Mathematical Induction

Let's begin with an example.

Example: A Sum Formula

Theore. For any positive integer n, 1 + 2 + ... + n = n(n+1)/2.

Proof. (Proof by Mathematical Induction) Let's let P(n) be the statement "1 + 2 + ... + n = (n (n+1)/2." (The idea is that P(n) should be an assertion that for any n is verifiably either true or false.) The proof will now proceed in two steps: the initial step and the inductive step.

Initial Step. We must verify that P(1) is True. P(1) asserts "1 = 1(2)/2", which is clearly true. So we are done with the initial step.

Inductive Step. Here we must prove the following assertion: "If there is a k such that P(k) is true, then (for this same k) P(k+1) is true." Thus, we assume there is a k such that 1 + 2 + ... + k = k (k+1)/2. (We call this the inductive assumption.) We must prove, for this same k, the formula 1 + 2 + ... + k + (k+1) = (k+1)(k+2)/2.

This is not too hard: 1 + 2 + ... + k + (k+1) = k(k+1)/2 + (k+1) = (k(k+1) + 2 (k+1))/2 = (k+1)(k+2)/2. The first equality is a consequence of the inductive assumption.

The Math Induction Strategy

Mathematical Induction works like this: Suppose you want to prove a theorem in the form "For all integers n greater than equal to a, P(n) is true". P(n) must be an assertion that we wish to be true for all n = a, a+1, ...; like a formula. You first verify the initial step. That is, you must verify that P(a) is true. Next comes the inductive step. Here you must prove "If there is a k, greater than or equal to a, for which P(k) is true, then for this same k, P(k+1) is true."

Since you have verified P(a), it follows from the inductive step that P(a+1) is true, and hence, P(a+2) is true, and hence P(a+3) is true, and so on. In this way the theorem has been proved.

Example: A Recurrence Formula

Math induction is of no use for deriving formulas. But it is a good way to prove the validity of a formula that you might think is true. Recurrence formulas are notoriously difficult to derive, but easy to prove valid once you have them. For example, consider the sequence a₀, a₁, a₂, ... defined by a₀ = 1/4 and a_n+1 = 2 a_n(1-a_n) for n ≥ 0.

Theorem. A formula for the sequence a_n defined above, is a_n = (1 - 1/2^2ⁿ)/2 for all n greater than or equal to 0.

Proof. (By Mathematical Induction.)

Initial Step. When n = 0, the formula gives us (1 - 1/2^2ⁿ)/2 = (1 - 1/2)/2 = 1/4 = a₀. So the closed form formula ives us the correct answer when n = 0.

Inductive Step. Our inductive assumption is: Assume there is a k, greater than or equal to zero, such that a_k = (1 - 1/2^{2^k})/2. We must prove the formula is true for n = k+1.

First we appeal to the recurrsive definition of a_k+1 = 2 a_k(1-a_k). Next, we invoke the inductive assumption, for this k, to get

a_k+1 = 2 (1 - 1/2^{2^k})/2 (1 - (1 - 1/2^{2^k})/2) = (1 - 1/2^{2^k})(1 + 1/2^{2^k})/2 = (1 - 1/2^{2^k+1})/2. This completes the inductive step.

Exercises

Prove each of the following by Mathematical Induction.

For all positive integers n, 1² + 2² + ... + n² = (n)(n+1)(2n+1)/6.
Define a sequence a₀, a₁, a₂ by the recurrsive formula a_n+1 = 2 a_n - a_n². Then, a closed form formula for a_n is a_n = 1 - (1 - a₀)^2ⁿ for all n = 0, 1, 2, ....

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