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!
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 $(0, 1) \sim (1, 0) \sim (0, -1) \sim (-1, 0)$.
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 $1^2$.
$$[(0,1)] ~ \equiv ~ \{(p,q) : p^2+q^2 = 1^2\}$$
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.
