Show if $0 \le a <b$ implies $0 \le a^{\frac{1}{n}}<b^{\frac{1}{n}}$

Question

Given that $0\le a<b$ show that $0\leq a^{1/n}<b^{1/n}$

Is this proof by induction?

Show it's correct for $n=1$

Assume true for $n=k$, then $0\leq a^{1/k}<b^{1/k}$ holds for some $k$,

Show that this holds for $0\leq a^{1/{k+1}}<b^{1/{k+1}}$ , but I can't get it to show this.

Answer 1

Can you show that if $a,b \ge 0$ then $b \le a$ implies $b^n \le a^n$?

Its contrapositive is the statement $a^n < b^n$ implies $a < b$.

Answer 2

$x\mapsto x^{1/n}$ is increasing, therefore the claim follow.