Showing that a function is one-to-one How can I show if the following function is one-to-one? $$y = x^2 - x \ (x \ge \frac{1}{2})$$
A function is one-to-one if $f(x_1) = f(x_2)$ then $x_1 = x_2$. So I put in $x_1$ and $x_2$ in the function and tried to see if $x_1(x_1 - 1) = x_2(x_2 - 1)$ made sense. However, I am stuck right here: how do I know that $x_1 = x_2$ in this situation? I think I might be missing something quite simple...please help!

From $x_1^2-x_1=x_2^2-x_2$ we get $x_1^2-x_2^2=x_1-x_2$, hence $(x_1-x_2)(x_1+x_2)=x_1-x_2$.
Now assume that $x_1 \ne x_2$. Then it follows that $x_1+x_2=1$. Since $x_1,x_2 \ge 1/2$, we derive $x_1=x_2 =1/2$. Conclusion ?
