Mathematical Induction I Lecture 19 Section 4.2 Mon, Feb 14, 2005

The Principle of Mathematical Induction Let P(n) be a predicate defined for integers n. Let a be an integer. If the following two statements are true P(a) For all integers k  a, if P(k), then P(k + 1) then the statement For all integers n  a, P(n) is true.

The Principle of Mathematical Induction Symbolically, this can be written as P(a)P(a)  k  Z with k  a, P(k)  P(k + 1)  n  Z with n  a, P(n).

Proof by Mathematical Induction Basic Step Choose a starting point a (typically 0 or 1). Prove P(a). Inductive Step Suppose P(k) for some arbitrary integer k  a. Prove P(k + 1), using the assumption P(k). Conclude P(n) for all n  a.

Example: Mathematical Induction Theorem: For any integer n  4, we can obtain n¢ using only 2¢ and 5¢ coins. Let P(n) be the predicate “we can obtain n¢ using only 2¢ and 5¢ coins.” Proof: Basic Step: (Start at a = 4.) P(4) is true since 4¢ = 2¢ + 2¢.

Proof continued Inductive Step Suppose that P(k) is true for some k  4. We must show that P(k + 1) is true. Consider two cases: Case 1: k¢ uses a 5¢ coin. Case 2: k¢ does not use a 5¢ coin.

Proof concluded Case 1: k¢ uses a 5¢ coin. Then remove it and replace it with three 2¢ coins, thereby obtaining (k + 1)¢. Case 2: k¢ does not use a 5¢ coin. Then it must use at least two 2¢ coins. Replace two 2¢ coins with one 5¢ coin, thereby obtaining (k + 1)¢.

Proof concluded Therefore, P(k + 1) is true. Therefore, P(n) is true for all n  4.

Example: Mathematical Induction Theorem: Σ i = 1..n 1/(i(i + 1)) = n/(n + 1). Proof: Basic Step P(1) is true since Σ i = /(i(i + 1)) = 1/(1.2) = 1/2 = 1/(1 + 1).

Proof continued Inductive Step Suppose P(k) is true for some k  1. Then Σ i = 1..k+1 1/(i(i + 1)) = (Σ i = 1..k 1/(i(i + 1)) ) + 1/((k + 1)(k + 2) = k/(k + 1) + 1/((k + 1)(k + 2) = (k + 1)/(k + 2). Therefore, P(k + 1) is true. Therefore, P(n) is true for all n  1.

Example: Mathematical Induction Theorem: Σ i = 1..n i = n(n + 1)/2. Proof: Basic Step P(1) is true since Σ i = 1..1 i = 1 = 1  (1 + 1)/2.

Proof continued Inductive Step Suppose P(k) is true for some k  1. Then Σ i = 1..k+1 i = (Σ i = 1..k i ) + (k + 1) = k(k + 1)/2 + (k + 1) = (k + 1)(k + 2)/2. Therefore, P(k + 1) is true. Therefore, P(n) is true for all n  1.

Sums of Powers of Integers We can also prove by induction that Σ i = 1..n i 2 = n(n + 1)(2n + 1)/6. Σ i = 1..n i 3 = n 2 (n + 1) 2 /4. Σ i = 1..n i 4 = n(n + 1)(2n + 1)(3n 2 + 3n – 1)/30. Σ i = 1..n i 5 = n 2 (n + 1) 2 (2n 2 + 2n – 1)/12.

Putnam Question B-1 (1981) Find lim n  [(1/n 5 )  h = 1..n  k = 1..n (5h 4 – 18h 2 k 2 + 5k 4 )]. Solution Express  h  k 5h 4,  h  k 18h 2 k 2, and  h  k 5k 4 as polynomials in n. Simplify the polynomials. Divide by n 5. Take the limit as n  . The answer is -1.

Mathematical Induction Mathematical induction requires that we “know” the answer in advance. The method verifies the answer. How would we come up with the guess that Σ i = 1..n i 2 = n(n + 1)(2n + 1)/6?

Finding the Formula We might conjecture that the answer is a cubic polynomial in n. That is, Σ i = 1..n i 2 = an 3 + bn 2 + cn + d for some real numbers a, b, c, and d.

Finding the Formula Then substitute n = 0, n = 1, n = 2, and n = 3 into the equation to get a system of four equations. n = 0: 0 = d. n = 1: 1 2 = 1 = a + b + c + d. n = 2: = 5 = 8a + 4b + 2c + d. n = 3: = 14 = 27a + 9b + 3c + d.

Finding the Formula Solve the system d = 0, a + b + c + d = 1, 8a + 4b + 2c + d = 5, 27a + 9b + 3c + d = 14. for a, b, c, and d. a = 1/3, b = 1/2, c = 1/6, d = 0. Then verify using mathematical induction.

Let’s Play “Find the Flaw” Theorem: For every positive integer n, in any set of n horses, all the horses are the same color. Proof: Basic Step. When n = 1, there is only one horse, so trivially they are (it is) all the same color.

Find the Flaw Inductive Step Suppose that any set of k horses are all the same color. Consider a set of k + 1 horses. Remove one of the horses from the set. The remaining set of k horses are all the same color.

Find the Flaw Replace that horse and remove a different horse. Again, the remaining set of k horses are all the same color. Therefore, the two horses that were removed are the same color as the other horses in the set. Thus, the k + 1 horses are all the same color.

Find the Flaw Thus, in any set of n horses, the horses are all the same color.