Proof of the image of a set $\mathbb R$ be the set of real numbers.
I have a set A = {$x$ ∈ $\mathbb R$ | $0< x$} . A function g : A --> $\mathbb R$ is defined by 
$f(x) = x^2
How do i prove whether the Image(A) = $\mathbb R$
 A: Let $r$ be any real number. Consider $g(x) = r$. So: $\dfrac{2x-1}{2x - 2x^2} = r$. So: $2x - 1 = 2rx - 2rx^2$, and $2rx^2 + 2(1 - r)x - 1 = 0$. So: $ D = b'^2 - ac = (1 - r)^2 + 2r = r^2 + 1 > 0$ implying the equation always has 2 distinct real zeroes. Let's look at one of the zeroes $x = \dfrac{r - 1 + \sqrt{r^2 + 1}}{2r}$. There are three cases to consider:
Case 1: $r < 0$. Then $r^2 - 2r + 1 > r^2 + 1$. So $(r - 1 - \sqrt{r^2 + 1})(r - 1 + \sqrt{r^2 + 1}) > 0$. This implies $r - 1 + \sqrt{r^2 + 1} < 0$. Hence: $\dfrac{r - 1 + \sqrt{r^2 + 1}}{2r} > 0$. Thus $x > 0$. Now $x - 1 = \dfrac{r - 1 + \sqrt{r^2 + 1}}{2r} - 1 = \dfrac{-1 -r + \sqrt{r^2 + 1}}{2r} < 0$ because $\sqrt{r^2 + 1} > 1 + r$ when $r < 0$ and this inequality is easily verified. So $x < 1$, and $0 < x < 1$. Thus if $r < 0$ we can always choose an $x$ such that $0 < x < 1$ and $g(x) = r$.
Case 2: $r > 0$. Observe that for this case $\sqrt{r^2 + 1} > 1 - r$. So $x > 0$, and $-1 - r + \sqrt{r^2 + 1} < 0$ because $\sqrt{r^2 + 1} < 1 + r$ when $r > 0$. So $x < 1$. Thus for this case we also have $0 < x < 1$ such that $g(x) = r$. 
Case 3: $r = 0$. Simply choose $x = \frac{1}{2}$, and we readily check that $g(\frac{1}{2}) = 0$.
So all three cases imply that Image($A$) = $\mathbb R$ as claimed.
A: Hint :
Take $g(x)=y$
$2xy-2x^2y=2x-1$
$(2y)x^2+x(2-2y)-1=0$.
Solve for $x$.
