If $F(x)$ is a continuous function in $[0,1]$ and $F(x) = 1$ for all rational numbers then $F(1/\sqrt2) = 1$. (True/ False)
I think the statement is true because since $F$ is continuous and $1/\sqrt2$ is in neighbourhood of some rational number so $F(1/\sqrt2)$ should be one so that limit can exist. Is this reason valid to prove the statement true.
On
The statement is true, although I would express it: there is a sequence $\{r_n\}_{n\in\mathbb{N}}$ of rational numbers in the interval $[0,1]$ such that $r_n \to 1/\sqrt{2}$, hence by continuity $$ F(1/\sqrt{2}) = F(\lim_{n\to\infty} r_n) = \lim_{n\to\infty} F(r_n) = \lim_{n\to\infty} 1 = 1 \,. $$
For every $n$, there exists a rational number $q_n$ such that $\mid \frac{1}{\sqrt{2}}-q_n\mid < \frac{1}{n}$. So $\displaystyle\lim_{n\rightarrow \infty}q_n$ = $\frac{1}{\sqrt{2}}$. Since F is continuous, 1 = $\displaystyle\lim_{n\rightarrow \infty}$F($q_n$) = F($\frac{1}{\sqrt{2}}$) .