prove that $\sqrt{2}$ is not periodic. If  I am asked to show that $\sqrt{2}$ does not have a periodic decimal expansion. 
Can I just prove that $\sqrt{2}$ is irrational , and since irrational numbers are don't have periodic decimal expansions then I am done?
Thank you. 

Suppose $\sqrt{2}$ has a periodic decimal expansion, i.e. it has the form
$$\sqrt{2} = 1.a_1a_2 \cdots a_m \overline{d_1 d_2 \cdots d_n}$$
where the overline indicates the repeating digits. Then
$$10^m \sqrt{2} = 1a_1a_2 \cdots a_m . \overline{d_1 d_2 \cdots d_n}$$
and
$$10^{m+n} \sqrt{2} = 1a_1 a_2 \cdots a_m d_1 d_2 \cdots d_n . \overline{d_1 d_2 \cdots d_n}$$
Subtracting the first from the second yields
$$(10^{m+n}-10^m)\sqrt{2} = 1a_1 a_2 \cdots a_m d_1 d_2 \cdots d_n - 1a_1a_2 \cdots a_m$$
and hence
$$\sqrt{2} = \frac{1a_1 a_2 \cdots a_m d_1 d_2 \cdots d_n - 1a_1a_2 \cdots a_m}{10^{m+n}-10^m}$$
This implies that $\sqrt{2}$ is rational, which is known to be false.
This argument generalises to show that all irrational numbers have nonperiodic expansions (in any number base).
