Past exam of the year 1 of the computer science course of the University of Oxford / 2022 / Discrete mathematics, continuous mathematics, and probability / Discrete mathematics / 3 / c

Write $q(x,y)=x^2+y^2$. Equality of $q$-values defines an equivalence relation induced by a function: equality is reflexive, symmetric, and transitive, so all three properties hold for $R$. The equivalence class of $(x,y)$ isFor $(x,y)\ne (0,0)$ this is the circle centred at the origin with radius $\sqrt{x^2+y^2}$; the remaining equivalence class is the singleton $\{(0,0)\}$.

This binary relation is not an equivalence relation, because it is not symmetric: $(0,0)R(0,1)$ since $0\le 1$, but $(0,1)R(0,0)$ is false. In fact the injective encoding $(a,b)\mapsto 10a+b$ gives a total order. There are no equivalence classes for this binary relation.

Every nonzero integer $a$ has $a^2>0$, proving reflexivity. Because $ab=ba$, the relation is symmetric. If $ab>0$ and $bc>0$, then $ac=(ab)(bc)/b^2>0$, proving transitivity. Thus this is an equivalence relation, whose two equivalence classes are the positive integers and the negative integers: