Problem:
Consider the relation R on $N$ defined by
$x$R$y$ iff $2$ divides $x + y$.
Prove that R is an equivalent relation
My work:
I know that to prove that a relation is an equivalent relation, I have to show that it's reflexive, symmetric and transitive.
So first reflexive. Suppose there is an integer $a$ in $N$. Then $a + a = 2a$.
$2$ divides $2a$ because $2a = (2)(a)$. Therefore this relation is reflexive because $(a,a)$ will be in the relation.
For symmetric - Suppose there is an ordered pair $(a, b)$ for which $a$ and $b$ are both in $N$. If
$2 \mid (a + b)$ then $2$ will also divide $b + a$ because of the commutative law of addition
Commutative Law. Therefore the ordered pair $(b, a)$ will also be in the relation and the relation will therefore be symmetric
For transitive,
Suppose $(a, b)$ and $(b, c)$ are in the relation. Now I have to show that $(a, c)$ will also be in the relation.
$a + b = 2k$ for some integer $k$, solving for $b$, we have
$b = 2k - a$.
$b + c = 2m$ for some integer $m$. Substituting for $b$, we have
$2k - a + c = 2m$
$c - a = 2(k + m)$
Algebraically I have shown the $2$ divides $c - a$. Is there some algebra step or trick I can use to show that $2$ will also divide $c + a$?
Start with the relationships, where $a,b,c,k,j \in \mathbb{Z}$: $$a R b, bRc \rightarrow 2|(a+b), 2|(b+c)$$ $$(a+b) = 2k, (b+c) = 2j$$ $$a+2b+c = 2(k+j)$$ $$a+c = 2(k+j)-2b$$ $$a+c = 2(k+j-b)$$
And thus $2$ divides $(a+c)$. Therefore, if $aRb,bRc$, then $aRc$, satisfying the definition of transitivity.