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Homework Help: Quick number theory clarification before exam

  1. Mar 11, 2014 #1
    Okay if a/b then doesn't a/nb for any integer n?
     
  2. jcsd
  3. Mar 11, 2014 #2
    Well maybe any non zero integer n; but maybe that would make the theorem kind of useless in proofs?
     
  4. Mar 11, 2014 #3

    Dick

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    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?
     
  5. Mar 11, 2014 #4
    Are you asking me to prove that?
     
  6. Mar 11, 2014 #5

    Dick

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    I'm mostly asking if you understand what it means. If you do, then showing a|b implies a|nb shouldn't be difficult.
     
  7. Mar 11, 2014 #6
    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.
     
  8. Mar 11, 2014 #7

    Dick

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    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.
     
  9. Mar 11, 2014 #8
    I give man, can't figure it out.
     
  10. Mar 11, 2014 #9

    Dick

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    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.
     
  11. Mar 11, 2014 #10
    ak=ak+ak+ak which would result in 1+1+1 which is an int. so concluded?
     
  12. Mar 11, 2014 #11
    Sorry to be negative but this kind of math is beyond frustrating to me. It is like opposite of what I am used to.
     
  13. Mar 11, 2014 #12
    I don't see myself being really creative, maybe this math is not for me.
     
  14. Mar 11, 2014 #13

    Dick

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    That's meaningless, sorry.
     
  15. Mar 11, 2014 #14
    I don't like puzzles either.
     
  16. Mar 11, 2014 #15

    Dick

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    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.
     
  17. Mar 11, 2014 #16
    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.
     
  18. Mar 11, 2014 #17
    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.
     
  19. Mar 11, 2014 #18
    Okay nb = n(ka)

    ak= a(nk)

    a is a number times nk a number so proved?
     
  20. Mar 11, 2014 #19

    Dick

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    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.
     
  21. Mar 11, 2014 #20
    Look at my last comment, pretty sure it is proved.
     
  22. Mar 11, 2014 #21
    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.
     
  23. Mar 11, 2014 #22
    Also, sorry for being negative; I don't feel good and have an exam tomorrow.
     
  24. Mar 11, 2014 #23
    Okay nb = n(ka)

    ak= a(nk)

    a is a number times nk a number so proved?
     
  25. Mar 11, 2014 #24
    so a int times an int equals an int solved?
     
  26. Mar 11, 2014 #25
    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.
     
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