Asked by 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 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!
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!
Answers
Answered by
Steve
well, a^n = a^(m)*a^(n-m)
n-m > 0, so a^(n-m) < 1
n-m > 0, so a^(n-m) < 1
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.