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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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)....
View Full Document

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.

Page1 / 6

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online