Prove $f(x) = 0$ for all rational numbers implies $f(x)=0$ for all reals. 
Let $F:\mathbb{R}\to\mathbb{R} $ be a continuous function such that $F(x)=0$, $\forall x\in \mathbb{Q}$.
Show that $F(x)=0$, $\forall x\in \mathbb{R}$.

I said: Let $k\in \mathbb{Q}$, $\varepsilon,\delta\in\mathbb{R}\setminus \mathbb{Q}$ such that $k-\varepsilon\le k \le k+\delta$
Since $F(k)=0$, then $F(k-\varepsilon)>0$, $F(k+\delta)<0$ (or $F(k-\varepsilon)<0$, $F(k+δ)>0$).
Then, I thought that before $k-\varepsilon$ there's another rational number and between them is another. If it's a rational, then there's another irrational between the two quotients etc. etc. and my mind explodes. Any hints?
 A: From density of rationals in $\Bbb{R}$ we have that:
If $x \in \Bbb{R}$ then exists $q_n \in \Bbb{Q}$ such that $q_n \to x$.
Thus $$0=f(q_n) \to f(x)$$ from continuity.
From uniqueness of limit we have that $f(x)=0$
Thus $f(x)=0,\forall x\in \Bbb{R}$.
A: Let $x\in \mathbb{R}$. Fix an $\epsilon>0$, we can find a $\delta>0$ with
$$
|f(x)-f(y)|<\epsilon 
$$
for any $y$ with $|y-x|<\delta$. But density insures we can find a $q$ rational with $|x-q|<\delta$. Then we have 
$$
|f(x)-f(q)|=|f(x)|<\epsilon
$$
and so $f(x)=0$ on the reals as well.
A: Suppose $f(x)\ne0$ for some $x$; then either $f(x)>0$ or $f(x)<0$.
Let's do the case $f(x)>0$; by continuity there exists $\delta>0$ such that, for $|y-x|<\delta$,
$$
|f(y)-f(x)|<f(x)/2
$$
(taking $\varepsilon=f(x)/2$ in the definition of continuity).
Can you see a contradiction? Hint: there's a rational number $y$ such that $x-\delta<y<x+\delta$.
Can you do the case $f(x)<0$?
A: Recall that $F:\mathbb{R}\to\mathbb{R}$ is continuous if for every sequence $(x_n)$ converges to $x,$ then the sequence $(f(x_n))$ will converge to $f(x).$
Fix $x\in\mathbb{R}\setminus\mathbb{Q}.$
Let $(x_n)$ be a sequence of rational numbers which converges to $x$ (the sequence $(x_n)$ can be obtained by taking $x_n$ to be truncated digit of $x$ up to $n$th digit. For example, if $x=\sqrt{2},$ then $x_1=1.4, x_2=1.41,x_3=1.414,$etc.)
By assumption, $f(x_n)=0$ for all natural number $n.$
By continuity of $f,$ we must have $f(x)=0.$
Since $x$ is arbitrary, therefore $f(x)=0$ for all real number $x.$
A: Left to show:
$F(a) = 0$,  $a \in \mathbb{R}$ \ $\mathbb{Q}.$
$\mathbb{Q}$ is dense in $\mathbb{R}:$
There exists sequence   $ x_n \in \mathbb{Q}$ with 
$\lim_{n \rightarrow \infty} x_n =a.$
Since $F$ is continuous at $a$ this implies:
$\lim_{n \rightarrow \infty} F(x_n) =F(a).$
For $\epsilon\gt 0$  there exists a $n_0$ , such that for $n \ge n_0$,  we have:
$|F(x_n) - F(a)| \lt \epsilon$,
and with $F(x_n)=0$:
$|F(a)| \lt \epsilon. $
Since $\epsilon$ can be chosen arbitrarily, this implies 
$F(a) =0.$
A: If $F(i) > 0$ for some irrational $i$ then, by continuity, there will be some neighbourhood of $i$, say $(i-\alpha,i+\alpha)$ in which $F(x)>0$ for every $x \in  (i-\alpha,i+\alpha)$, but, every neighbourhood of $i$ contains rationals. Same if $F(i) < 0$.
