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I am having a bit of trouble with the algebra for proving that the function is injective.

Basically I set $f(a)=f(b)$ for $a,b\in[0,\infty)$ and $a,b\neq 1$.

$\frac{a^2}{1-a}=\frac{b^2}{1-b}\rightarrow a^2(1-b)=b^2(1-a)\rightarrow a^2-b^2+b^2a-a^2b=0\rightarrow (a-b)(a+b)-ab(a-b)=0\rightarrow (a-b)(a+b-ab)=0$.

It is here that I am stuck. Basically I want to prove that either $(a-b)$ or $(a+b-ab)$ must be $0$ and if that was the case, $a=b$. However, I don't know how to prove that especially for $(a+b-ab)$.

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    $\begingroup$ Don’t you mean the domain to be $[0,1\rangle$? $\endgroup$
    – Lubin
    Apr 25, 2014 at 1:52

3 Answers 3

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Let's find out $x$ such that $f(x)=m$, ($m>0$).

$$\frac{x^2}{1-x}=m \Rightarrow x^2+mx-m=0 \Rightarrow$$ $$\Rightarrow x_1=\frac{-m-\sqrt{m^2+4m}}{2} \quad \mathrm{or} \quad x_2=\frac{-m+\sqrt{m^2+4m}}{2}$$ Note that $x_1$ is always negative and $x_2$ is in the domain, hence the function is injective for $m>0$.

If $m=0$, there is only a possibility $x=0$, and hence the function is injective for $m \ge 0$.

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Suppose $f(a) = f(b)$. In order to show $f$ is injective, we must prove that $a = b $. Suppose, to reach a contradiction, that this is not the case. Then, either $a > b $ or $b > a $. Assume first $a > b $, then $a^2 > b^2$ since $x^2$ is an increasing function. Furthermore, notice $ a > b \implies -a < - b \implies 1 - a < 1 - b \implies \frac{1}{1-a} > \frac{1}{1-b} $. Hence

$$ 1 = \frac{f(a)}{f(b)} = \frac{\frac{a^2}{1-a}}{\frac{b^2}{1-b}} > \frac{\frac{b^2}{1-b}}{\frac{b^2}{1-b}} = 1 \implies 1 > 1$$

This is a contradiction, hence $a > b$ must be false. By the same argument, you should be able to show that $b > a$ is also false. Hence, $a = b$. In other words, $f$ must be injective.

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Are you willing to use Calculus? The derivative is $$ \frac{(1-x)\cdot2x+x^2}{(1-x)^2}=\frac{x(2-x)}{(1-x)^2}\,, $$ in which every factor is positive on the interior $\langle0,1\rangle$ of the domain. So the derivative is positive, the function is increasing, thus one-to-one within its domain. But since the function starts at $f(0)=0$ and clearly goes to $\infty$ as $x\to1$, continuity now implies that the range is all of $[0,\infty\rangle$.

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