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Inductive step: Suppose that B(n) holds. Then
12 + 22 + ... + n2 + (n+1)2= n(n+1)(2n+1)/6 + (n+1)2
Factoring out (n+1) on the right hand side yields
(n+1)(n(2n+1)+6(n+1))/6 = (n+1)(2n2 +7n+6)/6
One easily verifies that this is equal to
(n+1)(n+2)(2(n+1)+1)/6
Thus, B(n+1) holds.
Therefore, the proof follows by induction on n.
8 Tip
How can you verify whether your algebra is
correct?
Use http://www.wolframalpha.com [Not allowed in any exams, though. Sorry!]
9 What’s Wrong? 10 Billiard Balls
“Theorem”: All billiard balls have the same color.
Proof: By induction, on the number of billiard balls.
Induction basis:
Our theorem is certainly true for n=1. What’s wrong? Induction step:
Assume the theorem holds for n billiard balls. We prove it
for n+1. Look at the first n billiard balls among the n+1. By
induction hypothesis, they have the same color. Now look at
the last n billiard balls. They have the same color. Hence all
n+1 billiard balls have the same color.
11 Weird Properties of Positive Integers “Theorem”: For all positive integers n, we have n=n+1.
“Proof”: Suppose that the claim is true for n=k. Then
k+1 = (k) + 1 = (k+1) + 1
by induction hypothesis. Thus, k+1=k+2. What’s wrong? Therefore, the theorem follows by induction on n. 12 Maximally Weird!
“Theorem”: For all positive integers n, if a and b are positive
integers such that max{a,b}=n, then a=b.
Proof: By induction on n. The result holds for n = 1, i.e., if max
{a, b} = 1, then a = b = 1.
Suppose it holds for n, i.e., if max {a,b} = n, then a = b. Now
consider max {a, b} = n + 1.
Case 1: a − 1 ≥ b − 1. Then a≥b. Hence a=max{a,b}=n+1.
Thus a − 1 = n and max {a − 1, b − 1} = n.
By induction hypothesis,...
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 Fall '11
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