Equivalence relation

a~b iff a-b (is an element of) Z

To prove a~b is symmetric, I showed that:
b-a = -(a-b) is an element of Z, so hence b~a

Is this correct?

--------------------------
a~b iff a+b (is an element of) Z

To prove a~b is symmetric, I showed that:
b+a = -(-a-b) is an element of Z, so hence b~a

Is this correct?

Because I think Im stuck on the 2nd part, im not sure if i did it correctly..

Reply 1

nota bene

rpan161

a~b iff a-b (is an element of) Z

To prove a~b is symmetric, I showed that:
b-a = -(a-b) is an element of Z, so hence b~a

Is this correct?

Yes. You may want to say something about that if c in Z then -c in Z, but it's probably not necessary.

Have you shown the other properties?

a~a as a-a=0 which is in Z

and transitivity

--------------------------
a~b iff a+b (is an element of) Z

To prove a~b is symmetric, I showed that:
b+a = -(-a-b) is an element of Z, so hence b~a

Is this correct?

Because I think Im stuck on the 2nd part, im not sure if i did it correctly..

Well, b~a is b+a, but you know a~b, which is a+b and as a+b=b+a by field axoims of R then b~a.

Although I guess saying b+a=-(-(a+b))=a+b works as well, but it seems a bit odd to me.

Reply 2

rpan161

oh thanks, yes i have shown the other properties, i just have trouble with symmetry.
I am not sure how to do this one:

a~b iff 3a+b (is an element of) Z

To prove a~b is symmetric, how would I show it?
3b+a = insert something here? is an element of Z, so hence b~a

How would i show 3b+a? Because I dont think showing 3b+a=a+3b works?

Reply 3

nota bene

It isn't true, so try to find a counterexample.

Spoiler

Reply 4

rpan161

Thanks, does this means that a~b iff 3a+b (is an element of) Z is not an equivalence relation?

Reply 5

nota bene

Yes, because it doesn't satisfy the symmetric property (in fact nor transitive as below) an E.R. has to have...

(a=1/27 b=8/9 then a~b, with same b and c=1/3 then b~c but a doesn't squiggle c as 1/9+1/3=4/9 which isn't an integer)