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# Induction

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A presentation on Mathematical Induction

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### Induction

1. 1. INDUCTION As soon as an Analytical Engine exists, it will necessarily guide the future course of the science. Whenever any result is sought by its aid, the question will then arise – By what course of calculation can these results be arrived at by the machine in the shortest time? - Charles Babbage, 1864
2. 2. Overview <ul><ul><li>Numerical Iteration </li></ul></ul><ul><ul><li>Structural Induction </li></ul></ul><ul><ul><li>Double Induction </li></ul></ul>03/12/11
3. 3. Numerical Iteration
4. 4. Numerical Iteration <ul><li>Iteration means the act of repeating a process usually with the aim of approaching a desired goal or target or result. Each repetition of the process is also called an &quot;iteration&quot;, and the results of one iteration are used as the starting point for the next iteration. </li></ul><ul><li>Iteration in computing is the repetition of a process within a computer program. It can be used both as a general term, synonymous with repetition, and to describe a specific form of repetition with a mutable state. </li></ul>03/12/11
5. 5. <ul><li>var i, a := 0 // initialize a before iteration </li></ul><ul><li>for i from 1 to 3 // loop three times </li></ul><ul><li>{ </li></ul><ul><ul><li>a := a + i // increment a by the current value of I </li></ul></ul><ul><ul><li>} </li></ul></ul><ul><ul><li>print a // the number 6 is printed </li></ul></ul>03/12/11
6. 6. Recursive Fibonacci <ul><li>procedure fibonacci (n: nonnegative integer) </li></ul><ul><li>if n = 0 then fibonacci (0) :=0 </li></ul><ul><li>else if n =1 then fibonacci (1) := 1 </li></ul><ul><li>else fibonacci (n): = fibonacci (n-1) + fibonacci (n-2) </li></ul>What’s the “problem” with this algorithm? 03/12/11
7. 7. Iterative Fibonacci <ul><li>procedure iterativefibonacci (n: nonnegative integer) </li></ul><ul><li>if n=0 then y:= 0 </li></ul><ul><li>else </li></ul><ul><li>begin </li></ul><ul><li>x := 0 </li></ul><ul><li>y := 1 </li></ul><ul><li>for i := 1 to (n-1) </li></ul><ul><li>begin </li></ul><ul><li>z := x + y </li></ul><ul><li>x := y </li></ul><ul><li>y := z </li></ul><ul><li>end </li></ul><ul><li>end </li></ul><ul><li>{y is the nth Fibonacci number} </li></ul>03/12/11
8. 8. Structural Induction
9. 9. Structural Induction <ul><li>Structural induction can be used to prove that all members of a set constructed recursively have a particular property. </li></ul><ul><li>Basic Step: Show that the result holds for all elements specified in the basis step of the recursive definition to be in the set. </li></ul><ul><li>Recursive step: Show that if the statement is true for each of the elements used to construct new elements in the recursive step of the definition, the result holds for these new elements. </li></ul>03/12/11
10. 10. Validity of Structural Induction follows Mathematical Induction ( for the nonnegative integers) <ul><li>P(n) the claim is true for all elements of the set that are generated by n or fewer applications of the rules in the recursive step of the recursive definition. </li></ul><ul><li>So, we will do induction on the number of rules applications. </li></ul><ul><li>We show that P(n) is true whenever n is a nonnegative integer. </li></ul>03/12/11
11. 11. Validity of Structural Induction follows Mathematical Induction ( for the nonnegative integers) <ul><li>Basis case - we show that P(0) is true (i.e., it’s true for the elements specified in the basis step of recursive definition). </li></ul><ul><li>From recursive step, if we assume P(k), it follows that P(k+1) is true. </li></ul><ul><li>Therefore when we complete a structural induction proof we have shown that P(0) is true, and that P(k)  P(k+1). </li></ul><ul><li>So, by mathematical induction P(n) follows for all nonnegative numbers. </li></ul>03/12/11
12. 12. Well-Formed Formulas T is a wff F is a wff p is a wff for any propositional variable p If p is a wff, then (  p) is a wff If p and q are wffs, then (p  q) is a wff If p and q are wffs, then (p  q) is a wff For example, a statement like ((  r)  (p  r)) can be proven to be a wff by arguing that (  r) and (p  r) are wffs by induction and then applying rule 5. Note: we have three recursive/construction rules to create new elements. 03/12/11 Basic Cases Recursive Step
13. 13. Structural induction --- illustrative example <ul><li>Well Formed Formulae for compound statement Forms We can define the set of well formed formulae for compound statement forms involving T,F, Propositional variables and operators from the set {  ,  ,  ,  ,  ,}. </li></ul>03/12/11
14. 14. Structural induction --- illustrative example <ul><li>Basis Step --- </li></ul><ul><ul><li>T,F, and s, where s is a propositional variable , are well formed formulae </li></ul></ul><ul><li>Recursive Step: </li></ul><ul><ul><li>If E and F are well formed formulae, then (  E),(E  F), (E  F). (E  F) and (E  F) are well formed formulae. </li></ul></ul>03/12/11
15. 15. Double Induction
16. 16. Double Induction 03/12/11
17. 17. Double Induction <ul><li>Double induction: There are cases where the proof of the inductive step requires, in its own right, an inductive argument. The following examples illustrate this point. </li></ul><ul><li>Sometimes we would like to prove a proposition P(m, n) involving two natural numbers m by iterating the inductive process. This is done by performing an induction on one of the variables, say m , and then an induction on n . This strategy is called double induction . </li></ul>03/12/11
18. 18. Steps of Double Induction <ul><li>following steps. </li></ul><ul><ul><li>Prove P(1,1) is true </li></ul></ul><ul><ul><li>Prove P(m,1)  P(m+1,1) </li></ul></ul><ul><ul><li>Prove P(m, n )  P(m,n+1) for all natural numbers </li></ul></ul>03/12/11
19. 19. Example of Double Induction <ul><li>Example Show that for each n∈ N, 2 ⋅ 7n + 3 ⋅ 5n – 5 is a multiple of 24. </li></ul><ul><li>Solution </li></ul><ul><ul><li>Writing an = 2 ⋅ 7n + 3 ⋅ 5n – 5, the claim is clear for n = 1. </li></ul></ul><ul><ul><li>Assuming it true for n = k then as ak+1 = 7 ⋅ ak – 6⋅ 5k + 30, the inductive argument would be complete if we could prove that 6⋅ 5k – 30 is always a multiple of 24. We can now start a new </li></ul></ul><ul><ul><li>inductive argument to prove the last statement, an easy task left to the reader </li></ul></ul>03/12/11
20. 20. Thank you for your co-operation