Let a be a real number and let m, n be natural numbers such that m < n. Prove that if 0 < a < 1 --> a^n < a^m < 1.

Question

So, what I thought to start was: since 0 < a
0 < a^n Also a < 1, a^n < 1. By the property of real #s where if a < b there is a real number t such that a < t < b, since a^n < 1, a^n < a^m < 1. But I don't think that is enough to justify the problem. HELP!

Answer 1

well, a^n = a^(m)*a^(n-m)
n-m > 0, so a^(n-m) < 1