Alternative Proof to show any integer multiplied with 0 is 0

1. Jan 31, 2014

Seydlitz

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. Feb 1, 2014

jgens

That is essentially the same proof as the one given in Spivak. I have no idea what simplification you think it affords.

3. Feb 1, 2014

Seydlitz

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.

4. Feb 1, 2014

jgens

Closure is not necessary in this argument.