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Problem:

Prove that if $n < m$ with $n,m \in \mathbf{Z}$ and $x>1$, then $x^n < x^m$.

Proof:

Since $ x > 0$, we know $x^n > 0$.

Also, since $m > n$, it follows that $m - n \geq 1$. Thus $x^{m-n} > 1$.

Finally, $x^n(x^{m-n} -1) > 0$. Rearranging yields the desired result: $x^n < x^m$.

Question: I hope my proof is correct. I am actually curious whether it is possible to prove this result using induction? Could you somehow map $n,m$ from $\mathbf{Z}$ to the positive integers?

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    Your proof needs better justification because the step $x^{m-n}>1$ only works for $x>1$. I also prefer writing using $\log$: $$ x>1\implies\log(x)>0\implies(m-n)\log(x)>0\implies m\log(x)>n\log(x)\implies x^m>x^n. $$ It's clear here where the assumptions are used.2017-01-14
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    We are assuming that $x>1$, no? But I agree, the proof using log(x) is very clear.2017-01-14
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    Yes I know you're assuming $x>1$. It's just that with the way your current proof is written, it's not clear where that assumption is used. The assumption is very important of course because without it, your inequality may flip direction. Cheers.2017-01-14

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Induction proof(for positive $n$ and $m$):

As we have : $m > n > 0$. Let's prove the base : $m = 2, n = 1 : x^2-x > 0 $ - true for $x > 1$.

Now assume that's true for $n = 1$ and $m > n$, so let's prove it for $m = m+1$ :

$x^{m+1} - x > x^m - x > 0$ - true.

Now let's assume that's true for $n = n$ and $m > n$, let's prove it for $m = m+1$.

$x^{m+1}-x^n > x^m-x^n>0$ - true.


This proof could be expand for negative $m$ and $n$. Just use the same idea.


If $m =0$ or $n = 0$ then the proof reduced to compare $1$ and $x^{n}$ or $x^{m}$.

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    @RobArthan this proof could be expand for negative $m$ and $n$2017-01-14
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    It's easy to extend to the case of negative $m$ or $n$: suppose $n < m$ with $n$ nonpositive, and note that $x^n < x^ m$ is equivalent to $x < x^{m + 1 - n}$. Here both exponents are positive, and $1 < m - n + 1$, so the initial argument applies.2017-01-14
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    @RobArthan of course, thanks for remark!2017-01-14