Equivalence Relation...

Question

I am trying to complete my homework based on equivalence relation and I don't seem to understand it properly so I need help !

My question is that do all the elements in my set must satisfy all the three conditions then I can say there is an equivalence relation or I can say there is an equivalence relation of just some of the elements satisfy the three conditions and does the order matter do I have to compare the element from the left to the right or can I pick randomly e.g. my set=( 0,2,4,6) relation defined as a~b if a+b>2 so if I start by oder I can say that 0 is not related to 2 because 0+2=2 but if I pick randomly I can state that 6~0 because 6+0>2 and u can see that some of the elements satisfy the conditions but not all of them so do I still say there is an equivalence relation ?? I am really confuse

Thank you

Answer 1

If I understand you correctly, you have a relation $\sim$ defined on the set $A=\{0,2,4,6\}$ by $a\sim b$ if and only if $a+b>2$, and the question is whether $\sim$ is an equivalence relation.

Start with the definition: in order to be an equivalence relation, the relation $\sim$ must be reflexive, symmetric, and transitive. If you can’t already define those three properties, you should make a point of learning them as soon as possible, because you’ll need to know them.

• Reflexivity: The relation $\sim$ is reflexive if $a\sim a$ for each $a\in A$. Go back to the definition of $\sim$: $a\sim a$ if and only if $a+a>2$, so $\sim$ is reflexive if and only if $a+a>2$ for each $a\in A$. But $0+0=0$ is not greater than $2$, so $\sim$ is not reflexive. The fact that $2\sim 2$ (since $2+2=4>2$), $4\sim 4$ (since $4+4=8>2$), and $6\sim 6$ (since $6+6=12>2$) is irrelevant: all it takes is the one failure to show that $\sim$ is not reflexive.

That already shows that $\sim$ cannot be an equivalence relation, but let’s check the other two properties of an equivalence relation just for the practice.

• Symmetry: The relation $\sim$ is symmetric if whenever $a\sim b$ holds, then $b\sim a$ also holds. If $a\sim b$ holds, then by definition $a+b>2$; but $b+a=a+b$, so $b+a>2$ as well, and therefore by definition $b\sim a$ holds. Thus, this relation is symmetric.

• Transitivity: The relation $\sim$ is transitive if whenver $a\sim b$ and $b\sim c$ hold (where $a,b$, and $c$ are not necessarily distinct elements of $A$), then $a\sim c$ also holds. Since $0+4=4>2$, we know that $0\sim 4$. We also just saw that $\sim$ is symmetric, so $4\sim 0$. Taking $a=0,b=4$, and $c=0$, we see that we have an example here in which $a\sim b$ and $b\sim c$, but $a\not\sim c$: $0+0=0\not>2$, so $0\not\sim 0$. Thus, the relation is not transitive.

Answer 2

It's not the "elements of the set" that must satisfy the conditions, it's the particular set of ordered pairs that must satisfy the three conditions.

You are correct saying $0$ is not related to $2$ and that $6$ is related to $0$, so the pair $(6,0)$ is in the relation but $(0,2)$ is not.

So write down more of the ordered pairs that you know do and don't represent related pairs and see if you can prove that the set of things that do satisfy the three conditions.