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I have to define whether the following relation is symmetric, reflexive, and transitive.

Define a relation R on Z as follows: (x, y) ∈ R if and only if x = |y|:

This is my answer so far:

Is R symmetric? No

Is R reflexive? No

Is R transitive? No

I am not exactly confident with this answer because I am not sure if we should check these properties based on x, y for all x, yR or based on x, y for all x, yZ

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Note that $R$ is defined on $\mathbb Z$, so we assume that $x,y \in \mathbb Z$.


You're correct that $R$ is not symmetric:

  • Take $x = 3 \in \mathbb Z$ and $y = -3 \in \mathbb Z$. Notice that $3 = |-3|$ but $-3 \neq 3 = |3|$. So $(x,y) \in R$ but $(y,x) \notin R$.

You're correct that $R$ is not reflexive:

  • Take $x = -3 \in \mathbb Z$. Notice that $-3 \neq 3 = |-3|$. So $(x,x) \notin R$.

However, $R$ is in fact transitive:

  • Choose any $x,y,z \in \mathbb Z$ such that $(x,y),(y,z) \in R$ so that $x = |y|$ and $y = |z|$. Then by substituting, notice that: $$ x = |y| = ||z|| = |z| $$ So $(x,z) \in R$, as desired.
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  • $\begingroup$ As far as I know, picking values for (x,y),(y,z)∈R requires us to pick 3 distinct values for x, y, and z. However, if we want to satisfy the equation x=|y| and y=|z|, we can only choose at the most 2 distinct values. Or it does not matter? $\endgroup$ – amundi12 Dec 10 '14 at 21:08
  • $\begingroup$ The three values need not be distinct. Indeed, the identity relation (which only contains ordered pairs of the form $(a,a)$) is transitive. $\endgroup$ – Adriano Dec 10 '14 at 21:11

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