Show bijection from (0,1) to R I have to show that the following function $f: (0,1) \rightarrow \mathbb{R}$.
I will use this function: $f(x)=\frac{1}{x}+\frac{1}{x-1}$.
To show 1-1, I am using $f(x_1)=f(x_2) \Rightarrow x_1=x_2$, where $x_1,x_2 \in (0,1)$.
To start off:
$\frac{1}{x_1}+\frac{1}{x_1-1}=\frac{1}{x_2}+\frac{1}{x_2-1}$.
Then, I get: $\frac{(x_1-1)+x_1}{x_1(x_1-1)}=\frac{(x_2-1)+x_2}{x_2(x_2-1)}$.
This leads to: $\frac{2x_1-1}{x_1(x_1-1)}=\frac{2x_2-1}{x_2(x_2-1)}$.
Even after I cross-multiply after this step, I don't see where I can conclude that $x_1=x_2$.
Now,to show surjective, I'm a little confused as well, since rearranging the function to solve for $x$ like in simple cases is not working. Any hints/clues/examples will be appreciated. Thanks!
 A: Calculus methods as suggested in other answers are good.  If you want to do it without calculus, here is a possibility.
Continuing your working leads easily to
$$2x_1x_2^2-2x_1^2x_2-x_2^2+x_1^2+x_2-x_1=0\ .$$
Now factorise the LHS.  This might sound difficult, but remember that you want to show $x_1=x_2$, so you would hope that $x_1-x_2$ is a factor.  And then you should notice that indeed if you take the first two terms, the next two and the last two, $x_1-x_2$ is a factor in every case.  So we get
$$(x_1-x_2)(-2x_2x_1)+(x_1-x_2)(x_1+x_2)-(x_1-x_2)=0$$
which then becomes
$$(x_1-x_2)(-2x_1x_2+x_1+x_2-1)=0\ .$$
To conclude for certain that $x_1=x_2$ you need to explain why the second factor is not zero.  Hint.  We have
$$-2x_1x_2+x_1+x_2-1=-\frac12\bigl((2x_1-1)(2x_2-1)+1\bigr)\ ,$$
and both $x_1$ and $x_2$ are in $(0,1)$.
To prove the function is surjective, try to solve $f(x)=a$ for any given real $a$.  This simplifies to
$$ax^2-(a+2)x+1=0\ .$$
Now call the LHS $q(x)$ and imagine the graph of $y=q(x)$.  This is a continuous curve; since $q(0)=1$ it is above the $x$-axis when $x=0$; since $q(1)=-1$ it is below the axis when $x=1$; so it must cross the axis somewhere between $0$ and $1$, which gives the solution you need.
A: All you have to show is that, given $a \in \mathbb R$, the equation $f(x) = a$ has exactly one solution for $x \in (0,1)$.
This equation can be rearranged as $ax^2-(a+2)x+1 = 0$.


*

*If $a = 0$, then it becomes $-2x+1 = 0$, which has the unique root $x=\frac12$ in $(0,1)$.

*If $a \ne 0$, we have a quadratic equation that takes the value $1$ at $x=0$, and the value $-1$ at $x=1$. So again it has exactly one root in $(0,1)$.

A: There is another proof: Since the derivative is negative then the function is decreasing strictly, and since the limits at the boundary points (0 and 1)  are +infinity and -infinity, then the function  f is one-to-one and onto from (0,1) to R.
