Suppose a,b are real numbers. I'm trying to prove that
$\forall n\geq 1$ ( $0<a<b$ entails $0 <a^{1/n}<b^{1/n}$ )
with the method of Induction.
P.S : I already know how to prove by contradiction.
$n=1$
This is easy to prove.
Inductive Hypothesis
Assume $0 <a^{\frac{1}{k}} <b^{\frac{1}{k}}$ .
I need to reach $0 <a^{1/(k+1)}<b^{1/(k+1)}$ .
I've found that $\frac{1}{k+1} = \frac{1}{k} - \frac{1}{k(k+1)} $.
So, if i can find the following inequation :
$0 <a^{\frac{1}{k(k+1)}} <b^{\frac{1}{k(k+1)}} $
We could divide the inductive hypothesis by it and reach the conclusion ... but i'm not sure this is the way to go ( i can't derive that one either ).
Thanks in advance.