[proofplan] We prove the no-zero-divisors condition directly. The key ordered-ring fact is that a positive integer is at least $1$, and products of positive integers are positive. Applying this to the absolute values $|a|$ and $|b|$ shows that if both integers were nonzero, then $|ab|=|a||b|$ would be positive, contradicting $ab=0$. [/proofplan] [step:Reduce the conclusion to showing that two nonzero integers have nonzero product] Let $a,b \in \mathbb{Z}$ and assume \begin{align*} ab = 0. \end{align*} If $a = 0$, then the desired conclusion holds. It remains to consider the case $a \ne 0$ and prove that $b = 0$. We argue by contradiction. Assume also that $b \ne 0$. Define the absolute value map \begin{align*} |\cdot|: \mathbb{Z} &\to \mathbb{Z}_{\ge 0} \end{align*} by $|n| = n$ if $n \ge 0$ and $|n| = -n$ if $n < 0$. Since $a \ne 0$ and $b \ne 0$, trichotomy for the order on $\mathbb{Z}$ gives \begin{align*} |a| > 0 \end{align*} and \begin{align*} |b| > 0. \end{align*} [/step] [step:Show that the product of two positive integers is positive] We use the ordered-ring properties of $\mathbb{Z}$ from [citetheorem:9702] and [citetheorem:9705]. Since $|a|$ and $|b|$ are positive integers, the least-positive-integer property gives \begin{align*} 1 \le |a| \end{align*} and \begin{align*} 1 \le |b|. \end{align*} Apply order compatibility with multiplication to $1 \le |a|$ and the nonnegative integer $|b|$. This gives \begin{align*} |b| \le |a||b|. \end{align*} Since $1 \le |b|$, transitivity of the order gives \begin{align*} 1 \le |a||b|. \end{align*} Hence $|a||b| > 0$. [guided] We need a precise reason that multiplying two nonzero integers cannot give zero. The order supplies that reason. Since $|a|$ and $|b|$ are positive integers, [citetheorem:9702] says that $1$ is the least positive integer in $\mathbb{Z}$. Therefore \begin{align*} 1 \le |a| \end{align*} and \begin{align*} 1 \le |b|. \end{align*} Now we use the compatibility of order with integer multiplication from [citetheorem:9705]. The theorem applies because $|b| \ge 0$. Multiplying the inequality $1 \le |a|$ by the nonnegative integer $|b|$ gives \begin{align*} 1 \cdot |b| \le |a||b|. \end{align*} Since $1 \cdot |b| = |b|$, this is \begin{align*} |b| \le |a||b|. \end{align*} Combining this with $1 \le |b|$ gives \begin{align*} 1 \le |a||b|. \end{align*} Thus $|a||b|$ is a positive integer, so it cannot be equal to $0$. [/guided] [/step] [step:Use multiplicativity of absolute value to obtain the contradiction] For integer absolute value, multiplication satisfies \begin{align*} |ab| = |a||b|. \end{align*} From the previous step, \begin{align*} |a||b| > 0. \end{align*} Therefore \begin{align*} |ab| > 0. \end{align*} But the hypothesis $ab = 0$ gives \begin{align*} |ab| = |0| = 0, \end{align*} a contradiction. Hence the assumption $b \ne 0$ is false. Therefore $b = 0$. We have shown that whenever $a,b \in \mathbb{Z}$ and $ab = 0$, then $a = 0$ or $b = 0$. Since $\mathbb{Z}$ is a commutative unital ring under its standard operations, this is exactly the no-zero-divisors condition in the definition of an [integral domain](/page/Integral%20Domain). Thus $\mathbb{Z}$ is an integral domain. [/step]