Prove √ is not a rational number

Homework Statement

"Prove m√n is not a rational number for any natural numbers with n,m > 1, where n is not an mth power"

Homework Equations

Natural numbers for us start at 1.
Since we know n is not an mth power, then n $\neq$ km 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 am=nbm for some a,b $\in$ $Z$, b $\neq$ 0, a,b are irreducible.
Then am 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 am 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 $\equiv$ 1 (mod 2), then a2 $\equiv$ 12 = 1 $\equiv$ 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?

Answers and Replies

Related Calculus and Beyond Homework Help News on Phys.org
jgens
Gold Member
You can do this proof much more easily just by counting prime factors.