# 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:

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

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

Transitive:

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

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

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

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 $\displaystyle \sim$ is symmetric and transitive.

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

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

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

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!

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