# Showing final part of equivalence relation

• Oct 5th 2010, 02:20 PM
kathrynmath
Showing final part of equivalence relation
Let S be the set of all ordered pairs (m,n) of positive integers. For (a1,a2) in S and (b1,b2) in S, define (a1,a2)~(b1,b2) if a1+b2=a2+b1.
I need to show we have an equivalence relation.

The first two parts I have already determined to be true, so i will omit my work on that.
For the third part I need to show if for all a,b,c in S if (a,b) is in ~ and (b,c) is in ~ then (a,c) is in ~.
We have:
a1+b2=a2+b1
b1+c2=b2+c1
We want to show:
a1+c2=a2+c1
I'm unsure of how to get there.
I tried using b1=a1-a2+b2 and b2=a2-a1+b1
a1-a2+b2+c2=a2-a1+b1+c1
That didn't really get me anywhere, though.
• Oct 5th 2010, 03:36 PM
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Quote:

Originally Posted by kathrynmath
For the third part I need to show if for all a,b,c in S if (a,b) is in ~ and (b,c) is in ~ then (a,c) is in ~.
We have:
a1+b2=a2+b1
b1+c2=b2+c1

If you simply add these two equations together, b1+b2 will cancel from both sides and you will be left with the desired result.
• Oct 6th 2010, 04:52 AM
HallsofIvy
By the way- this equivalence relation can be used to define the set of all integers from the set of positive integers. The integers are the equivalence classes with this equivalence relation. We then define addition and multiplication by using "representative": if $\displaystyle \alpha$ and $\displaystyle \beta$ are equivalence classes we define: choose one pair, $\displaystyle (x_1, y_1)$, of set $\displaystyle \alpha$ and one pair, (x_2, y_2), of $\displaystyle \beta$. Then $\displaystyle \alpha+ \beta$ is the equivalence class containing $\displaystyle (x_1+ x_2, y_1+ y_2)$ and $\displaystyle \alpha\beta$ is the equivalence class containing $\displaystyle (x_1x_2+y_1y_2, x_1y_2+ x_2y_1)$. Of course, you need to show that these are "well defined"- that is, that if you chose different "representatives" of each equivalence class, while $\displaystyle (x_1+ x_2, y_1, y_2)$ and $\displaystyle (x_1x_2+ y_1y_2, x_1y_2+ x_2y_1)$ would be different, they would still be in the same equivalence classes.

It is fairly easy to show that there exist a class consisting of all those pairs in which the two numbers are equal. That is, that (x, x) and (y, y) are equivalent and that if (x, x) was equivalent to (y, z) then y= z. It is also easy to show, then, that that equivalence class acts like the "0" element.

If X is an equivalence class containing (x, y) and 0 is the equivalence class containing (a, a), then X+ 0 is the equivalence class containing (x+ a, y+ a). But (x+ a)+ y= (y+a)+ x so this is still equivalent to (x, y) and is in the same equivalence class: X+ 0= X.

Also, if X is an equivalence class containing (x, y) and 0 is the equivalence class containing (a, a) then X*0 is the equivalence class containing (xa+ ya, ax+ ay) and, of course, xa+ ya= ax+ ay. That is, this pair is in the class 0: X*0= 0.

Similarly, in the set of all pairs, (x, y) where x is any integer and y is any integer except 0, we can define an equivalence relation by "$\displaystyle (x_1, y_1)~ (x_2, y_2)$ if and only if $\displaystyle x_1y_2= x_2y_1$ and get the rational numbers as the equivalence classes.