Which of the following relations is an equivalence relation?

Question

I'm struggling a bit with this question. Here are the choices

• R1:={(a,b)∈Z×Z | 8 divides (a−b)}
• R2:={(a,b)∈Z×Z | 8 divides (a+b)}.
• R3:={(a,b)∈Z×Z | a⋅b is divisible by 25 but a,b are not}.
• R4={(a,b)∈Z×Z | a⋅b is divisible by 36 but a,b are not}
• R5={(a,b)∈Z×Z | |a−b| is a prime number}

Thanks for the feedback. I've edited the original question below:

I know that equivalence relations are reflexive, symmetric and transitive. This I went through each option and followed these 3 types of relations.

For R1 I considered a - b such that I get a multiple of 8. For example (24, 16) (16,8) (8,8). these seem to work so I assumed it's equivalence.

For R2 I followed the same approach. And found that it, too, was an equivalence relationship.

I followed the same process for the rest of the options. And found them to be equivalent except for the last one (which is not)

For R5 aRb would not work for a reflexive relationship since if a-b then 0, which is not prime.

MY CONFUSION: I get confused because it seems that there are certain pairs inside each relationship that follow the properties for equivalence and some that don't. For a relation to be an equivalence relation do all pairs have to meet the criteria or just some?

Thanks, K

Answer 1

I get confused because it seems that there are certain pairs inside each relationship that follow the properties for equivalence and some that don't. For a relation to be an equivalence relation do all pairs have to meet the criteria or just some?

Neither.

To prove reflexivity all pairs of the form $(a,a)$ most have the relationship.

Th prove symmetry absolutely no pairs have to have the relationship. But if the pair $(a,b)$ has the relationship then $(b,a)$ must also have it. But no arbitrary pair needs to.

Transitive. Again no pair needs to. But if the two pairs $(a,b)$ and $(b,c)$ do. Then the pair $(a,c)$ must.

So which pairs need to in order to be an equivalence. Only the pairs $(a,a)$ have to. The other conditions only say IF some pairs have the relationship, which other pairs most also.

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Rather than go example by example I ll go property by property

Reflexive/

1) The is reflexive if $8|a-a$ for all $a \in \mathbb Z$. $a-a = 0$ and $8|0$ so this is true. Reflexivity holds.

2) This is reflexive if $8|a+a$ for all $\in \mathbb Z$. If $a =1$ then $a + a=2$ and $8\not \mid 2$. Reflexivity fails.

Can you continue?

Symmetry

1) This is symmetric if when ever $8|a-b$ the $8|b-a$. As $b-a = -(a-b)$ and if $8|a-b$ there is an integer $k$ so that $a-b = 8k$ then $b-a =8(-k)$ so yes $8|a-b \implies 8|b-a$. So symmetry holds.

2) This is symmetric if whenever $8|a+b$ then $8|b+a$. As $b+a = a+b$ anything we say about $a+b$ will be true about $b+a$ as they are the same thing. Symmetry holds.

can you continue.

Transitivity

1) This is transitive if whenever $8|a-b$ and $8|b-c$ then $8|a-c$.

If $8|a-b$ then thre is a $k$ so that $a-b = 8k$ and if $8|b-c$ there is a $j$ so that $b-c = 8j$. So $a = 8k +b$ and $c = b-8j$ and $a -c = (8k+b)-(b-8j) = 8k + 8j = 8(k+j)$. So $8|a-c$. Transitivity holds.

Can you continue?

Answer 2

It seems that $(1,1)\notin R_2$, $(0,0)\notin R_3$, $(0,0)\notin R_4$ and $(0,0)\notin R_5$ so $R_2$, $R_3$, $R_4$ and $R_5$ are not equivalence relations.

Clearly, $(a,a)\in R_1$ and if $(a,b)\in R_1$ then $(b,a)\in R_1$. Moreover, if $(a,b)\in R_1$ and $(b,c)\in R_1$, then $8\mid a-b$ and $8\mid b-c$ so $8\mid a-b+b-c$, that is $8\mid a-c$ and $(a,c)\in R_1$. We deduce that $R_1$ is an equivalence relation.