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Prove √ is not a rational number

  • Thread starter polatinte
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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 [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?
 

Answers and Replies

  • #2
jgens
Gold Member
1,580
49
You can do this proof much more easily just by counting prime factors.
 

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