Transitivity relation in the set of Integers

Question

Prove or disprove that R is transitive, where $R=\{ (a,a^2)| a \in \Bbb Z \}$ is a relation on $\Bbb Z$

By definition: $R$ is transitive $iff$

$$ (a,b)\in R \wedge (b,c) \in R\implies (a,c) \in R $$

If we pick two numbers $a$ and $b$, such that $ a\neq b$. $a^2$ will never equal $b^2$ (excluding 0 and 1)

So by the definition, we can not find a $c$ such that $a=b=c$.

Because if $b = a^2$ and $c = b^2$ then $c \neq a^2$ by the fact that $a \neq b$

if we let $a$ and $b$ equal the exclusions $1$ and $0$, respectively, once again a value of $c$ that satisfies the definition does not exist.

Therefore, this relation is not transitive.

Is this a valid proof?

Answer 1

Your definition is missing a crucial part. It should be

$R$ is transitive iff for all $a,b,c\in\mathbb Z$ it holds that $$ (a,b)\in R \wedge (b,c) \in R\implies (a,c) \in R $$

When you want to prove that a "for all" statement is false, it suffices to give a single counterexample.

For example, one counterexample would be $a=3, b=9, c=81$, because then $(3,9)$ and $(9,81)$ are both in $R$, but $(3,81)$ is not.

This, in itself, shows that $R$ is not transitive.

Answer 2

Hint: Suppose that $b = a^2$ and $c = b^2$, where $a, b, c \in \mathbb{Z}$. Does this imply that $c = a^2$?

Note that you cannot make the assumption that $a \neq b$.