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Help with proofs

  1. Feb 8, 2005 #1

    Nx2

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    i just started my second semester with geomtry and am having difficulties with these proofs. i am stuck on this one question which asks:

    prove that if n is an odd positive integer, then one of the numbers n+5 or n+7 is dividsible by 4.

    so this is what i came up with:

    let n = 2k+1

    f(n)= n+5
    = (2k+1)+5
    = 2k+6
    = 2(k+3)

    So we know that 2 is divisible by 2 and now im guessing i have to prove that (k+3) is divisible by 2 as well. then by using the factor tree thing we can say that since the 2 is divisible by 2 and (k+3) is divisible by 2, f(n) must be divisible by 4, no? but i dont get how to do this... am i doing something wrong?

    i did the same exact method with f(n)= n+7 and ended up with f(n)= 2(k+4).

    i just dont get all this proving stuff.

    Any help would be appreciated, thanks.

    - Tu
     
  2. jcsd
  3. Feb 8, 2005 #2

    dextercioby

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    You're basically "home",because you have to prove that one of the 2 no. K+3 or K+4 is divisible with 2,thing which is trivial.

    Daniel.
     
  4. Feb 8, 2005 #3

    Nx2

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    so how would i show that (k+3) or (k+4) is divisible by 2... that what i dont understand. it looks to me that not all cases will be divisible by 2 making the statement false, but my teacher says that none of them are false. unless im doing this question totaly wrong... any ideas?

    thnx.

    - Tu
     
  5. Feb 8, 2005 #4

    dextercioby

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    Can u show that from two consecutive natural numbers (as is the case with K+3 & K+4),one & only one is divisible by 2...??

    Daniel.
     
  6. Feb 8, 2005 #5
    yea tu it sucks balls (this is phillips)
     
  7. Feb 8, 2005 #6

    Nx2

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    oooo... omg i cant beleive i didnt see that... so (k+3) and (k+4) are consecutive number, which mean one of them must be even, making one of them divisible by 2 correct?

    - Tu
     
  8. Feb 8, 2005 #7

    dextercioby

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    That's right... :smile:

    Daniel.
     
  9. Feb 8, 2005 #8
    yes
    also you didnt need to substitute if you think about it, you already had the consecutive pairs

    n, n+5, and n+7
    n being odd, you are adding 1 to make it even, than another 4
    if that isnt divisible, the other pair adds another 2, so it must be divisible if the other wasnt.
     
  10. Feb 8, 2005 #9

    Nx2

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    Omg man thnx alot... i apreciate it man

    - Tu

    lol, hey wats up Philips, didnt know u go on here lol
     
  11. Feb 8, 2005 #10

    Nx2

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    ooo i got yea philips... i should have seen that comming... man i hate this proof stuf... brutal...
     
  12. Feb 8, 2005 #11
    useful forum
    hey daniel while you are here
    how would you do

    prove that n^5 - 5n^3 + 4n is always divisible by 120 when n is greater than or equal to 3

    and prove that there are no integer solutions to the equation 2x + 4y = 5
    since both 2 and 4 are even numbers, does that alone prove there are no solutions?
    thanks!
     
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