How do I prove if a relations is symmetric,transitive or reflexive?

Question

I have no idea how to start this problem. It is asking to prove if the following relation R on the set of all integers where $(x,y) \in R$ is reflexive, symmetric and/or transitive.

1) $(x, y)\in R \iff x\neq y$

2) $(x, y)\in R\iff x = y + 1 \;\text{ or }\; x = y - 1.$

Answer 1

Hints:

$(1)$

• (Reflexive?) Can it ever be the case that $x \neq x$? No: no ordered pair of the form $(x, x)$ can be in the relation R.

• (Symmetric?) For all $x, y \in \mathbb Z$, does $x\neq y $ imply $y\neq x$? Of cource it does.

• (Transitive?) Suppose $x \neq y$ and $y\neq z$. Does this necessarily mean that $x\neq z$? Hint: let $x = z = 1, y = 2$. So $(1, 2) \in R$ and $(2, 1) \in R$, but clearly, $(1, 1)\notin R$ because $1 = 1$.

Answer 2

Hint

• $(1,1)\not\in \mathcal R$ since $1=1$ hence $\mathcal R$ isn't reflexive.
• If $x\ne y$ then $y\ne x$ so...
• $(1,2)\in\mathcal R$ and $(3,1)\in\mathcal R$ but $(1,1)\not\in \mathcal R$ so...

I leave the second relation for you.