Let
$$
(x_0,x_1,\cdots,x_n)\in S=\mathbf{Q}\backslash \{(0,0,\cdots,0\}
$$
Let $[(x_0,x_1,\cdots,x_n)]$ denote the equivalence class of $(x_0,x_1,\cdots,x_n)$ in $P^n(\mathbf{Q})$.Prove that there exists exactly two elements $(a_0,a_1,\cdots,a_n)$ and $(b_0,b_1,\cdots,b_n)$in $S$ such that the numbers $a_0,\cdots,a_n$ are relatively prime integers,the numbers $b_0,\cdots,b_n$ are relatively prime integers ,and
\begin{equation}
[(x_0,x_1,\cdots,x_n)]=[(a_0,a_1,\cdots,a_n)]=[(b_0,b_1,\cdots,b_n)]\in P^n(\mathbf{Q})
\end{equation}
Moreover,
\begin{equation}
(b_0,b_1,\cdots,b_n)=-(a_0,a_1,\cdots,a_n)
\end{equation}
Proof:
\begin{align*}
\begin{cases}
a_0=tx_0\\
a_1=tx_1\\
\vdots\\
a_n=tx_n\\
\end{cases}
\end{align*}
Let $x_i=\frac{a_i}{b_i}(b_i\neq 0),(a_i,b_i)=1$.Let the least common multiple of $b_0,b_2,\cdots,b_n$ be $k$.Let the greatest common divisor of $a_0,a_1,\cdots,a_n$ be $m$.Then let
\begin{equation}
t=\pm \frac{k}{m}
\end{equation}
Done.