# Rational number proof

• Sep 21st 2013, 09:33 AM
euphony
Rational number proof
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. log(m)/log(n) = a/b implies that m^b = n^a.
But I don't know where to go from there.
• Sep 21st 2013, 10:49 AM
HallsofIvy
Re: Rational number proof
Quote:

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.

Quote:

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).
• Sep 21st 2013, 11:07 AM
euphony
Re: Rational number proof
Quote:

and since a and b have no common factors, so is n^(1/b)
I don't quite understand this.
• Sep 21st 2013, 02:40 PM
Plato
Re: Rational number proof
Quote:

Originally Posted by euphony
I don't quite understand this.

You you understand why $\displaystyle m=n^{\frac{a}{b}}~?$

If so, $\displaystyle m$ is an integer, therefore $\displaystyle n^{\frac{a}{b}}$ is also an integer.

But because $\displaystyle \frac{a}{b}}$ is reduced form that means $\displaystyle n^{\frac{1}{b}}$ must an integer.

Let $\displaystyle k=n^{\frac{1}{b}}$ so $\displaystyle m=k^a~\&~n=k^b$.
• Sep 22nd 2013, 06:48 AM
euphony
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.
• Sep 22nd 2013, 09:02 AM
Plato
Re: Rational number proof
Quote:

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

Oh come on. If you are going to work at this level then think at this level.

It is a matter of standard practice. If $\displaystyle \rho$ is a rational number then $\displaystyle \exists\{a,b\}\subset\mathbb{Z}$ such that $\displaystyle \text{GCD}(a,b)=1$ and $\displaystyle \rho=\frac{a}{b}$.
• Sep 22nd 2013, 09:28 AM
johng
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 pk exactly dividing c, b divides k. From this it immediately follows that if a and b are relatively prime with na/b an integer, then n1/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.