Mathematical induction and divisibility rules are methods for proving statements about numbers.
Mathematical induction has two steps: 1) proving the statement is true for the base case, usually n=1. 2) Assuming the statement is true for n=k, proving it is true for n=k+1. Divisibility rules transform numbers into smaller ones while preserving divisibility by certain divisors. Rules exist to test for divisibility by 2, 3, 4, 5, 6, 7, 9, 10 and 13.
The 1741 Goldbach [1] made his most famous contribution to mathematics with the conjecture that all even numbers can be expressed as the sum of the two primes (currently Conjecture) referred to as “all even numbers greater than 2 can be expressed as the sum-two primes” (DOI:10.13140/RG.2.2.32893.69600/1)
The 1741 Goldbach [1] made his most famous contribution to mathematics with the conjecture that all even numbers can be expressed as the sum of the two primes (currently Conjecture) referred to as “all even numbers greater than 2 can be expressed as the sum-two primes” (DOI:10.13140/RG.2.2.32893.69600/1)
this is a presentation on a a number theory topic concerning primes, it discusses three topics, the sieve of Eratosthenes, the euclids proof that primes is infinite, and solving for tau (n) primes.
Now we have learnt the basics in logic.
We are going to apply the logical rules in proving mathematical theorems.
1-Direct proof
2-Contrapositive
3-Proof by contradiction
4-Proof by cases
this is a presentation on a a number theory topic concerning primes, it discusses three topics, the sieve of Eratosthenes, the euclids proof that primes is infinite, and solving for tau (n) primes.
Now we have learnt the basics in logic.
We are going to apply the logical rules in proving mathematical theorems.
1-Direct proof
2-Contrapositive
3-Proof by contradiction
4-Proof by cases
How to Split Bills in the Odoo 17 POS ModuleCeline George
Bills have a main role in point of sale procedure. It will help to track sales, handling payments and giving receipts to customers. Bill splitting also has an important role in POS. For example, If some friends come together for dinner and if they want to divide the bill then it is possible by POS bill splitting. This slide will show how to split bills in odoo 17 POS.
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
Students, digital devices and success - Andreas Schleicher - 27 May 2024..pptxEduSkills OECD
Andreas Schleicher presents at the OECD webinar ‘Digital devices in schools: detrimental distraction or secret to success?’ on 27 May 2024. The presentation was based on findings from PISA 2022 results and the webinar helped launch the PISA in Focus ‘Managing screen time: How to protect and equip students against distraction’ https://www.oecd-ilibrary.org/education/managing-screen-time_7c225af4-en and the OECD Education Policy Perspective ‘Students, digital devices and success’ can be found here - https://oe.cd/il/5yV
This is a presentation by Dada Robert in a Your Skill Boost masterclass organised by the Excellence Foundation for South Sudan (EFSS) on Saturday, the 25th and Sunday, the 26th of May 2024.
He discussed the concept of quality improvement, emphasizing its applicability to various aspects of life, including personal, project, and program improvements. He defined quality as doing the right thing at the right time in the right way to achieve the best possible results and discussed the concept of the "gap" between what we know and what we do, and how this gap represents the areas we need to improve. He explained the scientific approach to quality improvement, which involves systematic performance analysis, testing and learning, and implementing change ideas. He also highlighted the importance of client focus and a team approach to quality improvement.
How to Create Map Views in the Odoo 17 ERPCeline George
The map views are useful for providing a geographical representation of data. They allow users to visualize and analyze the data in a more intuitive manner.
The Roman Empire A Historical Colossus.pdfkaushalkr1407
The Roman Empire, a vast and enduring power, stands as one of history's most remarkable civilizations, leaving an indelible imprint on the world. It emerged from the Roman Republic, transitioning into an imperial powerhouse under the leadership of Augustus Caesar in 27 BCE. This transformation marked the beginning of an era defined by unprecedented territorial expansion, architectural marvels, and profound cultural influence.
The empire's roots lie in the city of Rome, founded, according to legend, by Romulus in 753 BCE. Over centuries, Rome evolved from a small settlement to a formidable republic, characterized by a complex political system with elected officials and checks on power. However, internal strife, class conflicts, and military ambitions paved the way for the end of the Republic. Julius Caesar’s dictatorship and subsequent assassination in 44 BCE created a power vacuum, leading to a civil war. Octavian, later Augustus, emerged victorious, heralding the Roman Empire’s birth.
Under Augustus, the empire experienced the Pax Romana, a 200-year period of relative peace and stability. Augustus reformed the military, established efficient administrative systems, and initiated grand construction projects. The empire's borders expanded, encompassing territories from Britain to Egypt and from Spain to the Euphrates. Roman legions, renowned for their discipline and engineering prowess, secured and maintained these vast territories, building roads, fortifications, and cities that facilitated control and integration.
The Roman Empire’s society was hierarchical, with a rigid class system. At the top were the patricians, wealthy elites who held significant political power. Below them were the plebeians, free citizens with limited political influence, and the vast numbers of slaves who formed the backbone of the economy. The family unit was central, governed by the paterfamilias, the male head who held absolute authority.
Culturally, the Romans were eclectic, absorbing and adapting elements from the civilizations they encountered, particularly the Greeks. Roman art, literature, and philosophy reflected this synthesis, creating a rich cultural tapestry. Latin, the Roman language, became the lingua franca of the Western world, influencing numerous modern languages.
Roman architecture and engineering achievements were monumental. They perfected the arch, vault, and dome, constructing enduring structures like the Colosseum, Pantheon, and aqueducts. These engineering marvels not only showcased Roman ingenuity but also served practical purposes, from public entertainment to water supply.
2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
Instructions for Submissions thorugh G- Classroom.pptxJheel Barad
This presentation provides a briefing on how to upload submissions and documents in Google Classroom. It was prepared as part of an orientation for new Sainik School in-service teacher trainees. As a training officer, my goal is to ensure that you are comfortable and proficient with this essential tool for managing assignments and fostering student engagement.
2. UNIVERSITY OF AZAD JAMMU & KASHMIR
MUZAFFARABAD
Department of CS & IT
Dawood Faheem Abbasi 05
3. Mathematical induction
INTRODUCTION:
The principal of mathematical induction is useful tool for proving that a
certain predicate is true for all natural numbers.
It can not be used to discover theorems.
In mathematical induction, we try to prove that L.H.S is equal to R.H.S
It has only 2 steps
Step 1. Show it is true for first one.
Step 2. Show that if any one is true then the next one is true..
Then all are true.
4. Have you heard of “dominos effect”?
Step1. The first domino falls.
Step2. When any domino falls, the next domino falls.
So.. All dominos falls.
5. In the world of numbers we say;
Step 1. Show that it is true for n=1.
Step 2. Show that if n=k is true then n =k+1 is also true.
How to do it?
Step 1 is usually easy, we just have to prove it is true for n=1.
Step 2 is done by;
Assume it is true for n=k.
Prove it is true for n=k+1
6. Types of questions solve by induction
Statements giving expressions about summation or multiplication of
special series
Statements to show divisibility of an expression by a certain natural
number.
Statements containing signs of inequality.
7.
8. Adding up all odd numbers.
1+3+5+……….+(2n-1)=n2
Step 1. show it is true for n=1
1=12 is true
Step 2. assume it is true for n=k
1+3+5+…..+(2k-1)=k2 is true
Now, prove it is true for “k+1”
1+3+5+…..+(2k-1)+(2(k+1)-1)=(k+1) 2
We know that 1+3+5+…(2k-1)=k2 so,
K2+(2(k+1)-1)=(k+1) 2
Expanding;
k2+2k+2-1=k2+2k+1
K2+2k+1=k2+2k+1 they are same! So it is true.
9.
10.
11. Prove that 2n>n for all positive integers n
Let p(n)be the given statement
P(n):2n>n
step1. when n=1
21>1
Hence , p(1)is true.
Step 2. assume that p(k) is true for any positive integer k i.e
2k>k
We shall now prove that pk+1is true.
Multiplying both sides of 1 by 2.
2.2k>2k i.e
2k+1>2k
K+k>k+1
Therefore, pk+1 is true when p(k+1) is true.
So, P(n) is true for every positive integer n.
12. n < 2n
for all positive integers n.
Solution: Let P(n) be the proposition that n < 2n.
BASIS STEP: P(1) is true, because 1 < 21 = 2. This completes the basis step.
INDUCTIVE STEP: We first assume the inductive hypothesis that P(k) is true for anarbitrary
positive integer k. That is, the inductive hypothesisP(k) is the statement thatk < 2k.To complete
the inductive step, we need to show that if P(k) is true, then P(k + 1), which is the statement
that k + 1 < 2k+1, is true. That is, we need to show that ifk < 2k, then k + 1 < 2k+1. To show
320 5 / Induction and Recursion
that this conditional statement is true for the positive integer k, we first add 1 to both sides of
k < 2k, and then note that 1 ≤ 2k. This tells us that
k + 1
IH
<2k + 1 ≤ 2k + 2k = 2 · 2k = 2k+1.
This shows that P(k + 1) is true, namely, that k + 1 < 2k+1, based on the assumption that P(k)
is true. The induction step is complete.
Therefore, because we have completed both the basis step and the inductive step, by
the principle of mathematical induction we have shown that n < 2n is true for all positive
integers n.
13. Solution: To construct the proof, let P(n) denote the proposition: “7n+2 + 82n+1 is divisible by
57.”
BASIS STEP: To complete the basis step, we must show that P(0) is true, because we want
to prove that P(n) is true for every nonnegative integer. We see that P(0) is true because
70+2 + 82·0+1 = 72 + 81 = 57 is divisible by 57. This completes the basis step.
INDUCTIVE STEP: For the inductive hypothesis we assume that P(k) is true for an arbitrary
nonnegative integer k; that is, we assume that 7k+2 + 82k+1 is divisible by 57. To complete the
inductive step, we must show that when we assume that the inductive hypothesis P(k) is true,
then P(k + 1), the statement that 7(k+1)+2 + 82(k+1)+1 is divisible by 57, is also true.
The difficult part of the proof is to see howto use the inductive hypothesis.To take advantage
of the inductive hypothesis, we use these steps:
7(k+1)+2 + 82(k+1)+1 = 7k+3 + 82k+3
= 7 · 7k+2 + 82 · 82k+1
= 7 · 7k+2 + 64 · 82k+1
= 7(7k+2 + 82k+1) + 57 · 82k+1.
We can now use the inductive hypothesis, which states that 7k+2 + 82k+1 is divisible by
57. We will use parts (i) and (ii) of Theorem 1 in Section 4.1. By part (ii) of this theorem, and
the inductive hypothesis, we conclude that the first term in this last sum, 7(7k+2 + 82k+1), is
divisible by 57. By part (ii) of this theorem, the second term in this sum, 57 · 82k+1, is divisible
by 57. Hence, by part (i) of this theorem, we conclude that 7(7k+2 + 82k+1) + 57 · 82k+1 =
7k+3 + 82k+3 is divisible by 57. This completes the inductive step.
Because we have completed both the basis step and the inductive step, by the principle of
mathematical induction we know that 7n+2 + 82n+1 is divisible by 57 for every nonnegative
integer n.
14. Strong Induction
Before we illustrate how to use strong induction, we state this principle again.
STRONG INDUCTION To prove that P(n) is true for all positive integers n, where P(n)
is a propositional function, we complete two steps:
BASIS STEP: We verify that the proposition P(1) is true.
INDUCTIVE STEP: We show that the conditional statement [P(1) ∧ P(2) ∧ · · · ∧
P(k)] → P(k + 1) is true for all positive integers k.
15. Note that when we use strong induction to prove that P(n) is true for all positive integers n,
our inductive hypothesis is the assumption that P(j) is true for j = 1, 2, . . . , k. That is, the
inductive hypothesis includes all k statements P(1), P(2), . . . , P (k). Because we can use all k
statements P(1), P(2), . . . , P (k) to prove P(k + 1), rather than just the statement P(k) as in a
proof by mathematical induction,
strong induction is a more flexible proof technique. Because
of this, some mathematicians prefer to always use strong induction instead of mathematical
induction, even when a proof by mathematical induction is easy to find.
You may be surprised that mathematical induction and strong induction are equivalent.
That is, each can be shown to be a valid proof technique assuming that the other is valid.
18. Introduction:
The rules given below transform a given number into a generally
smaller number, while preserving divisibility by the divisor of
interest. Therefore, unless otherwise noted, the resulting number
should be evaluated for divisibility by the same divisor.
19. Divisibility by 2
First, take any number (for this example it will be 376) and note the last digit
in the number, discarding the other digits. Then take that digit (6) while
ignoring the rest of the number and determine if it is divisible by 2. If it is
divisible by 2, then the original number is divisible by 2.
20. Examples
376 (The original number)
37 6 (Take the last digit)
6 ÷ 2 = 3 (Check to see if the last digit is divisible by 2)
376 ÷ 2 = 188 (If the last digit is divisible by 2, then the whole number is
divisible by 2)
21. Divisibility by 3 or 9
First, take any number (for this example it will be 492) and add
together each digit in the number (4 + 9 + 2 = 15). Then take
that sum (15) and determine if it is divisible by 3. The original
number is divisible by 3 (or 9) if and only if the sum of its digits
is divisible by 3 (or 9).
If a number is a multiplication of 3 consecutive numbers then
that number is always divisible by 3. This is useful for when the
number takes the form of (n × (n − 1) × (n + 1))
22. Example
492 (The original number)
4 + 9 + 2 = 15 (Add each individual digit together)
15 is divisible by 3 at which point we can stop. Alternatively we can continue
using the same method if the number is still too large:
1 + 5 = 6 (Add each individual digit together)
6 ÷ 3 = 2 (Check to see if the number received is divisible by 3)
492 ÷ 3 = 164 (If the number obtained by using the rule is divisible by 3, then
the whole number is divisible by 3)
23. Divisibility by 4
The basic rule for divisibility by 4 is that if the number formed by the last two
digits in a number is divisible by 4, the original number is divisible by 4; this is
because 100 is divisible by 4 and so adding hundreds, thousands, etc. is
simply adding another number that is divisible by 4. If any number ends in a
two digit number that you know is divisible by 4 (e.g. 24, 04, 08, etc.), then
the whole number will be divisible by 4 regardless of what is before the last
two digits.
Alternatively, one can simply divide the number by 2, and then check the
result to find if it is divisible by 2. If it is, the original number is divisible by 4.
In addition, the result of this test is the same as the original number divided
by 4
24. Example
1720 (The original number)
1720 ÷ 2 = 860 (Divide the original number by 2)
860 ÷ 2 = 430 (Check to see if the result is divisible by 2)
1720 ÷ 4 = 430 (If the result is divisible by 2, then the original number is
divisible by 4)
25. Divisibility by 5
Divisibility by 5 is easily determined by checking the last digit in the number
(475), and seeing if it is either 0 or 5. If the last number is either 0 or 5, the
entire number is divisible by 5.
26. Conti….
If the last digit in the number is 5, then the result will be the remaining digits
multiplied by two (2), plus one (1). For example, the number 125 ends in a 5,
so take the remaining digits (12), multiply them by two (12 × 2 = 24), then add
one (24 + 1 = 25). The result is the same as the result of 125 divided by 5
(125/5=25
27. Example
If the last digit in the number is 5, then the result will be the remaining digits
multiplied by two (2), plus one (1). For example, the number 125 ends in a 5,
so take the remaining digits (12), multiply them by two (12 × 2 = 24), then
add one (24 + 1 = 25). The result is the same as the result of 125 divided by
5 (125/5=25).
28. If the last digit is 5
85 (The original number)
8 5 (Take the last digit of the number, and check if it is 0 or 5)
8 5 (If it is 5, take the remaining digits, discarding the last)
8 × 2 = 16 (Multiply the result by 2)
16 + 1 = 17 (Add 1 to the result)
85 ÷ 5 = 17 (The result is the same as the original number divided by 5)
29. Divisibility by 6
Divisibility by 6 is determined by checking the original number to see if it is both
an even number (divisible by 2) and divisible by 3. This is the best test to use.
If the number is divisible by six, take the original number (246) and divide it by
two (246 ÷ 2 = 123). Then, take that result and divide it by three (123 ÷ 3 = 41).
This result is the same as the original number divided by six (246 ÷ 6 = 41)
30. Example
324 (The original number)
324 ÷ 3 = 108 (Check to see if the original number is divisible by 3)
324 ÷ 2 = 162 OR 108 ÷ 2 = 54 (Check to see if either the original number or the
result of the previous equation is divisible by 2)
324 ÷ 6 = 54 (If either of the tests in the last step are true, then the original
number is divisible by 6. Also, the result of the second test returns the same
result as the original number divided by 6)
31. Divisibility by 7
Divisibility by 7 can be tested by a recursive method. A number of the form 10x + y
is divisible by 7 if and only if x − 2y is divisible by 7. In other words, subtract twice
the last digit from the number formed by the remaining digits. Continue to do this
until a number known to be divisible by 7 is obtained. The original number is
divisible by 7 if and only if the number obtained using this procedure is divisible by
7. For example, the number 371: 37 − (2×1) = 37 − 2 = 35; 3 − (2 × 5) = 3 − 10
= −7; thus, since −7 is divisible by 7, 371 is divisible by 7
32. Divisibility by 13
Remainder Test 13 (1, −3, −4, −1, 3, 4, cycle goes on.) If you are not comfortable
with negative numbers, then use this sequence. (1, 10, 9, 12, 3, 4)
Multiply the right most digit of the number with the left most number in the
sequence shown above and the second right most digit to the second left most
digit of the number in the sequence. The cycle goes on
33. Example
Example: What is the remainder when 321 is divided by 13?
Using the first sequence,
Ans: 1 × 1 + 2 × −3 + 3 × −4 = −17
Remainder = −17 mod 13 = 9
312 is divisible by 13?
Using first sequence
2 x 1+ 1 x -3+ 3x-4= -13 => 0
34. Divisor Divisibility Condition Examples
21 Subtract twice the last digit from the rest 168: 16 − (8×2) = 0, 168 is divisible.
1050: 105 − (0×2) = 105, 10 − (5×2) = 0, 1050 is
divisible.
23 Add 7 times the last digit to the rest. 3128: 312 + (8×7) = 368, 368 ÷ 23 = 16.
25 The number formed by the last two digits is divisible
by 25
134,250: 50 is divisible by 25.
27 Sum the digits in blocks of three from right to left. If
the result is divisible by 27, then the number is
divisible by 27.
2,644,272: 2 + 644 + 272 = 918 = 27×34.
35. Divisor Divisibility condition Examples
29 Add three times the last digit to the rest. 261: 1×3 = 3; 3 + 26 = 29
31 Subtract three times the last digit from the rest 837: 83 − 3×7 = 62
32
The number formed by the last five digits is divisible
by 32.[1][2] 25,135,520: 35,520=1110×32
32
If the ten thousands digit is even, examine the
number formed by the last four digits.
41,312: 1312.
33
Add 10 times the last digit to the rest; it has to be
divisible by 3 and 11
627: 62 + 7 × 10 = 132,
13 + 2 × 10 = 33.
36. Divisor Divisibility Condition Examples
35 Number must be divisible by 7 ending in 0 or 5. 700 is divisible by 7 ending in a 0
37 Take the digits in blocks of three from right to left and
add each block, just as for 27.
2,651,272: 2 + 651 + 272 = 925. 925 = 37×25.
39 Add 4 times the last digit to the rest 351: 1 × 4 = 4; 4 + 35 = 39
41 Subtract 4 times the last digit from the rest 738: 73 − 8 × 4 = 41
37. Divisor Divisible Condition Example
43 Add 13 times the last digit to the rest. 36,249: 3624 + 9 × 13 = 3741,
374 + 1 × 13 = 387,
38 + 7 × 13 = 129,
12 + 9 × 13 = 129 = 43 × 3
45 The number must be divisible by 9 ending in 0 or 5 495: 4 + 9 + 5 = 18, 1 + 8 = 9;
(495 is divisible by both 5 and 9.)
47 Subtract 14 times the last digit from the rest. 1,642,979: 164297 − 9 × 14 = 164171,
16417 − 14 = 16403,
1640 − 3 × 14 = 1598,
159 − 8 × 14 = 47.
49 Add 5 times the last digit to the rest. 1,127: 112+(7×5)=147.
147: 14 + (7×5) = 49
38. Divisor Divisible Condition Example
50 The last two digits are 00 or 50 134,250: 50.
51 Subtract 5 times the last digit to the rest 204: 20-(4×5)=0
53 Add 16 times the last digit to the rest. 3657: 365+(7×16)=477 = 9 × 53
55 Number must be divisible by 11 ending in 0 or 5. 935: 93 − 5 = 88 or 9 + 35 = 44
39. Divisor Divisible Condition Example
71 Subtract 7 times the last digit from the rest. 852: 85-(2×7)=71
81 Subtract 8 times the last digit from the rest 162: 16-(2×8)=0
100 Ends with at least two zeros 900 ends with 2 zeros
1000 Ends with at least three zeros. 2000 ends with 3 zeros