Past exam of the mathematics course of the University of Cambridge / 2020 / ia / Paper 2 / 2D (Numbers and Sets)

An equivalence relation is reflexive, symmetric, and transitive. HereThe relation is reflexive and symmetric, but it is not transitive. For example, takeThe real parts of $x\overline{y}$ and $y\overline{z}$ are positive, whereas that of $x\overline{z}$ is negative. Hence this is not an equivalence relation.

By the equality condition in the complex triangle inequality,for nonzero $x,y$ exactly when $x/y$ is a positive real number. Having the same complex argument is reflexive, symmetric, and transitive, so this is an equivalence relation. Its classes are the open rays from the origin.

This relation is reflexive because choosing $n=1$ gives $|x/x|=1$. It need not be symmetric: take positive real numbers $y=e$ and $x=e^2$. Then $|x/y^2|=1$ is rational, so $x\sim y$. Butis transcendental, hence irrational, for every integer $m\ge 1$. Thus $y\nsim x$, and the relation is not an equivalence relation.

Set $F(x)=|x^3-x|$. The relation is exactlyIt is therefore the equivalence relation induced by a function: equality of the two function values is reflexive, symmetric, and transitive. Hence this is an equivalence relation.