Mathematical Induction?

Question

Prove by mathematical induction that (10^n - 1) is divisible by 9 for all positive integers n.

Answer 1

The first step is to check that 10^n - 1 is divisible by 9 when n equals one.

10^1 - 1 = 10 - 1 = 9 Yup.

The second step is to prove that if it works for any positive integer k, then it will also work for the next positive integer, k + 1

Assume that 10^k - 1 is divisible by 9, that is,
10^k - 1 = 9m for some positive integer m

Now start playing with 10^(k+1) - 1

10^(k+1) - 1 =
10*10^k -1=
10*10^k - 10 + 9 =
10(10^k - 1) + 9 =
10(9m) + 9 =
9(10m + 1)

Therefore, since 10^(k+1) can be written as a product of 9 and some positive integer, it is divisible by 9.

Therefore it is true that for any positive integer n, the expression
10^n - 1 is divisible by 9.

Answer 2

First, test for n=1:
10^1 - 1 is divisible by 9.

Next, assume true for n=k:
10^k - 1 is divisible by 9 for all poisitive integers k.

Finally, prove true for k+1:
10^(k+1) - 1 = 10^k + 8 = (10^k - 1) + 9

By the principle of mathematical induction, 10^n - 1 is divisible by 9 for all positive integers n.