4
$\begingroup$

I just started my abstract algebra class and I am struggling with the concept of equivalence relations. I know that in order to prove equivalence relations, I have to prove the reflexive, symmetric, and transitive properties. However, I don't know how to go about starting the actual proof or solution. I have these examples and any help would be appreciated.

I have to show which of the following are equivalence relations on the set of real numbers and, if they are not, why.

  1. $a\sim b$ iff $|a|=|b|$
  2. $a\sim b$ iff $a\leq b$
  3. $a\sim b$ iff $|a-b| \leq 1$

Thank you for any help!

$\endgroup$
9
$\begingroup$

As you say, we check whether or not they're reflexive, symmetric, and transitive.

  1. We define $a \sim b$ if $|a|=|b|$ for $a,b \in \mathbb{R}$. So we check:

    • Reflexive: Is $a \sim a$ for all $a \in \mathbb{R}$? Yes, because $|a|=|a|$.
    • Symmetric: If $a,b \in \mathbb{R}$ and $a \sim b$, does it follow that $b \sim a$? Yes, because $|a|=|b|$ implies $|b|=|a|$.
    • Transitive: If $a,b,c \in \mathbb{R}$ and $a \sim b$ and $b \sim c$, does it follow that $a \sim c$? Yes, because $|a|=|b|$ and $|b|=|c|$ implies $|a|=|c|$.

    Hence this is an equivalence relation.

  2. We define $a \sim b$ if $a \leq b$ for $a,b \in \mathbb{R}$. So we check:

    • Reflexive: Is $a \sim a$ for all $a \in \mathbb{R}$? Yes, because $a \leq a$.
    • Symmetric: If $a,b \in \mathbb{R}$ and $a \sim b$, does it follow that $b \sim a$? Not in general, because e.g. $2 \leq 3$ but $3 \not\leq 2$.
    • Transitive: If $a,b,c \in \mathbb{R}$ and $a \sim b$ and $b \sim c$, does it follow that $a \sim c$? Yes, because $a \leq b$ and $b \leq c$ implies $a \leq c$.

    We conclude that $\leq$ is not an equivalence relation (since it's not symmetric).

And so on.

$\endgroup$
  • $\begingroup$ (If you're still around) Could you elaborate on 1. I don't understand the symmetric and transitive proofs. Aren't you just assuming the definition of symmetric to be true a priori ?. Why do we, for example, assume that $|a| = |b| \implies |b| = |a|$ $\endgroup$ – Scb Oct 21 '19 at 23:04
0
$\begingroup$

Reflexive:- (a,a)belongs to 'R' all 'a' belongs to'A' Symmetric:- (a,b) belongs to 'R' for all 'a' belongs to 'A' Transitive:- if (a,b) belongs to 'R' (b,c) belongs to 'R' then (c,a) belongs to 'R' for all (a,b,c) belongs to 'A' Antisymmetric :- 'a' relation 'b' and 'b' relation 'a' IF a=b for all (a,b) belongs total 'A'

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.