[Math] Proving equivalence relations

Question

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!

Answer 1

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.