Equivalence Relations and Classes 3 I am studying for a discrete math exam that is tomorrow and the questions on equivalence classes are not making sense to me.
Practice Problem: Let $\sim$ be the relation defined on set of pairs $(x, y) \in R^2$ such that $(x, y) \sim (p, q)$ if and only if $x^2 + y^2 = p^2 + q^2$. Find three elements in the equivalence class $[(0, 1)]$
The example solution shows $(0,1),(1,0),(-1,0),$ can somebody explain why those solutions hold true for this equivalence class? Thank you!
 A: So this says two pairs of numbers are equivalent if the sum of their squares is equal. In this case, $$1 = 0^2 + 1^2 = 1^2 + 0^2 = 0^2 +(-1)^2 = (-1)^2 + 0^2,$$ so $(0, 1) \sim (1, 0) \sim (0, -1) \sim (-1, 0)$.
A: 

Practice Problem: Let $\sim$ be the relation defined on set of pairs $(x, y) \in \Bbb R^2$ such that $(x, y) \sim (p, q)$ if and only if $x^2 + y^2 = p^2 + q^2$. Find three elements in the equivalence class $[(0, 1)]$

The example solution shows $(0,1),(1,0),(-1,0),$

Elements of an equivalent class are all equivalent (by the given relation).
$$\begin{align}(0,1)\sim(0,1) ~\iff & ~ 0^2+1^2 = 0^2+1^2\\[1ex] (0,1)\sim(1,0) ~ \iff & ~ 0^2+1^2 = 1^2+0^2 \\[1ex] (0,1)\sim(-1,0) ~ \iff & ~ 0^2+1^2 = (-1)^2 + 0^2 \\[2ex] \ddots ~ & ~ \textsf{et cetera} \end{align}$$
Basically the equivalence classes of this relation ($\sim$) are all points of the circumference of a circle of a certain radii.   In this case that is $1^2$.
$$[(0,1)] ~ \equiv ~ \{(p,q) : p^2+q^2 = 1^2\}$$
A: The equivalence relation is defined such that $(x,y) \sim (p,q)$ if $x^2 + y^2 = p^2 + q^2$.
If $(x,y) \in[(0,1)]$, then $(x,y) \sim (0,1)$, so it must satisfy
$$
x^2 + y^2 = 0^2 + 1^2 = 1
$$
Which is the equation of the unit circle, which all three of those points are a part of.
