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Alternative Proof to show any integer multiplied with 0 is 0

  1. Jan 31, 2014 #1
    In his book, Spivak did the proof by using the distributive property of integer. I am wondering if this, I think, simpler proof will also work. I want to show that ##a \cdot 0 = 0## for all ##a## using only the very basic property (no negative multiplication yet).

    For all ##a \in \mathbb{Z}##, ##a+0=a##.

    We just multiply ##a## again to get ##a^2+(a \cdot 0) = a^2##. Then it follows ##a \cdot 0 = 0##. (I remove ##a^2## by adding the additive inverse of it on both side)
     
  2. jcsd
  3. Feb 1, 2014 #2

    jgens

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    That is essentially the same proof as the one given in Spivak. I have no idea what simplification you think it affords.
     
  4. Feb 1, 2014 #3
    I'm glad then that it's the same. Because I thought it's fallacious because I haven't showed if the integers are closed with multiplication, and Spivak's proof is the more appropriate one.
     
  5. Feb 1, 2014 #4

    jgens

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    Closure is not necessary in this argument.
     
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