[proofplan] We prove both inclusions. First, if $n$ is a unit, then there is an integer $m$ with $nm=1$. Passing to absolute values gives a product of positive integers $|n||m|=1$, and the order properties of $\mathbb{Z}$ force $|n|=1$. Conversely, $1$ and $-1$ each multiply with themselves to give the multiplicative identity, so both are units. [/proofplan] [step:Show that every integer unit has absolute value $1$] Let $n \in \mathbb{Z}^{\times}$. By the definition of a unit in a unital ring, there exists $m \in \mathbb{Z}$ such that \begin{align*} nm=1. \end{align*} Since $nm=1$, neither $n$ nor $m$ is $0$. Define the nonnegative integers $a:=|n|$ and $b:=|m|$. Then $a,b \in \mathbb{Z}$ and \begin{align*} ab=|n||m|=|nm|=|1|=1. \end{align*} Because $a$ and $b$ are positive integers, the least-positive-element property of $\mathbb{Z}$ from [citetheorem:9702] gives \begin{align*} 1 \leq a \end{align*} and \begin{align*} 1 \leq b. \end{align*} Since $0 \leq a$ and $1 \leq b$, order compatibility with multiplication in $\mathbb{Z}$ from [citetheorem:9705] gives \begin{align*} a \leq ab. \end{align*} Using $ab=1$, we obtain \begin{align*} a \leq 1. \end{align*} Together with $1 \leq a$, antisymmetry of the order gives $a=1$. Hence \begin{align*} |n|=1. \end{align*} [guided] Let $n \in \mathbb{Z}^{\times}$. The notation $\mathbb{Z}^{\times}$ denotes the set of units of the ring $\mathbb{Z}$, so by definition there is an integer $m \in \mathbb{Z}$ such that $m$ is a multiplicative inverse for $n$. Thus \begin{align*} nm=1. \end{align*} We want to turn this equation into a statement about the size of $n$. The clean way is to use absolute values. Since $nm=1$, neither $n$ nor $m$ can be $0$. Define \begin{align*} a:=|n| \end{align*} and \begin{align*} b:=|m|. \end{align*} Then $a$ and $b$ are positive integers. Taking absolute values in the equation $nm=1$ gives \begin{align*} ab=|n||m|=|nm|=|1|=1. \end{align*} Now we use the order structure of $\mathbb{Z}$. By the least-positive-element property in [citetheorem:9702], every positive integer is at least $1$. Therefore \begin{align*} 1 \leq a \end{align*} and \begin{align*} 1 \leq b. \end{align*} The [second inequality](/theorems/2136) tells us that multiplying by $b$ should not decrease a nonnegative integer. We justify this by [citetheorem:9705]: since $1 \leq b$ and $0 \leq a$, multiplication by the nonnegative integer $a$ preserves the inequality. Hence \begin{align*} a \leq ab. \end{align*} But we already computed $ab=1$, so \begin{align*} a \leq 1. \end{align*} Combining $1 \leq a$ and $a \leq 1$, antisymmetry of the order on $\mathbb{Z}$ gives $a=1$. Since $a=|n|$, we have proved \begin{align*} |n|=1. \end{align*} [/guided] [/step] [step:Convert absolute value $1$ into the two possible units] From $|n|=1$, the definition of absolute value on $\mathbb{Z}$ gives \begin{align*} n=1 \end{align*} or \begin{align*} n=-1. \end{align*} Thus every unit of $\mathbb{Z}$ belongs to $\{-1,1\}$. [/step] [step:Verify that $1$ and $-1$ are units] The integer $1$ is a unit because \begin{align*} 1 \cdot 1=1. \end{align*} The integer $-1$ is a unit because \begin{align*} (-1)(-1)=1. \end{align*} Therefore $\{-1,1\} \subseteq \mathbb{Z}^{\times}$. Combining this inclusion with the previous step gives \begin{align*} \mathbb{Z}^{\times}=\{-1,1\}. \end{align*} This proves the theorem. [/step]