I have few questions

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.

 

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.

 

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.

 

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.

 

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.

 

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.

 

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.

 

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.