Mathematical induction

Maths · Mathematical induction and binomial theorem. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

Mathematical induction is a method of proving that a statement S(n) is true for every positive integer n. Checking a few values is not a proof, so the method uses two conditions.

Condition 1: show that the statement is true for n = 1. Condition 2: assume it is true for n = k, and use this to prove it is true for n = k + 1. Then the truth passes from 1 to 2, from 2 to 3, and so on without end.

Some statements are false for the first few values, such as n! > 2n. In the extended form you start at another integer i in place of 1, and the result is then proved for all integers n ≥ i.

Rules to remember

Common mistake

Students leave out condition 1, or assume S(k + 1) in place of proving it. Both conditions are needed: condition 2 alone proves nothing if the statement is false at the starting value.

Worked example

Show by mathematical induction that 9n − 1 is divisible by 8 for every positive integer n.

  1. n = 1: 91 − 1 = 8, which is divisible by 8, so condition 1 holds.
  2. Assume it is true for n = k: 9k − 1 = 8m for some integer m.
  3. For n = k + 1: 9k+1 − 1 = 9 × 9k − 1 = 9(9k − 1) + 8 = 9(8m) + 8.
  4. = 8(9m + 1), which is divisible by 8, so condition 2 holds.

Answer: Both conditions hold, so 9n − 1 is divisible by 8 for all positive integers n.

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