# Homework Help: 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.

21. Mar 11, 2014

### Tsunoyukami

Seeing proofs for the first time can be daunting (I know it was for me!) but you can learn how to do proofs and how to approach proofs. It takes a lot of work but you'll get it if you work at it.

I would suggest focusing your attention on understanding the definitions of the concepts you're required to know. For example, suppose you were given this problem on your test: "Prove that if a|b, then a|nb for any integer n." This is gibberish unless you understand what the definition " | " means. You need to have internalized this definition and translate it into something that you understand. So the statement a | b means a divides b, which means b is equal to a times some integer which can be written b = a*k where k is an integer.

Do you see what I did there? I took the befuddling definition and I wrote it out in English and translated it back into something I could work with.

Let's try giving you a step by step guide to this problem - but you'll have to do the work yourself.

"Prove that if a|b, then a|nb for any integer n."
1) Assume a|b.
2) Use the definition of a|b to write this in a form that is more useful (hint: I already did this for you above).
3) Try to manipulate your expression from (2) by multiplying by n and then use the definition of a|nb to come to the conclusion a|nb.

22. Mar 11, 2014

### chimath35

Also, sorry for being negative; I don't feel good and have an exam tomorrow.

23. Mar 11, 2014

### chimath35

Okay nb = n(ka)

ak= a(nk)

a is a number times nk a number so proved?

24. Mar 11, 2014

### chimath35

so a int times an int equals an int solved?

25. Mar 11, 2014

### Tsunoyukami

You're getting close but try to be a bit more precise and remember to show all your work explicitly.

"If a|b, then a|nb for any integer n."

Proof
Assume a|b. Then, by the definition of a|b, b=ak for some integer k. Multiplying this equation by n we find nb=n(ak)=a(nk)=al where l = nk is an integer since both n and k are integers. (This is where you got to. You should now have a concluding sentence.)

You were close to having a complete proof but you should strive to be more precise in presenting your solution and make sure that you conclude with the statement that you wanted to show.