Proof that Log2 of 5 is irrational

[Pascal's Pal]

Homework Statement

Prove that log2 of 5 is irrational.

Homework Equations

None.

The Attempt at a Solution

I just had a glimpse of the actual solution, but I'm wondering if mine would work too.

2^(a/b) = 5

square both sides...

2^(2a/b) =25

2 = 25^(b/2a)

(b/2a) = log25 of 2

b = 2aLog25 of 2

b is even...

and through a similar process...by taking the square root of both sides of "2^(a/b) = 5" you can arrive at a being even too. So how can they both be even etc etc.

 

[Kummer]

$$\log_2 5 = a/b$$ so $$2^{a/b} = 5 \implies 2^a = 5^b$$. Now use unique factorization.

 

[Pascal's Pal]

$$\log_2 5 = a/b$$ so $$2^{a/b} = 5 \implies 2^a = 5^b$$. Now use unique factorization.

But does mine work?

 

[rock.freak667]

$$\log_2 5 = a/b$$ so $$2^{a/b} = 5 \implies 2^a = 5^b$$. Now use unique factorization.

What is unique factorization ?