Equivalence Relation IFF (a+b)\2

Question

Let R be a relation on $Z$ defined as follows: for $a, b ∈ Z, a R b$ if and only if $2$ divides $a + b.$ Is R an equivalence relation? Prove your answer.

New to thinking about this. Not sure about the IFF part.

Is the relation than $$\frac{a}{2}=-\frac{b}{2}$$

Reflexive :$$aRa \Rightarrow \frac{a}{2}=\frac{a}{2}$$ Symmetric: $$aRa \rightarrow bRa \rightarrow -\frac{b}{2}R\frac{a}{2} \rightarrow -\frac{-b}{2}=\frac{a}{2}$$

Not sure i understand the concept.It seems like if they are elements of integers they will fail to hold equality. Help in the right direction is appreciated.

Answer 1

It is better if you work with your definition: $aRB\Leftrightarrow 2$ divides $ a+b$.

Since $a+a=2a$, and $2$ divides $2a$ for every $a$, you have that $aRa$ for every $a$.

If $2$ divides $a+b$, it surely divides $b+a$, therefore $aRb\Rightarrow bRa$.

And now, for the transitive property, let's assume $aRb$ and $bRc$. That is, $2$ divides $a+b$ and $b+c$. Then, 2 divides $(a+b)-(b+c)=a-c$. Now, since $2$ divides $a-c$ and $2c$, then it divides $(a-c)+2c=a+c$, so $aRc$.

Answer 2

Hint:

• Reflexive: $aRa$ means that $2$ divides $a+a$.
• Symmetric: $aRb\iff bRa$ means that $2$ divides $a+b$ if and only if $2$ divides $b+a$.
• Transitive: $aRb$ and $bRc\Rightarrow aRc$ means that if $2$ divides both $a+b$ and $b+c$ then $2$ also divides $a+c$.

Are all of them true?