Properties of a relation on $\mathbb Z\times\mathbb Z$ and $\mathbb Q \times\mathbb Q$.

Question

Is my proof of the following excersize sufficient? I am given two relations and i am supposed to show whether they are reflexive, transitive and symmetric.

Given $ R\subseteq\mathbb Z \times\mathbb Z $

1. $xRy \iff x= \sqrt{y^2} $

reflexivity: i) show a counter example that if $ x = -5 $ it follows $ -5 \neq 5 $ Thus R is not Reflexive.

symetric property: Here i am not sure if this is a proof or not, my idea is that, it follows from the observation that only non negative numbers are in relation and only with itself. Therefore,

$\implies $ $x = \sqrt{y^2} $ $\implies $ $y = \sqrt{x^2} $ and thus the relation R is symmetric.

transitivity: From the definition of transitivity of a relation, if $ $ $x = \sqrt{y^2} $ $\land$ $ $$ y = \sqrt{z^2}$ $\implies $ $x = \sqrt{z^2} $ Again, we observed how the relation looks like, so when we choose $z=y=x$ the relation is Transitive.

Given $ S\subseteq ℚ \times ℚ $

2. $xSy \iff \exists z $$\;\in$$\; ℚ $: $x\le z\le y$

reflexivity: My observation is that we can not find a $z:\ $z $ \neq$ 3 $\ $between for example $3\le z\le 3$ thus S is not reflexive.

symetric property: In order for S to be symmetric we need $x\le z\le y$ $\implies$ $y\le z\le x$ I show a counter example: $xSy$ for $(x,y) = (3,5)$ and $z=4$, but $ySx$ does not hold. therefore not symmetric.

transitivity: $x\le z\le y \quad \land \quad y\le z\le w$ $\implies$ $x\le z\le w$ if we choose $y=z$ it holds, so S is transitive.

Is this a valid solution?

Answer 1

Concerning $R\subset\mathbb Z\times\mathbb Z$. I would rewrite the relation by $xRy\iff x=|y|$.

• Correct, the relation is not reflexive since $-5\neq|-5|$.
• The relation is not symmetric. We have e.g. $5=|-5|$ but not $-5=|5|$.
• The relation is transitive: $x=|y|\wedge y=|z|\implies x=||z||$ and we have $||z||=|z|$.

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Concerning $S\subset\mathbb Q\times\mathbb Q$.

• The relation is reflexive since $x\leq x\leq x$ is true for every $x\in\mathbb Q$. Your reasoning makes no sense because it is not requested that $z\neq3$
• Correct, not symmetric.
• Correct, transitive.

Answer 2

$5R(-5)$ but as you already showed, $-5$ is not related to $5$. So it is not symmetric.