Proving operations are well-defined

[Raziel2701]

Homework Statement

Let X= N x N. Define a relation R on X by (x,y)R(z,w) if xw=yz. Then R is an equivalence relation.

Define "+" and "x"(multiplication) on the set X/R of equivalence classes by:

[x,y] "+" [z,w] = [xw + yz,yw]

[x,y] "x" [z,w] = [xz.yw].

Prove that each of the operations is well-defined, independent of choice of representatives.

Homework Equations

The Attempt at a Solution

The way R is defined: (x,y)R(z,w) if xw=yz, can be rewritten as \frac{x}{y}=\frac{z}{w}

I said that [x,y] is equivalent to \frac{x}{y}, and thus I did the following:

\frac{x}{y} +\frac{z}{w}= \frac{xw+yz}{yw} which is equivalent to [xw+yz,yw], thus the operation is well-defined.

Is this correct?

I did more or less the same process for the "x" operation.

 

[micromass]

No, well defined means the following. Take [x,y]=[x^\prime,y^\prime] and [z,w]=[z^\prime,w^\prime].

For + to be well defined, you'll need to show that [x,y]+[z,w]=[x^\prime,y^\prime]+[z^\prime,w^\prime].

For . to be well defined, you'll need to show that [x,y].[z,w]=[x^\prime,y^\prime].[z^\prime,w^\prime].

 

[Raziel2701]

But the problem statement defines the operations differently, shouldn't I have to work with the definitions I was already given?

 

[micromass]

I don't understand. Yes, you'll need to work with the operations given to you...

 

[Raziel2701]

Well, the thing is that I already have what [x,y] + [z,w] equals to from the problem statement, so trying to do [x',y']+ [z',w'] seems to me to be doing something else.

Does that make sense?

 

[micromass]

The problem is that sometimes two things are equal when they don't appear to be. Like [2,4]=[1,2]. In the left-hand side you have the representatives 2 and 4. And in the right-hand side the representatives are 1 and 2.

When you use the formula [x,y].[z,w]=[xz,yw], you're using the representatives x,y and z,w. But there could be other representatives.

For example: [1,2][z,w]=[z,2w] and [2,4][z,w]=[2z,4w]. Since you multiplicated the same elements, you'll need to get the same answer. So you'll need to see wheter it is true that [z,2w]=[2z,4w]. If this is not true, then multiplicating thesame element yields a different answer! The multiplication is then not well-defined.

 

[Raziel2701]

I think if I do what you suggest, I'd be forgetting to utilize the definition of the equivalence relation. I don't know, I'm not convinced that your suggestion is correct mainly because yours seems to be a way that's not taking into account all these tidbits of information embedded in the relation, and in the manner the operations are defined.

 

[micromass]

Alright what does the question mean to you then? What does the sentence "independent of choice of representatives" mean then??

 

[Raziel2701]

So independent of choice goes right along with your suggestion because it doesn't matter what numbers we pick, the operation should hold, however, don't I have to incorporate the relation into my proof somehow?

It'd be helpful if I could first understand what I'm doing wrong in my attempt at a solution.

 

[micromass]

So take [x,y]. Let's say that there are other representatives [x',y']. This means that [x,y]=[x',y'] (where the equality is in X/R, the set of equivalence classes). This means in X that (x,y)R(x',y'). This is how you incorporate the relation.

 

[Raziel2701]

[x,y]+[z,w]=[x^\prime,y^\prime]+[z^\prime,w^\prime] Implies [x,y]=[x^\prime,y^\prime]

Which means (x,y)R(x',y'), therefore xy'=yx', thus \frac{x}{y}=\frac{x^\prime}{y^\prime} which only holds if x=x' and y=y'.

Now this seems to be similar to what I had posted on my first post. Am I on the right path?

 

[micromass]

I don't see how you did that. You have that [x,y]=[x',y'] and [z,w]=[z',w'] (thus (x,y)R(x',y') and (z,w)R(z',w') ).

Now you need to show that [x,y]+[z,w]=[x',y']+[z',w']. I suggest applying the definition of the addition (your definition) and then see if the two couples you receive are in relation with each other)...

 

[Raziel2701]

[xw+yz,yw]=[x'w'+y'z',y'w']

Now if these two things are equal, then (xw+yz,yw)R(x'w'+y'z',y'w'), but from there I get that
(xw+yz)(y'w')=(x'w'+y'z')(yw)

And then I can do a bunch of algebra to get nowhere...sigh

What am I missing?

 

[micromass]

Yes, your on the right path. But you'll need to use that (x,y)R(x',y') (thus xy'=x'y) and (z,w)R(z',w') (thus zw'=z'w). Apply this to xwy'w'+yzy'w'=x'w'yw+y'z'yw ...

 

[Raziel2701]

Ok, so I get to this step after manipulating this: (xw+yz)(y'w')=(x'w'+y'z')(yw)

I get w'wy'x +zw'yy' = yx'ww' + z'wyy'

And since we know that y'x=yx' and zw'=z'w then I can make substitutions to make the expressions on both sides of the equal sign to be exactly the same. Is that the end of the proof?

 

[micromass]

Yes, that's it. The multiplication is the same thing...

 

[Raziel2701]

Phew, got it, thanks a lot micromass!