Expert: Sherry Wallin - 12/18/2008

Question
A relation R on the set  Z of integers given by R={(x,y):2|(x-y)} is an equvalance relation.

Soluction: R is reflexive ,since 2|(a-a),for all a є R
R is symmetric .since 2|(a-b)

⟹                                    a-b=2k,    kєZ
⟹                                    b-a=2(-k)
⟹                                    2|b-a                       i.e.,(a,b) єR ⟹(b,a) єR
Further        
                                        (a,b) є  R⟹2|a-b
                 (b,c) єR   ⟹2|bic
Now a-c=(a-b)+(b-c),which is even i.e., divisible by 2.
⟹         2|a-c
Therefore                          2|a-b,2|b-c⟹2|a-c
i.e.,         (a,b) єR,(b,c) єR⟹(a,c) єR

therefore  R is transitive
hence .R is an equivalence relation on Z.


my problem is I cloud understand this avove example.
Please explain the avobe example step by step
Please help me.it is important for me.please friend help me.  

Answer
Hi Pratap~
    There are three things you need to show in order to say something is an equivalence relation. The reflexive, symmetric, and transitive properties.

Reflexive says that the relation is a reflection of itself. A good example is "=" because any two things, thing 1 and thing 2 that are equal you have thing 1 = thing 2 and thing 2 = thing 1. This is essentially what needs to be shown for reflexivity. Now the relation that you have is that given an ordered pair (just like a point in the x-y coordinate system), that their difference is divisible by 2. Anything divisible by 2 is even so it means that if you take the ordered pair and subtract one from the other that the result is even. For reflexivity give me any number and take it away and you get 0 which is an even number so it is divisible by 2. (By the way this long winded explanation can be considered an alternate proof). Wherever you got the proof you presented they used another way to prove it. They used the definition of the relation (that 2 divides the difference 2|a-a which is the same thing as saying 2|0 so the relation is reflexive).

I'm not entirely comfortable with the proof of symmetry because it is only symmetric if the first condition is met and the writer has not assumed it is already met. You should always assume the first condition is met, i.e., assume that for a,b in R that 2|(a-b). This says if 2 divides the difference between the coordinates of the ordered pair then it can be shown that their difference in the opposite order is also divisible by 2. A concrete example is say your ordered pair is (2,4) their difference (In one direction) is 2-4 = -2 which is an even number hence divisible by 2 and also 4-2= 2 is even and therefore divisible by 2. So this is the way the proof should read:

Suppose a,b in R and 2|(a-b). This means a-b = 2k for some integer k. (this is the exact definition of what it means to divisible). So since a-b is divisible by 2, you need to show that b-a is also divisible by 2. This is simple because if you multiply by -1 you get b-a = -2k which can always be written as 2(-k), therefore the relation is symmetric.

The 3rd part you have written is about transitivity and basically says if the relation works in two ordered pairs and one of the entries in the ordered pair is the same as one other that the first is and the 3rd are still even. Again, this is written a bit better and perhaps acceptable by some since at least they let a,b,c be in R but you STILL NEED TO ASSUME IF a,b in R AND IF b,c in R THEN you can say (a,b) in R implies 2|(a-b) and b,c in R implies 2|(b-c) [and then translate those expressions using the definition into two equations and try to tie them together], there exists a "k" in the integers such that a-b = 2k and there exists a "j" in the integers such that b-c = 2j. [To combine these two so you can show the first implies the 3rd solve for b in the 2nd equation and substitute it into the first equation]. a-(2j+c) = 2k which implies a-c = 2k+2j = 2(k+j)which says that a-c is divisible by 2 and hence a~c.

Now that we've shown reflexivity and symmetry and transitivity you can conclude that this relation is an equivalence relation.

I hope this made what you need to do clearer.

Math Prof