Prove that for all $x \in R$ at least one of $\sqrt{3}-x$ or $\sqrt{3}+x$ is irrational

Question

Prove that for all $x \in R$ at least one of $\sqrt{3}-x$ or $\sqrt{3}+x$ is irrational.

Could you not just prove this using contradiction by plugging in $1$ for $x$?

I'm having difficulty even starting this proof.

Answer 1

If for any $x \in \Bbb R$,

$\sqrt 3 + x, \; \sqrt 3 - x \in \Bbb Q, \tag 1$

then

$2\sqrt 3 = (\sqrt 3 + x) + (\sqrt 3 - x) \in \Bbb Q, \tag 2$

whence

$\sqrt 3 \in \Bbb Q, \tag 3$

contradicting the irrationality of $\sqrt 3$; thus, (1) is false.