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**Kazy****Member**- Registered: 2006-01-24
- Posts: 37

I need to prove that multiplication on Q is well defined. I have no idea really what well defined even means. From what I understand from reading, it means that multiplication on Q does not depend on the choices of integers to represent the equivalence classes. However, I have no idea how to go about proving that. Anyone have any ideas?

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

English version:

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Think of well defined as defined well. When is something defined well? When you know precisely what it is by the definition!

So the definition we have is that a * b = c, where a, b, c ∈ Q. If we know exactly what a and b are, do we know exactly what c is? Or is it a little fuzzy?

Here's any example: y = x². If I tell you what y is, can you tell me what x is? Well, no, not exactly. If I say y is 9, x could be 3 or it could be -3. See what I mean by fuzzy? It's not defined well, you don't know exactly what x is.

On the contrary, if I tell you y = x, and then tell you what y is, you could tell me what x is every time. That's pretty well defined, right?

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Math (words) version:

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Well defined simply just means that if two elements of a set are eqaul, and there is a function that maps these elements to another set, then their mapped versions will also be equal.

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Math (symbol) version:

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A map f is well defined when f:A⇒B, for all a1,a2∈A if a1 = a2 then f(a1) = f(a2)

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I did that mostly to show that mathimaticans have such a great language that something which takes paragraphs in english can be shortned down to one sentence in pure mathish.

What you need to show is that if a = b and c = d, then a * c = b * d. Try it, and see what you can come up with.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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