would like some feedback(not the answer) on an equivalence relation proof

Question

asnaseer · Answer

go ahead...

anonymous · Answer

ok

anonymous · Answer

sorry still not used to the site since the changes

anonymous · Answer

ok so i'm supposed to either prove or disprove the following:

anonymous · Answer

A relation R is defined on Z by aRb if a is congruent to b(mod2) and a is congruent to b(mod3) . . .

anonymous · Answer

So I figured out that if 2|a-b and 3|a-b then a-b = 2k and a-b= 3k . . .

anonymous · Answer

so a = 2k + b and a = 3k + b which means the only integer for which this can actually be a relation would be when k = 0

anonymous · Answer

which would then mean that a would have to equal b for this to even be considered to be a relation

asnaseer · Answer

sorry mathgirl but this is beyond my knowledge domain - sorry :-(

asnaseer · Answer

there is guy called jamesj on this site who is VERY good at this type of stuff - if you can find hime he would be your best option.

anonymous · Answer

yeah, i know. . . he always helps me out.  . . if he's on, i'm sure he'll help out

jamesj · Answer

Ok.  So, so far so good:
"So I figured out that if 2|a-b and 3|a-b"

jamesj · Answer

and that means that 
a - b = 2k for some k 
a - b = 3j for some j

Note that k need not be j.  This is important.

jamesj · Answer

Intuitively if a and b differ by a multiple of 2 AND a multiple of 3, they must in fact differ by a multiple of 6.  

And that is indeed the case, because if
a - b = 2k for some k 
a - b = 3j for some j

then 2k = 3j
which implies 3 | k and 2 | j.  So for example, we can write k = 3m, for some m.
Therefore
a - b = 2k = 2(3m) = 6m

Therefore what we've shown is that 
 $ a \equiv b $ (mod 2) and $ a \equiv b $ (mod 3) $ \implies $ $ a \equiv b $ (mod 6)

jamesj · Answer

Now what's the question?  Is the question: 
"If aRb is defined by a = b (mod 2) and a = b (mod 3), is R an equivalence relation?"

anonymous · Answer

Ok, so now based on showin that a≡b (mod 6) I can proceed with testing reflexivity, symmetry, and transitivity

jamesj · Answer

sorry, just saw your reply.   Yes.

jamesj · Answer

It is the case that a = b (mod n) for any n is an equivalence relationship; and you should have no trouble showing that this is the case for n = 6.