Question

How can you prove that `log_10 2` is irrational ?

Answer 1

We can prove this by contradiction;

Explanation:

if `x=QQ` then `x=p/q` for `p,q in ZZ`

so hence let, `log_10 2 =p/q`
then `2 = 10^(p/q)`

so hence `2^q = 10^p`

`=> 2^q = 5^p 2^p`

`=> 2^(q-p) = 5^p`

we know `log_10 1 < log_10 2 < log_10 10`

so hence `0 < log_10 2 < 1`

so hence `0 < p/q < 1`

`p < q`

so `q-p > 0` and let `m=q-p`

But we can see that `2^m` is even and `5^p` is odd, hence there are no values for `p and q` that satisify this

`=>` Hence proven by contradiction