Example Proofs: Induction
Proofs Workshop
Note: the proofs in this handout are not necessarily in the same form as they were presented at the workshop. In particular, any errors you spot here are entirely accidental, not deliberate. Theorem Let { f n } be the sequence of integers satisfying: Prove that 1 + f1 + f3 + … + f2n 1 = f2n −
f1 = 1, f2 = 2, and fr = fr-1 + fr-2 for all r ≥ 3 .
for all positive integers n.
Proof: Let P(n) be the proposition 1 + f1 + f3 + … + f2n-1 = f2n .
Reminder: P(k+1) is the statement:
Initial step:
1 + f1 + f3 + … + f2(k+1) 1 = f2(k+1) i.e. 1 + f1 + f3 + … + f2k+1 = f2k+2
P(1) is true, since 1 + f1 = 1 + 1 = 2 = f2 (= f2n).
−
Inductive step: Suppose P(k) is true for some integer k ≥ 1, so 1 + f1 + f3 + … + f2k 1
= f2k
−
Then
1 + f1 + f3 + … + f2k-1 + f2k+1 = f2k + f2k+1 (adding f2k+1 to both sides) = f2k+2
(by definition, since 2k + 2 ≥ 3)
and this is the statement P(k + 1). So we have shown that for all integers k ≥ 1, whenever P(k) is true, then P(k+1) is also true. So P(n) is true for all positive integers n by the principle of mathematical induction.
Theorem Prove that
d n x = nx n −1 using the product rule, where n is a positive integer. dx
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Proof: Let P(n) be the proposition
d n x = nx n−1 dx
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Initial step: Let n = 1 , then
d n d x = (x ) = 1 = 1⋅ x 0 = 1⋅ x1−1 = nx n −1 . Therefore P(1) is true. dx dx
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Inductive step: Assume the proposition is true when n = k , where k ≥ 1, i.e.
d k +1 d x = x ⋅ xk dx dx d (x ) ⋅ x k + x ⋅ d x k = dx dx k k −1 = 1 ⋅ x + x ⋅ kx
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(
)
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k
= x + kx
(by the product rule)
d k x = kx k −1. Then for n = k+1, dx
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Reminder: P(k+1) is the statement:
(by assumption)
k
= (k + 1)x k This shows that if the proposition is true for n = k , k ≥ 1 , then it is true for n = k + 1. We have shown that P(1) is true, and that whenever P(k) is true, where k ≥ 1, then P(k+1) is true. Hence by the principle of mathematical induction, P(n) is true for all n ≥ 1. Therefore
d n x = nx n −1 for all positive integers n. dx
Shirleen Stibbe
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http://www.shirleenstibbe.co.uk1