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I believe this is an induction problem.

Let $a, b$ be positive integers with $a < b$. Prove that for any natural number $n$, $a^n < b^n$.

I feel I should start with a base case $n = 1$ which yields true since $a$ is already less than $b$.

Next I would implement the induction hypothesis, but I'm kinda shaky on what that is.

After that I would check the $n + 1$ case.

Could someone check and verify what I'm doing?

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    Minor point, post should be edited to make sure negatives are not allowed. For $-10\lt -2$, but $(-10)^2\gt (-2)^2$.2012-11-28

2 Answers 2

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For $n\in\Bbb Z^+$ let $P(n)$ be the statement that $a^n. You get the induction off the ground by showing that $P(1)$ is true; indeed that’s simply the original hypothesis, that $a. The induction step is to show that if $P(n)$ is true for some positive integer $n$, then $P(n+1)$ is also true. Thus, your induction hypothesis is $P(n)$: $a^n. From this assumption you want to prove $P(n+1)$, i.e., that $a^{n+1}. You can do this in two steps. First multiply your induction hypothesis by $a$ to conclude that $a^{n+1}. Then multiply the inequality $a by $b^n$, and put the pieces together to get $a^{n+1}.

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    @blutuu: You’re welcome; I’m glad it clicked.2012-11-28
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You're off to a good start. For the inductive step, assume that $a^k < b^k$ for some $k \geq 1$.

Then, what can you conclude?

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    Looking at Brian Scott's answer I would multiply both sides by a and then by b^n. I'm not sure about it on my own.2012-11-28