[proofplan] First clear denominators to replace the given rational homogeneous coordinates by a nonzero integer tuple. Then divide by the positive greatest common divisor of its coordinates to obtain a primitive integer tuple representing the same projective point. For uniqueness, compare two primitive integer representatives by the rational scalar relating them in projective space; writing that scalar in lowest terms forces both numerator and denominator to divide all coordinates of primitive tuples, hence both have absolute value $1$. [/proofplan] [step:Clear denominators to obtain an integer representative] For each $i \in \{1,\dots,n\}$, write $q_i = u_i/v_i$ with $u_i \in \mathbb Z$, $v_i \in \mathbb N$, and $v_i>0$. Define \begin{align*} D := \prod_{i=1}^{n} v_i \in \mathbb N. \end{align*} For each $i \in \{1,\dots,n\}$, define \begin{align*} c_i := Dq_i = u_i \prod_{\substack{1 \le j \le n \\ j \ne i}} v_j \in \mathbb Z. \end{align*} Since not all $q_i$ are zero and $D \ne 0$, the tuple $c := (c_1,\dots,c_n) \in \mathbb Z^n$ is nonzero. Also $c_i = Dq_i$ for every $i$, so multiplication by the nonzero rational scalar $D \in \mathbb Q^\times$ gives \begin{align*} [c_1:\dots:c_n] = [q_1:\dots:q_n] \end{align*} in $\mathbb P^{n-1}(\mathbb Q)$. [/step] [step:Divide by the common gcd to make the representative primitive] Let \begin{align*} g := \gcd(c_1,\dots,c_n) \in \mathbb N \end{align*} denote the positive greatest common divisor of the integer coordinates of $c$. Since $c$ is nonzero, $g \ge 1$. For each $i \in \{1,\dots,n\}$, define \begin{align*} a_i := \frac{c_i}{g} \in \mathbb Z. \end{align*} Then $a := (a_1,\dots,a_n) \in \mathbb Z^n$ is nonzero. Since $c_i = g a_i$ for every $i$, the tuples $a$ and $c$ differ by the nonzero rational scalar $g$, and therefore \begin{align*} [a_1:\dots:a_n] = [c_1:\dots:c_n] = [q_1:\dots:q_n]. \end{align*} By construction, every common divisor of $a_1,\dots,a_n$ multiplied by $g$ is a common divisor of $c_1,\dots,c_n$. Since $g$ is the greatest positive common divisor of the coordinates of $c$, the greatest positive common divisor of $a_1,\dots,a_n$ is $1$. Hence \begin{align*} \gcd(a_1,\dots,a_n)=1. \end{align*} [/step] [step:Compare two primitive representatives by a reduced rational scalar] Let $b := (b_1,\dots,b_n) \in \mathbb Z^n$ be another nonzero tuple such that \begin{align*} [b_1:\dots:b_n] = [a_1:\dots:a_n] \end{align*} and $\gcd(b_1,\dots,b_n)=1$. By equality in projective space over $\mathbb Q$, there exists $\lambda \in \mathbb Q^\times$ such that \begin{align*} b_i = \lambda a_i \end{align*} for every $i \in \{1,\dots,n\}$. Write $\lambda = m/\ell$ with $m \in \mathbb Z$, $\ell \in \mathbb N$, $\ell>0$, and $\gcd(m,\ell)=1$. Then \begin{align*} \ell b_i = m a_i \end{align*} for every $i \in \{1,\dots,n\}$. Since $\gcd(m,\ell)=1$, the divisibility relation $\ell \mid m a_i$ implies $\ell \mid a_i$ for every $i$. Thus $\ell$ is a positive common divisor of $a_1,\dots,a_n$. Because $\gcd(a_1,\dots,a_n)=1$, we get $\ell=1$. With $\ell=1$, the relation becomes \begin{align*} b_i = m a_i \end{align*} for every $i \in \{1,\dots,n\}$. Hence $m$ divides every coordinate $b_i$. Since $\gcd(b_1,\dots,b_n)=1$, this forces $|m|=1$. Therefore $m \in \{-1,1\}$, and setting $\varepsilon := m$ gives \begin{align*} b_i = \varepsilon a_i \end{align*} for every $i \in \{1,\dots,n\}$. This proves uniqueness up to multiplication by $-1$. [/step]