1. ## Equivalence relation?

Reflexive: I'd say it's reflexive since x*x would stiil belong to the rational numbers, right?
But is it enough to just write that, or do I need to do something more?

Symmetric: Again I'd say it is symmetric because multiplication is commutative.
Is that enough? xD

Transitive: No idea, I'm so bad at transitive, so any help would be appreciated, thanks!

2. Your idea for the reflexivity of the relation is good

$xTx = \{x\in\\Q$ ^ ${x^2}\in\\Q\}$

which is obvious if we let $x = a/b\rightarrow\\{x^2} = {a^2}/{b^2}$. which will be in Q

Symmetric:

$xTy = \{x\in\\Q$ ^ ${x*y}\in\\Q\}\rightarrow\{y\in\\Q$ ^ ${y*x}\in\\Q\} = yTx$

since multiplication in Q is communative, i.e., $xy = yx$.
If $x = a/b$ and $xy = c/d$, since both x and xy are rational then it follows there must be a y in Q such that $xy = {a/b}*{y}= c/d$. Apply the same argument but for x and you show $xTy = yTx$

Transitive:

if $xTy$ and $yTz$ then $xTz$

by $xTy$, we know x and xy are in Q, and by $yTz$ we know y and yz are in Q. So we want to know if xz is in Q. It follows if y and yz are in Q then z must be in Q, by the same argument for the Symmetric. So if x is in Q and z is in Q, then xz must be in Q.