Mathematical Induction
Tan Conghui
Department of Systems Engineering and Engineering Management,
The Chinese University of Hong Kong
September 15, 2016
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Outline
1
Review of Basic Concepts
2
Examples
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Review of Basic Concepts
Outline
1
Review of Basic Concepts
2
Examples
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Review of Basic Concepts
Definition of Predicate
A predicate is a statement whose value depends on one or more
variables.
Example 1
The following two statements are both predicates depends on n (n is a
nonnegative integer):
n is a perfect square,
Pn
n(n+1)
,
i=1 i =
2
However, the former is only true for some particular choices of n,
while the latter is true for all n.
Usually, we use P(n) to denote a predicate concerning the variable n.
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Review of Basic Concepts
Principle of (Weak) Mathematical Induction
Let P(n) be a predicate. If
P(0) is true, and
P(n) implies P(n + 1) for all nonnegative integers n,
then P(n) is true for all nonnegative integers n.
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Review of Basic Concepts
A Template for Induction Proofs
1
2
3
4
5
State that the proof uses induction.
Define an appropriate predicate P(n).
Prove that P(0) is true (basic step).
Prove that P(n) implies P(n + 1) for every nonnegative integer n
(inductive step).
Invoke induction. Given these facts, the induction principle allows
you to conclude that P(n) is true for all nonnegative n.
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Review of Basic Concepts
Principle of Strong Mathematical Induction
Let P(n) be a predicate. If
P(0) is true, and
P(0), P(1), . . . , P(n) together implies P(n + 1) for any
nonnegative integer n,
then P(n) is true for all nonnegative integers n.
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Examples
Outline
1
Review of Basic Concepts
2
Examples
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Examples
Problem 1
Question: Prove by induction that if 0 < a < 1, then (1 − a)n ≥ 1 − na
for all n = 0, 1, . . . .
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Examples
Problem 1
Question: Prove by induction that if 0 < a < 1, then (1 − a)n ≥ 1 − na
for all n = 0, 1, . . . .
Solution: We can prove it by induction. Let P(n) be the predicate
“(1 − a)n ≥ 1 − na”.
1
2
(Basic Step) When n = 0, LHS = (1 − a)0 = 1 = 1 − 0 × a = RHS, so
P(0) is true.
(Inductive Step) For any nonnegative integer n, assume P(n) is true, i.e.,
(1 − a)n ≥ 1 − na. Then we have
(1 − a)n+1 = (1 − a)n (1 − a)
≥ (1 − na)(1 − a)
(from P(n) and the fact 1 − a ≥ 0)
2
= 1 − (n + 1)a + na
> 1 − (n + 1)a,
which means P(n + 1) is true.
Hence, by the principle of induction, P(n) is true for n = 0, 1, 2, . . . .
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Examples
Problem 2
Question: Let T(n) be the number of comparisons needed to sort n cards
by merge sort. We already know that T(n) has recursive relationship
T(n) = 2T( n2 ) + n − 1 and T(1) = 0. Prove that T(n) ≤ n log2 n for all
n = 2k , where k = 0, 1, 2, . . . .
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Examples
Problem 2
Solution: We prove it by induction. Let P(k) be the predicate
“T(n) ≤ n log2 n holds where n = 2k ”.
1 (Basic step) When k = 0, then n = 2k = 1 and therefore
T(n) = 0 = 1 × log2 1 = n log2 n, so P(0) is true.
2 (Inductive Step) For any nonnegative integer k, assume P(k) is true, i.e.,
T(n) ≤ n log2 n where n = 2k . Note that 2k+1 = 2n, so we need to prove
T(2n) ≥ 2n log2 (2n) for P(n + 1):
T(2n) = 2T(n) + 2n − 1
(from the recursive relationship of T(2n))
≤ 2n log2 n + 2n − 1 (by inductive hypothesis)
= 2n(log2 n + 1) − 1
= 2n log2 (2n) − 1 ≤ 2n log2 (2n),
thus P(k + 1) is also true.
Remark: log2 n + 1 = log2 n + log2 2 = log2 (2n)
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Examples
Problem 3
Question: Show that any integer n (n ≥ 2) can be written as the product
of primes, e.g., 20 = 2 × 2 × 5.
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Examples
Problem 3
Solution: We prove it by strong mathematical induction. Let P(n) be the
predicate “n can be written as the product of primes”.
1
2
(Basic Step) For n = 2, since 2 itself is a prime number, thus P(2) holds.
(Inductive Step) Assume P(k) holds for all 2 ≤ k ≤ n. Now let’s
consider n + 1:
If n + 1 is not divisible by any positive integer greater than 1 and except
itself, then n + 1 itself is a prime number and therefore P(n + 1) is
obviously true;
Otherwise, then there must exist integers s and t such that n + 1 = s · t and
1 < s, t < n + 1. By the inductive hypothesis, we known both s and t can
be written as the product of primes, i.e., s = s1 · s2 · · · sl and
t = t1 · t2 · · · tm , where si and tj are all primes. Thus,
n + 1 = s · t = s1 · s2 · · · sl · t1 · t2 · · · tm and therefore P(n + 1) is true.
Hence, by the principle of strong MI, P(n) is true for n = 0, 1, 2, . . . .
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Examples
Problem 4
Question: Recall that the Fibonacci numbers are defined recursively by:

if n = 0
 0
1
if n = 1 .
Fn =

Fn−1 + Fn−2 if n ≥ 2
√ Let φ be the number 1 + 5 /2 (note it satisfy φ2 = 1 + φ). Prove
Fn ≤ φn−1 for all nonnegative integer n by weak mathematical induction.
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Examples
Problem 4
Solution: Let P(n) be the predicate “Fn ≤ φn−1 and Fn+1 ≤ φn ”.
(Basic Step): For n = 0, because F0 = 0 ≤ φ−1 and F1 = 1 ≤ 1 = φ0 ,
P(0) is true.
(Inductive Step): Assume all P(n) is true, i.e., Fn ≤ φn−1 and
Fn+1 ≤ φn . Now let us look into Fn+2 :
Fn+2 = Fn + Fn+1
(from def. of Fn+2 )
≤ φn−1 + φn
(by IH)
=φ
n−1
(1 + φ)
(because φ2 = 1 + φ).
= φn+1
Together with Fn+1 ≤ φn (by IH directly), we know P(n + 1) is also true.
Obviously, Fn ≥ φn−1 for all nonnegative integer n.
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Examples
Thank You!
If you have any questions, feel free to contact me.
My Email is [email protected].
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