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Hello Mathematics Community,

currently I am trying to prove that the Relation:

$R = \{ (a,b) \in \mathbb{R} \times \mathbb{R} : |a|=|b| \}$

is reflexive, symmetric and transitive.

I know the definitions but I don't know how to prove it formally. Hope somebody can give me a lead.

sincerely, M.Hisoka

William Elliot
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M.Hisoka
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2 Answers2

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More generally if $f:X\to Y$ denotes a function then the relation: $$R:=\{\langle a,b\rangle\mid f(a)=f(b)\}$$can easily be shown (see the answer of DanG) to be an equivalence relation (i.e. it is reflexive, symmetric and transitive).

If conversely $R$ denotes an equivalence relation on $X$ then we have the natural function $\nu:X\to X/R$ prescribed by $a\mapsto[a]$ where $[a]$ denotes the equivalence class of $a$, and we can write:$$R=\{\langle a,b\rangle\mid \nu(a)=\nu(b)\}$$

drhab
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Reflexive: $|a|=|a|$

Symmetric: If $|a|=|b|$, then $|b|=|a|$

Transitive: If $|a|=|b|$ and $|b|=|c|$, then $|a|=|c|$.

Either this is formal enough, or you must show these for all cases of positive and negative values of a,b, and c.

DanG
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  • Now it makes sense, was confused how to write it formally. Thank you. – M.Hisoka Oct 30 '17 at 09:25
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    This is formal enough. Splitting up in cases after all would only violate its elegance. Here we are dealing with function $a\mapsto|a|$ but actually this works for every function $a\mapsto f(a)$. – drhab Oct 30 '17 at 10:52