Is my proof for (log10 a)/(log10 b) being irrational correct?

[Thinker8921]

I can never seem to create proofs the way it is shown in every textbook I've seen. To be honest, I don't really know how to write the proofs correctly. I've seen sometimes my reasons are flawed and other times I go around aimlessly and get home after some unnecessary steps. So I would just like some feedback here:

Homework Statement

Prove that (log₁₀ a)/(log₁₀ b) is not rational if a and b are relatively prime and both are greater than 1.

Homework Equations

The Attempt at a Solution

Suppose that (log₁₀ a)/(log₁₀ b) is rational, then they are expressed as fraction:
(log₁₀ a)/(log₁₀ b) = p/q, where p and q are integers. q not 0. p not 0 because a is greater than 1.
Let 10^x = a
and 10^y = b
x and y cannot be zero because a and b cannot be 1.
Now if a and b are relatively prime, ie they have only 1 as their common factor then x and y too are relatively prime.

Reason:
Suppose that x and y share a common factor greater than 1, then:
let x = kc
and y = kz
where k is the common factor, which is not negative or 0 because a and b are greater than 1 and c and z are also not 0 or negative.
So then, 10^kc = a
and 10^kz = b
x = log₁₀ a
y = log₁₀ b
So, kc = log₁₀ a
and kz = log₁₀ b
But then, (10^k)^c = a
and (10^k)^z = b
And this contradicts the fact that a and b are relatively prime. So x and y are also relatively prime.

So now we can write:
x/y = p/q
i.e, xq = py
This cannot be true because x and y are relatively prime. This means x/y is not rational.
And since x = log₁₀ a
and y = log₁₀ b
(log₁₀ a)/(log₁₀ b) is not rational.

Thank-you

 

[tiny-tim]

Hi Thinker8921!

hmm … this is really complicated

you're introducing x and y for no particular reason.

Just do …

Suppose that (log₁₀ a)/(log₁₀ b) is rational, then they are expressed as fraction:
(log₁₀ a)/(log₁₀ b) = p/q, where p and q are integers. q not 0. p not 0 because a is greater than 1.

… then write that as q(log₁₀ a) = p(log₁₀ b), and carry on from there.

(or better still use log_b a, if you know how to do that)