Define a relation $∼$ on $Z$ by $a ∼ b$ if and only if $a − b$ is even. Show that it is reflexive, symmetric and transitive.
a) Reflexive: $a \sim a$ holds because $a-a = 0$ and $0$ is an even number, so $\sim$ is reflexive.
b) Symmetric: We need to show if $a \sim b$ then $b \sim a$. If $a,b$ ∈ $Z$ then $a \sim b$ then $a-b = 2n$, from which we get $a= 2n + b$. To show $b \sim a : b-a = b – (2n +b) = -2n$. So $b-a$ is an even number therfore $b\sim a$ holds, so $\sim$ is symmetric.
c) Transitive: We need to show if $a \sim b$ and $b \sim c$ then $a \sim c$. For $a \sim b: a-b = 2n$, then $b = a-2n$. For $b \sim c: b-c = a-2n -c = 2n$ , the $2n$'s cancel out so we get $a-c = 0$ , so $a=c$ and therfore for $a \sim c a=c$ so its just $a-a = 0$ which is even, thefore $a \sim c$ holds and we can say $\sim$ is transitive.
Is this correct?
Best Answer
There is one small problem in the proof of transitivity. You know that $a\sim b$. Hence there is an $n$ such that $a-b=2n$. And, you know that $b\sim c$. Thus, there is an $m$ such that $b-c=2m$. HOWEVER, you in your proof, you assume that $m=n$ and you are not allowed to do this. To complete the proof, you should now do something like:
$b=2n-a$ and $b=2m-c$, hence, $2n-a=2m-c$ and, thus, $a-c=2n-2m=2(n-m)$.