Which of these subsets of $\mathbb R$ is uncountable?

Question

Which of the following sets are uncountable?

1. $\{x \in\mathbb R:$ integral part of $x$ is $1\}$.

2. $\{x \in\mathbb R: x^2$ is rational$\}$.

3. $\{x\geq 0: \sqrt x$ is rational$\}$.

4. $\{x\geq 0: x + \sqrt x$ is an integer$\}$.

Clearly, 1st option is true. Since it is $[1,2)$

For second option, such set contains all rationals ( which are countably infinite), and all the numbers of the form $\sqrt p$ (also countably infinite), where $p$ is prime. So, it is not uncountable.

For 3rd option, we can get at most countably infinite such $x$ for which $\sqrt x$ is rational. So, it should be countably infinite.

I'm clueless about 4th. Any hint would be helpful. And please correct me if my reasonings are wrong about first three options.

Answer 1

$x + \sqrt x$ is strictly increasing, so for each integer there can be at most one $x$ such that $x+\sqrt x$ is that integer.

Answer 2

Since $4)$ is a subset of algebraic real numbers, it is countable

In more detail: Fix $n\in\mathbb{N}$ and $x+\sqrt{x}=n$ Then ${(n-x)}^2=x$ and x is algebraic

So if $A_n= \{ x\geq 0: x+\sqrt{x}=n \} $, therefore $A_n$ is a subset of the algebraic real numbers and therefore countable

Hence $ \{x\geq 0, x+\sqrt{x} $ x is an integer $\}=\cup \{ A_n\}$ and so it is countable.

The same argument applies to $2)$ and $3)$.