I have tried to solve these exercises that require proving if a certain relation is an equivalence relation, but I have no solutions, are my solutions right?
Note: I'll use $\rho$ to refer to the relation.
$$ x \rho y \iff x^2=y^2 \\ \textbf{resolution:} \\ \text{reflexive:} \ x^2= x^2 \ (true) \\ \text{symmetric:} \ x^2=y^2 \iff y^2=x^2 (true) \\ \text{transitive:} \ x^2=y^2, y^2 = z^2 \implies z^2 = x^2 (true)$$ and so is an equivalence relation.
$$ x \rho y \iff x-y\in \mathbb{Z} \\ \textbf{resolution:} \\ \text{reflexive:} \ x-x= 0 \in \mathbb{Z} \ (true) \\ \text{symmetric:} \ x-y \iff y-x \in \mathbb{Z} \ (true) \\ \text{transitive:} \ x-y \in \mathbb{Z}, y-z \in \mathbb{Z} \implies x-z \in \mathbb{Z} (true) \text{ because} \ x-y+y-z \in \mathbb{Z}$$ and so is an equivalence relation.
$$ (x,y) \rho (z,t) \iff x+y = z+t \\ \textbf{resolution:} \\ \text{reflexive:} \ x+y = y+x \ (true) \\ \text{symmetric:} \ x+y = z+t \implies z+t = x+y \ (true) \\ \text{transitive:} \ x+y = z+t, z+t = a+b \implies x+y = a+b \ (true) \text{ because} \ a+b = z+t \ $$ and so is an equivalence relation.
$$ A=\{f: \mathbb{R \to R}\}\\ f \rho g \iff f(1) = g(1) \\ \textbf{resolution:} \\ \text{reflexive:} \ f(1) = f(1) \ (true) \\ \text{symmetric:} \ f(1) = g(1) \implies g(1) = f(1) \ (true) \\ \text{transitive:} \ f(1) = g(1), g(1) = h(1) \implies f(1)=h(1) \ (true) \text{ because } h(1) = g(1)$$ and so is an equivalence relations.