# Thread: stumped on this equivalence relation

1. ## stumped on this equivalence relation

define ~ on Z by a~b iff 3a+b is a mulitple of 4.

so far i only have reflexive...

reflexive: 3a+a=4a

how can i prove that its symmetric and transitive?

2. Originally Posted by glopez09
define ~ on Z by a~b iff 3a+b is a mulitple of 4.

so far i only have reflexive...

reflexive: 3a+a=4a

how can i prove that its symmetric and transitive?
Symmetric:

$a\sim b\implies 3a+b=4n$,

But $3a+b=4n\implies 9a+3b=4(3n)\implies3b+a=4(3n)-8a$ $\implies 3b+a=4(3n-2a)\implies b\sim a$

Transitive:

I'll start this, and then I'll leave it for you to finish.

$a\sim b\implies 3a+b=4n$ and $b\sim c\implies {\color{red}3b+c}=4k$

Now, $3a+b=4n\implies 3a+(b+3b)+c-({\color{red}3b+c})=4n\implies\dots$

Can you finish this?

3. i am not understanding either of those...?

4. Hello glopez09
Originally Posted by glopez09
i am not understanding either of those...?
You need to work out what it is you need to prove, in order to show that $\sim$ is symmetric and transitive.

First it's symmetric if $a\sim b \Rightarrow b\sim a$.

Now $a\sim b$ means that $3a+b$ is a multiple of $4$. And if we're going to show that $b\sim a$, we shall need to show that this will mean that $3b+a$ is also multiple of $4$. So, read carefully through Chris L T521's answer. He has clearly shown that if $3a+b$ is a multiple of $4$ then $3b+a$ is also a multiple of $4$.

Then, what does it mean to prove that $\sim$ is transitive? It is this: if $a\sim b$ and $b\sim c$, then we must prove that $a\sim c$. Translating this into multiples of $4$, this is: if $3a+b$ is a multiple of $4$, and $3b+c$ is a multiple of $4$, then we must prove that $3a+c$ is also a multiple of $4$. Chris L T521's answer has shown you how to build up an expression for $3a+c$ starting with $3a+b$ and $3b+c$. Study it again, and make sure that you can see how this shows that $3a+c$ is a multiple of $4$.

5. OK. i got some time to work on it. and i now understand the symmetry. but that -(3b+c) is throwing me through a loop?

this was my attempt without that.

3a+b+3b+c = 4n+4k

3a+4b+c = 4n+4k

3a+c = 4n+4k-4b

3a+c = 4(n+k-b)

would that work?

6. Originally Posted by glopez09
OK. i got some time to work on it. and i now understand the symmetry. but that -(3b+c) is throwing me through a loop?

this was my attempt without that.

3a+b+3b+c = 4n+4k

3a+4b+c = 4n+4k

3a+c = 4n+4k-4b

3a+c = 4(n+k-b)

would that work?
Perfect!

In the transitive case, I had to add zero:

$3a+b=4n\implies 3a+b+{\color{blue}0}=4n\implies3a+b+(3b+c){\color{ red}-(3b+c)}=4n$ $\implies 3a+4b+c-4k=4n$...

Then your result follows. Does this clarify things more?

7. ahhh ok thanks so much.