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Prove that sum of two even integers is even

  1. Aug 9, 2007 #1
    Okay, I know that this is probably super easy. This is not homework, I just grabbed tis book at the library today and am trying to get familiar with the subject (Abstract Algebra). The book is hella old and doesn't have many of the solutions, especially if the author regarded the solution as easy as this (it's the first problem!).

    My main problem is that I do not know how the proof should look (formally). In the past, I think I recall that a proof must work both ways (forwards and backwards). But I do not know how to start these things.

    1. The problem statement, all variables and given/known data An integer [tex]n[/tex] is defined to be even if [tex]n=2m[/tex] for some integer [tex]m[/tex] It is a theorem that the sum of two even integers is even. The definition of an even integer must be used to prove this theorem.


    2. Relevant equations
    [tex]n[/tex] is defined to be even if [tex]n=2m[/tex]


    3. The attempt at a solution
    I don't know where to start..I mean on which side of the equivalency [tex]n=2m[/tex]

    Should I start like: [tex]n[/tex] is defined to be even if [tex]n=2m[/tex]
    [tex]n+n=2n=4n[/tex]

    ....I feel like I am just babbling here....can someone start me off?

    Thanks,
    Casey

    p.s. sorry if this is the wrong forum.
     
    Last edited: Aug 9, 2007
  2. jcsd
  3. Aug 9, 2007 #2
    Let m and n be two even numbers. m = 2p and n = 2q, where p and q are integers. If you have either assumed or proved certain properties of integers, proving the sum of m and n is even is quite easy.
     
  4. Aug 9, 2007 #3

    Gib Z

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    As long as you know that natural numbers p and q add to some other natural number. No way to actually "prove" that, it comes of Peano's axioms (well, sort of).
     
  5. Aug 9, 2007 #4

    HallsofIvy

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    You problem involves two even integers- as neutrino suggested, try calling them 2n and 2m. Now, what is their sum?

    I have no idea what you mean by n+ n= 2n = 4n! Even if you mean 4m, you are not using two different numbers. You would only have proved that adding a number to itself gives an even number.
     
  6. Aug 9, 2007 #5
    Good point. Like I said, I felt like I was just rambling like a D-bag....

    Anyway. Given two even numbers m=2p and n=2q where p and q are both integers. 2p+2q=2(p+q)...now what, I assume that I include somewhere that the sum of two integers p and q = an integer, thus giving back my deinition of an even number..?
     
  7. Aug 9, 2007 #6

    HallsofIvy

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    Exactly! 2p+ 2q= 2(p+q): 2 times a whole number and so even.
     
  8. Aug 9, 2007 #7
    Nice! Thanks.
     
  9. Aug 9, 2007 #8

    HallsofIvy

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    New exercise: show that the product of any two odd numbers is odd!~
     
  10. Aug 9, 2007 #9
    Another excercise: Show that the sum of odd many odd numbers is odd.
     
  11. Aug 9, 2007 #10
    Definition: an integer s is defined to be odd if s=2w+1 for some integer w.

    s and t are both odd integers where s=2w+1 and t=2x+1 where w and x are integers.
    st=(2w+1)(2x+1)
    =2xw+2w+2x+1
    =2(xw+w+x)+1
    ummm..let me think....
     
  12. Aug 9, 2007 #11
    Seriously though, I don't call it an exercise to prove results that stem from associativity.
     
  13. Aug 9, 2007 #12
    How about expressing two odd number as x + 1 and y + 1 where x and y are even numbers? What does (x + 1)(y + 1) develops into?
     
  14. Aug 9, 2007 #13
    I will try it, howeber I was going by the definitions that the book uses (or the types that it uses)
     
  15. Aug 9, 2007 #14
    Simpler still, if x and y are two odd numbers, xy = x(y-1) + x. Since y - 1 is an even number, so is x(y-1). An even number plus an odd number gives an odd number.
     
  16. Aug 10, 2007 #15

    HallsofIvy

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    I'm sure it just personal preference but I would prefer:
    (2n+1)(2m+1)= 4mn+ 2n+ 2m+ 1= 2(2mn+ n+ m)+ 1 and so is an odd number.
     
  17. Aug 10, 2007 #16
    Yeah I forgot to distribute that 2...wow:redface:

    I'd like to have a go at this one:
    I need to first transpose that statement into something quantitative.....
     
  18. Aug 11, 2007 #17
    [tex]\sum_{k=1}^{n+1}{(a_k + 1)},[/tex]

    where n is even and so is every [itex]a_k[/itex]
     
  19. Aug 11, 2007 #18

    CompuChip

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    Do you know about induction? This seems like a good way to prove that statement.
    Obviously, the sum of one odd number is odd. Now assume that the sum of 2n - 1 odd numbers is odd and add two odd numbers. Can you prove it's odd?
     
  20. Aug 11, 2007 #19
    I don't see the need to complicate things by adding a factor of 2.
     
    Last edited: Aug 11, 2007
  21. Aug 13, 2007 #20

    CompuChip

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    And now we are supposed to ask: "Why not?", right? Because just giving the reasons in your first post would not increase our post counts, right?
    OK, so here goes:

    Werg22, why not?
     
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