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Prove that if m, n are natural, then the root |
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| Nov4-07, 05:38 AM | #1 |
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Prove that if m, n are natural, then the root
Hi,
I've encountered this exercise which I'm having a hard time proving. It goes like this: Prove that if m and n are natural, then the nth root of m is either integer or irrational. Any help would be greatly appreciated. Thanks. |
| Nov4-07, 07:07 AM | #2 |
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I would imagine you would start by proving it for m= p, a prime number. That would make it easy to prove "If integer k is not divisible by p then neither is kn" so you could mimic Euclid's proof that [itex]\sqrt{2}[/itex] is irrational.
After that, look at products of prime. |
| Nov4-07, 05:15 PM | #3 |
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if p is prime and divides m^n, p must divide m...
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