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If $c>1$ Prove that $c^m>c^n$ if and only if $m>n$. I don't know how to apply mathematical induction to this statement. Can anyone help?

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    Assuming $c,m$ and $n$ are all able to be arbitrary real numbers, induction won't work. – 2017-02-26

2 Answers 2

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I'll assume c $\in \mathbb{Z}$ and $m,n >1$

Check if true for $c= 2$:

Note that $2^m, 2^n>0$

We have that: $2^m - 2^n> 0 \Leftrightarrow 2^{m} > 2^{n}\Leftrightarrow 2^{m-n} > 1 \Leftrightarrow m-n > 0 \Leftrightarrow m>n$

There it's true for $n=1$ as each statement was an if and only if one.

Now we assume true for $c$:

$c^{m} > c^{n} \Leftrightarrow m>n$

Then we must prove it's also true for $c+1$:

$(c+1)^{m} > (c+1)^{n} \Leftrightarrow (c+1)^{m-n} > 1 \Leftrightarrow m>n $

Hence true $\forall c \in \mathbb{Z}^{+}$

Note that we could have proved this without induction, using the proof we did for $c+1$.

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    Where is the flaw in my logic? – 2017-02-26
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    I demonstrated there (with some inbetween iff steps) that $2^{m} > 2^{n} \Leftrightarrow m>n$ which is what is required to the OP's statement to be true for c =2 – 2017-02-26
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    In fact, my statement $2^{n} ,2^{m} > 0$ is unnecessary, the following line is self-sufficient - I suppose to make it clearer, I should say "Note that", instead of "since" – 2017-02-26
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    Sorry for the previous rushed comment. Here's the problem that persists after your clarification. When you say $2^{m-n} > 1 \Leftrightarrow m - n > 0$ you are begging the question. While it is not explicitly stated what properties of powers may be used, it's reasonable to assume that $2^{m+n} = 2^m \cdot 2^n$ is known to be true, and that $2^{m-n} > 1 \Leftrightarrow m-n > 0$ may not be used, even though it's not a generic $c$, but a specific $2$. We also know $2>1$, and that's the key to the base step of the induction. – 2017-02-26
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    @Fabio Somenzi Ok let me phrase it like this: Suppose $2^{m} > 2^{n}$, then I prove that $m>n$. Now suppose $m>n$, then I prove that $2^{m} > 2^{n}$. Then I have proved from for c =2. Assuming you're ok with this, then what I have written is exactly equivalent to this. – 2017-02-26
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    I'm not OK with that. I understand the chain of "if and only if." What I'm saying is that the step I mentioned in my previous comment consists of assuming the result you want to prove holds in the proof itself. – 2017-02-26
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    @Fabio Somenzi I don't see how it does. Take any A iff B statement, and in order to prove it, you have to suppose A then prove B and vice versa. – 2017-02-26
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    Don't look at the whole chain of $\Leftrightarrow$ steps. Focus on this: $2^{m-n} > 1 \Leftrightarrow m-n > 0$. This is the statement of the theorem, and you are using it to prove the theorem itself. This is what is technically called "begging the question" or "petitio principii," or "circular reasoning." – 2017-02-26
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    That's the thing though, I haven't used the statement of theorem. I started with $2^{m}>2^{n}$, proved a result m>n and showed that if I assume m>n, I can show $2^{m}>2^{n}$. Unless you're saying my steps wouldn't be sufficient to constitute a proof (i.e. I haven't shown that $2^{m} > 2^{n}$ implies $ m>n$) – 2017-02-26
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    Exactly. You have not shown that $2^m > 2^n$ is equivalent to $m > n$, because you've shown that $2^m > 2^n$ if and only if $2^{m-n} > 1$ (using the basic laws of powers) and also proved that $m-n>0$ if and only if $m > n$. That's all well and good, but the important proof obligation is to prove that $2^{m-n} > 1$ if and only if $m-n>0$, which you take for granted, but is effectively *the theorem*. – 2017-02-26
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    I understand you now, admittedly I would have included more but the OP's post wasn't in latex and I kind of guessed that he wasn't at a high enough level in math for a more rigorous proof to be required(note we are not told what definitions can be used in the question - or what c,m and n are). – 2017-02-26
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    Agreed. But the suggestion that induction should be used should not be ignored. I don't think there's a need for the special case $c=2$ if things are done properl. I have to leave SE now, but I'm glad we came to an agreement. – 2017-02-26
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$\dfrac{c^m}{c^n}=c^{m-n} \gt 1 \text{ for } m\gt n $

$\implies c^m \gt c^n \text{ for } m\gt n $