Quick number theory clarification before exam

[chimath35]

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

 

[chimath35]

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

 

[Dick]

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

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?

 

[chimath35]

Are you asking me to prove that?

 

[Dick]

Are you asking me to prove that?

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

 

[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
proof class. I might get a bad grade on my exam.

 

[Dick]

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
proof class. I might get a bad grade on my exam.

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.

 

[chimath35]

I give man, can't figure it out.

 

[Dick]

I give man, can't figure it out.

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.

 

[chimath35]

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

 

[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.

 

[chimath35]

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

 

[Dick]

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

That's meaningless, sorry.

 

[chimath35]

I don't like puzzles either.

 

[Dick]

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

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.

 

[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.

 

[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.

 

[chimath35]

Okay nb = n(ka)

ak= a(nk)

a is a number times nk a number so proved?

 

[Dick]

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

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.

 

[chimath35]

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

 

[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.

 

[chimath35]

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

 

[chimath35]

Okay nb = n(ka)

ak= a(nk)

a is a number times nk a number so proved?

 

[chimath35]

so a int times an int equals an int solved?

 

[Tsunoyukami]

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

Here's how I would go about this problem.


"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.

 

[chimath35]

3 Z multipied together are a Z by def. solved

 

[Dick]

so a int times an int equals an int solved?

That's the basic idea of the proof. If you can state that in the form of a proof as Tsunoyukami outlined, that would do it.

 

[Dick]

3 Z multipied together are a Z by def. solved

Now that's just annoying.

 

[chimath35]

Ya, I was trying to say that; thanks. How did you get good at proofs? Intro proof class?

 

[chimath35]

Now that's just annoying.

I was being serious.