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Can I prove $a^n<b^n$ when $0<a<b$ by inducton?

It's easy to see when $n=1$: $a^n<b^n$ is true. my attempt is from

$a^{n+1} = (a^n)*a$

$b^{n+1}=b^n*b$

then don't know what to do next.

It look so easy but I just can't say it in mathematical language, will you help me? thank you in advance.

Daniel Fischer
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user401653
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2 Answers2

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Suppose that $a^n<b^n$, you need to prove that $a^{n+1}<b^{n+1}$, since $a^n<b^n$ and $a>0$ , multiplying both sides by $a$ we get that $$a^{n+1}<b^na$$ since $a<b$ and $b>0$, multiplying both sides by $b^n$: $$ab^n < b^{n+1}$$ Now, using transitivity: $$a^{n+1}<ab^n< b^{n+1} \rightarrow a^{n+1}<b^{n+1}$$

Dieter Enns
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By induction for $n=1$ we have that $a<b \Rightarrow a^1<b^1 \Rightarrow a<b$

Now, suppose that $a<b\Rightarrow a^n<b^n$, this would be our induction hypothesis. We have to prove that $a<b\Rightarrow a^{n+1} <b^{n+1}$

Then, $a<b\Rightarrow a^n<b^n \Rightarrow a^na<b^na \Rightarrow a^{n+1}<b^n a$

We also have that $a<b\Rightarrow ab^n<bb^n\Rightarrow ab^n<b^{n+1}$

Now, $a^{n+1}<ab^n<b^{n+1} \Rightarrow a^{n+1}<b^{n+1}$

$\therefore a<b\Rightarrow a^n<b^n$