Equivalence Relations: $(a,b)\sim(c,d)$ if and only if $a^2+b^2=c^2+d^2$

Question

I need someone to check my proof.

Question: On the set $\mathbb{R}^2$ of ordered pairs define the 2-plane relation $\sim$ as follows $(a,b)\sim(c,d)$ if and only if $a^2+b^2=c^2+d^2$. Prove that $\sim$ is an equivalence relation on R2

My Answer: We must show reflexivity, symmetry, and transitivity.

Symmetry: $(a,b)\sim(b,a) \Rightarrow a^2+b^2=b^2+a^2$

Transitivity: Suppose $(a,b)\sim(c,d)$ and $(c,d)\sim(e,f)$ Then $a^2+b^2=c^2+d^2$ and $c^2+d^2=e^2+f^2$ Therefore $a^2+b^2=e^2+f^2$

Reflexivity: I am not sure - might need a little help with this.

Answer 1

For reflexivity, it is enough to write that for all $(a,b) \in \mathbb{R}^2$, $a^2+b^2=a^2+b^2$...

Answer 2

In this cant we say that (a,b)~(c,d) if they are on the same circle about the center because the radius square is equal. So this must be a equivalence relation with the classes being the circles about the origin.

Answer 3

Your transitivity reasoning is correct, but for symmetry, as was commented, you rather need $(a,b)\sim (c,d) \implies (c,d)\sim (a,b)$.

For reflexivity, you need $(a,b)\sim (a,b)$, which is even more trivial than the others.

Answer 4

It is also useful to note that if $A$ is a set, and we have a relation $\equiv$ such that $a\equiv b$ iff $f(a)=f(b)$ (Where $f$ is a function whose domain is $A\times A$), then $\equiv$ is an equivalence relation. In this case $f$ is the function that sends $(a,b)$ to $a^2+b^2$