Let Z be the set of integers. Show that the relation

Let Z be the set of integers. Show that the relation
R = {(ab) : ab ∈ Z and a + b is even}
is an equivalence relation on Z.


We observe the following properties of R.


Let $a$ be an arbitrary element of $Z$. Then,

$a \in R$

Clearly, $a+a=2 a$ is even for all $a \in Z$.

$\Rightarrow(a, a) \in R$ for all $a \in Z$

So, $R$ is reflexive on $Z$.


Let $(a, b) \in R$

$\Rightarrow a+b$ is even


$\Rightarrow b+a$ is even

$\Rightarrow(b, a) \in R$ for all $a, b \in Z$

So, $R$ is symmetric on $Z$.


Let $(a, b)$ and $(b, c) \in R$

$\Rightarrow a+b$ and $b+c$ are even

Now, let $a+b=2 x$ for some $x \in Z$

and $b+c=2 y$ for some $y \in Z$

Adding the above two, we get

$a+2 b+c=2 x+2 y$

$\Rightarrow a+c=2(x+y-b)$, which is even for all $x, y, b \in Z$

Thus, $(a, c) \in R$

So, $R$ is transitive on $Z$.

Hence, R is an equivalence relation on Z


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