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$S$ 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 $T$.

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Hint: construct the inverse function. –  Levon Haykazyan Oct 24 '11 at 20:48
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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

1 Answer 1

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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 $(x_1,y_1)=(x_2,y_2)$

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

then

$x_1y_2=x_2y_1$ 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 $T$

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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

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