Show that the function $f(x)=x+\sqrt{x}$ is one-to-one Show that the function $f(x)=x+\sqrt{x}$ is one-to-one.
I know that for showing that a function is one-to-one I have to prove that if $f(a)=f(b)$ then $a=b$.
Then I'm trying that in here but I get stuck.
$$f(a)=f(b)$$
$$a+\sqrt{a}=b+\sqrt{b}$$
$$a-b=\sqrt{b}-\sqrt{a}$$
How to do I show from here that $a=b$?
I've tried square both sides, completing the square and haven't worked. :(
I will appreciate a detail to understand, thanks in advance.

Picking up from your last step, assuming $ a, b \neq 0$, we have
\begin{align*}
a - b = \sqrt{b} - \sqrt{a} & \iff (a - b)(\sqrt{b} + \sqrt{a}) = b - a \\
& \iff (a - b)(\sqrt{a} + \sqrt{b} + 1) = 0 \\
& \iff a - b = 0 \quad \text{or} \quad \sqrt{a} + \sqrt{b} + 1 = 0 \\
& \iff a = b \quad \text{(Since $ \sqrt{a} + \sqrt{b} + 1 \neq 0 $ for any $ a $ or $ b $)}
\end{align*}
So all in all we have $ f(a) = f(b) $ if and only if $a = b $ or $ a = b = 0$, or we can absorb the latter condition into the former one and says $ f(a) = f(b) $ if and only if $ a = b$.
