Proof that $\mathbb{Z}$ has no zero divisors

Solution 1:

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:

1. For all $a,b,c\in R$, if $a\neq 0$ and $ab=ac$, then $b=c$.
2. 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

Solution 2:

Hint: Show that $\mathbb{Z}$ has characteristic 0 and note that 1 generates $\mathbb{Z}$ as an additive group.

Solution 3:

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, $\beta = 0$. $\Box$

Solution 4:

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 $a\times b\neq 0$.

1. 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).
2. 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.