show that the relation give by $\vert x \vert=\vert y \vert$ is an equivalence relation

Question

For the relation $$\vert x \vert = \vert y \vert$$

this consists of all the pairs $(\vert -a \vert,\vert a \vert)$

Clearly $$\vert x \vert = \vert x \vert$$ and $$\vert y \vert = \vert y \vert$$

for example $\vert 2 \vert = \vert 2 \vert$ and say $\vert 4 \vert = \vert 4 \vert$

Thus the relation is reflexive

Clearly if $\vert x \vert = \vert y \vert$ then $\vert y \vert = \vert x \vert$

for example $\vert 3 \vert =\vert -3 \vert$ or $\vert -3 \vert = \vert 3 \vert$

Thus the relation is symmetric

If $\vert x \vert = \vert y \vert$ and $\vert y \vert = \vert z \vert$

Clearly $\vert x \vert = \vert z \vert$

Thus the relation is transitive.

Not sure how to show an example for the transitive property. I feel like this relation is pretty much self explanatory. Am I right that this consists of the pairs $(\vert -a \vert,\vert a \vert)$? Are my proofs without the examples acceptable? Are my examples correct? Can I improve my answer at all?

Answer 1

For the transitive property you could show that
Since $|x|=|y|$ then $|x|-|y|=0$.
Since $|y|=|z|$ then $|y|-|z|=0$.
Then $$0=\\0+0=\\(|x|-|y|)+(|y|-|z|)=\\|x|-|z|$$ Since $0=|x|-|z|$ then $|x|=|z|$

Answer 2

You don't need to make examples: you just need to prove the property.

Suppose $|x|=|y|$ and $|y|=|z|$. Then $|x|=|z|$.

That's all.

More generally, if you have a function $f\colon X\to Y$ and define the relation $\sim_{f}$ on $X$ by $$ a\sim_{f}b \quad\text{if and only if}\quad f(a)=f(b) $$ then $\sim_f$ is an equivalence relation:

1. $a\sim_f a$ follows from $f(a)=f(a)$
2. if $a\sim_f b$, then $f(a)=f(b)$, so $f(b)=f(a)$ and therefore $b\sim_fa$
3. if $a\sim_f b$ and $b\sim_f c$, then $f(a)=f(b)$ and $f(b)=f(c)$, so $f(a)=f(c)$ and therefore $a\sim_f c$

Your case is $f\colon\mathbb{R}\to\mathbb{R}$, $f(a)=|a|$.

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You seem to be baffled by the fact that there are no distinct $a,b,c$ such that $|a|=|b|$ and $|b|=|c|$. That's completely irrelevant for the task of proving transitivity.

Actually, having proved reflexivity, you can, in this particular case, even dispense from proving transitivity, because it cannot be false. But there's no need to go this way.