Proving relation $\{(x,y)\in \Bbb R\times\Bbb R\mid y = x^2\}$ is not a transitive relation

Question

I have trouble proving how the following statement is false:

The relation $g = \{\,(x,y)\in \Bbb R\times\Bbb R\mid y = x^2\,\}$ is transitive.

I know you have to use $yRx$, $zRy$, and $xRz$, but I'm at a loss.

Also.. What's an example of a transitive function that isn't the equality relation ($y=x$)?

Thank you for help.

Answer 1

If you want an example of a transitive relation, a classic is $x\leq y$

If your relation is given by elements in $g$, then $(x, y) \in g \implies y = x^2$, and $(y, z) \in g \implies z = y^2$. For the relation to be transitive, we would need to have that $(x, z) \in g$, i.e., we would need to have that $z = x^2$ whenever $(x, y), (y, z) \in g$.

Can you think of a counterexample for values of $x, y, z$ such that $y = x^2$, and $z = y^2$, but $z \neq x^2$?

Take $x = 2$, $y = 4$, $z = 16$: then we have $y = x^2, \; z = y^2, \; \text{ but} \; z \neq x^2,\;$ since $16 \neq 4$