Proove that

Question

log₃² is not a rational number

Answer 1

This is log base 3 of 2, right?

 

Suppose it were a rational number.  That means you could write it as some fraction x/y, where x and y are integers.  So you'd have:

 

log₃ 2 = x/y

 

Now raise both sides to powers of 3, in other words:

 

3^{log₃ 2} = 3^x/y

2 = 3^x/y

 

Now raise both sides to the y power:

 

2^y = [3^(x/y)]^y = 3^(x/y)y = 3^x

 

Let's call 2^y M.  So M the prime factoring of M would be 2^y.  There are no other integers in this factoring other than 2.

 

But if this is also equal to 3^x, then you could also factor M as 3^x, which has no other integers in it than 3.  That means that M would have two completely different prime factorings, which is not possible.

 

Therefore, our original assumption, that log₃ 2 can be expressed as a rational number, must not be true.