Mirie

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. 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...

Prove: [1/sqrt(2)] [sqrt(a) + sqrt(b)] <= sqrt(a + b) <= sqrt(a) + sqrt(b) for all non-negative real numbers a and b.