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Divisibility property

  1. Jul 31, 2010 #1
    1. The problem statement, all variables and given/known data

    prove the following

    if a is an odd integer, then, 24 l a(a2-1)

    (i'm not familiar with modulo yet, i think it can help, but let don't use it yet ;P)

    2. Relevant equations

    n/a

    3. The attempt at a solution

    i stumbled when using 2n+1=a for all integer n, because i will only get (2n)(4n2+4n), i can't see how it can divided by 24, should i think of any other equivalent classes of odd integer?? or any other clue? T_T help
     
  2. jcsd
  3. Jul 31, 2010 #2
    For 24 to divide a(a2-1), convince yourself that this happens exactly when 8 and 3 divide evenly.
     
  4. Jul 31, 2010 #3
    i don't understand T_T, do you mean i need to find 8 l "something" , 3 l "something"

    like 8 l a and 3 l a2-1 then 24 l a(a2-1), something like that??
     
  5. Jul 31, 2010 #4
    I meant 24 | a(a2-1) if and only if 8 | a(a2-1) and 3 | a(a2-1).
     
  6. Jul 31, 2010 #5
    hmm, if thats the case i can proof 3 l a(a2-1)
    , 4 l a(a2-1) , and 2 l a(a2-1)
    will imply 8 l a(a2-1) and 3 l a(a2-1), then 24 l a(a2-1) for all odd integer a

    but now this troubled me how to proof a l d and c l d then ac l d , wait let me think, don't go offline, hoho
     
  7. Jul 31, 2010 #6
    Well, proving for 4 and 2 doesn't imply 8. Why not?
     
  8. Jul 31, 2010 #7
    i'm sorry, i'm not really good with english

    before i continue, "Well, proving for 4 and 2 doesn't imply 8. Why not?" i'm not sure if this a question, or you implying that 4 and 2 does imply 8,

    if you're implying, i follow your logic, then 4 and 2 does imply 8,

    but now i want to verify is all a,b,c,d in Z follow this logic

    if a l d and c l d then ac l d

    i cannot proof this T_T, should i proof by contradiction maybe, wait, i'll try ;P
     
  9. Jul 31, 2010 #8
    wait i'm confused with myself
     
  10. Jul 31, 2010 #9
    No, I mean 4 and 2 actually don't imply 8. Can you think of the reason why not?
     
  11. Jul 31, 2010 #10
    ugh, i can't find counter example (should i use proof by contradiction?),

    and you said 8 and 3 does imply 24, should i use (8 and 3 and (not 24)), is a contradiction??
     
  12. Aug 1, 2010 #11

    Dick

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    You really haven't explained why you think anything divides a(a^2-1). Why do you think 3 divides it? Why do you think 4 divides it? Let's start this over again.
     
  13. Aug 1, 2010 #12
    lets lets start again,

    i mentioned

    4 l a(a^2-1) for all odd integer

    proof,

    a=2n+1 for all integer n , then,

    (2n+1)((2n+1)^2-1 = 8n^3 + 12n^2 + 4 n = 4(2n^3+3n^2+1), and it can divides by 4,

    hmm, and i don't remember how i got 3 l a(a^2-1),
     
  14. Aug 1, 2010 #13

    Dick

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    Here's a hint. (a^2-1)=(a+1)(a-1). So a(a^2-1)=(a-1)a(a+1). Now rethink this. You don't even need a lot of technical stuff to answer this.
     
  15. Aug 1, 2010 #14
    yea, just now i got it like this,

    when a=3n , 3(n(3n+1)(3n-1))

    when a=3n-1 , 3(n(3n-1)(3n-2))

    when a=3n+1 , 3(n(3n+1)(3n+2)) for all integer n

    so 3 l a(a^2-1), for all integer a, hence, 3 l a(a^2-1) for all odd integer,

    is this alright?
     
  16. Aug 1, 2010 #15
    hmm , i just noticed that

    4 l a(a^2-1) for all integer,

    anyway, what next thing to do?
     
  17. Aug 1, 2010 #16

    Dick

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    I think it's ok. But maybe you are making this too hard. a-1,a,a+1 are three consecutive integers. Of course one is divisible by three. Now why is the product divisible by 8 if a is odd. Just say it in words.
     
  18. Aug 1, 2010 #17
    more clue please ;P
     
  19. Aug 1, 2010 #18
    hmm, when a is odd,

    even*odd*even is even?
     
  20. Aug 1, 2010 #19

    Dick

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    Ok, but you shouldn't need this. If a is odd then (a-1) and (a+1) are two consecutive even numbers. Might one be divisible by 4?
     
  21. Aug 1, 2010 #20
    i can see that it divisible by 4, but if you asking "why", i don't know how to answer T_T
     
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