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I'm going through Lang's "Basic Mathematics", and one of the exercises is to prove that if $0 < a < b$ for $a, b \in \Bbb{R}$ then $a^{1\over{n}} < b^{1\over{n}}.$

Earlier we were asked to prove that $a^{n} < b^{n}$ with the same assumptions. I've proven this by induction.

Basis: for $n = 1$ we have $a^{n} = a < b = b^{n}.$

Inductive step: assume $a^{k} < b^{k}.$ We multiply by $a$ on both sides, and we get $a^{k+1} < ab^{k}.$ Then we multiply $a < b$ by $b^{k}$ on both sides to get $ab^{k} < b^{k+1}.$ Since we're dealing with positive numbers, it's safe to do the multiplications. So now from our inequalities we get $a^{k+1} < b^{k+1},$ and we're done.

Now for the main problem. We start by taking $r$ and $s$ such that $r^n = a$ and $s^n = b$---we're told earlier it's OK to assume the existence of real roots. Then we have $a < b \implies r^n < s^n.$ and using the result we've just proved, we get $r < s \implies a^{1\over{n}} < b^{1\over{n}}.$

My question is: is it OK use the proposition we've proved inductively like this. It seems fine to me, because we've proven it for all $n.$ So, even if we've used it to go downards, we're still OK.

NplusOne
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  • I'm confused how exactly did you "use the result you've just proved". You've proved $a<b\implies a^n<b^n$. Where is this used? To fix this you can first prove $a<b\Leftarrow a^n<b^n$ (for example by using the contrapositive) and then use this for your main problem. – Benjamin Wang Oct 29 '23 at 22:06
  • That's the bit I wasn't sure about. I proved $a<b \implies a^n<b^n$ by inductions. Then I intended to use $a^n < b^n \implies a < b$ because it was proved by induction, and was wondering if that is valid. – NplusOne Oct 29 '23 at 22:38
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    Welcome to MSE! <> It seems to me you're not using induction, but contraposition: If $0 < b \leq a$, then by what you've proven, $0 < b^n \leq a^n$. Contrapositively, if $a$ and $b$ are positive and the conclusion is false, i.e., $a^n < b^n$, then the hypothesis is false, i.e., $a < b$. – Andrew D. Hwang Oct 29 '23 at 23:11
  • Thanks, Andrew :) I see how I could have used the contrapositive to get $a^n < b^n \implies a < b.$ – NplusOne Oct 30 '23 at 08:03

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