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Unformatted text preview: a−1=b−1. Hence a=b.
Case 2: b − 1 ≥ a − 1.
Same argument.
13 Maximally Weird!! Fallacy: In the proof we used the inductive hypothesis to
conclude max {a − 1, b − 1} = n a − 1 = b − 1.
However, we can only use the inductive hypothesis if a − 1 and
b − 1 are positive integers. This does not have to be the case
as the example b=1 shows. 14 More Examples 15 Factorials
Theorem. n
i=0 i(i!) = (n + 1)! − 1. By convention: 0! = 1
Induction basis:
Since 0 = 1 − 1, the claim holds for n = 0.
Induction Step:
Suppose the claim is true for n.Then
n+1
n
i(i!) = (n + 1)(n + 1)! +
i(i!)
i=0 i=0 =
=
= (n + 1)(n + 1)! + (n + 1)! − 1 by ind. hyp.
(n + 2)(n + 1)! − 1
(n + 2)! − 1
16 Divisibility
Theorem: For all positive integers n, the number
7n2n
is divisible by 5.
Proof: By induction.
Induction basis. Since 72=5, the theorem holds
for n=1. 17 Divisibility
Inductive step:
Suppose that 7n2n is divisible by 5. Our goal is to show that
this implies that 7n+12n+1 is divisible by 5. We note that
7n+12n+1 = 7x7n2x2n= 5x7n+2x7n2x2n = 5x7n +2(7n2n).
By induction hypothesis, (7n2n) = 5k for some integer k.
Hence, 7n+12n+1= 5x7n +2x5k = 5(7n +2k), so
7n+12n+1 =5 x some integer.
Thus, the claim follows by induction on n. 18 Strong Induction 19 Strong Induction
Suppose we wish to prove a certain assertion concerning
positive integers.
Let A(n) be the assertion concerning the integer n.
To prove it for all n >= 1, we can do the following:
1) Prove that the assertion A(1) is true.
2) Assuming that the assertions A(k) are proved for all
k<n, prove that the assertion A(n) is true.
We can conclude that A(n) is true for all n>=1.
20 Strong Induction Induction basis:
Show that A...
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This note was uploaded on 03/24/2014 for the course CSCE 222 taught by Professor Math during the Fall '11 term at Texas A&M.
 Fall '11
 math

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