$xTy\iff |x^{2}-2|=|y^{2}-2|$
I'd love to solve it but we don't deal in absolutes. Joke aside I have no clue where to start with this one, it has me stumped.
If I had been provided the answer to the problem I could just bump around until I got there but its not the case.
$x^{2}-y^{2}=0 \rightarrow (x+y)(x-y)=0 \rightarrow x=y \lor x=-y$
Is the only thing that I can scavenge from this, but I doubt it covers all the classes, thus I can't even begin to partition the set.
$cl(x)=\{x\in R:x,-x,x-2,2-x\}$ But I'm not sure x-2 and 2-x counts in general I think it's only valid if x=0.
Well if we fix $k \in \mathbb R$ then
the equivalence class $[k]$ is by definition $\{x\in \mathbb R: |x^2 -2| = |k^2 - 2|\}$.
Now if $|x^2 - 2| = |k^2 - 2|$ then a) $x^2 -2 = k^2-2$ or b) $x^2 - 2 = -(k^2 -2) = 2-k^2$.
a) $x^2 - 2 = k^2 -2\implies x^2 = k^2 \implies x =\pm k$.
b) $x^2 - 2 = 2-k^2 \implies x^2 = 4 -k^2\implies |k| \le 2$ and $x=\pm \sqrt{4-k^2}$.
So $[k]=\{x\in \mathbb R: |x^2 -2| = |k^2 - 2|\}=$
$=\{k, -k\}$ if $|k| > 2$ and $= \{k,-k, \sqrt{4-k^2}, -\sqrt{4-k^2}\}$ if $|k| \le 2$
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Note: All of this assumes $T$ is an equivalence relation in the first place.