Proof of Equivalence relation ~ on $\mathbb{Z}\times\mathbb{Z}$ such that $(a,b)$~$(c,d)$ when $a+b=c+d$

Question

Define a relation ~ on $\mathbb{Z}\times\mathbb{Z}$ such that $(a,b)$~$(c,d)$ when $a+b=c+d$.

To prove ~ is an equivalence relation, I need to prove reflexive, symmetric, transitive, right? For the reflexive, $(a,a)$~$(a,a)$, that is $a+a=a+a$. For the symmetric, if $(a,b)$~$(c,d)$, then, $(c,d)$~$(a,b)$. That is If $a+b=c+d$, then $c+d=a+b$?? For the transitive, I need help!

Answer 1

To show it's transitive, you need to show that if $a + b = c+d$ and $c+d = e+f$ then $a+b = e+f$.

Answer 2

In general, if $f\colon X\to Y$ is a map, the relation $\sim_f$ defined on $X$ by $$ a\sim_f b\qquad\text{if and only if}\qquad f(a)=f(b) $$ is an equivalence relation. All three properties are very easy to verify.

Your relation is a special case of this one, by considering the function $f\colon\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}$ defined by $$ f\colon (a,b)\mapsto a+b $$

If you don't want to use this general fact, you have to do the checks right: the reflexive property means that, for all $(a,b)\in\mathbb{Z}\times\mathbb{Z}$ it holds $$ (a,b)\sim (a,b) $$ Remember it should hold for every element of the set. Similarly, the symmetry is $$ \text{if $(a,b)\sim (c,d)$, then $(c,d)\sim (a,b)$} $$ and the transitivity is $$ \text{if $(a,b)\sim(c,d)$ and $(c,d)\sim(e,f)$, then $(a,b)\sim(e,f)$.} $$ Just apply the definition of $\sim$.