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Math Help - Prime numbers

  1. #1
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    Prime numbers

    The sum of 2 positive prime numbers is a prime number and 3 times that sum is an odd number.

    Question: Why does the the prime number 2 need to be one of the original prime numbers?
    How do I show this?
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  2. #2
    Moo
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    Hello,

    A prime number has to be odd, except 2.
    So if you add two prime numbers, nor of them being 2, since they're both odd, their sum will be an even number. So it couldn't be a prime number !
    Hence 2 has to be one of the original prime numbers.
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  3. #3
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    I'm sorry, I'm still confused, umm..is there a more formal proof of this?? Can maybe someone restate? It's seems like hocus pocus to me..
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  4. #4
    Moo
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    The only even prime number is 2 (otherwise, it's divisible by 2).

    So let's suppose 2 is not among the original prime numbers.
    They're both odd. So if you add them, the sum of two odd numbers being even, their sum will be even. But their sum wouldn't be a prime number (unless it's equal to 2, which is not possible) since it's even, that is divisible by 2.

    So the original assumption of 2 not being among the original prime numbers is false.

    If you want to do it formally, you have to use a proof by contradiction.
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  5. #5
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    The answer to your question involves Parity. Parity is a fancy word for two properties of numbers: even or odd. Notice that an odd number plus another odd number is always even. An even + even is also always even. Since we are told the sum of two primes is another prime, we know that this sum results in an odd number. The only way for a sum to result in an odd number is for one of the terms to be even and one to be odd.

    even + even = even number
    odd + odd = even number
    odd + even = odd number

    2 is the only even prime, therefore 2 must be one of the primes.


    EDIT: Sorry Moo, I didn't see your post.
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  6. #6
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    Here is a rigorous proof that the sum of two odd integers is even (just in case you care):

    Let m and n be odd integers. Then there are integers k and r such that m=2k+1 and n=2r+1. So,
    m+n=(2k+1)+(2r+1)=2k+2r+2=2(k+r+1). Since k+r+1 is an integer, m+n is even.
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