"Prove m√n is not a rational number for any natural numbers with n,m > 1, where n is not an mth power"
Natural numbers for us start at 1.
Since we know n is not an mth power, then n [itex]\neq[/itex] 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 [itex]\in[/itex] [itex]Z[/itex], b [itex]\neq[/itex] 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 [itex]\equiv[/itex] 1 (mod 2), then a2 [itex]\equiv[/itex] 12 = 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?