Induction

# Induction - Examples of Proof by Induction Introduction:...

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Unformatted text preview: Examples of Proof by Induction Introduction: “A Journey of a thousand miles begins with a single step” This phrase rather nicely sums up the core idea of proof by induction where we attempt to demonstrate that a property holds in an infinite, but countable, number of cases, by extrapolating from the first few. More specifically, we have the following: 1) The “base” case : Our property holds for some initial case(s). 2) The induction step: Show if the property holds for some n th case, it must hold for the next, the “n+1” case. 3) The conclusion: Once we have the first case holds, from the induction step we know if any step holds, the next step holds. So the first implies the 2 nd , implies the third etc., etc. Thus our property holds for all cases. ( Note : technically, this is what is known as weak induction . This differs in step 2 from the other version, called strong induction. With strong, we assume, in the induction step that all earlier cases have the requisite property, whereas here we only required the n th step to work .) Generally it's much easier to see, if we work through some examples. Example #1 Solution: Well, we can see this is obviously true, even without induction, by deconstructing it into the expression: Σ( i + 1) = Σ i + Σ 1, and using the relations from Appendix E. But let's work this through now with induction. 1) Base case : The simplest possible sum occurs at n = 1. Which fits the equation as predicted. Page #1 of 6 ( 29 1 ( 3) Show, using induction, that 1 2 n i n n i = + + = ∑ ( 29 1 1 1 1(1 3) ( 3) 1 (1 + 1) = 2 = 2 2 i n n n i = = + + + = = ∑ From here, we could check n = 1, n = 2, n = 3 etc. and each of these would fit our formula. But how do we know that n = 100, n = 10000, n = 10 100 works? That's why we need step #2. 2) Induction Step: Here we assume our relation holds for the sum up to n = k : If it's true for n = k, is it true for the “next” expression, n = k + 1? That is, can we say that: Well, let's see. We can split up our new k +1 th sum in terms of the previous sum, which we know worked (by our assumption)....
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## This note was uploaded on 12/12/2011 for the course MATH 1ZA3 taught by Professor Childs during the Spring '11 term at McMaster University.

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Induction - Examples of Proof by Induction Introduction:...

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