# Homework Help: Equivalence relation with the Cartesian product of a set

1. May 8, 2013

### unawareness

1. The problem statement, all variables and given/known data
Let A be the set that contains all rational numbers, but not zero. Let (a,b),(c,d) $\in$ A×A. Let (a,b)$\tilde{}$(c,d) if and only if ad = bc. Prove that $\tilde{}$ is an equivalence relation on A×A.

2. Relevant equations

3. The attempt at a solution
The solution just needs to show reflexivity, symmetry, and transitivity. My problem is I don't get how to go between elements of A and elements of A×A.

The solution, according to my teacher, is:

Reflexivity - (a,b) $\tilde{}$ (a,b) $\Leftrightarrow$ ab = ba

I don't understand this. Is it because (I will probably unsatisfactorily define this) from the information in the problem, all cartesian products follow: (x$_{1}$,y$_{1}$) $\tilde{}$ (x$_{2}$,y$_{2}$) $\Leftrightarrow$ x$_{1}$y$_{2}$ = y$_{1}$x$_{2}$ ?

So, the relation is the mapping of two elements of A×A to a product of elements from A? I don't understand what the relation is. Is it a mapping? I'm very confused how I'm supposed to get from (a,a) $\tilde{}$ (a,a) to aa = aa. a is an element of A, but (a,a) is an element of A×A. I can't... even... comprehend... anything about how these two things are related. I understand that if A = {a,b} then A×A = {(a,a),(a,b),(b,a),(b,b)} (i think). But I don't get how elements of A and A × A can imply things about each other. I don't know how to work with elements between two sets of different dimension, I guess, is what I'm saying. I'm assuming A is one dimensional and A×A is two dimensional. I mean... I can't do things like a$\bullet$(b,b) to get something? Or is that equal to (a$\bullet$b,a$\bullet$b)? This is assuming there's a distributive property available. I don't know how it just automatically would be available, though. Like I guess if it's like matrices and scalars. But how does (a,a) get represented in terms of non-ordered pair, a?

Symmetry - (a,b) $\tilde{}$ (c,d) $\Leftrightarrow$ ad = bc $\Leftrightarrow$ bc = ad $\Leftrightarrow$ (c,d) $\tilde{}$ (a,b)

I don't follow this, either. I feel like nothing happened to prove anything. If it's this simple I don't know why it needs to be proved. I don't know why he elaborated anything.

Maybe, is it because of something like this:

(a,b) $\tilde{}$ (c,d) $\Leftrightarrow$ ad = bc , information from problem
ad = bc = bc = ad $\Rightarrow$ (ad = bc) = (bc = ad) , i don't know... this step is weird to me
(c,d) $\tilde{}$ (a,b) $\Leftrightarrow$ bc = ad , I'm assuming this information is given from my example with x's and y's which is equivalent to the information given in the problem, but easier for me to visualize.

With all that information, then I could feel more comfortable with the solution. I just don't quite understand why some of the things are just assumed to be equal to each other. I don't really understand why ab = bc can imply bc = ad. I don't understand the equals sign, I guess. Is it just something like these two things will always be each other. They will never not be each other?

The bigger problem is why can (c,d) $\tilde{}$ (a,b) $\Leftrightarrow$ bc = ad? Is this just something I'm assuming, so I can prove this is an equivalence relation? The problem only says (a,b) $\tilde{}$ (c,d) $\Leftrightarrow$ ad = bc. Again it seems my confusion is arising from when it is appropriate to say two things are equal to each other. So, (c,d) $\tilde{}$ (a,b) $\Leftrightarrow$ bc = ad, is just an assumption? Why am I assuming this particular thing? Why can I not assume like: (c,d) $\tilde{}$ (a,b) $\Leftrightarrow$ ab = cd ? What if for some strange reason, that's how it worked? Is it possible for it to work like that? It doesn't seem consistent with anything, but I guess what I'm asking is why is it impossible?

Transitivity - (a,b) $\tilde{}$ (c,d) and (c,d) $\tilde{}$ (e,f)
$\Rightarrow$ ad = bc and cf = de
$\Rightarrow$ adf = bcf and bcf = bde
$\Rightarrow$ adf = bde $\Rightarrow$ af = be $\Rightarrow$ (a,b) $\tilde{}$ (e,f)
d $\neq$ 0

Okay... he's just introducing another element of A×A, (e,f). I guess I'm supposed to assume (c,d) $\tilde{}$ (e,f) $\Leftrightarrow$ cf = de. Probably because that's how (a,b) $\tilde{}$ (c,d) works. Then there's the concept of right and left multiplication. I don't know why proof is necessary, though. I mean... if you just assume everything works the same way as (a,b) $\tilde{}$ (c,d) , then you just should already know that (a,b) $\tilde{}$ (e,f) will work out to be af = be. I don't get why all that work is necessary. If you're just operating under the assumption that this is how everything works, then everything will work this way. I feel like I'm missing some sort of concept. Also, why does right multiplication by f work the same for ad and bc? What if, for some reason, adf = adf, but bcf = a rabbit, and adf $\neq$ a rabbit. Is this possible somehow?

Thanks to anyone who dares try to help me. Lol.

Last edited: May 8, 2013
2. May 9, 2013

### Dick

The relation isn't a mapping from AxA to A. It's really a mapping from (AxA)x(AxA)->(true,false). (a,b)~(b,a) is true if ab=ba by the definition of the relation, and since ab=ba is always true for rationals (i.e. elements of A) so (a,b)~(b,a) is always true. I'm not sure why you are so severely confused.