Can we prove that $1/\sqrt2\in$ Cantor set on $[0,1]$? Today, I was trying to prove Cantor set is uncountable and I completed it just a while ago.
So, I know that the end-points of each $A_n$ are elements of $C$ and those end-points are rational numbers. But since $C$ is uncountable, $C$ must contain uncountable numbers of irrational numbers. Then, is their a way to prove that, a specific irrational number (say $1/\sqrt2$ or $1/4\pi$) belongs to the set $C$ or not?
(Description of notation can be found in the link given above or here)
Lets say,

Prove or disprove that $1/\sqrt2\in$ Cantor set on $[0,1]$.

Can we do that? Or is their a way to solve such problem?

Any number in $[0, 1]$ which has (at least one) base $3$ expansion without a $1$ will be in the Cantor set. More precisely,
$$\mathcal{C} = \left\{ \sum_{n=1}^{\infty} \frac{c_n}{3^n} : (\forall n)( c_n \in \{ 0, 2 \} ) \right\}.$$
It is known that a number as above is irrational if and only if the expansion is non-recurring. So for example $0{.}202202220222202222202\ldots$ will be an irrational number in the Cantor set.
But you seem to be looking for an irrational number expressed by radicals, for example $\frac{\sqrt{7-\sqrt{3}}}{4}$. This appears to be difficult, since there is no obvious connection between an expression of an irrational number by radicals and it's base $3$ expansion.
As others have indicated, $\frac{1}{\sqrt{2}} \notin \mathcal{C}$. This is quite understandable, because it's like a random guess, and since the Cantor set has measure $0$, it's difficult to hit the Cantor set with such.
