Relations exercise 3.2 Charles C. Pinter set theory

Question

Each of the following describes a relation in the set Z of the integers. State, for each one, whether it has any of the following properties: reflexive, symmetric, antisymmetric or transitive. Determine whether it is an equivalence relation, an order relation, or neither. Prove your answer in each case.

$$ G=\{(x,y):x+y <3 \}$$

Reflexivity:

$$\text{let}\hspace{0.2cm} x \in\mathbb Z,\hspace{0.2cm}x+x <3,\hspace{0.2cm}2x<3 $$

$$ x=2\hspace{0.5cm} 2(2)>3\hspace{0.5cm} (2,2)\notin G$$

So G isn't reflexive

Symmetric:

$$\text{let}\hspace{0.2cm} x,y \in\mathbb Z,\hspace{0.2cm}x+y <3,\hspace{0.2cm}y<3-x $$

$$(x,3-x)\in G \Rightarrow (3-x,x) \in G $$

So G is symmetric

Could you help me with these demonstrations? also I would like check your answers. Thanks

Answer 1

$G:=\{(x,y)\in\Bbb Z^2:x+y<3\}$ so $(x,y)\in G\equiv x+y<3$

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Reflexivity [$\forall x\in\Bbb Z~(x+x<3)$] is falsified by witnessing a counter example.

• such as yours: $2\in\Bbb Z$ and $2 + 2\not< 3$. $\color{green}\checkmark$

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Symmetry [$\forall x\in\Bbb Z~\forall y\in\Bbb Z~(x+y<3\to y+x < 3)$] is verified by arguing that if an arbitrary pair is in the relation, then its inverse must be too. A valid reason for this needs to be supplied.

• You merely argued for arbitrary $x$, and specific $y:=3-x$.
• Worse, $(x,3-x)$ will never be in $G$, since $x+3-x\not< 3$, and while $3<3\to 3<3$ is vacuously true, this does not verify symmetry.
• Rather consider that $+$ is commutative over $\Bbb Z$ (in short: $x+y=y+x$ so ...).

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Transitivity [$\forall x\in\Bbb Z~\forall y\in\Bbb Z~\forall z\in\Bbb Z~(x+y<3\land y+z<3\to x+z<3)$] and antisymmerty [$\forall x\in\Bbb Z~\forall y\in\Bbb Z~((x+y<3)\land (y+x<3)\to (x=y))$] are checked similarly.

Both can be falsified.