Recent content by UofC

After a lot of thinking and an allnighter, think I got it, allong with the rest of my problem set. Now I turn it in and :zzz: Thanks guys.

Homework Statement So I have to prove that addition and multiplication in Zn are well defined. Homework Equations The Attempt at a Solution I have no idea where to start.

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.

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.

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...

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.

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...

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.

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...

Homework Statement Prove cancellation axiom using the properties of addition, multiplication and the order axioms. This is cancellation axiom in the integers. Homework Equations 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...