Proof that $\mathbb{Z}$ has no zero divisors 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: $\forall a,b,c\in\mathbb{Z}:$


*

*$a+0=a,\;\;\;\; a\cdot 1=a$

*$0<1$

*$1+(-1)=0,\;\;\;\; 0-a=(-1)\cdot a$

*$a+b=a+c \Rightarrow b=c$

*$a\cdot b=a\cdot c,a\neq 0\Rightarrow b=c$

*$0 < a \Rightarrow a\neq 0$

*$a<b\wedge b<c\Rightarrow a<c$

*$a<b\rightarrow a+c<b+c$

*$a<b\wedge 0<c\Rightarrow a\cdot c<b\cdot c$

*$a<b\wedge c < 0\Rightarrow b\cdot c < a\cdot c$

 A: Hint: Show that $\mathbb{Z}$ has characteristic 0 and note that 1 generates $\mathbb{Z}$ as an additive group.
A: The rules you provide are incorrect: your fifth "rule" currently reads:
$$a\cdot b = a\cdot c \Rightarrow b=c.$$
This is not a valid rule of multiplication in $\mathbb{Z}$: after all, $0\cdot 1 = 0\cdot 0$, but we do not have $1=0$.
The correct cancellation rule is:
$$\Bigl(a\cdot b = a\cdot c \land a\neq 0\Bigr) \Rightarrow b=c.$$
But this is equivalent to the fact that there are no zero divisors.
Theorem. Let $R$ be a ring. Then the following are equivalent:


*

*For all $a,b,c\in R$, if $a\neq 0$ and $ab=ac$, then $b=c$.

*For all $x,y\in R$, if $xy=0$ and $x\neq 0$, then $y=0$.
Proof. $(1)\Rightarrow (2)$: Let $x$ and $y$ be such that $xy=0$ and $x\neq 0$. Then $xy=0 = x0$, so by (1) (with $a=x$, $b=y$, $c=0$) we conclude $y=0$.
$(2)\Rightarrow (1)$: Let $a,b,c\in R$ be such that $a\neq 0$ and $ab=ac$. Then $ab-ac = 0$, so $a(b-c)=0$. Since $a\neq 0$, then by (2) (with $x=a$ and $y=b-c$) we conclude that $b-c=0$, hence $b=c$. QED
A: Suppose $\alpha = 0$ then there is nothing left to prove. So suppose that $\alpha \neq 0$. Suppose for contradiction that $\beta \neq 0$. Since $\alpha\beta = 0$, and $\beta$ is non-zero it follows that either $\alpha + \alpha + ... +\alpha = 0$,  or $-(\alpha + \alpha + ... +\alpha) = 0$ for some finite positive number, $\beta$, of $\alpha$'s. However, this is a contradiction since in $\mathbb{Z}$ no element can be added to itself indefinitely to reach $0$. Hence, $\beta = 0$. $\Box$ 
A: HINT $\: $ For $\rm\ h(n) = a\:n\:,\:$ $\rm\ ker\ h = 0\ \Leftrightarrow\ h\:$ is $1$ to $1\:,\:$ i.e. $\:$ non-zero-divisor $\:\Leftrightarrow\:$ cancellable.  
Said more simply, specialize $\rm\:c =0\:$ in your fifth rule: $\rm\:a\ne 0,\ a\:b = a\:c\:\Rightarrow\:b=c\:.$ 
A: From the rules you're provided (plus a very little more), you can give a fairly snazzy induction proof:
Theorem. Suppose $a\gt 0$; then $\forall b\gt 0$, $a\times b\neq 0$.
In fact, we can prove a little more; we'll prove that $a\times b\gt 0$, and your sixth rule then implies that $a\times b\neq 0$.


*

*For $b=1$, $a\times b = a\times 1 = a$ (by your first rule), and $a\gt 0$, so $a\neq 0$ (by your sixth rule).

*Assume that $a\times b\gt 0$. Then $a\times (b+1) = (a\times b) + (a\times 1)$ (by distribution of multiplication, which isn't on your list but should be 'basic') = $(a\times b) + a$ (by the first rule).  Now, $a\times b \gt 0$ (induction hypothesis) so your eighth rule gives $\bigl((a\times b) + a\bigr) \gt a$, and then the seventh rule along with the hypothesis that $a\gt 0$ lets us conclude that $\bigl((a\times b) + a\bigr) \gt 0$, so $a\times(b+1)\gt 0$; this induction step then gives the result for all $b$.


The other cases ($b\lt 0$, etc.) can be handled straightforwardly, although you'll also need the law of the excluded middle (in the form that $a\neq 0$ implies either $a\lt 0$ or $a\gt 0$ holds) to get the final result.
A: We may assume that $\alpha$ and $\beta$ are positive. Then $\alpha \beta = (\alpha \beta) \cdot 1 = 1+1+\cdots+1\space$ ($\alpha \beta$ times). Now $1>0$ and so $0<1+1<1+1+1<\cdots<1+1+\cdots+1$. So, no positive multiple of $1$ can be zero.
