Prove if $f(x) = g(x)$ for each rational number x and $f$ and $g$ are continuous, then $f = g$ $f,g: \mathbb{R} \to \mathbb{R}$
I'd like to see a sketch for this proof.
[sorry for posting errors, I am on a cell phone]

Suppose $h(x) = f(x) - g(x)$. Then $h(x)$ is continuous and $h(x) = 0$ if $x \in \mathbb{Q}$. Let $\alpha$ be irrational. If $h(\alpha) > 0$ then since it is continuous there is a neighborhood $I$ of $\alpha$ in which $h(x)$ is positive. Clearly this neighborhood also includes rational numbers at which $h(x) = 0$. This contradiction shows that we can't have $h(\alpha) > 0$. Similarly we can't have $h(\alpha) < 0$. Thus $h(\alpha) = 0$. So $h(x) = 0$ for all $x$.
