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What's going on in this proof about integers?

  1. Jul 6, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove that for every integer n>=8, there exists nonnegative integers a and b, such that n =3a+5b

    2. Relevant equations



    3. The attempt at a solution
    I'm trying to understand the proof of this. It goes as follows:
    OtysN97.jpg

    I am having a hard time figuring out what is going on in any of the three cases.

    Firstly, why did they pick 3q, 3q+1, and 3q+2?
    They could do this proof using n = 5q, 5q+1, 5q+2, 5q+3, 5q+4 also correct? It would just have more cases, and therefor, be more work?

    Then, about the cases.

    Case I: n = 3q
    I see what they did here. Since they picked to represent their integer as 3q, they supposed that b = 0, and then it's easy to show there exists an a, since a = q at that point. And a >=3 because n = 3q needs to be greater than 8.

    Case II: n = 3q+1
    Here is where I am troubled.
    How do they make the jump to claim that n = 3(q − 3) + 10 ??
    What is the thought process behind this?

    I would appreciate any explanation that you could provide. Thanks!
     
  2. jcsd
  3. Jul 6, 2013 #2

    Dick

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    n=3q+1, 3(q-3)+10=3q-9+10=3q+1=n. It's not as big a jump as you think.
     
  4. Jul 6, 2013 #3
    Ah I see. So they are still playing the make one side look like the other game. Thanks again!
     
  5. Jul 6, 2013 #4

    Dick

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    That's a good way to phrase it. They are just splitting up into something divisible by 3 plus something divisible by 5.
     
  6. Jul 6, 2013 #5
    Yeah.

    That "add 0 to an equation by adding some quantity x and then subtracting x again" thing always catches me off guard. I'm very familiar with that trick from various calc II integrals,


    but it still always hits me like a truck.
     
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