I need to prove that the function $f: \mathbb R_{>0}\times\mathbb R \to \mathbb R^2$ , $f(x,y)=(xy, x^2-y^2)$ is injective. I know I have to show that $f(x,y)=f(a,b)$ implies $x=a$ and $y=b$ but I have no idea how to prove it. Could you give me a hint?
Injectivity in two variables
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multivariable-calculus
3 Answers
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If you are given $xy$ and $x^2 - y^2$ and you know that $x > 0$, you can determine all of the following: $$ \begin{align*} (x^2 + y^2)^2 &= x^4 + y^4 + 2(xy)^2 = (x^2 - y^2)^2 +4 (xy)^2\\ x^2 + y^2 &= \sqrt{(x^2 + y^2)^2} \\ x^2 &= \frac{(x^2 - y^2) + (x^2 + y^2)}{2} \\ x &= \sqrt{x^2} \\ y &= \frac{xy}{x}. \end{align*} $$
So $(x, y)$ is a function of $(xy, x^2 - y^2)$ for $(x, y) \in \Bbb{R}_{>0}\times\Bbb{R}$.
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if $f(x,y) = f(a,b)$ you have two equations
$(1)\;\; xy = ab$
$(2)\;\; x^2 - y^2 = a^2 - b^2$.
Since $x > 0$, from (1) you get $y = ab/x$. Sub such $y$ on equation (2) and simplify to get
$$x^4 - x^2 (a^2 - b^2) - (ab)^2 = 0$$
Use the quadratic equation and the fact that $x,a > 0$ to conclude that $x = a$.
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So: $xy = ab, x^2-y^2 = a^2-b^2\implies x^2-a^2=y^2-b^2$. We should first make the added condition that $x,y,a,b$ be positive reals to make the result valid. Thus with this in mind, if $x > a > 0 \implies y > b\implies xy > ab$. Thus $x = a$, and then $y = b$.
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0The question gives you that $x > 0$ and $a > 0$ and with that information the sign of $y$ is the same as the sign of $xy$ (and similarly for $a$ and $b$). So your "added condition" is not necessary. – 2017-01-19