Let R be the following relation on the set of pairs of integers:

Question

Let $R = \left\{\bigl((a, b), (c, d)\bigr) \in \mathbb{Z}^2 \times \mathbb{Z}^2; a + d = b + c\right\}$.

1. Prove that $R$ is an equivalence relation.
2. Find the equivalence class of the pair $(0, 0)$.

Answer 1

To show that something is an equivalence relation, we need to show 3 properties:

1. $R$ is reflexive: $aRa$ for all $a$.
2. $R$ is symmetric: $aRb\implies bRa$ for all $a,b$.
3. $R$ is transitive: $aRb,bRc\implies aRc$ for all $a,b,c$.

For reflexive: $$a+b=a+b\implies(a,b)R(a,b)$$

For symmetric: $$(a,b)R(c,d)\implies a+d=b+c\implies c+b=d+a\implies (c,d)R(a,b)$$

For transitive: $$(a,b)R(c,d),(c,d)R(e,f)\implies a+d=b+c,c+f=d+e\implies a+d+c+f=b+c+d+e\implies a+f=b+e\implies(a,b)R(e,f)$$

Notice that $(a,b)R(0,0)\implies a+0=b+0\implies a=b$. Hence the equivalence class is $\{(k,k)\mid k\in\mathbb Z\}$.

Answer 2

The relation $R$ is

• reflexive. For all $(a,b)\in\mathbb{Z}^2$, we have $(a,b) R (a,b)$ because $a+b = a+b$.
• symetric. Let $(a,b)\in\mathbb{Z}^2$ and $(c,d)\in\mathbb{Z}^2$ such that $(a,b) R (c,d)$. We have $a + d = b + c$ or equivalently $c + b = d + a$, that is $(c,d) R (a,b)$.
• transitive. Let $(a,b)\in\mathbb{Z}^2$, $(c,d)\in\mathbb{Z}^2$ and $(e,f)\in\mathbb{Z}^2$ such that $$ (a,b) R (c,d) \quad\text{and}\quad (c,d) R (e,f), $$ then $$ \begin{align*} a + d &= b + c && (1) \\ c + f &= d + e. && (2) \end{align*} $$ We add $(1)$ and $(2)$: $$ a + d + c + f = b + c + d + e, $$ and now we substract $c + d$ from each side of the above equality. We get $$ a + f = b + e, $$ that is $$ (a,b) R (e,f). $$

The equivalence class of $(0,0)$ is the set $$ \left\{ (a,b)\in\mathbb{Z}^2 : a + 0 = b + 0 \right\} = \left\{ (a,a) : a\in\mathbb{Z}\right\}. $$