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)
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)
0 commentaires:
Enregistrer un commentaire