Everyone knows the rules of zero divisors like $\forall \alpha,\beta\in\mathbb{R}\;:\;\alpha\cdot\beta = 0\Rightarrow\alpha=0\vee \beta=0.$ But how can I prove it for $\mathbb{Z}$? My first try was this one: For $\alpha\cdot \beta=0$ and $\alpha\neq 0$ let $0=\alpha^{-1}\cdot 0=\alpha^{-1}\cdot (\alpha\cdot\beta)=(\alpha^{-1}\cdot \alpha)\cdot\beta = \beta = 0\Rightarrow \beta = 0;$ and the same for $\beta\neq 0\Rightarrow \alpha=0$, however i realized that the multiplicative inverse of a number $\alpha\in\mathbb{Z}$ is not defined in $\mathbb{Z}$ (because $(\mathbb{Z},\cdot)$ is not a multiplicative group). What now?
Furthermore the information: it's about basic multiplication and I should prove this via the basic "rules" neutrality of 0 and 1, comparability of 0 and 1, commutativity, associativity, distributivity, irreflexivity or transitivity. Group-theory should not be mentioned in the solution as out instructors don't want us to use these "advanced techniques"!
The rules (to use some from) are the following: $\forall a,b,c\in\mathbb{Z}:$
- $a+0=a,\;\;\;\; a\cdot 1=a$
- $0<1$
- $1+(-1)=0,\;\;\;\; 0-a=(-1)\cdot a$
- $a+b=a+c \Rightarrow b=c$
- $a\cdot b=a\cdot c,a\neq 0\Rightarrow b=c$
- $0 < a \Rightarrow a\neq 0$
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