# Quick number theory clarification before exam

1. Mar 11, 2014

### chimath35

Okay if a/b then doesn't a/nb for any integer n?

2. Mar 11, 2014

### chimath35

Well maybe any non zero integer n; but maybe that would make the theorem kind of useless in proofs?

3. Mar 11, 2014

### Dick

If you mean a|b (i.e. a divides b) then sure, a|nb for any integer n. How would you prove that? What does '|' mean?

4. Mar 11, 2014

### chimath35

Are you asking me to prove that?

5. Mar 11, 2014

### Dick

I'm mostly asking if you understand what it means. If you do, then showing a|b implies a|nb shouldn't be difficult.

6. Mar 11, 2014

### chimath35

ac=b

then

ac=b+b+b....+bn

but if you broke it up into cases a goes into b clean so repeatedly you are adding integers so bn must
be an Z thus we have ac=b

I am not good at proofs, having very hard time; never did or saw them before this class, no intro to

7. Mar 11, 2014

### Dick

Maybe. That's not at all clear. Try this. The definition of a|b is that b=ka for some integer k. Now can you show that if a|b then a|nb? Try that to practice at proofs.

8. Mar 11, 2014

### chimath35

I give man, can't figure it out.

9. Mar 11, 2014

### Dick

You give up pretty easily. If a|b then b=k*a for some integer k by the definition of '|'. Then nb=n(ka). Why? So nb=n(ka)=(nk)a. To show a|nb you want to find an integer j such that nb=ja. What's good choice for j? Just read that through a few times.

10. Mar 11, 2014

### chimath35

ak=ak+ak+ak which would result in 1+1+1 which is an int. so concluded?

11. Mar 11, 2014

### chimath35

Sorry to be negative but this kind of math is beyond frustrating to me. It is like opposite of what I am used to.

12. Mar 11, 2014

### chimath35

I don't see myself being really creative, maybe this math is not for me.

13. Mar 11, 2014

### Dick

That's meaningless, sorry.

14. Mar 11, 2014

### chimath35

I don't like puzzles either.

15. Mar 11, 2014

### Dick

You seem to be more interested in expressing your frustration with the course than in listening to advice. Which is ok, but I don't see how I can help with that.

16. Mar 11, 2014

### chimath35

That is because I like math a lot just got an A in calc three and I sit for hours with this stuff and have yet to solve ONE proof by myself.

17. Mar 11, 2014

### chimath35

It is like crazy thinking and time with these problems. Yet I still I am unable to solve a problem. Also I have no idea where to go with your proof. I appreciate you trying, I just give don't know the answer.

18. Mar 11, 2014

### chimath35

Okay nb = n(ka)

ak= a(nk)

a is a number times nk a number so proved?

19. Mar 11, 2014

### Dick

Congrats on calc 3! But so far you haven't listened to a single thing I've said. I thought you wanted help, not a forum to express how negative you are about the course. I can't help with that.

20. Mar 11, 2014

### chimath35

Look at my last comment, pretty sure it is proved.