Past exam of the mathematics course of the University of Cambridge / 2003 / ia / Paper 4 / 2C (Numbers and Sets)

An equivalence relation on a set $X$ is a binary relation that is reflexive, symmetric and transitive: every $x\in X$ satisfies $x\sim x$; $x\sim y$ implies $y\sim x$; and $x\sim y$, $y\sim z$ imply $x\sim z$. Each of these quantifiers ranges over the entire specified set. The separate entries below test these three properties.

This is an equivalence relation. The difference $x-x=0$ is an even integer, proving reflexivity. If $x-y=2k$ then $y-x=-2k$, proving that the relation is symmetric. If $x-y=2k$ and $y-z=2\ell$ then $x-z=2(k+\ell )$, proving transitivity. Its equivalence classes are the translates $x+2\mathbb{Z}$; two real numbers need not be integers themselves to belong to the same class.

For nonzero complex numbers, $x\overline{y}=(x/y)|y{|}^2$. Since $|y{|}^2$ is positive real, the condition is equivalent to $x/y\in \mathbb{R}\setminus \{0\}$. This is an equivalence relation. The ratio $x/x=1$ gives reflexivity; the reciprocal of a nonzero real ratio shows that the relation is symmetric; and $(x/y)(y/z)=x/z$ gives transitivity. An equivalence class is a real line through zero with zero removed, or equivalently all nonzero complex numbers with the same argument modulo $\pi$.

This is not an equivalence relation. For $x=1/2$, $x\overline{x}=1/4\notin \mathbb{Z}$, so reflexivity fails. It is symmetric, because $y\overline{x}$ is the conjugate of $x\overline{y}$, and an integer is real. It also fails transitivity: take $x=z=1/2$ and $y=2$. Then $x\overline{y}=y\overline{z}=1$ are integers but $x\overline{z}=1/4$ is not. A single failure would suffice, but these examples identify exactly which properties break.

This is not an equivalence relation. With the usual convention that zero is a perfect square, $x^2-x^2=0$ gives reflexivity, and changing the sign of $x^2-y^2$ shows that the relation is symmetric. However $3\sim 5$ because $3^2-5^2=-16=-4^2$, and $5\sim 13$ because $5^2-13^2=-144=-12^2$, whereas $3^2-13^2=-160$ is neither a positive nor a negative perfect square: $12^2<160<13^2$. Thus transitivity fails. If “perfect square” were restricted to positive squares, reflexivity would fail as well, leaving the same negative classification.