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.