Re: Rational number proof

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Originally Posted by

**euphony** I'm supposed to prove that log(*m*)/log(*n*) is rational if and only if there is some integer *k* such that *m* and *n *(which are integers) are powers of *k*.

It's an "if and only if" proof, so there are two parts:

a) Show that the existence of said integer *k* implies that log(*m*)/log(*n*) is rational. I did this, but I don't understand how to do the second part:

b) Show that log(*m*)/log(*n) *being rational implies that there is some integer *k* such that *m* and *n* are powers of *k*.

If log(*m*)/log(*n*) is rational, it can be represented as *a*/*b* where *a* and *b* are integers and *b* is not 0.

And, a/b is reduced to lowest terms. That is, a and b have no factors in common.

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log(*m*)/log(*n*) = *a/b *implies that *m^b = n^a*.

But I don't know where to go from there.

m^b= n^a is the same as saying m= n^(a/b). Since m is an integer, so is n^(a/b) and since a and b have no common factors, so is n^(1/b). Let k= n^(1/b).

Re: Rational number proof

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and since a and b have no common factors, so is n^(1/b)

I don't quite understand this.

Re: Rational number proof

Re: Rational number proof

I don't understand why a and b having no common factors implies that n^(1/b) has to be an integer.

Re: Rational number proof

Re: Rational number proof

Hi,

Maybe what you're missing is this fact: if b and c are integers greater than 1, then the bth root of c is integral iff for any prime p with p^{k} exactly dividing c, b divides k. From this it immediately follows that if a and b are relatively prime with n^{a/b} an integer, then n^{1/b} is an integer.

By the way, your original problem statement should have included the statement that m > 1, otherwise of course it's false. I think you should foster the habit of specifying the exact hypotheses of a statement.