- #1

polatinte

- 1

- 0

## Homework Statement

"Prove

^{m}√n is not a rational number for any natural numbers with n,m > 1, where n is not an m

^{th}power"

## Homework Equations

Natural numbers for us start at 1.

Since we know n is not an m

^{th}power, then n [itex]\neq[/itex] k

^{m}for an arbitrary integer k.

## The Attempt at a Solution

I believe this would follow a similar argument for the classic proof that √2 is irrational.

We set up the contradiction that

^{m}√n is actually rational. Then a

^{m}=nb

^{m}for some a,b [itex]\in[/itex] [itex]Z[/itex], b [itex]\neq[/itex] 0, a,b are irreducible.

Then a

^{m}is a multiple of n, so a is a multiple of n.

I'm assuming I need to prove this as a lemma.

I've attempted proving this by modular arithmetic.

Prove by contrapositive (If a is not a multiple of n, then a

^{m}is not a multiple of n). But then there would be n-1 cases to check (infinitely many), so I follow with induction.

Base case: n=2, a [itex]\equiv[/itex] 1 (mod 2), then a

^{2}[itex]\equiv[/itex] 1

^{2}= 1 [itex]\equiv[/itex] 1 (mod 2).

Base case checks out, so we can assume the inductive hypothesis that this is true for all n.

To show show n+1, the (mod n) turns into (mod n+1), so how do I use the inductive hypothesis on this?

This is the point where I get lost. Any suggestions?