stumped on this equivalence relation

**glopez09** · September 26th 2009, 08:44 PM

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?

**Chris L T521** · September 26th 2009, 08:56 PM

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?

**glopez09** · September 26th 2009, 09:54 PM

i am not understanding either of those...?

**Grandad** · September 27th 2009, 06:31 AM

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 $4$ .

Grandad

**glopez09** · September 27th 2009, 05:14 PM

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?

**Chris L T521** · September 27th 2009, 05:21 PM

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?

**glopez09** · September 27th 2009, 05:23 PM

ahhh ok thanks so much.

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