I have few questions

1. Apr 7, 2007

UofC

1. The problem statement, all variables and given/known data
Prove cancellation axiom using the properties of addition, multiplication and the order axioms. This is cancellation axiom in the integers.

2. Relevant equations

3. The attempt at a solution

I basically said that if ab=ac, then ab-ac=0, so a(b-c)=0, and if "a" does not equal to 0, then b-c=0, hence b=c.

I just started working with proofs and I have no idea if this is correct, and I didn't have the time to go to TA session so now I'm on my own.

2. Apr 7, 2007

Dick

Clearly, to show this you need to show that if a and b are non-zero then a*b is non-zero in the integers. It's not clear exactly what axioms you are starting with, but I have a hunch that something along the lines of an 'order axiom' might help you.

3. Apr 7, 2007

UofC

So I need to prove that if a,b in Z, "a" not equal to 0, and ab=ac, then b=c. So if I say that if ab=ac, then ab-ac=o. Then by the distributive law a(b-c)=0. So if a(b-c)=0, and a does not equal to 0, then b-c=0, hence b=c. I don't understand why do I need to show that (ab) is non zero? Also what do you mean by "hunch"? Sorry, I'm not a native spreaker of English.

4. Apr 7, 2007

Dick

Sorry, I didn't mean to confuse you. I'll switch letters. Do you already know that if x and y are non-zero then x*y is non-zero? In that case your proof is already correct and complete. To prove that x*y is non-zero given that x and y are non-zero would probably involve something like an order axiom. 'Hunch' just means 'guess'.

5. Apr 7, 2007

UofC

Yes I know that and I think that I don't need to prove it. Professor said that whatever we prove in class or for hw, or we use as an assumption before, can be used without proving it in the future problems. So I guess that one is done...Man these proofs are frustrating.

6. Apr 7, 2007

UofC

Hmmm...I have another one. So the problems says: Let X be a nonempty set and R=power set of X. Show that R with symmetric difference as addition and interstection as multiplication is a commutative ring with 1.

So, I guess I can say that the symmetric difference is commutative because...(A-B)U(B-A)=(B-A)U(A-B), and that intersection is commutative since...A intersection B= B intersection A. But I have no idea if there is any logical flow in this one.

7. Apr 7, 2007

UofC

Ooooops...It is commutative ring with 1, so I'll have to prove it for all 5 properties of addition, all 4 of multiplication and distributive one also.

8. Apr 8, 2007

Dick

True, but I don't think any of them are particularly difficult, are they?

9. Apr 8, 2007

UofC

Yeah I think I got that one, however...

The problem says:

If c|a and c|b, then c|(sa+tb), for s,t in Z.

Now I tried to say that if c|a and c|b then c|(a+b), and if c|a then c|sa and if c|b then c|tb, so if c|(a+b) and c|sa and c|tb, then c|(sa+tb)...So yeah I'm basically stuck on this one.

10. Apr 8, 2007

Dick

Stuck where? If c|a and c|b then a=i*c and b=j*c. So sa+tb=s*i*c+t*j*c=(s*i+t*j)*c which looks divisible by c to me. I thought that was basically what you were saying?

11. Apr 8, 2007

UofC

I was...sorta. I see why this is true. How do I show that if c|a and c|b, then c|(a+b)? Or that if c|a, then c|sa? I do understand why is it true but can't think of the way to show it.

12. Apr 8, 2007

matt grime

Dick explained that to you. He even showed you the proof.

13. Apr 8, 2007

UofC

Oh I see now. Since sa+tb=(s*i+t*j)*c, c is a divisor of sa+tb. Oh, oh...I see now. Thanks a lot guys.