Relations on $\mathbb{R}^2$ I am asked to define a relation $R$ on $\mathbb{R}^2$ by $(x,y)R(u,v)$ iff $x^2+y^2=u^2+v^2$.
How do I show that $R$ is an equivalence relation? The notation here is confusing me.
To show reflexivity, I know to show that for all $a\in R, aRa$ or $(a,a)\in R$.
So am I supposed to let $a=(x,y), (u,v)$. Then we have reflexivity if $a^2+a^2=a^2+a^2$, which is obviously reflexive. Am I presenting the argument here correctly?
Likewise, to show symmetry, am I to show that for all $a,b\in R$, $a^2+b^2$ is the same as $b^2+a^2$?
And finally, to show transitivity, am I to show that for all $a,b,c\in R$, if $a^2+a^2=b^2+b^2$ and $b^2+b^2=c^2+c^2$, then $a^2+a^2=c^2+c^2$?
 A: Actually, you're kind of messed up, but that's good. You showed your work (thank you), and now I have a better idea of how to respond to you.

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*Reflexive: $(\forall a,b \in \Bbb R), \quad a^2+b^2 = a^2 + b^2 \implies (a,b)R(a,b)$


*Symmetric: $(\forall a,b,c,d \in \Bbb R), \quad (a,b)R(c,d) \implies
    a^2+b^2 = c^2 + d^2 \implies c^2+d^2 = a^2 + b^2 \implies (c,d)R(a,b)$


*Transitive : $(\forall a,b,c,d,e,f \in \Bbb R), \quad (a,b)R(c,d) \wedge (c,d)R(e,f)
    \implies a^2+b^2 = c^2 + d^2 \wedge c^2+d^2 = e^2+ f^2 
    \implies a^2+b^2 = e^2+ f^2 \implies (a,b)R(e,f).$
A: 
To show reflexivity, I know to show that for all a∈R,aRa or (a,a)∈R. So am I supposed to let a=(x,y),(u,v). Then we have reflexivity if a2+a2=a2+a2, which is obviously reflexive. Am I presenting the argument here correctly?

You've gone a level too deep.
An element of $\mathbb R^2$ is of the form $(x,y)$.  And an element of $R \subset \mathbb R^2 \times \mathbb R^2$  will be of the form $((x,y), (u,v))$.
You need to show for all $(x,y) \in \mathbb R^2$ (not for every $a \in R$) then $(x,y)R(x,y)$ in other words $((x,y),(x,y)) \in R$.  Which is true if and only if $x^2 + y^2 = x^2 + y^2$.

So your job:  For all $(x,y) \in \mathbb R^2$ prove that $x^2 +y^2 = x^2 + y^2$.

To show symmetry you have to show if $((x,y), (u,v)) \in R$ (or in other words $(x,y)R(u,v)$; which only happens if $x^2 +y^2 = u^2 + v^2$) so if that is true, that $x^2 +y^2 = u^2 + v^2$ you must show that $u^2 + v^2 = x^2 + y^2$ (which means $(u,v)R(x,y)$ and $((u,v), (x,y)) \in R$.

SO your job: For any $(x,y)$ and $(u,v) \in \mathbb R^2$ where $x^2 + y^2 = u^2 + v^2$, you most prove $u^2 + v^2 = x^2 + y^2$.

And to show transitivity you have to show for any $((x,y),(u,v))\in R$ (so that means $(x,y)R(u,v)$ and that $x^2 + y^2 = u^2 + v^2$) and for any $((u,v),(a,b))\in R$ (so that means $(u,v)R(a,b)$ and that $u^2 + v^2=a^2 + b^2$, then you must show that $x^2 + y^2 = a^2 + b^2$ (which means $(x,y)R(a,b)$ and $((x,y),(a,b)) \in R$.

SO your jog:  For any $(x,y), (u,v), (a,b)\in \mathbb R^2$ where $x^2 + y^2 = u^2 +v^2$ and $u^2 + v^2 = a^2 + b^2$, you must prove $x^2 + y^2 = a^2 + b^2$.

That's your job.
And it's pretty trivial.
