# Proving $f(x,y,z)=\big(\frac{x}{a+x+y+z}, \frac{y}{a+x+y+z}, \frac{z}{a+x+y+z} \big)$ is injective

Let $$a\in \mathbb{R}$$, $$a\neq 0$$, and $$E=\{(x,y,z)\in \mathbb{R}^3:a+x+y+z\neq 0\}$$ and $$f:E\rightarrow \mathbb{R}^3$$ defined by $$f(x,y,z)=\big(\frac{x}{a+x+y+z}, \frac{y}{a+x+y+z}, \frac{z}{a+x+y+z} \big)$$ Show that $$f$$ is injective.

I've been trying to prove it directly by showing that if $$f(x_1,y_1,z_1)=f(x_2,y_2,z_2)\Rightarrow$$ $$x_1=x_2, y_1=y_2,z_1=z_2$$. But I get a system of equations which I'm find difficult to deal with. Any help would be greatly appreciated.

• If you would insist, the polynomial equations obtained easily imply either $a+x+y+z=0$, or $a=0$ or $x_i=y_i$ for $i=1,2,3$. So we are done. – Dietrich Burde Nov 5 '18 at 12:37

Adding the three components of the equation $$f(x_1,y_1,z_1)=f(x_2,y_2,z_2)$$ gives $$\frac{x_1+y_1+z_1}{a+x_1+y_1+z_1} = \frac{x_2+y_2 + z_2}{a+x_2+y_2+z_2}$$ which implies $$x _1+y_1+z_1 = x_2+y_2 + z_2 \, .$$ Then consider the equation $$f(x_1,y_1,z_1)=f(x_2,y_2,z_2)$$ again, and conclude that $$(x_1,y_1,z_1)=(x_2,y_2,z_2)$$.
• Why is that last part true? if $x_1 = x_2-1, y_1=y_2+1$, and $z_1=z_2$ then wouldn't that equation still be true and the conclusion false? – Joe Man Analysis Nov 5 '18 at 12:29
• @JoeManAnalysis: Use the equation $f(x_1,y_1,z_1)=f(x_2,y_2,z_2)$ as well! You have that $\frac{x_1}{a+x_1+y_1+z_1} = \frac{x_2}{a+x_2+y_2+z_2}$. If the denominators are equal, $x_1 = x_2$ follows. – Martin R Nov 5 '18 at 12:31
$$x_1f_2(x_1,y_1,z_1)=y_1f_1(x_1,y_1,z_1)$$
$$x_1f_2(x_2,y_2,z_2)=y_1f_1(x_2,y_2,z_2)$$
$$x_1y_2=x_2y_1$$