Is $\mathbb{N}\cup \big\{\sqrt{2}\big\}$ an uncountable set? 
Is $\mathbb{N}\cup \big\{\sqrt{2}\big\}$ an uncountable set?

I think it is.
 A: No, it is not. Since we can create bijection between $\mathbb{N} \cup \{ \sqrt{2}\}$ and $\mathbb{N}$.
Let $f: \mathbb{N} \cup \{ \sqrt{2}\} \to \mathbb{N}$ be given by:
$f(1) = \sqrt{2}$ and $f(n) = n-1$ for $n \ge 2$.
It is clear that $f$ is a bijection, which means the two sets must be equal in cardinality. Therefore both sets are countable.
Note that the set that results from adding any finite number of points (or even a countable number of points) to a countable set is still countable.
A: No, it isn't. You can start counting from $\big\{\sqrt{2}\big\}$ and then proceed to counting $\mathbb{N}$ 
A: It is countable. Here is an explicit enumeration of its elements:
$a_1 = \sqrt{2}$
$a_2 = 1$
$a_3 = 2$
$a_4 = 3$
$a_5 = 4$
$\cdots$
$a_n = n-1$ (for $n>1$)
$\cdots$
A: If you add countably many elements to a countable set then the resulting set is also countable. Here $\Bbb N$ is countable and we are adding a single element $\sqrt 2$ to it. Thus the resulting set is countable.
A: Consider : $A:=$ {$\mathbb{N}$}$\bigcup ${$√2$}  ;
$A$ as the union of 2 countable sets is countable.
