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Difference between any two odd numbers is even

  1. Jun 5, 2003 #1
    Is there a law or theorem somewhere that states the difference between any two odd numbers is even? Or the difference between 2 even numbers is even?
    Last edited by a moderator: Feb 5, 2013
  2. jcsd
  3. Jun 5, 2003 #2


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    It's trivially obvious. Why do you ask?
  4. Jun 5, 2003 #3
    Well its also trivially obvious that the sum of any two odd primes yields an even integer over 4, yet that has not been proven yet. I just wanted to make sure that these were because someone keeps telling me they aren't. Luckily in the meanwhile I came up with my own proof for them, so it is all good.
  5. Jun 5, 2003 #4


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    Let m= 2n+1
    Let r= 2s+1

    r and m are arbitrary odd numbers(1 greated then an even number)

    r+m = (2n +1)+ (2s+1)= 2n+2s+2 = 2(n+s+1)

    r+m is even.
    You can do something similar for 2 even numbers.
  6. Jun 5, 2003 #5
    Yes that is the proofs I came up with...different notation but same message.
  7. Jun 7, 2003 #6


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    What's not trivially obvious is what you MEAN by "an even integer over 4". If you mean, literally, "an even integer divided by 4" then it doesn't make sense because the sum of two odd primes certainly has to be an INTEGER, not a half. If you mean "divisible by 4" then it's not true: 3+ 7= 10.
    The fact that the sum of two odd PRIMES is even is "trivially obvious" because the sum of two odd numbers is always even. That certainly has been proven, in fact, I've seen it given as an exercise in high-school algebra texts.
  8. Jun 7, 2003 #7
    I forgot that mathematicians need to be exact here.

    Every even integer greater than 4 is the sum of two odd primes.

    It is indeed obvious the sum of any two odd primes is even (after all, they are just special odd numbers), but that seems to be obvious as well at first glance, but not proven. It is, after all, the Goldbach Conjecture.
  9. Jun 14, 2003 #8
    Actually, the sum of ANY two odd numbers is an even number (wether they are primes or not).
    Look at integral's proof.
    You can adjust Integral's proof to proove this.
    Let A be an odd number, B another odd number
    A = 2k + 1
    B = 2n + 1
    (where both k and n are integers)
    A - B = (2k + 1) - (2n + 1) = 2k-2n + (1-1) = 2(k+n)
    Since k and n are integers, k+n is an integer too, and A-B is even (since it can be expressed as 2*integer).
  10. Jun 14, 2003 #9
  11. Jun 14, 2003 #10


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    "It is, after all, the Goldbach Conjecture," is a great line.
    I suspect that you set up this thread with that line
    in mind from the start
    in order to have an opportunity to deliver it.

    Perhaps I'm easily amused today but find it difficult
    to stop chuckling at this thread.
  12. Jun 14, 2003 #11
    Hehe, I wish I was ingenious enough to have that planned from the start. The actual purpose was to make sure my arguments were correct. Of course the sum of any two odd primes will be even since the sum of any two odds is even. As I then explained to my friend, the other way...all even numbers are the sum of two primes, is much less obvious and so far not proven.
  13. Jun 21, 2003 #12
    Re: question

    The difference between the number êÄ and the number Ç{ is DEFINITELY ODD (don't you think?)
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