This document provides examples and explanations of set theory concepts including:
- Types of sets such as universal sets, disjoint sets, and subsets
- Set operations including intersection, union, and complement
- Relationships between sets such as subsets and disjoint sets
- Calculating quantities such as the number of elements in sets
It contains examples of sets of various items like fruits, numbers, playing cards, and fish to demonstrate set theory ideas and operations.
This document provides an overview of the syllabus for the course CS6303 - Computer Architecture. It covers the following key topics in 3 sentences or less:
- Components of a computer system including input, output, memory, datapath, and control. Instructions and their representation. Addressing modes for accessing operands.
- Eight major ideas in computer architecture: designing for Moore's law, using abstraction, optimizing common cases, performance via parallelism and pipelining, performance via prediction, hierarchy of memories, and dependability via redundancy.
- Evolution from uniprocessors to multiprocessors to address power constraints. Instruction formats, operations, logical and control operations, and different addressing modes for specifying operand locations
This document summarizes digital image processing techniques including algebraic approaches to image restoration and inverse filtering. It discusses:
1) Unconstrained and constrained restoration, with unconstrained having no knowledge of noise and constrained using knowledge of noise.
2) Inverse filtering which is a direct method that minimizes error between degraded and original images using matrix operations, but can be unstable due to noise or near-zero filter values.
3) Pseudo-inverse filtering which adds a threshold to the inverse filter to avoid instability, working better for noisy images by not amplifying high frequency noise.
The document discusses concurrency issues in operating systems and solutions to the critical section problem. It begins by introducing the critical section problem and describing software and hardware solutions. It then defines key concurrency concepts like critical sections, mutual exclusion, deadlocks, livelocks, race conditions, and starvation. Specific hardware approaches like interrupt disabling and test-and-set instructions are presented. Software approaches using semaphores are also introduced as a way for processes to signal each other and synchronize access to shared resources.
1) Amdahl's Law describes the theoretical speedup from parallel processors based on the proportion of a program that can be parallelized (1-B) versus the portion that must run serially (B).
2) The speedup formula is: Speedup = 1 / (B + (1-B)/Number of Processors). This shows diminishing returns as more processors are added.
3) A speedup curve based on Amdahl's Law will always be below the ideal linear speedup (S=N) line, showing the limits on parallelization from the serial components of a program.
This document discusses various methods for structuring fields and records in files. It compares methods such as fixing field lengths, using length indicators, delimiters, and self-describing fields. It also discusses organizing records using fixed lengths, predicable numbers of fields, length indicators, index files, and delimiters. One effective method discussed is using a length indicator to store variable length records by writing the size first followed by the record contents.
This document outlines a presentation on pipelining and data hazards in microprocessors. It begins with rules for participant questions and outlines the topics to be covered: what is pipelining, types of pipelining, data hazards and their types, and solutions to data hazards. It then defines pipelining as executing subsequent instructions before prior ones complete. Types of pipelining include control, data, and structure hazards. Data hazards occur if an instruction uses a value before it is ready, and their types are RAW, WAR, and WAW. Solutions involve forwarding newer register values to bypass stale values in the pipeline and prevent hazards.
The document discusses cache organization and mapping techniques. It describes:
1) Direct mapping where each block maps to one line. Set associative mapping divides cache into sets with multiple lines per set.
2) Replacement algorithms like FIFO and LRU that determine which block to replace when the cache is full.
3) Write policies like write-through and write-back that handle writing cached data back to main memory.
This document provides an overview of digital image fundamentals including:
- The electromagnetic spectrum and how light is sensed and sampled by sensor arrays to create digital images.
- Common sensor technologies like CCD and CMOS sensors and how they work.
- How digital images are represented through spatial and intensity discretization via sampling and quantization.
- Factors that affect image quality like spatial and intensity resolution.
- Concepts like aliasing, moire patterns, and their relationship to sampling rates.
- Basic image processing techniques like zooming, shrinking, and relationships between pixels.
This document provides an overview of the syllabus for the course CS6303 - Computer Architecture. It covers the following key topics in 3 sentences or less:
- Components of a computer system including input, output, memory, datapath, and control. Instructions and their representation. Addressing modes for accessing operands.
- Eight major ideas in computer architecture: designing for Moore's law, using abstraction, optimizing common cases, performance via parallelism and pipelining, performance via prediction, hierarchy of memories, and dependability via redundancy.
- Evolution from uniprocessors to multiprocessors to address power constraints. Instruction formats, operations, logical and control operations, and different addressing modes for specifying operand locations
This document summarizes digital image processing techniques including algebraic approaches to image restoration and inverse filtering. It discusses:
1) Unconstrained and constrained restoration, with unconstrained having no knowledge of noise and constrained using knowledge of noise.
2) Inverse filtering which is a direct method that minimizes error between degraded and original images using matrix operations, but can be unstable due to noise or near-zero filter values.
3) Pseudo-inverse filtering which adds a threshold to the inverse filter to avoid instability, working better for noisy images by not amplifying high frequency noise.
The document discusses concurrency issues in operating systems and solutions to the critical section problem. It begins by introducing the critical section problem and describing software and hardware solutions. It then defines key concurrency concepts like critical sections, mutual exclusion, deadlocks, livelocks, race conditions, and starvation. Specific hardware approaches like interrupt disabling and test-and-set instructions are presented. Software approaches using semaphores are also introduced as a way for processes to signal each other and synchronize access to shared resources.
1) Amdahl's Law describes the theoretical speedup from parallel processors based on the proportion of a program that can be parallelized (1-B) versus the portion that must run serially (B).
2) The speedup formula is: Speedup = 1 / (B + (1-B)/Number of Processors). This shows diminishing returns as more processors are added.
3) A speedup curve based on Amdahl's Law will always be below the ideal linear speedup (S=N) line, showing the limits on parallelization from the serial components of a program.
This document discusses various methods for structuring fields and records in files. It compares methods such as fixing field lengths, using length indicators, delimiters, and self-describing fields. It also discusses organizing records using fixed lengths, predicable numbers of fields, length indicators, index files, and delimiters. One effective method discussed is using a length indicator to store variable length records by writing the size first followed by the record contents.
This document outlines a presentation on pipelining and data hazards in microprocessors. It begins with rules for participant questions and outlines the topics to be covered: what is pipelining, types of pipelining, data hazards and their types, and solutions to data hazards. It then defines pipelining as executing subsequent instructions before prior ones complete. Types of pipelining include control, data, and structure hazards. Data hazards occur if an instruction uses a value before it is ready, and their types are RAW, WAR, and WAW. Solutions involve forwarding newer register values to bypass stale values in the pipeline and prevent hazards.
The document discusses cache organization and mapping techniques. It describes:
1) Direct mapping where each block maps to one line. Set associative mapping divides cache into sets with multiple lines per set.
2) Replacement algorithms like FIFO and LRU that determine which block to replace when the cache is full.
3) Write policies like write-through and write-back that handle writing cached data back to main memory.
This document provides an overview of digital image fundamentals including:
- The electromagnetic spectrum and how light is sensed and sampled by sensor arrays to create digital images.
- Common sensor technologies like CCD and CMOS sensors and how they work.
- How digital images are represented through spatial and intensity discretization via sampling and quantization.
- Factors that affect image quality like spatial and intensity resolution.
- Concepts like aliasing, moire patterns, and their relationship to sampling rates.
- Basic image processing techniques like zooming, shrinking, and relationships between pixels.
The document describes the basic processing unit. It discusses how (1) the processor fetches and executes instructions one at a time from memory, (2) an instruction is executed by performing more basic operations like register transfers, arithmetic/logic operations, and memory access, and (3) the processor uses control signals to coordinate the execution of instructions step-by-step. It also introduces hardwired control and microprogrammed control as two approaches to generate the necessary control signals.
Concurrency: Mutual Exclusion and SynchronizationAnas Ebrahim
This document discusses concurrency and synchronization in operating systems. It covers mutual exclusion and how it must be enforced to prevent interference between concurrent processes accessing shared resources. Various synchronization mechanisms are described, including semaphores, mutexes, monitors and event flags. The producer-consumer problem is presented and solutions shown using semaphores to ensure processes can access shared resources like buffers safely. Implementation of semaphores also discussed, needing atomic operations.
basic computer programming and micro programmed controlRai University
The document discusses microprogrammed control unit implementation. It describes that a microprogrammed control unit uses microinstructions stored in read-only control memory to generate control signals for executing microoperations. Each computer instruction is mapped to a routine in control memory containing a sequence of microinstructions. The microinstructions include fields that specify microoperations to perform and the address of the next microinstruction. A control address register holds the address of the current microinstruction, and a next address generator determines the next address based on branching conditions.
Memory is organized in a hierarchy with different levels providing trade-offs between speed and cost.
- Cache memory sits between the CPU and main memory for fastest access.
- Main memory (RAM) is where active programs and data reside and is faster than auxiliary memory but more expensive.
- Auxiliary memory (disks, tapes) provides backup storage and is slower than main memory but larger and cheaper.
Virtual memory manages this hierarchy through address translation techniques like paging that map virtual addresses to physical locations, allowing programs to access more memory than physically available. When data is needed from auxiliary memory a page fault occurs and page replacement algorithms determine what data to remove from main memory.
This document outlines the course syllabus for Digital Image Processing (DIP). It includes 5 units covering key topics in DIP like digital image fundamentals, image enhancement, restoration and segmentation, wavelets and compression, and image representation and recognition. The syllabus allocates 45 class periods to cover these units in depth. Recommended textbooks and references for the course are also provided.
The document discusses pipeline hazards including structural, data, and control hazards. It provides details on how each hazard can occur in a 5-stage pipeline and techniques to resolve them, including forwarding, stalling, and compiler scheduling. Data hazards are classified as RAW, WAW, and WAR. Control hazards from branches are reduced by computing the branch target and outcome earlier in the ID phase to minimize stalls.
In non-preemptive scheduling once a process has been allocated the CPU it runs uninterrupted until it finishes execution.
On the other hand, in preemptive schedule algorithms, the running process may be interrupted by a higher priority process in between its execution.
Whenever a process gets into ready state or the currently running finishes execution, the priority of the ready state process is checked against that of the running process.
If the priority of the ready process is more it is allowed to be allocated to the CPU.
Therefore, in these schemes, the CPU is allocated to the process with the highest priority all the time.
This gives rise to frequent context switching, which can become very costly in terms of CPU time wasted in switching. In the following sections we will explore some of such scheduling algorithms
This document discusses loop parallelization and pipelining as well as trends in parallel systems and forms of parallelism. It describes loop transformations like permutation, reversal, and skewing that can be used to parallelize loops. It also discusses parallelization conditions, wavefront transformations for fine-grained parallelism, and tiling to improve data locality. The document then covers software pipelining of loops to reduce execution time. Finally, it discusses trends in parallel computing and different forms of parallelism like instruction-level, data, and task parallelism.
This document discusses region-based image segmentation techniques. Region-based segmentation groups pixels into regions based on common properties. Region growing is described as starting with seed points and grouping neighboring pixels with similar properties into larger regions. The advantages are it can correctly separate regions with the same defined properties and provide good segmentation in images with clear edges. The disadvantages include being computationally expensive and sensitive to noise. Region splitting and merging techniques are also discussed as alternatives to region growing.
This presentation discusses array processors, which are parallel computers composed of multiple identical processing elements that can operate simultaneously. The presentation covers the history of array processors, how they work, classifications, architectures, performance and scalability. It explains that array processors are well-suited for tasks involving repetitive arithmetic operations on large datasets, as they can improve performance for such workloads, but may not provide benefits for operations with data dependencies or decisions based on computations.
This document discusses the evolution of computer systems from early relay-based computers to modern parallel processing systems. It covers the progression from vacuum tubes to integrated circuits, increasing computer speeds and capabilities over generations. The key aspects covered are:
1. Computer components including the CPU, memory, and I/O have advanced significantly from early electromechanical to modern integrated systems.
2. Parallel processing has increased from basic multiprocessing to finer-grained instruction-level parallelism using pipelining and multiple functional units.
3. Uniprocessor computers exploit parallelism through techniques like overlapping I/O and CPU operations, hierarchical memory systems, and multiprogramming.
Comparison between Lossy and Lossless Compressionrafikrokon
This presentation compares lossy and lossless compression. It discusses the group members, topics to be covered including definitions of compression, lossless compression, and lossy compression. It explains that lossless compression allows exact recovery of original data while lossy compression involves some data loss. Lossy compression removes non-essential data and has data degradation but is cheaper and requires less space and time. Lossless compression works well with repeated data and allows exact data recovery but requires more space and time. The presentation discusses uses of each compression type and their advantages and disadvantages.
The DBMS schema defines the overall logical structure and description of a database. A sub-schema allows different personalized views of the same underlying data. A DBMS instance refers to the specific set of data stored in a database at a particular moment in time.
The document outlines the main stages of image processing which include image acquisition, restoration, enhancement, representation and description, segmentation, object recognition, color processing, compression, and morphological operations. It describes each stage in detail, explaining their purposes and some common techniques used. The overall stages take a raw image and perform various operations to extract useful information and simplify analysis for applications like object identification and extraction.
The document discusses arithmetic pipelines used in computers. It describes how pipelines divide sequential processes into sub-processes that execute concurrently. As an example, it explains the 4-part floating point addition and subtraction process, including comparing exponents, aligning mantissas, adding or subtracting mantissas, and producing the result. Registers are used to store intermediate results between operations. The document also briefly mentions that RISC pipelines typically have 2-3 segments - one to fetch instructions, one for ALU execution, and optionally one for storing results.
This document discusses parallel processing and pipelining. It describes different levels and types of parallel processing including job level, task level, inter-instruction level, and intra-instruction level parallelism. It also covers Flynn's classification of parallel computers as SISD, SIMD, MISD, and MIMD based on the number of instruction and data streams. Pipelining is defined as decomposing a process into sub-operations that execute concurrently. The key benefits of pipelining are that multiple computations can progress simultaneously through different pipeline stages.
This document provides an introduction to First Order Predicate Logic (FOPL). It discusses the differences between propositional logic and FOPL, the parts and syntax of FOPL including terms, atomic sentences, quantifiers and rules of inference. The semantics of FOPL are also explained. Pros and cons are provided, such as FOPL's ability to represent individual entities and generalizations compared to propositional logic. Applications include using FOPL as a framework for formulating theories.
The document provides an overview of key concepts in set theory, relations, and graph theory that are important foundations for computer science. It defines what a set is, and methods for describing sets. It discusses operations on sets like unions, intersections, complements and differences. It presents results on the number of elements in sets and provides examples to illustrate set concepts. Graph theory concepts like graphs, subgraphs, trees and their applications in computer science are also introduced. The document aims to build awareness of these fundamental mathematical concepts used extensively in computer languages and their utility.
important Questions class 11 chapter 1 setsakstudy1024
This document contains solutions to multiple choice and multi-part questions about sets from a Maths Class 11 Chapter on sets. Some key details:
- Questions cover topics like roster form, set-builder form, union, intersection, complement and power sets of sets.
- Solutions provide step-by-step workings and explanations for finding subsets, set operations and properties of sets.
- One question involves a Venn diagram to determine the number of individuals exposed to certain chemicals.
- Another question uses set operations and properties to calculate the number of families buying different newspaper combinations in a town.
The document describes the basic processing unit. It discusses how (1) the processor fetches and executes instructions one at a time from memory, (2) an instruction is executed by performing more basic operations like register transfers, arithmetic/logic operations, and memory access, and (3) the processor uses control signals to coordinate the execution of instructions step-by-step. It also introduces hardwired control and microprogrammed control as two approaches to generate the necessary control signals.
Concurrency: Mutual Exclusion and SynchronizationAnas Ebrahim
This document discusses concurrency and synchronization in operating systems. It covers mutual exclusion and how it must be enforced to prevent interference between concurrent processes accessing shared resources. Various synchronization mechanisms are described, including semaphores, mutexes, monitors and event flags. The producer-consumer problem is presented and solutions shown using semaphores to ensure processes can access shared resources like buffers safely. Implementation of semaphores also discussed, needing atomic operations.
basic computer programming and micro programmed controlRai University
The document discusses microprogrammed control unit implementation. It describes that a microprogrammed control unit uses microinstructions stored in read-only control memory to generate control signals for executing microoperations. Each computer instruction is mapped to a routine in control memory containing a sequence of microinstructions. The microinstructions include fields that specify microoperations to perform and the address of the next microinstruction. A control address register holds the address of the current microinstruction, and a next address generator determines the next address based on branching conditions.
Memory is organized in a hierarchy with different levels providing trade-offs between speed and cost.
- Cache memory sits between the CPU and main memory for fastest access.
- Main memory (RAM) is where active programs and data reside and is faster than auxiliary memory but more expensive.
- Auxiliary memory (disks, tapes) provides backup storage and is slower than main memory but larger and cheaper.
Virtual memory manages this hierarchy through address translation techniques like paging that map virtual addresses to physical locations, allowing programs to access more memory than physically available. When data is needed from auxiliary memory a page fault occurs and page replacement algorithms determine what data to remove from main memory.
This document outlines the course syllabus for Digital Image Processing (DIP). It includes 5 units covering key topics in DIP like digital image fundamentals, image enhancement, restoration and segmentation, wavelets and compression, and image representation and recognition. The syllabus allocates 45 class periods to cover these units in depth. Recommended textbooks and references for the course are also provided.
The document discusses pipeline hazards including structural, data, and control hazards. It provides details on how each hazard can occur in a 5-stage pipeline and techniques to resolve them, including forwarding, stalling, and compiler scheduling. Data hazards are classified as RAW, WAW, and WAR. Control hazards from branches are reduced by computing the branch target and outcome earlier in the ID phase to minimize stalls.
In non-preemptive scheduling once a process has been allocated the CPU it runs uninterrupted until it finishes execution.
On the other hand, in preemptive schedule algorithms, the running process may be interrupted by a higher priority process in between its execution.
Whenever a process gets into ready state or the currently running finishes execution, the priority of the ready state process is checked against that of the running process.
If the priority of the ready process is more it is allowed to be allocated to the CPU.
Therefore, in these schemes, the CPU is allocated to the process with the highest priority all the time.
This gives rise to frequent context switching, which can become very costly in terms of CPU time wasted in switching. In the following sections we will explore some of such scheduling algorithms
This document discusses loop parallelization and pipelining as well as trends in parallel systems and forms of parallelism. It describes loop transformations like permutation, reversal, and skewing that can be used to parallelize loops. It also discusses parallelization conditions, wavefront transformations for fine-grained parallelism, and tiling to improve data locality. The document then covers software pipelining of loops to reduce execution time. Finally, it discusses trends in parallel computing and different forms of parallelism like instruction-level, data, and task parallelism.
This document discusses region-based image segmentation techniques. Region-based segmentation groups pixels into regions based on common properties. Region growing is described as starting with seed points and grouping neighboring pixels with similar properties into larger regions. The advantages are it can correctly separate regions with the same defined properties and provide good segmentation in images with clear edges. The disadvantages include being computationally expensive and sensitive to noise. Region splitting and merging techniques are also discussed as alternatives to region growing.
This presentation discusses array processors, which are parallel computers composed of multiple identical processing elements that can operate simultaneously. The presentation covers the history of array processors, how they work, classifications, architectures, performance and scalability. It explains that array processors are well-suited for tasks involving repetitive arithmetic operations on large datasets, as they can improve performance for such workloads, but may not provide benefits for operations with data dependencies or decisions based on computations.
This document discusses the evolution of computer systems from early relay-based computers to modern parallel processing systems. It covers the progression from vacuum tubes to integrated circuits, increasing computer speeds and capabilities over generations. The key aspects covered are:
1. Computer components including the CPU, memory, and I/O have advanced significantly from early electromechanical to modern integrated systems.
2. Parallel processing has increased from basic multiprocessing to finer-grained instruction-level parallelism using pipelining and multiple functional units.
3. Uniprocessor computers exploit parallelism through techniques like overlapping I/O and CPU operations, hierarchical memory systems, and multiprogramming.
Comparison between Lossy and Lossless Compressionrafikrokon
This presentation compares lossy and lossless compression. It discusses the group members, topics to be covered including definitions of compression, lossless compression, and lossy compression. It explains that lossless compression allows exact recovery of original data while lossy compression involves some data loss. Lossy compression removes non-essential data and has data degradation but is cheaper and requires less space and time. Lossless compression works well with repeated data and allows exact data recovery but requires more space and time. The presentation discusses uses of each compression type and their advantages and disadvantages.
The DBMS schema defines the overall logical structure and description of a database. A sub-schema allows different personalized views of the same underlying data. A DBMS instance refers to the specific set of data stored in a database at a particular moment in time.
The document outlines the main stages of image processing which include image acquisition, restoration, enhancement, representation and description, segmentation, object recognition, color processing, compression, and morphological operations. It describes each stage in detail, explaining their purposes and some common techniques used. The overall stages take a raw image and perform various operations to extract useful information and simplify analysis for applications like object identification and extraction.
The document discusses arithmetic pipelines used in computers. It describes how pipelines divide sequential processes into sub-processes that execute concurrently. As an example, it explains the 4-part floating point addition and subtraction process, including comparing exponents, aligning mantissas, adding or subtracting mantissas, and producing the result. Registers are used to store intermediate results between operations. The document also briefly mentions that RISC pipelines typically have 2-3 segments - one to fetch instructions, one for ALU execution, and optionally one for storing results.
This document discusses parallel processing and pipelining. It describes different levels and types of parallel processing including job level, task level, inter-instruction level, and intra-instruction level parallelism. It also covers Flynn's classification of parallel computers as SISD, SIMD, MISD, and MIMD based on the number of instruction and data streams. Pipelining is defined as decomposing a process into sub-operations that execute concurrently. The key benefits of pipelining are that multiple computations can progress simultaneously through different pipeline stages.
This document provides an introduction to First Order Predicate Logic (FOPL). It discusses the differences between propositional logic and FOPL, the parts and syntax of FOPL including terms, atomic sentences, quantifiers and rules of inference. The semantics of FOPL are also explained. Pros and cons are provided, such as FOPL's ability to represent individual entities and generalizations compared to propositional logic. Applications include using FOPL as a framework for formulating theories.
The document provides an overview of key concepts in set theory, relations, and graph theory that are important foundations for computer science. It defines what a set is, and methods for describing sets. It discusses operations on sets like unions, intersections, complements and differences. It presents results on the number of elements in sets and provides examples to illustrate set concepts. Graph theory concepts like graphs, subgraphs, trees and their applications in computer science are also introduced. The document aims to build awareness of these fundamental mathematical concepts used extensively in computer languages and their utility.
important Questions class 11 chapter 1 setsakstudy1024
This document contains solutions to multiple choice and multi-part questions about sets from a Maths Class 11 Chapter on sets. Some key details:
- Questions cover topics like roster form, set-builder form, union, intersection, complement and power sets of sets.
- Solutions provide step-by-step workings and explanations for finding subsets, set operations and properties of sets.
- One question involves a Venn diagram to determine the number of individuals exposed to certain chemicals.
- Another question uses set operations and properties to calculate the number of families buying different newspaper combinations in a town.
The document contains questions related to trigonometric functions, sets, relations and functions, complex numbers, and sequences and series. Some questions ask students to prove trigonometric identities, find sets operations, determine if relations are functions, solve complex equations, and evaluate infinite geometric series. The document provides hints for many questions and includes the answers for some questions.
The document defines and describes sets. Some key points include:
- A set is a collection of well-defined objects. Sets can be described in roster form or set-builder form.
- There are different types of sets such as the empty/null set, singleton sets, finite sets, and infinite sets.
- Set operations include union, intersection, difference, symmetric difference, and complement. Properties like subsets and the power set are also discussed.
- Cardinality refers to the number of elements in a set. Formulas are given for finding the cardinality of sets under different operations.
This document contains notes on additional mathematics including topics on progression, linear laws, integration, and vectors. Some key points:
- It discusses arithmetic and geometric progressions, defining the terms and formulas for finding terms and sums. Examples are worked through finding terms, sums, and differences between sums.
- Linear laws are explained including lines of best fit, converting between linear and non-linear forms using logarithms, and working through examples of finding equations from graphs.
- Integration techniques are outlined including formulas for integrals of powers, areas under and between curves, volumes of revolution, and the basic rules of integration. Worked examples find areas and volumes.
- Vectors are introduced including addition using the triangle
The document defines and provides examples of key concepts related to sets:
1) A set is a collection of distinct objects that can be defined using a description, set notation, or a Venn diagram.
2) Sets have elements and can be equal, subsets, or complements of one another.
3) Common set operations include intersection, union, and complement which combine elements of sets in different ways.
4) Venn diagrams can visually represent relationships between sets and their operations.
The document provides definitions and examples related to set theory concepts that are important for business mathematics. It defines what a set is, the different types of sets, set operations like union, intersection, complement and difference. It also discusses subsets, universal sets, disjoint sets, and the power set. Further concepts covered include Cartesian products of sets, Venn diagrams and solving word problems using set concepts.
The document provides information about sets, relations, and functions. It defines what a set is and how sets can be described in roster form or set-builder form. It discusses different types of sets such as the empty set, singleton sets, finite sets, and infinite sets. It also covers topics like subsets, power sets, operations on sets such as union, intersection, difference, complement, and applications of these concepts to solve problems involving cardinality of sets.
This document introduces some basic concepts of set theory:
1. Sets are collections of well-defined objects that can be represented using capital letters. Elements of a set are denoted using symbols like ∈ and ∉.
2. Important sets in number systems include real numbers (IR), positive/negative reals (IR+/IR-), integers (Z), positive/negative integers (Z+/Z-), rational numbers (Q), and natural numbers (N).
3. Sets can be specified using a roster method that lists elements or a set-builder notation that describes elements. Operations on sets include union, intersection, complement, and symmetric difference.
This document contains a presentation on the topic of sets. It includes definitions of key set concepts like unions, intersections, complements and Venn diagrams. It also provides examples to demonstrate these concepts, such as representing categorical data from a survey using a Venn diagram. The presentation covers fundamental topics in sets including types of sets, operations on sets and applications of sets in mathematics. It concludes by acknowledging the teacher and principal for providing the opportunity to create this project and thanks others who provided assistance.
- A set is a well-defined collection of objects. The empty set is a set with no elements. A set is finite if it has a definite number of elements, otherwise it is infinite.
- Two sets are equal if they have exactly the same elements. A set A is a subset of set B if every element of A is also an element of B.
- The power set of a set A is the collection of all subsets of A, including the empty set and A itself. It is denoted by P(A).
This document provides an introduction to set theory. It defines what a set is and provides examples of common sets. A set can be represented in roster form by listing its elements within curly brackets or in set builder form using a characteristic property. There are different types of sets such as finite sets, infinite sets, empty sets, singleton sets, and power sets which contain all subsets. Set operations like union, intersection, difference and symmetric difference are introduced. Important concepts like subsets, equivalent sets, disjoint sets and complements are also covered.
- The document discusses integers, adding and subtracting positive and negative numbers using a number line. It provides examples of adding, subtracting and using integers to solve temperature problems.
- It also discusses squares, square roots, and finding the square of numbers using a calculator or by multiplying a number by itself. Examples of finding the square root with and without a calculator are provided.
- Factors and primes are defined, with examples of finding factors and prime factors of different numbers. The highest common factor and lowest common multiple of sets of numbers are also explained.
- A set is a collection of distinct objects where order and repetition do not matter. Basic set operations include union, intersection, and difference.
- A set can be finite or infinite. Infinite sets can be countable, where elements can be paired with natural numbers, or uncountable. The set of real numbers is uncountable.
- Key relationships between sets include subset, proper subset, power set, membership (∈), and equality determined by mutual inclusion. Venn diagrams can visualize set relationships.
This document contains a summary of a student's third semester examination in field theory. Some key points:
1) The exam had two parts - Part A covered electrostatics and Part B covered magnetostatics.
2) In Part A, the student was asked to define electric field intensity, derive Maxwell's first equation, find potential due to line and point charges, and solve Laplace's equation for different boundary value problems.
3) In Part B, the student was asked to derive expressions for magnetic field and force between current elements, define displacement current density, and derive Maxwell's equations for time-varying fields.
4) The final section covered electromagnetic wave propagation - including deriving the wave
The document provides information about sets including definitions of key terms like union, intersection, complement, difference, properties of these operations, and counting theorems. It discusses describing sets by explicitly listing members or through a relationship. Examples are provided to illustrate concepts like subsets, proper subsets, power sets, De Morgan's laws, and using Venn diagrams to solve problems involving sets. Counting theorems are presented to calculate the number of elements in unions, intersections, and complements of finite sets.
This document contains a series of exercises related to vectors in a plane. It begins with exercises involving vector operations like finding scalar multiples that satisfy equations and vector addition and subtraction. Later questions involve vector properties such as parallelism of vectors, orthogonality, vector lengths, and linear combinations of vectors. Geometric representations of vectors are also explored through problems finding points and line segments. The document aims to reinforce concepts of vector algebra and geometry through multiple practice problems.
1) The document discusses algebraic formulas and expressions. It provides examples of writing formulas based on word problems and situations.
2) Key terms discussed include: algebraic formula, algebraic expression, variables, operations, equations, and relating factors.
3) The document also contains exercises on writing formulas, expressing variables as subjects of formulas, evaluating formulas for given values, and solving word problems algebraically.
1. The document provides a sample paper for mathematics class 12 with questions covering various topics like vectors, matrices, integrals, probability, linear programming, etc. divided into three sections - short 1 mark questions, 4 mark questions and 6 mark questions.
2. It also includes the marking scheme with answers for all the questions.
3. The questions aim to test different cognitive levels from remembering to applications and higher order thinking skills.
1. The document provides a sample paper for mathematics class 12 with questions covering various topics like vectors, matrices, integrals, probability, linear programming, etc. divided into three sections - short 1 mark questions, 4 mark questions and 6 mark questions.
2. It also includes the marking scheme with answers for all the questions.
3. The questions assess different cognitive levels like remembering, understanding, application and higher order thinking skills.
This document contains 3 short entries dated October 06, 2014 that are all labeled "6th october 2014". The document appears to be a log or record with multiple brief entries made on the same date.
This document is a list of dates, all occurring on October 3rd, 2014. Each entry repeats the date and contains a page number. There are 9 total entries in the list, each with the same date but incrementing page numbers from 1 through 9.
This document appears to be a log of dates from October 1st, 2014. It contains four entries all with the date October 1st, 2014 listed. The document provides a brief record of dates but does not include any other contextual information.
This document is a series of 8 entries all with the date of September 30, 2014. Each entry contains only the date with no other text or information provided.
The document is dated September 25, 2014. It appears to be a brief one paragraph document that does not provide much context or details. The date is the only substantive information given.
The document is dated September 25, 2014. It appears to be a brief one paragraph document that does not provide much context or details. The date is the only substantive information given.
This document is a record of events from September 24, 2014. It consists of 7 entries all with the same date of September 24, 2014 listed at the top, suggesting some type of daily log or journal was being kept for that date.
This document is a series of 7 entries all dated September 23, 2014 without any other notable information provided. Each entry simply states the date of September 23, 2014.
This four sentence document repeats the date September 22, 2014 four times without providing any additional context or information. The document states the same date, September 22, 2014, in each of its four sentences without elaborating on the significance of the date or including any other details.
This document is a record of dates, containing six identical entries of "September 18, 2014" with no other text or context provided. Each entry is on its own line and labeled with "18th sept 2014" and a number.
This document is a log of dates from September 16, 2014. It contains 5 entries all with the same date of September 16, 2014 listed in various formats including 16th sept 2014 and September 16, 2014.
This 3 sentence document simply repeats the date "September 11, 2014" three times on three different lines. It does not provide any other context or information.
The document is a list of dates, all occurring on September 9th, 2014. Each entry repeats the date 10 times, once for each numbered line. The sole purpose of the document is to repeatedly record the same date, September 9th, 2014, across 10 lines.
This document is a series of 7 entries all dated September 23, 2014 without any other notable information provided. Each entry simply states the date of September 23, 2014.
This four sentence document repeats the date September 22, 2014 four times without providing any additional context or information. The document states the same date, September 22, 2014, in each of its four sentences without elaborating on the significance of the date or including any other details.
This document is a record of dates, containing six identical entries of "September 18, 2014" with no other text or context provided. Each entry is on its own line and labeled with "18th sept 2014" and a number.
The document is a record of dates from September 17, 2014. It contains 20 entries, each listing the date September 17, 2014. The document functions as a log or record of the single date of September 17, 2014 recorded 20 separate times.
This document is a log of dates from September 16, 2014. It contains 5 entries all with the same date of September 16, 2014 listed in various formats including 16th sept 2014 and September 16, 2014.
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DeFi represents a paradigm shift in the financial industry. Instead of relying on traditional, centralized institutions like banks, DeFi leverages blockchain technology to create a decentralized network of financial services. This means that financial transactions can occur directly between parties, without intermediaries, using smart contracts on platforms like Ethereum.
In 2024, we are witnessing an explosion of new DeFi projects and protocols, each pushing the boundaries of what’s possible in finance.
In summary, DeFi in 2024 is not just a trend; it’s a revolution that democratizes finance, enhances security and transparency, and fosters continuous innovation. As we proceed through this presentation, we'll explore the various components and services of DeFi in detail, shedding light on how they are transforming the financial landscape.
At Intelisync, we specialize in providing comprehensive DeFi development services tailored to meet the unique needs of our clients. From smart contract development to dApp creation and security audits, we ensure that your DeFi project is built with innovation, security, and scalability in mind. Trust Intelisync to guide you through the intricate landscape of decentralized finance and unlock the full potential of blockchain technology.
Ready to take your DeFi project to the next level? Partner with Intelisync for expert DeFi development services today!
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TrustArc Webinar - 2024 Global Privacy SurveyTrustArc
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The original Czech 🇨🇿 version of the presentation can be found here: https://www.slideshare.net/slideshow/hlavni-novinky-souvisejici-s-ccs-tsi-2023-2023-1695/269688092 .
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5th LF Energy Power Grid Model Meet-up SlidesDanBrown980551
5th Power Grid Model Meet-up
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Power Grid Model
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What to expect
For the upcoming meetup we are organizing, we have an exciting lineup of activities planned:
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Letter and Document Automation for Bonterra Impact Management (fka Social Sol...Jeffrey Haguewood
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We believe integration and automation are essential to user experience and the promise of efficient work through technology. Automation is the critical ingredient to realizing that full vision. We develop integration products and services for Bonterra Case Management software to support the deployment of automations for a variety of use cases.
This video focuses on automated letter generation for Bonterra Impact Management using Google Workspace or Microsoft 365.
Interested in deploying letter generation automations for Bonterra Impact Management? Contact us at sales@sidekicksolutionsllc.com to discuss next steps.
Ocean lotus Threat actors project by John Sitima 2024 (1).pptxSitimaJohn
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This talk will focus on how to collect data from a variety of sources, leveraging this data for RAG and other GenAI use cases, and finally charting your course to productionalization.
1. Chapter 3: Set Theory and Logic
L e s s o n 3.1: T y p e s of S e t s a n d S e t Notation,
page 154
1. a) e.g., Y e s , those explanations make sense.
b) T h e universal set is set C.
C = {produce}
O = {orange produce} = {oranges, carrots}
Y = {yellow produce} = {bananas, pineapple, corn}
G = {green produce} = apples, pears, peas, beans}
B = {brown produce} = {potatoes, pears}
c) e.g., S c F because all fruits you can eat without
peeling are also fruits. S c C because all fruits you can
eat without peeling are also produce.
d) Sets S and V are disjoint sets, as are F and V.
e) Yes. e. g., C = { F and V}, F = V.
f) n{V) = n{C)-n{F)
n{/) = 1 0 - 5
n{V) = S
g) oranges, pineapple, bananas, peas, corn, carrots,
beans, potatoes
2. a)
b) Sets E a n d S are disjoint sets, as are sets F a n d S.
c) i) True, e.g.. Multiples of 8 are also multiples of 4 .
ii) False, e.g.. Not all multiples of 4 are multiples of 8.
iii) True, e.g., All multiples of 4 are multiples of 4.
iv) False, e.g., F' = {all numbers from 1 to 4 0 that are not
multiples of 4}
v) True, e.g.. T h e universal set includes natural numbers
from 1 to 4 0 .
3. a)
.1. A A A A A A
l l i l S i i t l i l i l l S # f ^ ^
A A A A A A
A A i A ' . A iA . A - . A - ' A / A ,' -1 . .
;-;A>;'AIOAIA;,ia-'..A / •) lo j o
b) Subsets of set B: C c B a n d S c B
c) Subsets of set R: H c R anti D a R
d) Y e s , the sets S a n d C are disjoint, e.g., A card
cannot be both a spade and a club.
e) Y e s , the events in sets H and D are mutually
exclusive, e.g., Y o u cannot draw a card that is a
heart and a diamond at the s a m e time.
f) Y e s , that statement is correct, e.g., Because
these sets are disjoint, they contain no c o m m o n
elements. Therefore, w h e n the numbers of
elements in each set are added, no element will be
counted twice.
niS or D ) = n{S) + n ( D )
n{S or D ) = 13 + 13
n{S or D ) = 26
5. a) e.g., C = {all clothes}, S = { s u m m e r clothes},
W = {winter clothes}, H = { s u m m e r headgear}
b) e.g., In set C, but not in set S or set W, because
they would be w o r n year round.
c) N o , set S ' is not equal to set W. Set S ' includes
the jacket, but W does not.
d) Sets S and Ware disjoint sets. Sets H a n d W
are disjoint sets.
e) e.g., C = {clothes},
H = {headgear} = {cap, sunglasses, toque},
6 = {clothing for body} = {shirt, shorts, coat,
jacket}, F = {footwear} = {sandals, insulated boots}
6. n{X') = n{U) - n(X)
n{X')= 100 0 0 0 - 1 2
niX') = 99 988
7. Not possible; e.g., there m a y be s o m e elements
that are in both X and Y.
walleye northern pike 8. n(L/) = n(X) + n(X')
n{U) = 34 + 4 2
lake trout Arctic char
Arctic grdylinq lake whitefish
b) e.g., N a 7 m e a n s that aft the fish found in Nunavut
are also found in the Northwest Territories. Tct N m e a n s
that not all the fish found in the Northwest Territories are
found in Nunavut.
niU) = 76
9. a) S = {A, E, F, H, I, K, L, M, N, T, V, W , X, Y, Z}
C = {C, 0 , S}
b) False, e.g., B is not in S or C.
F o u n d a t i o n s of Mathematics 12 S o l u t i o n s Manual 3-1
2. 10. Let U represent ttie universal set. Let L represent ttie
set of land transportation. Let represent ttie set of
water transportation. Let A represent ttie set of air
transportation.
lu-.t .ill K>,il!(>cu!s '.".ai-kiny i.ikir,!;
•.kiinii riiiv'inij
pO'.M'l lUi.lt',
11. a)
1 2 .5 4 S lO 9 3 / h
b) Sets A and B are disjoint sets.
c) i) False, e.g., 1 is not in B.
ii) False, e.g., - 1 is not in A.
iii) False, e.g., 0 is in A' but not in B.
iv) True, e.g., n{A) = 10, n{B) = 10.
v) True, e.g., No integer from - 2 0 to - 1 5 is in U.
12. a) S = {4, 6, 9, 10, 14, 15, 2 1 , 22, 25, 26, 33, 34, 35,
38, 39, 46, 49}
W= { 1 , 2, 3, 5, 7, 8, 1 1 , 12, 13, 16, 17, 18, 19, 20, 23,
24, 27, 28, 29, 30, 3 1 , 36, 37, 40, 4 1 , 42, 43, 44, 45, 47,
48, 50}
b) e.g., £ = {even semiprime numbers}
E = {4, 6, 10, 14, 22, 26, 34, 38, 46}
c) n{W) =n{U)-n{S)
n{W) = 5 0 - 1 7
n{W) = 3 3
d) No, it is not possible to determine n{A). e.g., Ttiere is
an infinite number of prime numbers, so there is an
infinite number of semiprime numbers.
13. e.g.. Let U represent the universal set. Let E
represent the set of entertainment items. Let T represent
technology items.
equipment television
luniputv.'-
14. Agree; e.g., A (z B means that set A s a part of
set B, and it could be that set A and set B are
equal. f Ac: B, then set A will have the s a m e
number or fewer elements than set B. With
numbers, x < y m e a n s that x is less than or equal
to y. Or, or if y4 c B, then n{A) < n{B). The number
of elements in a subset must be equal to or less
than the number of elements in the set.
15. a) S = {x I - 1 0 0 0 < X < 1000, x e I }
r = { f | f = 2 5 x , - 4 0 < x < 4 0 , X G 1}
F = { f | f = 5 0 x , - 2 0 < x < 2 0 , x e 1}
Fez Td S
b)
16. a)U = { H H H , HHT, H T H , T H H , HTT, THT, T T H ,
TTT}
b) £ = { H T H , HTT, T T H , TTT}
c) n{U) = 8, n(£) = 4
d) Yes, e.g., because each element of £ is also an
element of U, and there are some elements of U
that are not elements of £ .
•^•'•'•'•^••IT THH IMI
e) For example, £ ' is the set of elements of U
where the second coin turns up heads.
n(£0 = n((y)-n(£)
n ( £ 0 = 8 - 4
n{E^ = 4
£ ' = { H H H , HHT, T H H , THT} and n(£') = 4
f) Yes. e.g., A coin cannot show both heads and
tails at the s a m e time.
17. a)
b) e.g., N' is the set of all non-natural numbers. W
is the set of all non-whole numbers. 1' is the set of
non-integer numbers. Q' is the set of numbers that
cannot be described as a ratio of two integers. Q
is the set of numbers that can be described as a
ratio of two integers.
3-2 C h a p t e r 3: S e t T h e o r y a n d L o g i c
3. Set C o m p l e m e n t
N W = {x 1 X € f?, X g N}
.1 / ' = { x 1 X e R , X « 1}
Q Q
Q Q
c) Sets N and Q are disjoint sets. Sets lA^and Q are
disjoint sets. Sets / and Q are disjoint sets. Sets Q and
Q are disjoint sets.
d) Yes. e.g., Q' is ttie set of numbers that cannot be
described as a ratio of two integers, which is the set of
irrational numbers.
e) W, I, Q, R
f) No. e.g., The area of a region in a V e n n diagram is not
related to the number of elements in the set.
18. a) S = { 1 , 4, 9, 16, 25, 36, 49, 64, 8 1 , 100, 1 2 1 , 144,
169, 196, 225, 256, 289}
n ( S ) = 17
E = {4, 16, 36, 64, 100, 144, 196, 256}
n{E) = 8
b) n ( S ) = 17, n(£) = 8
n(0) = n{S)-n{E)
n ( 0 ) = 1 7 - 8
n ( 0 ) = 9
c) n{U) = 300, n(S) = 17
n(S') = niU) - n{S)
n(S') = 3 0 0 - 17
n(S') = 283
19. a) e.g., /A c S if al! elements of A are also in B. For
example, all weekdays are also days of the week, so
w e e k d a y s is a subset of days of the week.
b) e.g., A consists of all the elements in the universal set
but not in A. For example, all days of the week that are
not w e e k d a y s are w e e k e n d days. So weekend days is
the complement of weekdays.
20. e.g., Disagree; since both the subsets are empty,
they both contain the same elements and are therefore
the s a m e subset.
L e s s o n 3.2: E x p l o r i n g R e l a t i o n s h i p s b e t w e e n
S e t s , p a g e 1 6 0
1 a )
u
IS n 10 14
b) i) n{A) = 5
ii) n{A but not S) = n{A) - n{A and B)
n(A but n o t e ) = 5 - 2
n{A but not S) = 3
iii) n{B) = 7
iv) n{B but not A) = n{B) - n{A a n d B)
n{B but not A) = 7-2
n{B but not yA) = 5
v) n{A and 6 ) = 2
vi) n{A or S) = n(A but not B) + n{A and 6 )
+ n{B but not yA)
n(>AorB) = 3 + 2 + 5
n{A or B) = 10
vii) n{A) = 5, therefore n{A') = 5
2. a) 8 students are in both the drama club and the
band.
b) 11 students are in the drama club only.
6 students are in the band only.
c) Drama: 1 1 + 8 = 19
Band: 8 + 6 = 14
d) Drama club or band: 1 1 + 8 + 6 = 2 5
e) 38 students in grade 12 - 25 in drama club or
band = 13 students in neither drama club nor band
3. a) hockey or soccer: 45 - 16 = 29
hockey and soccer: 20 + 14 = 34
overlap: 34 - 29 = 5
5 students like hockey and soccer.
b) only hockey: 20 - 5 = 15
only soccer: 1 4 - 5 = 9
15 + 9 = 24
24 students like only hockey or only soccer.
c)
4. a) ski or snowboard: 55 - 9 = 46
ski and snowboard: 25 + 32 = 57
Overlap: 57 - 46 = 11
11 guests plan to ski and snowboard.
b) only ski: 2 5 - 11 = 14
14 guests will only ski.
c) only s n o w b o a r d : 32 - 11 = 2 1
21 guests will only s n o w b o a r d .
5. a) n{U) - n{U but not A or B): 25 - 4 = 21
n{A) + n{B): 13 + 10 = 23
n{A and 6 ) : 2 3 - 2 1 = 2
n{A only): 1 3 - 2 = 11
n ( B only): 1 0 - 2 = 8
b)
2 8
F o u n d a t i o n s of Mathematics 12 S o l u t i o n s Manual 3-3
4. L e s s o n 3 . 3 : I n t e r s e c t i o n a n d U n i o n o f T w o S e t s ,
p a g e 1 7 2
1. a) ^ = { - 1 0 , - 8 , - 6 , - 4 , - 2 , 0, 2, 4, 6, 8, 10}
B = {0, 1,2, 3, 4, 5, 6, 7, 8, 9 , 1 0 }
AuB = { - 1 0 , - 8 , - 6 , - 4 , - 2 , 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,
10}
b) n{AuB)= 16
c) AnB = {0, 2 , 4 , 6, 8, 10}
d) n{A n B) = 6
2. a) Let A represent ttie universal set. Let N represent
ttie set of tundra animals. Let S represent the set of
southern animals.
N = {arctic fox, caribou, ermine, grizzly bear, muskox,
polar bear}
S = {bald eagle, Canadian lynx, grey wolf, grizzly bear,
long-eared owl, wolverine}
Nu S = {Arcticfox, caribou, ermine, muskox, polar bear,
grizzly bear, bald eagle, Canadian lynx, grey wolf, long-
eared owl, wolverine}
Tr, S = {grizzly bear}
b)
Arctii, fox bald eagle
tcuibou Canadian lynx
ermint; grizzly bear grey wolf
rTin^kox lonc]-eared owl
r^otar bear wolverine
3. a) / u C = { - 1 0 , - 8 , - 6 , - 4 , - 2 , 0, 2, 4, 6, 8, 10, 12,
14, 16}
niA u C) = 14
A nC = {2,4, 6, 8, 10}
n{A nC) = 5
b)
2 4
12 14
4. a) 7 u C = {half-ton trucks, quarter-ton trucks, vans,
SUVs, crossovers, 4-door sedans, 2-door coupes, sports
cars, hybrids}
b) n ( r u C ) = 9
c) T n C = {crossovers}
5. a) Let U represent the universal set. Let F
represent the set of African animals. Let S
represent the set of Asian animals
lion
qiraffe
hippo
i..Vni';i
elephant
tKjer
b) F = {lion, camel, giraffe, hippo, elephant}
S = {elephant, tiger, takin, camel}
F u S = {lion, giraffe, hippo, camel, elephant, tiger,
takin}
F n S = {camel, elephant}
6. a)
0 C!
6 12 3 I
15
b)AuB = { - 1 2 , - 9 , - 6 , -4, - 3 , - 2 , 0, 2, 3, 4, 6,
9, 10, 12, 15}
n{AuB)= 16
AnB = {-6, 0 , 6 , 12}
n{A n B) = 4
7. Let U represent the universal set. Let H
represent the set of people w h o liked Sherlock
Holmes. Let P represent the set of people w h o
liked Hercule Poirot.
n{H uP) = n{U) - n{(H u P)')
n{H u P) = 25 - 4
n{H u P) = 21
n{H nP) = n{H) + n(P) - n{H u P)
n ( H n P ) = 16 + 11 - 2 1
n{H n P) = 6
6 people like both detectives.
n{H only) = n{H) - n{H u P)
n ( H only) = 1 6 - 6
n ( H o n l y ) = 10
10 people liked Sherlock Holmes only.
n(P only) = n(P) - n{H u P)
n(P only) = 1 1 - 6
n{P only) = 5
5 people liked Hercule Poirot only.
3-4 C h a p t e r 3: S e t T h e o r y a n d L o g i c
5. 8. Let U represent ttie universal set. Let V represent ttie
set of people wtio liked vanilla ice c r e a m . Let C
represent ttie set of people wtio liked ctiocolate ice
cream.
n(C u VO = n{U) - n{(C u V)')
n ( C u VO = 80 - 9
n ( C u V0 = 71
n{C only) = n ( C u V) - n{ V only) - n ( C n VO
n ( C o n l y ) = 71 - 1 1 - 2 0
n{C only) = 40
40 people like chocolate ice cream only.
9. Let U represent the universal set. Let K represent the
set of people w h o like to ski. Let W represent the set of
people w h o like to swim.
niKuW) = n{U) - n{(K u W)')
n{KuW) = 26-5
n{Ku 140 = 21
n{KnW) = n{K) + n{W) -n{KuW)
n{Kn t V ) = 1 9 + 1 4 - 2 1
n ( K n MO =12
12 people like to ski and swim.
10. e.g., She could draw a V e n n diagram showing the
set of multiples of 2 and the set of multiples of 3. The
intersection of the sets would be the multiples of 6.
11. a) U = {all customers surveyed}
C = {customers ordering coffee}
D = {customers ordering donuts}
N = {customers ordering neither coffee nor doughnuts}
b) For the following Venn diagram:
The rectangular area labelled U represents the universal
set.
The shaded area labelled D represents the set of people
w h o ordered doughnuts.
The shaded area labelled C represents the set of people
w h o ordered coffee.
The shaded area labelled D n C represents the set of
people w h o ordered coffee and doughnuts.
The unshaded area labelled N represents those people
did not order coffee or doughnuts.
customers ordering botti coffee and a d o u g h n u t
I — ^ _
customers ordering neither
c) Determine n{D n C) using the information available.
niU) = 100, n(D) = 45, n(C) = 65, n{(D u C)') = 10
n{D u C) = n{U) - n{(D u C)')
n{DuC)= 1 0 0 - 1 0
n{D u C) = 90
Therefore,
/7(D nC) = n{D) + n(C) - n ( D u C)
n(D n C) = 4 5 + 65 - 90
n(D n C) = 20
There were 20 people w h o ordered coffee and a
doughnut.
12. Let U represent the universal set. Let T
represent the set of seniors w h o watch television.
Let R represent the set of seniors w h o listen to the
radio.
n{R only) = n ( L / ) - n ( T )
n ( R o n l y ) = 1 0 0 - 6 7
n{R only) = 33
33 seniors prefer to listen to the radio only.
13. Let U represent the universal set. Let C
represent the set of people w h o attended the
Calgary S t a m p e d e . Let P represent the set of
people w h o attended the P N E .
n(C uP) = n{U)- n{(C u P)')
n ( C u P ) = 5 6 - 1 4
n ( C o P) = 42
n{C n P) = n(C) + n(P) - n{C u P)
n ( C n P) = 30 + 22 - 42
n{CnP) = 10
10 people had been to both the Calgary S t a m p e d e
and the P N E .
14. Of the 54 people, 31 o w n their home, so
54 - 31 = 23 people rent their home. Of that 23, 9
rent their house, so 23 - 9 = 14 rent their
condominium. O f t h e 30 people w h o live in a
condominium, 14 rent, so 30 - 14 = 16 must o w n
the condominium in which they live.
15. Let U represent the universal set. Let R
represent the set of people w h o like reality shows.
Let C represent the set of people w h o like contest
shows.
n{C u R) = niU) - n{(C u P)')
n ( C u P ) = 3 2 - 4
n{C u P) = 28
n{C nR) = n{C u P ) - n{C only) - n{R only)
n ( C n P) = 2 8 - 1 3 - 9
n{C n P ) = 6
6 people like both type of shows.
16. No. e.g.. The three numbers do not add up to
48. There is an overlap between sets B and C, but
B<xC.
The s u m of the three values in the problem is 59.
59 - 48 = 11
11 students must drive a car and take a bus.
31 - 1 1 = 2 0
20 students drive a car but do not take a bus.
1 6 - 11 = 5
5 students take a bus but do not drive a car.
There are a total of 15 + 12 = 27 students w h o do
not take a bus.
F o u n d a t i o n s of Mathematics 12 S o l u t i o n s Manual 3-5
6. 17. a) Sets A and B are disjoint sets.
b) Sets /A and C intersect.
c) Yes; B and C; e.g., C intersecting A and /A and S
being disjoint says nottiing about the intersection, if any,
of 8 and C.
18. e.g.. The union of two sets is more like the addition of
two numbers because all t h e elements of each set are
counted together, instead of those present in both sets.
19. a) e.g., indoor, outdoor, races
b) e.g., U = {all sports}
/ = {indoor sports} = {badminton, basketball, curling,
figure skating, gymnastics, hockey, indoor soccer, speed
skating, table tennis, volleyball, wrestling, Arctic Sports,
Dene G a m e s }
O = {outdoor sports} = {alpine skiing, cross-country
skiing, freestyle skiing, s n o w s h o e biathlon, ski biathlon,
dog mushing, s n o w b o a r d i n g , snowshoeing. Dene
G a m e s }
R = {races} = {speed skating, alpine skiing, cross-country
skiing, biathlon, dog m u s h i n g , snowboarding,
snowshoeing}
c)
b^clrnii i k n i bdskotlj-all
curling hockuy
fuiure ^k:Jtln()
c)ynuui!>li(.h
iiKk>or sui.i,(.'r
'.vr-sliii!ij
Arc!i< Sp'.ifti
Dens?
Games
jlpin(> skiing
(ros'-'tountry skiiny
t'rcH'styk' skiiny
doq rTUishiny
snowlsoijrdiiiy
snowshoi'ifiy
Miowshoe bidtdlon
'ki ri.jirl'.•!:•,'•
d) Yes. e.g.. My classmate sorted the g a m e s as
individual, partner and t e a m games.
H i s t o r y C o n n e c t i o n , p a g e 1 7 5
A . e.g.. The "barber paradox" can be stated as follows:
Suppose there is one male barber in a small town, and
that every m a n in the town keeps himself clean-shaven.
S o m e do so by shaving thennselves and the others go to
the barber. So, the barber stiaves all the m e n w h o do not
shave themselves. Does the barber shave himself? The
question leads to a paradox: If he does not shave
himself, then he must abide by the rule and shave
himself. If he does shave himself, then according to the
rule he will not shave himself.
B. e.g.. O n e remarkable paradox that arises from
Cantor's work on set theory is the Banach-Tarski
theorem, which states that a solid, three-dimensional ball
can be split into a finite number of non-overlapping
pieces, which can then be p u t back together in a different
w a y to yield two identical copies of the original ball of the
same size.
M i d - C h a p t e r R e v i e w , p a g e 1 7 8
1. a) V c N, M d N, F c N, F cz M
b) e.g., N = {all foods}, V = {fruits and vegetables},
M = {meats}, F = {fish}
c) No. e.g.. Pasta is not part of M or V.
d) Sets V and M are disjoint, Sets V and F are
disjoint.
2. a)
') i-s M> .8 -lU
b) Sets F and S are disjoint sets.
c) i) False, e.g., 6 is in E but not F.
ii) True, e.g.. All elements of S are in E.
iii) False, e.g., 9 is not a multiple of 15.
iv) True, e.g., F = {15, 30}.
v) True, e.g., A set is a subset of itself
3. e.g., S = {summer sport equipment} = {baseball,
soccer ball, football, tennis ball, baseball glove,
volleyball net}
W = {winter sport equipment} = {hockey puck,
skates, skis}
B = {sports balls} = {baseball, soccer ball, football,
tennis ball, hockey puck}
E = {sports equipment worn on body} = {baseball
glove, skates, skis}
baseball
!;-o:h.:-l!
baseball glove
volleybdll net
Vl^ ki-y (ii.i k
'A.'tC'S
skis
4. a) beverage or soup: 40 - 5 = 35
beverage a n d soup: 34 + 18 = 52
overlap: 5 2 - 3 5 = 17
17 students bought a beverage and soup,
b) only beverage: 34 - 17 = 17
only soup: 1 8 - 1 7 = 1
18 students bought only a beverage or only soup.
3-6 C h a p t e r 3: S e t T h e o r y a n d L o g i c
7. c)
5. a) sunglasses or hat: 20 - 5 = 15
sunglasses and hat: 13 + 6 = 19
overlap: 1 9 - 1 5 = 4
4 students are wearing sunglasses and a hat.
b) only sunglasses: 1 3 - 4 = 9
9 students are wearing sunglasses but not a hat.
c) only hat: 6 - 4 = 2
2 students are wearing a hat but not sunglasses.
6. a) e.g., Tanya did not put any elements in the
intersection of A and B.
n{A u B) = n(L/) - n{{A u By)
n{AuB)= 4 0 - 8
n{A u 6 ) = 32
n{A nB) = n{A)+ n(B) - n{A u 6 )
n{AnB)= 1 6 + 1 9 - 3 2
n{A n 6 ) = 3
n{AB) = n{A)- n{A n B)
n{AB)= 1 6 - 3
n{AB)= 13
n ( B l 4 ) = n(B) - n{A n B)
n{BA) = 1 9 - 3
n{BA)= 16
b)
! S
7. Let ty represent the universal set. Let D represent the
set of students w h o have a dog. Let C represent the set
of students w h o have a cat.
n{C u D) = niU) - n{(C u D)')
n(C u D) = 20 - 4
n ( C u D ) = 16
n(C n D) = n(C) + n(D) - n{C u D)
n ( C n D) = 8 + 8 - 16
n(C n D) = 0
No students have a cat and a dog.
L e s s o n 3.4: A p p l i c a t i o n s o f S e t T h e o r y ,
p a g e 191
1. n(P) = p + 16, n(Q) = g + 2 1 , n(R) = r + 18
e.g., p Can be any number. Suppose p = 14. T h e n
n(P) = 30.
n(Q) = 30, so q = 30 - 21 or 9
n(P) = 30, so r 301 - 1 8 or 12
2. a) n ( ( F u M ) ^ ) = 9 + 1 5 + 8
n ( ( F u M ) ^ ) = 32
b) n{{A u F) A/f) = 9 + 11 + 7
n{{A uF)M) = 27
c) n((F u /) u (F u M))
= (9 + 11 + 7 + 9) + (15 + 8 + 4)
= 36 + 27
= 63
d) n{AFM) = 7
3. e.g., Staff could look at how many David Smiths
were on that bus route or they could look at the
books in the bag and see how many David Smiths
are taking courses that use those books.
4. P = {population surveyed}
n(P) = 641
L = {people wearing corrective lenses}
L' = {people not wearing corrective lenses}
n ( L ' ) = 167
G = {people wearing glasses}
C = {people wearing contact lenses}
n{L) = n{P)-n{n
n(L) = 641 - 1 6 7
n ( L ) = 474
n(G u C) = n{L)
n(G u C) = n{G) + n{C) - n(G n C)
474 = 442 + 83 - n{G n C)
51 = n(G n C)
51 people might make use of a package deal. This
51
is = 10.759...% or about 10.8% of all
574
potential customers.
5. e.g., "Canadian Rockies," "ski
accommodations," "weather forecast," "Whistler."
By combining two or more of these terms, Jacques
can search for the intersection of w e b pages
related to these terms. For example, "ski
accommodations" and "Canadian Rockies" is more
likely to give him useful information for his trip than
either of those terms on its o w n .
F o u n d a t i o n s of Mathematics 12 S o l u t i o n s Manual 3-7
8. 6. Using the principle of inclusion and exclusion for three
sets:
32 + 35 + 3 8 - ( 9 + x ) - ( 1 1 + x ) - ( 1 3 + x) + x = 5 8
1 0 5 - 9 - X - 1 1 - X - 1 3 - X + X = 5 8
7 2 - 2 x = 5 8
- 2 x = 5 8 - 7 2
- 2 x = - 1 4
X = 7
7 teens are training for the u p c o m i n g triathlon.
7. • s a m e numbers, s a m e s h a p e s , different shadings
• s a m e numbers, different s h a p e , different shading
• s a m e numbers, different s h a p e , different shading
o •
8. a) e.g., T h e dealer might use exterior colour, interior
colour, or year.
b) e.g.. The dealer might prioritize the search according
to options Travis wants or b y distance from where Travis
lives.
9. e.g., John a s s u m e d that 9 0 people ate at only one
restaurant for each of the 3 restaurants. He did not
calculate the correct n u m b e r of people eating at only one
of each of the 3 restaurants.
I defined these sets.
C = {students w h o like only Chicken and More
F = {students w h o like only Fast Pizza}
G = {students w h o like only Gigantic Burger}
I listed the values I knew a n d entered t h e m in a V e n n
diagram.
n ( C n P B ) = 37; n { C n e P ) = 19; n ( P n S C ) = 11
n{CnBnP)= 13
Chicken Fast Pizza
and More
Gigantic Burger
I used these figures and diagram to determine the
u n k n o w n values.
n(C B P) = 90 - n(C n P e ) - n{C n B P)
-n{CnBnP)
= 9 0 - 3 7 - 1 9 - 1 3
= 21
n(B P C) = 90-n{CnBP)-n{PnB C)
-n{CnBnP)
= 9 0 - 1 9 - 1 1 - 13
= 47
n ( P B C ) = 90-n{PnBC)-n{CnPB)
-n{CnBnP)
= 9 0 - 1 1 - 3 7 - 13
= 29
Fast Pizza
Gigantic Burger
n ( P ) = 21 + 29 + 4 7 + 37 + 19 + 11 + 13
n(R)= 177
177 students like at least one of these restaurants.
2 4 0 - 177 = 63
So, 63 students do not like any of the restaurants.
10. a) e.g., He can search for colleges and
(Calgary or Edmonton).
b) He should use "and" to connect the words.
c) He should use "or" to search for one or the other
city.
d) e.g., colleges and (Calgary or Edmonton) and
"athletics programs" -university
e) e.g., about 1500
f)
.iii',.ac'>
F..c!nionton
'Athle'if. I'rocjiain,')
3-8 C h a p t e r 3: S e t T h e o r y a n d L o g i c
9. 11. Set 1: different numbers, different colours, different
shading, different shape
Set 2: different numbers, different colours, different
shading, s a m e shape
Set 3: different numbers, different colours, different
shading, same shape
Set 4: different numbers, s a m e colour, same shading,
different shape
Set 5: different numbers, s a m e colour, different shading,
same shape
Set 6: s a m e number, different colours, different shading,
different shape
Set 1:
O O
14. e.g.. No, they did not get the s a m e results.
Elinor got all of James' results, plus others dealing
with either string or bean, but not both.
)i .-in ,' -.tiinn
15. a) and b) e.g..
Set 2: m
Set 3:
A A
Set 4:
A O O • :
Sets:
A A A
Set 6:
A A O
12. a) n(D), the total number of cards in the deck: there
are 3 shapes, 3 colours, 3 numbers, and 3 shadings, so
in total there are 3 3 3 3 or 81 cards.
b) n(T), the total number of triangle cards in the deck:
there are 3 colours, 3 numbers, and 3 shadings, so in
total there are 3 • 3 • 3 or 27 triangle cards.
c) n(G), the total number of green cards in the deck: 3
shapes, 3 numbers, and 3 shadings, so in total, there are
3 3 3 or 27 green cards.
d) n{S), the total number of cards with shading: there are
27 cards with striped shading and 27 cards with solid
shading, so there are 27 + 27 or 54 cards with shading.
e) n ( 7 u G): there are 27 triangle cards and 27 green
cards, but 9 triangle cards are also green, so there are
54 - 9 or 45 cards that have triangles or are green.
f) n{G n sy. there are 27 green cards. Since 2/3 of the
cards have either striped shading or solid shading, 18
cards are both green and have shading.
13. a) 36 sites would appear in a search for fishing
boats. There are 35 sites that involve boats, 20 of which
deal with fishing boats. 21 sites involve fishing, but these
sites include the 20 sites that deal with fishing boats.
b) e.g.. Because fishing and boats will turn up sites that
deal with boats and fishing, but not just fishing boats.
c) 20 o f t h e 21 fishing sites deal with fishing boats, so 1
site would not have boats.
c) e.g., 1 = A{Bu Cu D)
2 = B{Au Cu D)
3 = C{Au Bu D)
4 = D{Au Bu C)
'5 = iAnB){CuD)
6 = iAnC){Bu D)
7 = {AnD){BuC)
8 = iBnC){Au D)
9 = {BnD){Au C)
^0 = {CnD){Au B)
n ={AnBnC)D
^2 = {AnBnD)C
^3 = {AnCnD)B
U = {BnCnD)A
15 = AnBnCn D
16. e.g., Let B = {blue}, y = {yellow}, R = {red}, and
G = {green}. There is no area representing
( e n R) (G u Y) or (G n T) ( e u R).
F o u n d a t i o n s of Mathematics 12 S o l u t i o n s Manual 3-9
10. M a t h in A c t i o n , p a g e 1 9 4
e.g..
• I decided to researcti texting in relation to driving safely.
S e a r c h W o r d s Number of Hits
texting and driving 3 830 000
texting while driving 976 000
"texting while driving" 599 000
"texting while driving in
Canada"
8
• I had w a y too m a n y hits for texting and driving. I figured
out that the issue is texting while driving, so I tried that
combination. Putting quotes around it netted even fewer
results. Since I live in Canada, I w a s interested in what's
happening here, so I added Canada to my search. T h e n
I tried "texting while driving in Canada". That really cut
d o w n the hits to a manageable number.
Let T represent "texting while driving" sites, and C
represent Canada sites. The overlap of the two circles
represents the sites that contain both "texting while
driving" and C a n a d a .
• The search engine's A d v a n c e d Search feature allows
you to exclude any sites that contain certain words from
your search.
L e s s o n 3.5: C o n d i t i o n a l S t a t e m e n t s a n d T h e i r
C o n v e r s e , p a g e 2 0 3
1. a) Hypothesis, p = I a m swimming in the ocean.
Conclusion, q = I a m s w i m m i n g in salt water.
b) Yes, the conditional statement is true, because all
oceans contain salt water.
c) Converse: If I a m swimming in salt water, then I a m
swimming in the ocean.
The converse is false, because I could be swimming in a
salt-water pool, or a salt-water lake.
2. a) Yes, the conditional statement is true. Four is
divisible by 2, so any number that is divisible by 4 is also
divisible by 2.
b) Converse: If a number is divisible by 2, then it is
divisible by 4. The converse is false.
c) e.g., A counterexample o f t h e converse is the number
2, which is divisible by 2, but not 4.
3. a) If a triangle is equilateral, then it has 3 equal sides.
b) If a triangle has 3 equal sides, then it is equilateral.
c) Both statements are true, because the definition of an
equilateral triangle is a triangle that has 3 equal sides.
d) Yes, the statement is biconditional, because both the
conditional statement and its converse are true.
4. a) If w e cannot get what w e like, then let us like
what w e get.
b) Hypothesis: W e cannot get what w e like.
Conclusion: Let us like what w e get.
5. a) The statement is false. A counterexample is
the number 25. It is divisible by 5, but it does not
end in a 0.
b) If a number ends in a 0, then it is divisible by 5.
c) The converse is true. The V e n n diagram s h o w s
that all multiples of ten are also multiples of 5, but
not all multiples of 5 are multiples of 10.
I j>t Clltilt K 0
L'iViMt.iie by ;S
.... "3, i:x
6. a) The conditional statement is true, because
Canada is in North America. T h e converse is false.
Counterexample: You might live in Mexico and still
be in North America. The statement is not
biconditional.
b) The statement is true, because Ottawa is the
capital of Canada. The converse is also true.
Biconditional statement: Y o u live in the capital of
Canada if and only if you live in Ottawa.
7. Biconditional, e.g.,
X is not
V x ^ = X => X is
not negative
V A — A
negative
V x ^ = X => X is
not negative
true true true
false false true
false true true
true false false
Both the conditional statement and its converse
are always true, so the statement is biconditional.
T h e statement can be written as: V x ^ = x if and
only if X is non-negative.
8. a) Conditional statement: If a glass is half-
empty, then it is half full. This statement is true.
Converse: If a glass is half full, then it is half-
empty. The converse is true. The statement is
biconditional, because both the conditional
statement and its converse are true.
Biconditional statement: A glass is half-empty if
and only if it is half full,
b) Conditional statement: If a polygon is a
rhombus, then it has equal opposite angles. The
statement is true.
Converse: If a polygon has equal opposite angles,
then it is a rhombus. The converse is false.
Counterexample: A rectangle has equal opposite
angles. The statement is not biconditional.
3-10 C h a p t e r 3: S e t T h e o r y a n d L o g i c
11. c) Conditional statement: If a number is a repeating
decimal, then it can be expressed as a fraction. The
statement is true.
Converse: If a number can bo expressed as a fraction,
then it is a repeating decimal.
The converse is false. Counterexample; The decimal
number 0,3 can be expressed as the fraction ^
10
but 0.3
is not a repeating d e c i m a l
The statement is not biconditional
9. a | Conditional statement; If AB and CD are
parallel, then the alternate angles are e q u a l
Converse; If the alternate angles are e q u a l then and
CD are parallel
b| Proof of conditional statement:
I drew two lines crossed by a transversal and numbered
the angles as shown.
//
/
- /
^ i -
l_ _
AlU'rn.ii.!' rfi-fjiot.
eQy.ll.
It
h
(-iven.
Al'f-rnate angles.
Z 2 and ,^1 ciEc
supplementary.
niOy form a straight
line.
_ 4 A'^vl e
supplementary.
They form a straight
line.
Z I = Z 3 Supplements of equal
angles are also equal.
AB II CD Corresponding angles
are e q u a l
Therefore, the conditional statement is true.
Proof of converse:
1 used Inti i-iwn' di-it/.-iin
AB CO Gi'en.
Z I = , lines are parallel,
corresponding angles
are equal.
Z 2 and Z I are
supplementary.
They form a straight
line.
Z 4 and Z 3 are
supplementary.
They form a straight
line.
Z I = Z 3 Supplements of equal
angles are also equal.
' 1 and . 3 (Hi
alternate angles.
Therefore, the converse is true.
10, a) Converse; If your pet is a dog, then it barks.
The statement and its converse are true, so the
statement is biconditional
b) Converse; If your pet w a g s its tail, then it is a
dog.
The converse is false. A cat w a g s its t a i l The
statement is not biconditional
11. a) True.
x + y
X
b) True,
P - Q
P • Q + Q
P
= z
= z-y
= z-y
' ' 'I
= r+q
12. e.g., If a number appears in the same row,
column, or large square as the shaded square,
then it is not in the shaded square. The numbers 1,
4, 5, 6, and 8 must go in column 4, If I were to put
1, 4, 5, or 8 in this square, then I could not put 6 in
any other square in column 4, I conclude that 6 is
the only number that can go in this square. As a
result, 5 can only go in the square above. 4 can
only go in the square below. 8 can only go in the
top square, and 1 must go in the remaining square.
The numbers in the column should be. from top to
bottom: 8. 3, 9, 5, 6, 4. 1. 7, 2.
13. a) i) If a figure is a square, then it has four nght
angles.
li) If a figure has four nght angles, then it is a
square.
ill) The statement is true. The converse is false.
The figure could be a rectangle.
iv) The statement is not biconditional
b) i) If a triangle is a nght tnangle, then a' + b' = c
ii) If, for a triangle, + = c^, then it is a right
tnangle.
iii) The statement is true. The converse is true,
iv) A triangle is a right tnangle if and only if
+b' =
c) i) If a quadnlateral is a trapezoid, then it has two
parallel sides.
ii) If a quadrilateral has two parallel sides, then it is
a trapezoid.
iii) The statement is true. The converse is false, A
regular hexagon has two sides that are parallel
iv) The statement is not biconditional
c) The onginal statement is true, because both the
statement and the converse are true, so the statement is
biconditional
F o u n d a t i o n s of Mathematics 12 S o l u t i o n s Manual 3-11
12. 14. Use the finance application on a calculator. Note:
Mortgages are c o m p o u n d e d semi-annually in C a n a d a .
a) i) The number of payments is 25 • 12 or 300.
T h e interest rate is 6.5%.
T h e present value is $250 000.
The payment amount is unknown.
The future value is $0.
The payment frequency is 12.
T h e c o m p o u n d i n g frequency is 2.
They should pay $1674.559... or $1674.56 per month.
ii) The number of payments is 25 • 24 or 600.
The interest rate is 6.5%.
The present value is $250 000.
The payment amount is unknown.
T h e future value is $0.
The payment frequency is 24.
The compounding frequency is 2.
They should pay $836.163... or $836.16 bi-monthly.
b) 2 payments/week • 52 weeks/year = 104
The number of payments is unknown.
The interest rate is 6.5%.
The present value is $250 000.
The payment amount is $836.16 - 4 = $209.04.
The future value is $0.
The payment frequency is 104.
The c o m p o u n d i n g frequency is 2.
They will make 2164.088... or 2164 mortgage payments.
If Michelle and Marc m a k e one payment each month for
300 months, they will pay
$1674.56 300 = $502 368 in total.
If they pay two payments each month for 300 months,
they will pay
$836.16 600 = $501 696 in total.
If they make 2164 payments of $209.04, they will pay
2164 • $209.04 = $452 362.56 in total.
They will save
$502 368 - $452 362.56 = $50 005.44 by paying more
frequently, so they should do that if they can.
15. e.g., a) M: If it is December, then it is winter.
U: If a number is even, then it is divisible by 2.
b) Let W represent winter, and D represent December.
Let £ represent even numbers, and D represent being
divisible by 2.
c) e.g.. If the sets are the same (i.e., there is one area in
the V e n n diagram), then the converse is true. If there are
two or more areas in the V e n n diagram, then the
converse is false.
16. a) e.g.. If the first letter is a consonant, the
second letter is a vowel.
b) E is the most c o m m o n letter used in the English
language. X is very frequent in the puzzle.
Substitute
J = A and X = E.
E
K S Q Q S C A X H B M V T D T Y
A E E
D K J D C S N S A U C X X A
A A E
Q J T D J Y T C L V X
E E E
P S P X C D N X B S H X
A E
Y D J H D T C L D S T P Z H S W X
E
D K X Q S H V A .
A E A
- J C C X B H J C G
The last two words are someone's name. W h a t
name starts with A, has the same two letters, and
then ends with E? Anne. Substitute C = N.
N E
K S Q Q S C A X H B M V T D T Y
A N N E E
D K J D C S N S A U C X X A
A A N E
Q J T D J Y T C L V X
E N E E
P S P X C D N X B S H X
A N E
Y D J H D T C L D S T P Z H S W X
E
D K X Q S H V A .
A N N E A N
- J C C X B H J C G
W h a t could N E E _ be? Need. There are three
vowels left: I, O, and U. Two-letter words usually
have a I or an O. If T w a s a vowel, it would
probably be an I rather than an O. Substitute A = D
and try T = I.
N D E I I
K S Q Q S C A X H B M V T D T Y
A N D N E E D
D K J D C S N S A U C X X A
A I A I N E
Q J T D J Y T C L V X
E N E E
P S P X C D N X B S H X
A I N 1 E
Y D J H D T C L D S T P Z H S W X
E D
D K X Q S H V A .
A N N E A N
- J C C X B H J C G
3-12 C h a p t e r 3: S e t T h e o r y a n d L o g i c
13. i r-«i fi • I - < u s c d . Try " i f and
'•, ' . > . J ' r . * . i i | l - ; ) .sM-t { S
I T I S
V s . , I , x 1! ;: f.. y I D T Y
i , ^ . E D
I. • 1 ^ : !. • n i; f ... /.
••J ' i I* J Y I i i V >.
. F- • . : • ;-J X ( !
i ^ ' : E
/ .11 : i i : : L '•• I /: H s w X
r) i- > ••/ A
f '. ' s i ' . U • ; r .• :.•',( .-l^v ni'^/'S'- • Af S
t i - . ' V,",., I ^ < ; i , - ' .= i b - . | . t . ; t > - S = O and
K = H.
' • • • • ' ^. • I s
V :J l l (, A X i i b M ^/ 1 i Y
D K ,1 Li ( .> fl A U c. >, X h
0 J I D f f; L y /•
p f. p X c l l rj X b P. H X
V D J H n r f ! ri L. r p / n vv x
K X O h •/ A
J r c /. ^ , . .
H O ^ is probably "how," so ^ A I T is "wait." which makes
sense . J I -v'T uses the s a m e letter twice. All I can think
of is "moment." STA^TIN^^ probably ends in "ing." Try
Q = W, P = M <"rn t .
H O W W O N D E I T I S
K S O O S sX A X H B M V 1 D "1 Y
D K J U C S N S A U G X X A
O J 1 D J Y T C L V X
P S P X C D N X B S H X
S T A T I N G T O I y O E
Y D J H D T C L D S T P Z H S W X
D K X Q S H V A .
A N N E A N
- J C C X B H J C G
- > I M f I X r -Kihr;..- XX(. M A TING is
: ifiMHj H . H v V o N i i i _ : " l i i o De "wonderful."
X-iljs]i»i;l' X i P P I ! . M = U,
H O W W O N D E R F U L I T I S
K X .J (J ; (, A X H s M v i y T Y
h K .1 r t. X, /t pi c X X A
i j I I I ) .1 f f s . L V /
P X P X t; P N X H f. H >'
• • : • R O E
' P 1 P 1, i L P> G I P Z H S W X
P K X i } I , H V A
J { i ; X p, H J (;
F R A N ^ is "Frank." ^ E F O R E is "before." So
NO O D could be "nobody " W h a t letters are left?
C, J, P, Q, V, X. Z, IM^RC^ L could be "improve,"
"iutiolilut,; (J K.
N p. IJ V X - P and VV -
H O W W O N D E R F U L I T I S
K X (J u X X A X H B M ^ I D T Y
P P ; f: X. -. IJ i^, / 1^ < X < A
X» J r PI I V 1 C i V X
P X X C iJ X B S H X
' i ' I h i l I P L Is X T P Z H S W X
IJ K X u G ! l V A
K
J C (; X B H J C G
17. a) Use the finance application on a calculator.
Ihe number of payments is unknown.
T h e interest rate is 4 % .
The present value is $265 233.48.
The payment amount is $1400 + $250 or $1650,
T h e future value is $0,
T h e payment frequency is 12.
T h e compounding frequency is 12.
It will take them 230.631 or 231 months to pay off
the mortgage.
b) At $1400/month; $1400 • 300 = $420 000
At $1650/month: $1650 ^ 231 = $381 150
$420 000 - $ 351 150 = $38 850
They will save $38 850 over the life of the
mortgage by paying $1650 per month instead of
$1400.
F o u n d a t i o n s o f M a t h e m a t i c s 12 S o l y t i o n s M a n u a i 3-13
14. A p p l y i n g P r o b l e m - S o l v i n g S t r a t e g i e s , p a g e 2 0 7
C . e.g., In this solution, squares are numbered from 1 to
9, from the top left to the bottom nght, as on the numehc
pad of a telephone. First, I examined the coloured
squares. I know from the first clue that either square 5 or
6 must be blue. T h e second clue tells me that either
square 6 or 9 is blue. The fifth clue tells me that either
square 7 or 8 is blue. So I know that either 7 or 8 must
be blue, and one or two of 5, 6, and 9 must be blue.
Also, either square 4 or 5 is red, either square 6 or 9 is
red, and square 2 or 5 is yellow.
Since squares 6 and 9 are red or blue, they cannot be
yellow.
I know that there is a heart in either square 1 or 2, and
two hearts in either 2 and 4, or 3 and 5. The yellow heart
is in either 2 or 3. All of the hearts are in the first two
rows.
I decided to begin to place the colours and symbols,
knowing that I might need to move them around. I put
hearts in squares 1, 3, and 5.
Since the ttiree hearts must be three different colours, I
think that square 5 will be a blue heart, which makes
square 1 a red heart, because the blue colours appear to
be in the second and third rows.
red
heart
yellow
heart
blue
heart
red or
blue
red or
blue
I know that either square 4 or 5 must be red. Since there
is a blue heart in square 5, square 4 must be red, which
makes square 7 blue, according to the fifth clue.
The sixth clue indicates that either square 1 or 4 is a
star. Since I have a heart in square 1, this means that a
star must be in square 4; so, it is a red star.
I now have three red squares and three blue squares, so
the remaining two squares, square 2 and 7, must be
yellow.
red
heart
yellow
yellow
heart
red
star
blue
heart
red or
blue
blue yellow
red or
blue
The fourth clue tells me that either 3 or 6 is a
pentagon. Since I already have a heart in square
3, square 6 must contain the pentagon. This clue
also tells me that a star must be in square 8, and
since I know square 8 is yellow, it is a yellow star.
T w o yellow shapes have been placed. The yellow
pentagon belongs in square 2.
red
heart
yellow
pentagon
yellow
heart
red
star
blue
heart
red or
blue
pentagon
blue
yellow
star
red or
blue
I used the sixth clue to determine whether square
9 is red or blue. The star is in square 4, so square
9 must be red. T h e only missing red shape is the
pentagon, so it is a red pentagon, and square 6 is
the blue pentagon. Finally, the blue star belongs in
square 7.
red yellow yellow
heart pentagon heart
red blue blue
star heart p e n t a g o n
blue yellow red
star star pentagon
I double-checked my clues. My puzzle is correct.
Solution:
KV
yellow ?
, . _ J
^ yp||(..)vv /
/
A ,
hi
blue / { blue 1
J
X y e l l o v ^ '
y 'A
/
/
3-14 C h a p t e r 3: S e t T h e o r y a n d L o g i c
15. D. The solution is; Inverse; If a quadrilateral is not a square, then its
diagonals are not perpendicular,
Contrapositive; If the diagonals of a quadrilateral
are not perpendicular, then it is not a square,
cl) Converse: If 2n is an even number, then n is a
natural number.
Inverse; If n is not a natural number, then 2ri is not
an even number.
Contrapositive; If 2n is not an even number, then n
is not a natural number.
2. a) Converse: If an animal is a giraffe, then it has
a long neck.
Contrapositive; If an animal is not a giraffe, then it
does not have a long neck.
b) No. e.g., Ostriches and llamas have long necks,
so the contrapositive is not true.
3. a) Converse; If a polygon is a pentagon, then it
has five sides.
Inverse; If a polygon does not have five sides, then
it is not a pentagon.
b) Since pentagons are the only shapes with
5 sides, both of these statements are true. They
are logically equivalent.
4. a) I do not agree with Jeb. If / = 25, then
X = 5 or X = - 5 .
b) Converse; If x = 5. then x^ = 25, This statement
it true,
c) Inverse; if / ^ 25, then x ^ 5, This statement is
true,
d) Contrapos!t!ve:1f x t 5, then / # 25. This
statement is not true, because x could equal 5, and
/ would still equal 25.
F. To make the puzzle easier, I could give more clues
where both the colour and shape are given. Or, I could
give the shapes in a diagonal, or a group of colours or
shapes that would show one square in every row and
column
To make the puzzle harder. I could not give any clues
With both the colour and the shape, or I could make the
pieces smaller or without angles so there are more
possibilities for their location in the 3 by 3 grid.
L e s s o n 3.6: T h e I n v e r s e a n d t h e C o n t r a p o s i t i v e
o f C o n d i t i o n a l S t a t e m e n t s , p a g e 2 1 4
1. a) Converse: If you are looking in a dictionary, then
you will find success before work.
Inverse; If you do not find success before work, then you
are not looking in a dictionary.
Contrapositive: If you are not looking in a dictionary, then
you will not find success before work.
b) Converse: If you can drive, then you are over 16.
Inverse: If you are not over 16, then you cannot drive,
Contrapositive: If you cannot dnve, then you are not over
16.
c) Converse: If the diagonals of a quadnlateral are
perpendicular, then it is a square.
5. a) I) I ne statement is true.
ii) Converse; If you are in Northwest Terntones,
then you are in Hay River,
The converse is false. You could be in another city
or town in Northwest Territories, for example,
Yellowknife.
iii) Inverse: If you are not in Hay River, then you
are not in the Northwest Terntones.
The inverse is false. You could be in Norman
Wells, Northwest Territories for example.
iv) Contrapositive: If you are not in the Northwest
Terntones, then you are not in Hay River.
The contrapositive is true.
b) i) The statement is true. A puppy is either male
or female.
ii) Converse: If a puppy is not female, then it is
male.
The converse is true.
iii) Inverse: If a puppy is not male, then it is female.
The inverse is true,
iv) Contrapositive: If a puppy is female, then it is not
male.
The contrapositive is true.
c) i) The statement is true.
ii) Converse: If the Edmonton Eskimos are number 1
in the west, then they w o n every g a m e this season.
F o u n d a t i o n s of Mathematics 12 S o l u t i o n s Manual 3-15
16. T h e converse is false. T o be number 1, they must win
more g a m e s than the other western teams, but they do
not have to win t h e m all.
iii) Inverse: If the Edmonton Eskimos did not win every
g a m e this season, then they are not number 1 in the
The inverse is false. T h e y m a y have w o n more g a m e s
than the other western t e a m s and would still be number
1.
iv) Contrapositive: If the Edmonton Eskimos are not
number 1 in the west, then they did not win every g a m e
this season.
The contrapositive is true.
d) i) The statement is false. The integer could be 0. Zero
is neither negative nor positive.
ii) Converse: If an integer is positive, then it is not
negative.
The converse is true.
iii) Inverse: If an integer is negative, then it is not
positive.
The inverse is true.
iv) Contrapositive: If an integer is not positive, then it is
negative.
The contrapositive is false. The integer could be 0.
6.
Conditional Statement
Invers'
. C o n v e r s e _ _
C o n t r a p o s i t i v e
a) 1 b)
T
1
F 1 1
T j | T
c) ' d)
JY r f
F_ ^ f
F 1
7 a) If the - l a t o m o n ! is true, the contrapositive -s ahr.
hill- If th.; slateriHjnt is false, Ihr- contrapositive i-^- ak-(^
b) If the inverse is true, the converse is also true. If the
inverse is false, the converse is also false.
The pairs of statements are logically equivalent.
8. a) No, I cannot draw a conclusion about the
conditional statement and its converse. There is no
relationship between the two statements.
b) No, I cannot draw a conclusion about the inverse and
the contrapositive. There is no relationship between the
two statements.
9. a) Converse: If a polygon is a quadnlateral, then it is a
square.
Inverse: I f a polygon is not a square, then the polygon is
not a quadnlateral.
Contrapositive: If a polygon is not a quadnlateral. then it
is not a square.
b) The conditional statement is true. Every square is a
quadnlateral by definition.
The converse is false. A counterexample is a
parallelogram, which is not a square, but is a
quadnlateral.
The inverse is false. A counterexample is a rectangle,
which is a quadnlateral, but is not a square. The polygon
could be a rectangle, which is not a square, but is a
quadnlateral.
The contrapositive is true. If a polygon is not a
quadrilateral, then it cannot be a square.
10. a) Converse: If a line has a y-intercept of 2,
then the equation of this line is y = 5x + 2.
Inverse: If the equation of a line is not y = 5x + 2,
then its y-intercept is not 2.
Contrapositive: If a line does not have a y-intercept
of 2, then the equation of this line is not y = 5x + 2
b) The onginal statement is true, because the
y-intercept of that line is 2.
The converse is not true. If a line has a y-intercept
of 2. its equation could be y = 2. or infinitely other
equations.
The inverse is also not true. A line could not have
that equation, and still have a y-intercept of 2. For
example, the equation y = x + 2 has a y-intercept
of 2.
The contrapositive is true, because if a line does
not have a y-intercept of 2, it cannot have that
equation.
11. e.g.. If a conditional statement, its inverse, its
converse and its contrapositive are all true, I know
the conditional statement is biconditional.
12. a) i) e.g , The statement is true. Pins can burst
balloons.
ii) Converse: If a pin can burst the M o o n , then the
Moon is a balloon.
The converse is false, e g , The M o o n could be a
soap bubble, for example.
iii) Inverse: If the Moon is not a balloon, then a pin
cannot burst the Moon.
The inverse is false, e.g.. Again, the M o o n could
be a soap bubble.
iv) Contrapositive: If a pin cannot burst the Moon,
then the Moon is not a balloon, e g.. T h e
contrapositive is true.
b) i) The statement is true. The negative of a
negative number is a positive number,
ii) Converse: If x is a positive number, then x is a
negative number.
The converse is true. The negative of a positive
number is a negative number.
iii) Inverse: If x is not a negative number, then -x
is a not positive number. The inverse is true.
iv) Contrapositive: If - x is not a positive number,
then X IS not a negative number. The
contrapositive is true,
c) i) The statement is true.
ii) Converse: If a number is positive, then it is a
perfect square. The converse is false. 3 is a
positive number, but it is not a perfect square.
iii) Inverse: If a number is not a perfect square,
then it is not positive. The inverse is false. 3 is not
a perfect square, but it is positive.
iv) Contrapositive: If a number is not positive, then
it is not a perfect square.
The contrapositive is true. Negative numbers
cannot be perfect squares.
3-16 C h a p t e r 3: S e t T h e o r y a n d Logic
17. d) i) The statement is true.
ii) Converse: If a number can be expressed as a fraction,
then it can be expressed as a terminating decimal.
The converse is false. For example, - , written as a
decimal, is 0.333... This is a repeating decimal.
iii) Inverse: If a number cannot be expressed as a
terminating decimal then it cannot be expressed as a
fraction.
The inverse is false. For example, 0.333... is a repeating
decimal. It is also — .
3
iv) Contrapositive: I f a number cannot be expressed as a
fraction, then it cannot be expressed as a terminating
decimal. The contrapositive is true.
e) i) This statement is true
ii) Converse: If a graph is a parabola, then the equation
of this parabola is f{x) = 5x^ + 10x + 3.
This statement is false, because there are many
parabolas that do not have that equation, such as f(x) =
iii) Inverse: If the equation of a function is not
f(x) = 5x^ + 10x + 3, then its graph is not a parabola.
This statement is false. For example, a function can have
the equation f{x) = x^, yet it is a parabola.
iv) Contrapositive: If a graph is not a parabola, then the
equation of this parabola is not f(x) = 5x^ + lOx + 3.
This statement is true, because only a parabola can
have that equation.
f) i) This statement is false. For example, - 1 is an
integer, but not a whole number.
ii) Converse: If a number is a whole number, than it is an
integer.
This statement is true,
iii) Inverse: If a number is not an integer, than it is not a
whole number.
This statement is true.
iv) Contrapositive: If a number is not a whole number,
than it is not an integer.
This statement is false, for example - 1 is not a whole
number, but it is an integer.
13. a) e.g.. The contrapositive a s s u m e s as its hypothesis
that the original conclusion is false, which m e a n s that the
onginal hypothesis must also not be true. If the onginal
hypothesis is not true, then the conditional statement
must be false.
b) e.g., The converse of a conditional statement is
formed by stating the conclusion before the hypothesis.
The inverse is formed by negating the hypothesis and
conclusion of a conditional statement. Since negating
both parts of the statement is the same as reversing
them, the converse and inverse are logically equivalent.
The inverse of a statement is the contrapositive of the
statement's converse.
14. e.g., a) Conditional statement: If you are tall, then
you like chocolate.
Contrapositive statement: If you do not like chocolate,
then you are not tall.
Counterexample: I a m tall and do not like
chocolate. Both the conditional statement and the
contrapositive are false.
b) Conditional statement: If a traffic light is green, it
is not red. Contrapositive: If a traffic light is red, it is
not green. Both the conditional statement and the
contrapositive are true.
15. e.g., a) Conditional statement: If it is Saturday,
then it is the w e e k e n d .
Inverse: If it is not Saturday, then it is not the
w e e k e n d . The inverse is false. Counterexample: it
could be Sunday and be the w e e k e n d .
Converse: If it is the w e e k e n d , then it is Saturday.
The converse is false. Counterexample: it could be
the w e e k e n d and be Sunday.
b) Conditional statement: If a polygon has six
sides, then it is a hexagon.
Inverse: If a polygon does not have six sides, then
it is not a hexagon. The inverse is true by
definition.
Converse: If a polygon is a hexagon, then it has six
sides. The converse is true by definition.
C h a p t e r S e l f - T e s t , p a g e 2 1 7
1. Let L/ represent the universal set of writers. Let
P represent the set of poets. Let N represent the
set of novelists, and let F represent the set of
fiction whters.
Subset fiction writer, F = {Armand Ruffo, Richard
Van C a m p }
p i s i l ^ ^ M f l ^ l l i i i M
A j i n . i n d Rijfi;:
f;< b.-iid V.ui ,inip
2 ^
} 14 1 1- 17 •. 1^ -• 21 22. • : 24 i )
Set C is inside set A, therefore C czA.
b)A = { 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10, 1 1 , 12}
B = { 1 , 2 , 3 , 4 , 5, 6, 7, 8}
A u e = { 1 , 2 , 3, 4, 5, 6, 7 , 8 , 9, 10, 1 1 , 12}
F o u n d a t i o n s of Mathematics 12 S o l u t i o n s Manual 3-17
18. niAuB)=M
C = { 1 , 2 , 3 , 4 , 5, 6}
A n C = = { 1 , 2 , 3, 4, 5, 6}
n{A n C) = 6
c) A n B = { 1 , 2 , 3 , 4 , 5, 6, 7, 8}
AnBC = {7,8}
d) A u 8 u C = { 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
(yA u S u Cy is all the elements not in / u S u C.
( A u e u C ) ' = {13, 14, 15, 16, 17, 18, 19, 2 0 , 2 1 , 2 2 , 23,
24}
3. Let U represent the universal set. Let W represent the
students w h o dhnk bottled water. Let L represent the
students w h o follow a low fat diet. Let F represent the
students w h o eat fruit.
W e know 1 5 % of students d o all three, so that number is
the three-way intersection.
W e know 2 2 % dhnk bottled water and follow a low-fat
diet, so n(L u l ^ / F) = 22 - 15 = 7.
Similarly, n(lA^u F / L ) = 27 - 15 = 12 and
n ( L u F / l V ) = 2 3 - 1 5 = 8.
From here, I know 5 0 % of t h e students dhnk bottled
water, 5 6 % eat fruit, and 4 3 % follow a low-fat diet.
U ^ / L u F = 5 0 - 1 5 - 7 - 1 2 = 16
F / W u L = 5 6 - 1 5 - 1 2 - 8 ^ 2 1
L / l 4 ^ u F = 4 3 - 1 5 - 7 - 8 = 13
T o determine the percent of students w h o do not dhnk
bottled water, eat fruit or follow a low-fat diet w e need {W
u F u L ) ' .
( H / u F u L ) = 1 6 + 1 3 + 2 1 + 1 2 + 7 + 8 + 1 5 = 92
( W u F u L ) ' = 1 0 0 - 9 2 = 8
Therefore, 8 % of students d o not dhnk bottled water, eat
fruit or follow a low-fat diet.
4. a) Conditional statement: If you want to win an
election, then you must get t h e most votes.
Inverse: If you do not want t o win an election, then you
must not get the most votes.
The statement is not biconditional; e.g., in s o m e electoral
systems, you need a majority to win.
b) Conditional statement: If the planet is Earth, then it is
the third planet from the S u n .
Inverse: If the planet is not Earth, then it is not the third
planet from the Sun.
The statement and inverse a r e true so this is a
biconditional statement.
The planet is Earth if and o n l y if it is the third planet from
the Sun.
c) Conditional statement: If a number is between 1 and
2, then it is not a whole number.
Inverse: I f a number is not between 1 and 2, then it is a
whole number.
The statement is true but the inverse is false.
Counterexample: 0.75 is not a whole number but it is
less than one. The statement is not biconditional.
5. a) i) Conditional statement: If you are over 18, then
you are an adult. This statement if false, e.g., The age of
majority in British Columbia is 19.
ii) Converse: If you are an adult, then you are over
18. This statement is true.
iii) Inverse: If you are not over 18, then you are not
an adult. This statement is true.
iv) Contrapositive: If you are not an adult, then you
are not over 18. This statement is false, e.g., since
the age of majority in British Columbia is 19 the
statement would not hold true for an 18-year-old.
b) i) Conditional statement: If you are 16, then you
can drive. This statement is false, e.g., a 44-year-
old may know how to drive.
ii) Converse: If you can drive, then you are 16.
This is false.
iii) Inverse: If you are not 16, then you cannot
drive. This statement is also false.
iv) Contrapositive: If you cannot drive, then you
are not 16. This statement is false, e.g., a 16-year-
old may know how to drive.
C h a p t e r R e v i e w , p a g e 220
1 a )
• ' i! 1 ^ 1/ '9 u I
- : ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ :
b) Sets F and S are disjoints sets.
c) Yes, set F is a subset of set E.
d) S = {3, 6, 9, 12, 15, 18, 2 1 , 24, 27, 30}
S' = { 1 , 2, 4, 5, 7, 8, 10, 1 1 , 13, 14, 16, 17, 19, 20,
22, 23, 25, 26, 28, 29} = {natural numbers from 1
to 30 not divisible by 3}
Set S' is different from set E' because it includes
numbers that are not divisible by two and set E'
only includes numbers not divisible by two.
e) e.g., H = {multiples of 50}
2. a) black hair or blue eyes: 28 - 9 = 19
Since 19 students have black hair, all students with
blue eyes have black hair.
8 students have black hair and blue eyes.
b) only black hair: 1 9 - 8 = 11
11 students have black hair
c) only blue eyes: 8 - 8 = 0
No students have blue eyes but not black hair.
Z.a)A = { - 1 2 , - 9 , - 6 , - 3 , 0, 3, 6, 9, 12}
e = { x | - 1 2 < x < 1 2 , X € 1}
A u e = { x | - 1 2 < x < 12, X G 1}
= { - 1 2 , - 1 1 , - 1 0 , - 9 , - 8 , - 7 , - 6 , - 5 , - 4 , - 3 ,
- 2 , - 1 , 0 , 1,2, 3, 4, 5, 6, 7, 8, 9, 10, 1 1 , 12}
n{A u B) = 25
AnB = {-12, - 9 , - 6 , - 3 , 0, 3, 6, 9, 12}
niA n 6 ) = 9
3-18 C h a p t e r 3: S e t T h e o r y a n d L o g i c
19. b) Draw a V e n n diagram of these two sets.
10. 1 •
4. Number of people asked: 40
Number w h o like romance novels; 10
Number w h o like horror novels: 13
Number w h o do not like either: 18
Romance or horror or both: 40 - 18 = 22
Both romance and horror: 10 + 13 - 22 = 1
O n e person likes both romance and horror novels.
5. Set 1: Different numbers, same shape, different
colours.
Set 2: S a m e number, different shape, different colour.
Set 3: Different numbers, different shape, different
colour.
Set 4: S a m e number, different shape, different colour.
Set 5: Different number, s a m e shape, different colour.
Set 6: Different number, different shape, different colour.
S e t l :
Set 2:
Set 3:
Set 4:
Set 5;
Set 6:
H ^
O A
O mm
o A
o
b) Statement: If you live in Victoha, then you live
on Vancouver Island.
This statement is true. Victoha is located on
Vancouver Island.
Converse: If you live on Vancouver Island, you live
in Victoha.
The converse is false. Counterexample: There are
other parts of Vancouver Island you could live in,
for example, Port Hardy.
Since the converse is false, the statement is not
biconditional.
c) Statement: If xy is an odd number, then both x
and y are odd numbers.
This statement is true. If an odd number has two
factors, both factors are also odd.
Converse: If both x and y are odd numbers, then
xy is an odd number.
The converse is true. The product of two odd
numbers is odd.
Since both the statement and converse are true,
the statement is biconditional.
Biconditional statement: xy is an odd number if and
only if X and y are odd numbers.
d) Conditional statement: If two numbers are even,
then their s u m is even. This statement is true.
Converse: If the s u m of two numbers is even, then
the two numbers are even. The converse is false.
Counterexample: 5 + 7 = 12.
Since the converse is false, the statement is not
biconditional.
7. a) Use the finance application on a calculator.
The number of payments is 60.
The interest rate is 2.9%.
The present value is $24 729.56.
The payment amount is unknown.
The future value is $0.
The compounding frequency is 12.
The monthly payment is 443.259...
Serge's monthly payment will be $443.26.
b) Use the finance application on a calculator.
The number of payments is unknown.
The interest rate is 2.9%.
The present value is $24 729.56.
The payment amount is
443.259... + 100 = 543.259...
The future value is $0.
The compounding frequency is 12. Serge will pay
off the car in 48.28 months, or neady 1 year
sooner.
6. a) Conditional statement; If x is positive, then lOx > x.
This statement is true. A positive number multiplied by
ten will always be greater than the onginal number.
Converse: If lOx > x, then x is positive.
The converse is true. Ten times a number will be greater
than the onginal number if the number is positive.
Since both the statement and converse are true, the
statement is biconditional.
Biconditional statement: x is positive if and only if
1 0 x > x .
8. a) This statement is true.
Converse: If a number is not negative, then it is
positive.
The statement is false. The number could be 0.
Zero is neither negative nor positive.
Inverse: Contrapositive: If a number is not positive,
then it is negative.
This statement is false, because the number could
be 0, which is neither negative nor positive.
Contrapositive: If a number is negative, then it is
not positive.
F o u n d a t i o n s of IVIathematics 12 S o l u t i o n s Manual 3-19
20. This statement is t m e .
b) This statement is t m e . Converse: If it is a long
w e e k e n d , then M o n d a y is a holiday.
This statement is false, because it could be a long
w e e k e n d , but Friday is a holiday.
Inverse: If M o n d a y is not a holiday, then it is not a long
w e e k e n d . This is false, because Friday could be a
holiday white M o n d a y is a workday. This would still
create a long w e e k e n d .
Contrapositive: If it is not a long w e e k e n d , then M o n d a y
is not a holiday. This statement is true.
C h a p t e r T a s k , p a g e 2 2 1
A. I would organize most o f ttie animals according to
geographic region. I would u s e three sets: A m e h c a (A),
Afhca (F), and Asia/Australia (K). I would need to
consider both indoor and outdoor animals, as well as
animals that need room to r o a m or graze. I could put
birds, insects, and reptiles a s subsets of each
geographic region, or I c o u l d have a separate building for
t h e m and categohze t h e m inside this building. If I have
fish, it would make sense to put them all in an aquarium
building, rather than have several buildings. I would also
need water for birds w h o s w i m . I would need to have
enclosures for small a n i m a l s and for animals that require
controlled climate conditions. I could put predators and
prey in the s a m e areas, but not in the s a m e c o m p o u n d s .
B. and C. Set America (A) will have three subsets: North
America (N), South A m e r i c a (S), and Central America
(C). Central America will intersect sets North America
and South America. Set Africa (F) will have two subsets:
Savanna (V) and Rainforest (R). S o m e of the animals
from sets Africa a n d A m e r i c a will need roaming and
grazing room, so set Graze (G) will intersect both Africa
and America. The bears, lions, and wolves will need
large areas to roam and should be kept apart from the
other animals, so they will have a separate area (L). I a m
also going to separate the Australia/Asia (K) set of
animals as a special attraction area. This will have two
subsets: indoor (D) and outdoor (T).
I decided to put the reptiles a n d insects with the indoor
animals from Australia (H), since I thought they would
thrive best there. I decided t o put the birds in a separate
area. I could not show the intersection of the sets on m y
diagram, but I listed the birds. Also, I could not show the
intersection of the jaguar f r o m South America with the
area for the bears, lions, a n d wolves, but I put it in the
top area. My sets will contain ttie following animals:
America:
N = { moose, cougar, lynx, grizzly bear, polar bear,
bison, elk, raccoon, lynx, Arctic fox, Arctic wolf, snowy
owl, beaver, ferret, prairie d o g , flamingo, swan}
S = {tarantula, black w i d o w spider, blue poison dart frog,
boa constrictor, two-toed sloth, tamarin, marmoset,
jaguar, spider monkey, m a c a w , llama}
C = {boa constrictor, poison dart frog, ocelot,
jaguar, spider monkey}
I = {ferret, black widow spider, prairie dog,
burrowing owl, beaver, boa constrictor, blue poison
dart frog, marmoset, tamarin, tarantula, two-toed
sloth, macaw}
O = { moose, cougar, lynx, grizzly bear, polar bear,
bison, elk, raccoon, lynx, Arctic fox, Arctic wolf,
snowy owl, jaguar, spider monkey, llama}
Africa:
R = {tortoise, mandrill, pygmy hippopotamus, fruit
bat, gorilla}
V = { meerkat, elephant, zebra, lion, cheetah,
crane, stork, baboon, hippopotamus, ostrich,
hyena}
G = { moose, bison, elk, llama, elephant, zebra,
ostrich}
Australia:
D = { t r e e boa, frilled lizard, k o m o d o dragon,
bearded dragon, tree python, kookaburra, tree
kangaroo, sugar glider}
T = {kangaroo, wombat}
Reptile House
H = {tarantula, blue poison dart frog, boa
constrictor, black widow spider, tree boa, frilled
lizard, k o m o d o dragon, bearded dragon, tree
python, tree kangaroo, wombat, sugar glider}
D is now a subset of H.
Birds
Z = {snowy owl, flamingo, macaw, crane, stork,
ostrich, swan}
D.
My Zoo
Legend
path: ••«•"••.
A; America
N: North America
S: South America
C; Central America
F F: Africa
V: Savanna
R: Rainforest
G; Graze
K: Australasia
H: Reptile house
D: Indoor Australasia
T: Outdoor Australasia
L: Large Animals
Z: Bird house and Pond
E Yes, my zoo is easy to navigate. There is a
wide-open space at the entrance, with the feature
birds and pond as you go in. The animals are
categorized, so you can just walk around to see
the different continents.
The more dangerous animals are located together,
and there is a c o m m o n area for the roaming and
grazing animals. I think visitors will find my zoo
3-20 C h a p t e r 3: S e t T h e o r y a n d L o g i c
21. easy to navigate. Ttiey will be able to describe it to their
fhends a n d r e c o m m e n d it. Attendance will increase,
because their fhends will want to check it out for
themselves. Therefore, the conditional statement is true
for m y zoo.
To form the inverse of a conditional statement, I need to
negate the hypothesis and the conclusion. So, the
inverse of the conditional statement is this: If visitors do
not find the z o o easy to navigate, then they are not more
likely to r e c o m m e n d it to their friends, and attendance
will not increase. T h e inverse is true. If visitors find the
zoo confusing and awkward to navigate, they m a y not
return and they will not r e c o m m e n d the zoo to their
fhends. A s a result, the attendance will not increase, and
perhaps it will even decline.
C h a p t e r 3 D i a g n o s t i c T e s t , T R p a g e 1 9 9
1. A die has six possible outcomes: 1 , 2 , 3 , 4, 5, or 6.
Tossing a coin has two possible outcomes: heads (H) or
tails (T).
1 1 2 3 4 5 6
H H, 1 H, 2 H, 3 H, 4 H, 5 H, 6
T 1 T, 1
T, 2 T, 3 T, 4 T, 5 T, 6
2. e.g., long-necked birds = {flamingo, osthch, stork,
swan}
short-necked birds = {bluebird, duck, robin, sparrow}
3. e. g., shapes with curved sides:
The statement is false. Counterexample: the
product of 10 and 0.5 is 5.
7. Statement: Kara always sleeps in on Saturdays.
Today is Saturday. Conclusion: Kara will sleep in
today.
Review of T e r m s and C o n n e c t i o n s , T R
page 201
1. a) il) Venn diagram
X .•V
Y
/ /
( 1
I /' y
b) v) outcome table
Nickel
H T
H H) H T )
(T,H) (T,T)
c) iv) set-builder notation;
y = { x | 3 < x < 10, X G N}
d) iii) atthbutes; 3-D, cube, square faces
object, cube, square faces
shapes with straight sides:
4. Answers will vary. e. g., the, pen, red, s o n , o d d , boo,
nut, cat, dog, too, bib, tam, did, fog, mat
S = {words with t w o of the s a m e letters}
S = {odd, boo, too, bib, did}
V = {words with the letter 0 }
V= {son, o d d , boo, dog, too, fog}
The words belonging in both groups are: o d d , boo, too.
The words belonging in neither group are: the, p e n , red,
nut, cat, dog, tam, mat.
5. a) = {x I 1 < X < 7, X G N}
R = { 1 , 2 , 3, 4, 5, 6, 7}
b) T = { 3 x | - 3 < x < 2 , x e 1}
r = { - 9 , - 6 , - 3 , 0 , 3, 6}
6. a) Statement: T h e product of two odd numbers is o d d .
The statement is true.
b) Statement: If the product of two numbers is 5, then
one of the numbers is either 5 or - 5 .
e) i) set notation; A = {2, 4, 6, 8, 10}
a) Sort according to shading: Let H represent the
set of hollow shapes and S represent the set of
solid shapes.
H = { 1 , 2 , 4, 6} S = {3, 7}
b) Sort according to number of sides: Let O
represent the set of shapes with an o d d number of
sides and E represent the set of shapes with an
even number of sides.
0 = { 1 , 3 , 4} E = { 2 , 5, 6, 7}
3. a) 6.4 is a decimal, so it is in the rational number
system, Q and the real number system, R.
F o u n d a t i o n s of Mathematics 12 S o l u t i o n s Manual 3-21
22. b) V36 is a square root number. In this case, the
number is 6 which is a natural, whole, rational, real
integer. It belongs to N, W , I. Q, and R,
c) - 1 2 3 is a negative n u m b e r so it belongs to the integer
number system. I. It also belongs to the rational n u m b e r
system, Q, which includes all integers as well as the real
n u m b e r system, R
d) 8-5 IS a decimal, so it is in the rational n u m b e r
system, Q and the real n u m b e r system, R.
1
e) 72 is another w a y of wnting a square root, so the
number belongs to the irrational number system. Q and
the number system, R.
4. T={5, 6. 7, .... 97, 98. 99)
Let X represent the numbers in the set T.
T={K5<X<m,X€ 1}
5. Statement: If you know the length of two sides of a
tnangle, you can determine the length of the third side
using the Pythagorean theorem: a' + b'' = c^.
T h e statement is false. Counterexample: A tnangle can
be drawn with sides of length 4 c m . 6 c m , and 8 c m , but
4^ + 6^ IS not equal to 8^ (16 + 36 is equal to 52, but 8'' is
64).
6. Statement: If the temperature is greater than 0" C, any
snow on the ground will begin to melt. There is snow on
the ground. Today, the temperature is going up to 6" C.
Conclusion: T h e snow on the ground will begin to melt.
7. T h e outcomes nf a six sided die are- 1. 2, 3. 4, 5 or 6.
fti(,' (>un;(»rr-(-'s nf ,4 trn.r-sided die 1, 2, 3 ot 4.
1
o
6
3
9
10 J
8. e.g.. a) odd = {3, 9. 15, 2 1 , 27. 33}
even = {6. 12, 18. 24, 30. 36}
b) Yes, there is more than one solution. For example:
numbers w h o s e digits add to 9 = (9. 18. 27, 36}
numbers w h o s e digits do not add to 9 = {3, 6, 12, 15, 2 1 ,
24. 30, 33}
9. a) -789 is a negative number so it belongs to the
integer number system, I. It also belongs to the rational
number system. Q. which includes all integers. It also
belongs to the real numbers, R.
b) 62.3 is a decimal, so it is in the rational number
system. Q. It also belongs to the real numbers. R.
c) -981 is a decimal, so it is in the rational number
system. Q. It also belongs to the real numbers, R.
d) 2.349 583 430 723 4 2 3 4 4 5 4 2 9 743. .. is a non-
repeating, non-terminating decimal number, so it belongs
to the irrational number system. Q . It also belongs to the
real numbers. R.
e) V59 is a square root number, so it belongs to
the irrational n u m b e r system, Q . It also belongs to
the real numbers, R.
f) cos 116° = 0.971 ... which is a non-repeating,
non-terminating decimal number, so it belongs to
the irrational n u m b e r system. Q . It also belongs to
the real numbers, R.
g) 19 387 IS a whole, natural, rational integer, and
a real number, so it belongs to N, W , I, Q, and R.
h) tan 45° = 1, which is a natural number, so it I
belongs to the natural number system. N. It is also
in the whole (W), integer (I), rational (Q), and real
(R) n u m b e r systems, since all of these systems
include the natural numbers.
1§, a ) K = {a | - 3 < a < 5 , a e 1}
K = { - 3 , - 2 , - 1 , 0 . 1 , 2 , 3, 4, 5}
b) {2p I 1 < p < 4 , p c N}
M = {2. 4, 6. 8}
11. a ) Z = { x | x > 1 0 0 , XV N}
Z = {all natural n u m b e r s 100 or greater}
b) L = {x I X ^ , 1 < x < 10, X r N}
T h e range of x is the natural numbers from 1 to 10
inclusive. So, the values of y must be natural
n u m b e r s that are multiples of 4 that range from 4
to 40 inclusive.
For X = 1. 1 ^ , so y = 4, and so o n .
L = {all multiples of 4 from 4 to 40}
12. a) M/= {integers from -25 to 250}
V V = { x ! - 2 5 < x < 250, x f : 1}
b}E = {even positive numbers greater than 8}
E = {2x I X > 5. X c N}
13. a) Statement: The square of a number is
greater than or equal to the number itself.
This statement is false. Counterexample: The
square of 0.5 is 0.25.
b) Statement: If all three angles of a tnangle are
equal, then all three sides will be equal.
This statement is true. Tnangles with equal angles
are equilateral, meaning their sides are also equal.
14. a) Statement: The s u m of the three angles in a
triangle is 180". In A X Y Z , Z X = 40°, and / Y= 65".
Conclusion: ZZ = 75°
b) Statement: In the movie Field of Dreams, based
on the novel "Shoeless Joe", Ray hears a voice
whispenng. "If you build it, he will come." Ray
builds it. Conclusion: He came.
3-22 C h a p t e r 3: S e t T h e o r y a n d L o g i c
23. Chapter 3 Test, T R page 209
1. a) and b)
u
p N
!0„ 9. ;•
i f i i
f 2, 4, 6. N / .
7, 6, ' '1, i
0
c) Sets P and A/ are disjoint sets. Sets E and A/ are
disjoint sets.
d) Yes, Set E is a subset of set P, because set P
contains all the elements of set E.
e) No, set P ' d o e s not equal set hi because 0 does not
belong to set P or set H. Set P ' i n c l u d e s all the negative
numbers in set A/, plus 0.
2. a) n{A) = 7
b) n{B) = 2
c) n{A n e) = 1
d) n((A u 8)') = 7
e) n{U)= 15
3. >A = {x I X < 12, X is a phme number}
A = {2, 3, 5, 7, 11}
B = {x I 1 < X < 10, X is an even number}
8 = {2, 4, 6, 8, 10}
n{A) = 5
r?(B) = 5
n{A 8 ) = 1
n{A uB) = niA) + n{B) - niA n 8 )
r7(A u 8 ) = 5 + 5 - 1
niA u 8 ) = 9
4. Let C represent using a cellphone, and L represent
using a land line.
n(C) + n ( L ) = 152
This is 56 more than the number of people surveyed, so
56 people used both.
5. I drew a V e n n diagram, and wrote 20 where the sets
for canoeing and swimming ovedap. I wrote 11 outside
those sets, for those do not like either sport. Since 28
campers want to canoe, then there are 28 - 20 or 8
campers w h o only want to canoe. Since 45 campers
want to swim, then there are
45 - 20 or 25 campers w h o only want to swim. I added
the numbers in each region. There are 11 + 8 + 20 + 25
or 64 campers in all.
c uc WlSm
S
u
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11
6. 60 - 13 = 47 people had ice cream or chocolate
sauce.
Of these:
47 - 34 = 13 did not have vanilla ice cream.
47 - 28 = 19 did not have chocolate sauce.
4 7 - 1 3 - 1 9 = 15 had vanilla ice cream and
chocolate sauce.
19 I S 1
7. 12 students took all three sciences, and 27
students took physics and chemistry, so 27 - 12 =
15 students took physics and chemistry but did not
take biology. Similady, 15 students took physics
and biology, so 1 5 - 1 2 = 3 took physics and
biology, but did not take chemistry. A n d , 33
students took chemistry and biology, so 33 - 12 =
21 students took chemistry and biology, but did not
take physics. That means:
3 7 - 1 5 - 1 2 - 3 = 7 students took only physics.
6 2 - 1 5 - 1 2 - 2 1 = 1 4 students took only
chemistry.
6 8 - 3 - 1 2 - 2 1 = 3 2 students took only biology.
There were 104 grade 12 students.
F o u n d a t i o n s of Mathematics 12 S o l u t i o n s Manual 3-23
24. 8. a) Statement: If today is the longest day of the year,
then the s u m m e r solstice occurs today.
The s u m m e r solstice marks the first day of s u m m e r and
occurs on the day with the most daylight, Dominique's
statement is true.
b) Converse: If the s u m m e r solstice occurs today, then
today IS the longest day of the year
T h e converse is true. T h e s u m m e r solstice occurs on the
longest day of the year.
c) Biconditional statement: Today is the longest day of
the year if and only if the s u m m e r solstice occurs today.
9. a) Statement; If an integer is not negative, then it is
positive
The statement is false. The integer could be 0. Zero is
neither negative nor positive.
b) i) Converse; If an integer is positive, then it is not
negative. The converse is true.
ii) Inverse; If an integer is negative, then it is not positive.
T h e inverse is true.
iii) Contrapositive: If an integer is not positive, then it is
negative
The contrapositive is false. Counterexample; The integer
could be 0.
3-24 Chapter 3: Set Theory and Logic