For N, consider a relation R on N given by R = (x,y) : 3x+ y is a multiple of 7.

Question

For $\alpha\in\mathbb{N}$, consider a relation $R$ on $\mathbb{N}$ given by $R = \{(x,y) : 3x+\alpha y 	ext{ is a multiple of } 7\}$. The relation $R$ is an equivalence relation if and only if:

Accepted Answer

Reflexive: need $(3+\alpha)a \equiv 0\pmod7$ for all $a\in\mathbb{N}$, which forces
$$3+\alpha\equiv0\pmod7 \implies \alpha\equiv-3\equiv4\pmod7$$

Symmetric: with $\alpha=7k+4$, if $3a+\alpha b=7m$ then $3(a-b)+7kb = $ (rearranged) shows $3b+\alpha a$ is also a multiple of 7, so symmetry holds for this $\alpha$.

Transitive: if $(a,b),(b,c)\in R$, adding the two divisibility conditions and simplifying (using $\alpha\equiv4\pmod7$) shows $(a,c)\in R$ as well.

So $R$ is an equivalence relation exactly when $\alpha=7k+4$, i.e. 4 is the remainder when $\alpha$ is divided by 7.

Solutions · 1