Proving that 1/3 has no finite decimal representation

Question

There is a problem where i need to prove that 1/3 has no finite decimal representation

Here's my proof, can someone tell me if its valid?

Proof

Lets assume there is a decimal representation for $\frac{1}{3}$, Therefore:

$ \exists n,b \in \mathbb{N} $ : $ (\frac{b}{10^n}=\frac{1}{3}$) $ \land (\sum_{k=1}^n \frac{a_k}{10^k}=\frac{1}{3})$

By the theorem: $\frac{1}{3} = \frac{b}{10^n} $

b = $\frac{10^n}{3}$ = $\frac{(2 \times 5)^n}{3}$

Thats a contradiction ($b \notin \mathbb{N}$), Since that fraction is irreducible (Both 2,5,3 are prime numbers).

Is my proof valid? If not, Can someone explain what's wrong with it?

Thanks.

Answer 1

It is almost correct, but you should not write $2.5$ when what you mean is $2\times5$.

And, yes, $\frac{(2\times5)^n}3=\frac{2^n5^n}3$, which is indeed an irreducible fraction. You didn't say why it is irreducible, but it is easy: since $3$ is prime and $3\nmid2^n5^n$, $3$ and $2^n5^n$ are coprime and therefore, yes, the fraction is irreducible.