$P$ is a binary relation. $P ⊆ \mathbb{R}^2$. $P = \{(x,y): y = |x|\}$.
As I understand for relation to be transitive: $(a,b) \in P$ and $(b,c) \in P$ then $(a,c \in P)$
for this particular relation we can take $(a,b) = (-2,2)$ and $(b,c) = (2,2)$, then $(a,c) = (-2,2) \in P$, so it means that relation is transitive. Is this correct?
I'm not quite sure about antisymmetric property: for it to work for different a,b both $(a,b)$ and $(b,a)$ can't $\notin P$. For example $(-2,2) \in P$, but $(2,-2) \notin$ P. Which proves that relation is antisymmetric. Is this correct?
Have I answered questions correctly?
Update: for anti-symmetric:
By definition $\forall a,b \in P, aPb \land bPa => a=b$
let's assume that $(a,b) \in P$. Then $(b,a) \in P <=> a=b$
Relation is anti-symmetric.
Best Answer
You can disprove a universal statement by a single counter-example, but to prove a universal statement you need much more than a few examples. You need a guarantee that counter-examples do not exist.
The nature of a proposed counter example to a universal claim is found by negating the statement (using dual negation rules).
Reflexivity: $\forall x: x=\lvert x\rvert \\ \neg \exists x: x\neq x$
Symmetry: $\forall x\,\forall y: \big(y=\lvert x\rvert ~\to~ x=\lvert y\rvert\big)\\ \neg\,\exists x\,\exists y: \big(y=\lvert x\rvert \,\wedge\, x\neq \lvert y\rvert\big)$
Transitivity: $\forall x\,\forall y\,\forall z: \big(y=\lvert x\rvert\,\wedge\, z=\lvert y\rvert ~\to~ z=\lvert x\rvert\big) \\ \neg \exists x\,\exists y\,\exists z:\big(y=\lvert x\rvert\,\wedge\, z=\lvert y\rvert\,\wedge\, z\neq \lvert x\rvert\big)$
Anti-symmetry: $\forall x\,\forall y:\big(y=\lvert x\rvert\,\wedge\, x=\lvert y\rvert ~\to~ x=y\big) \\ \neg \exists x\,\exists y:\big(y=\lvert x\rvert\,\wedge\, x=\lvert y\rvert \,\wedge\, x\neq y\big)$