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.