I'll answer a slight generalization of your brief question. Although much of this information is already in the comments, I feel that putting it all in one place would be helpful.

The statement I'll prove is the following:
Suppose $\alpha$ is irrational (transcendental), then if $q\ne 0$ is rational, $q\alpha$ is also irrational (transcendental resp.)

I'll assume you're familiar with the facts that if $q,r$ are rational, then
$qr$ is rational, and when $q\ne 0$, $1/q$ is also rational.

Then suppose $\alpha$ is irrational, then if $q\alpha$ is rational, say $q\alpha = r$ for a rational $r$, then $\alpha = (1/q)r$, which is the product of two rational numbers, and therefore rational. This contradicts the irrationality of $\alpha$. Hence $q\alpha$ must be irrational.

Now suppose $\alpha$ is transcendental. If $q\alpha$ were not transcendental, i.e. algebraic, then we have a nonzero polynomial with rational coefficients $p(x)$ such that $p(q\alpha)=0$.
Let $$p(x)=a_nx^n + a_{n-1}x^{n-1}\cdots+a_1x +a_0.$$
Then
$$p(q\alpha)
=a_n(q\alpha)^n +a_{n-1}(q\alpha)^{n-1}+\cdots +a_0
=a_nq^n\alpha^n + a_{n-1}q^{n-1}\alpha^{n-1}+\cdots+a_0
=0,
$$
so if we let
$r(x)$ be the polynomial
$$r(x)=a_nq^nx^n+a_{n-1}q^{n-1}x^{n-1}+\cdots +a_1qx+a_0$$,
since each of the coefficients $a_iq^i$ is rational (because products of rational numbers are rational), this is a polynomial with rational coefficients. But by construction, we see that $r(\alpha)=0$, contradicting the fact that $\alpha$ is transcendental. Hence $q\alpha$ must in fact be transcendental.