Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$S$ is the set of ordered pairs $(x,y)$ such that $x,y \in \mathbb{R}^+$ and let $T$ be the set of all ordered pairs $(a,b)$ such that $ab<1$. Show that $$f(x,y) = \left(\frac{x}{y}, \frac{y}{x+1}\right)$$ defines a bijection from $S$ to $T$.

share|cite|improve this question
Hint: construct the inverse function. – Levon Haykazyan Oct 24 '11 at 20:48
To do this, you simply need to verify: (1.) For any $(x,y) \in S$, $f(x,y) \in T$. (2.) $f$ is injective, and (3.) $f$ is surjective. Which part is giving you trouble? – Srivatsan Oct 24 '11 at 20:50
up vote 1 down vote accepted

1/ $f$ is injective from $(\mathbb{R}^+)^2$ to $(\mathbb{R}^+)^2$

let $(x_1,y_1)$ and $(x_2,y_2)$ two elements in $(\mathbb{R}^+)^2$ such that $f(x_1,y_1)=f(x_2,y_2)$

we must show that $(x_1,y_1)=(x_2,y_2)$

$$\frac{x_1}{y_1}=\frac{x_2}{y_2}\text{ and }\frac{y_1}{x_1+1}=\frac{y_2}{x_2+1}$$


$x_1y_2=x_2y_1$ and $x_1*y_2=+x_1x_2*y_1+x_2$

so $x_1=x_2$ and $y_1=y_2$

2/ $f$ is subjective from $(\mathbb{R}^+)^2$ to $T$

let $(a,b) \in M$

let find $(x,y) \in (\mathbb{R}^+)^2$ such that $f(x,y)=(a,b)$

with simple calculation $x=\dfrac{ab}{1-ab}$ and $y=\dfrac{b}{1-ab}$

3/ So $f$ bijection from $(\mathbb{R}^+)^2$ to $T$

share|cite|improve this answer
I’m not sure what you were trying to say with $x_1*y_2=+x_1x_2*y_1+x_2$, so I had to leave it alone. Perhaps you should clarify by expanding the algebra: $y_1(x_2+1)=y_2(x_1+1)$, so $x_2y_1+y_1=x_1y_2+y_2$, but the first terms are equal, so $y_1=y_2$, and since they’re not $0$, it follows that $x_1=x_2$ as well. – Brian M. Scott Oct 24 '11 at 21:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.