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Alternative Proof to show any integer multiplied with 0 is 0

by Seydlitz
Tags: alternative, integer, multiplied, proof
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Seydlitz
#1
Jan31-14, 11:59 PM
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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)
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jgens
#2
Feb1-14, 12:04 AM
P: 1,622
That is essentially the same proof as the one given in Spivak. I have no idea what simplification you think it affords.
Seydlitz
#3
Feb1-14, 12:13 AM
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Quote Quote by jgens View Post
That is essentially the same proof as the one given in Spivak. I have no idea what simplification you think it affords.
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.

jgens
#4
Feb1-14, 12:26 AM
P: 1,622
Alternative Proof to show any integer multiplied with 0 is 0

Quote Quote by Seydlitz View Post
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.
Closure is not necessary in this argument.


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