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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- October 29th 2008, 09:25 PMkathrynmathClosed 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.

- October 29th 2008, 09:34 PMvagabond
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. - October 29th 2008, 09:50 PMkathrynmath
- October 30th 2008, 09:01 AMkathrynmath
- October 31st 2008, 06:39 AMHallsofIvy
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! - November 4th 2008, 07:31 AMkathrynmath