<s>[INST] Prove for all $x \in \mathbb{R}$ at least one of $\sqrt{3}-x$ and $\sqrt{3}+x$ is irrational I have attempted to do some work on this, but I'm not sure if I'm heading in the correct direction. If I am heading in the correct direction, then I'm not quite sure where to take it from here. 
Roughly what I have so far is:
We will use proof by contradiction
Assume $\sqrt{3}-x$ and $\sqrt{3}+x$ is rational
Then $\sqrt{3}-x=\frac{a}{b}$ for some $a,b \in \mathbb{R}$ and $\sqrt{3}+x=\frac{c}{d}$ for some $c,d \in \mathbb{R}$
Then $(\sqrt{3}-x)^2=\frac{a^2}{b^2}$ and $(\sqrt{3}+x)^2=\frac{c^2}{d^2}$
Then $(\sqrt{3}-x)(\sqrt{3}-x)=\frac{a^2}{b^2}$ and $(\sqrt{3}+x)(\sqrt{3}+x)=\frac{c^2}{d^2}$
Then $x^2-2\sqrt{3}x+3=\frac{a^2}{b^2}$ and $x^2+2\sqrt{3}x+3=\frac{c^2}{d^2}$
Then $x(x-2\sqrt{3})+3=\frac{a^2}{b^2}$ and $x(x+2\sqrt{3})+3=\frac{c^2}{d^2}$
That's as far as I have got with this. 
 [/INST] The sum of two rational numbers is rational, but
$$(\sqrt 3 -x)+(\sqrt 3+x)=2\sqrt 3$$
which is irrational. Therefore they cannot both be rational.
</s>