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Math Help - Closed Under Addition Proofs

  1. #1
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    Closed Under Addition Proofs

    I'm trying to prove x is a real number is closed under addition and 3Z is closed under addition, where Z is the set of integers.
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    To prove 3Z is closed under addition, we need to show 3*k_1 + 3*k_2 belongs to 3Z for arbitrary k_1 and k_2 in Z.

    Since an interger ring Z distributes multiplication over addition, 3*k_1 +3*k_2 = 3*(k_1 +k_2).

    Since k_1 and k_2 belongs to Z which is closed under addition, we can pick k_3 in Z satisfying k_3 = k_1 + k_2.

    Now, 3*k_1 +3*k_2 = 3*(k_1+k_2) = 3*k_3, where k_1, k_2, k_3 belongs to Z.

    Thus, 3Z is closed under addition.
    Last edited by vagabond; October 29th 2008 at 09:06 PM.
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  3. #3
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    Quote Originally Posted by vagabond View Post
    To prove 3Z is closed under addition, we need to check 3*k_1 + 3*k_2 belongs to 3Z for arbitrary k_1 and k_2 in Z.

    Since an interger ring Z distributes multiplication over addition, 3*k_1 +3*k_2 = 3*(k_1 +k_2).

    Since k_1 and k_2 belongs to Z which is closed under addition, we can pick k_3 in Z satisfying k_3 = k_1 + k_2.

    Now, 3*k_1 +3*k_2 = 3*(k_1+k_2) = 3*k_3, where k_1, k_2, k_3 belongs to Z.

    Thus, 3Z is closed under addition.
    Sorry, I forgot to include that 3Z = {3x such that x is an element of Z}. Is ther an easier way to do a formal proof now?
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  4. #4
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    Quote Originally Posted by vagabond View Post
    To prove 3Z is closed under addition, we need to show 3*k_1 + 3*k_2 belongs to 3Z for arbitrary k_1 and k_2 in Z.

    Since an interger ring Z distributes multiplication over addition, 3*k_1 +3*k_2 = 3*(k_1 +k_2).

    Since k_1 and k_2 belongs to Z which is closed under addition, we can pick k_3 in Z satisfying k_3 = k_1 + k_2.

    Now, 3*k_1 +3*k_2 = 3*(k_1+k_2) = 3*k_3, where k_1, k_2, k_3 belongs to Z.

    Thus, 3Z is closed under addition.
    Ok, I understand that, but what about proving that the set of real numbers is closed under addition. It's obvious to me that it's true, but proving it isn't so obvious.
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  5. #5
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    Quote Originally Posted by kathrynmath View Post
    Ok, I understand that, but what about proving that the set of real numbers is closed under addition. It's obvious to me that it's true, but proving it isn't so obvious.
    What DEFINITION of "real numbers" are you using, then, and what definition of addition of real numbers?

    That's a much more subtle question than you might think, but you surely can't prove something about addition in the real numbers without using the definitions!
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    What DEFINITION of "real numbers" are you using, then, and what definition of addition of real numbers?

    That's a much more subtle question than you might think, but you surely can't prove something about addition in the real numbers without using the definitions!
    Still a little confused. I don't remember ever using a definition for real numbers in the class.
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