Real Analysis Proof (Inequality)

[illjay7005]

If n belong to natural #'s, 0<a, 0<b, and a<b
then a^(1?n)<b^(1?n).

I am choosing to use induction.

base case- for n=1, we get a<b, so the claim is true for n=1.

inductive hypothesis- let n=k and assume that a^(1/k)<b^(1/k) for some k. We need to show that for all k, a^(1/k+1)<b^(1/k+1).

this is where i am having some trouble. i think this just comes down to algebraic manipulation, but i cannot see how to get from the assumption to the k+1. I know we know that 0<a, 0<b, and a<b so we know that 0<ab, 0<a+b, and 0<b-a. I tried using differnt manipulations of these to get what i need but could use some pointers if anyone would like to help.

thank you!

 

[mmm4444bot]

Given two distinct positive products, if each has its own set of n identical factors, then the smaller product must come from expanding the set of smaller factors because multiplication is nothing more than repeated addition.

Therefore, raising both sides of the given inequality to the nth power proves that the nth root of a is always less than the nth root of b, given that positive base a is less than positive base b.

Raising both sides of the given inequality to the nth power is an alternative to proof by induction.