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[SOLVED] Help with an Equivalence Relation?

katye333

New member
Jul 24, 2013
10
Hello all,


I have an equivalence relation that I need some help with. Normally I find these to be fairly simple, however I'm not sure if I'm over-thinking this one or if it's just tricky.

For the relation: aRb $\Longleftrightarrow$ |a| = |b| on $\mathbb{R}$ determine whether it is an equivalence relation.

Reflexive: Would it really be reflexive? If a = -2, then wouldn't |a| = +2?

Or would it be reflexive, since all a's are contained in a?
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Reflexive: Would it really be reflexive? If a = -2, then wouldn't |a| = +2?
Or would it be reflexive, since all a's are contained in a?
A relation $R$ is reflexive on a set $S$ if and only if $aRa$ for all $a\in S$. In our case $|a|=|a|$ for all $a\in \mathbb{R}$ i.e. $aRa$ for all $a\in \mathbb{R}$, which implies that $R$ is reflexive on $\mathbb{R}$.
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
You can prove a general theorem. Let $A$, $B$ be sets and $f:A\to B$ a total function. Let a relation $R$ on $A$ be defined by $a_1Ra_2\iff f(a_1)=f(a_2)$. Then $R$ is an equivalence relation. In this case $f:\mathbb{R}\to\mathbb{R}$ is the absolute value function. You can also apply this theorem to $f(x)=\mathop{\text{sgn}}(x)=\begin{cases}1&x>0\\ 0&x=0\\ -1&x<0\end{cases}$, $f(x)=\lfloor x\rfloor$, etc.

Note also that if we take not a function $f$, but an arbitrary relation $F\subseteq A\times B$ and define \[a_1Ra_2\iff \text{there exists a }b\in B\text{ such that }a_1Fb\text{ and }a_2Fb\] then the statement does not hold in general. (Which property of an equivalence relation gets violated?) For example, it does not hold if $A=B$ is a set of people and $aFb\iff b$ is a friend $a$.
 

katye333

New member
Jul 24, 2013
10
Thank you both for the responses.
I don't know why that one confused me, while none of the others did. :p