Injectivity proof

Question

Prove the injectivity(~strict monotony) of the following function, without utilising calculus:

$f:(0;∞)$ $\rightarrow$$(0;1]$ $f(x) = \sqrt{x + 1} - \sqrt{x} $

I have already determined the given codomain for the function to be surjective, as per the task,but this second task I haven't been able to demonstrate;I have observed by studying the table of values that it appears strictly decreasing,but I need a rigorous proof.

Answer 1

Notice that\begin{align}f(x)=f(y)&\iff \sqrt{x+1}-\sqrt x=\sqrt{y+1}-\sqrt y\\&\iff\frac1{\sqrt{x+1}+\sqrt x}=\frac1{\sqrt{y+1}+\sqrt y}\\&\iff\sqrt{x+1}+\sqrt x=\sqrt{y+1}+\sqrt y.\end{align}If $x\neq y$, the $x>y$ or $x<y$. If $x>y$, then $\sqrt x>\sqrt y$ and $\sqrt{x+1}>\sqrt{y+1}$. Therefore $\sqrt{x+1}+\sqrt x>\sqrt{y+1}+\sqrt y$ and, in particular, $\sqrt{x+1}+\sqrt x\neq\sqrt{y+1}+\sqrt y$. The case in which $x<y$ is similar.

Answer 2

HINT

$$f(x) = \sqrt{x + 1} - \sqrt{x}= \left(\sqrt{x + 1} - \sqrt{x}\right)\cdot \frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x + 1} + \sqrt{x}}=\frac{1}{\sqrt{x + 1} + \sqrt{x}}$$

Answer 3

Note that $$f(x) = \sqrt {x + 1} - \sqrt{x} = \frac {1}{\sqrt {x + 1} + \sqrt {x}}$$

As you notice the bottom is a strictly increasing function thus the fraction is strictly decreasing,which makes it one to one.

Answer 4

To prove that a function is injective, it is necessary to show that:

$f(x_1) = f(x_2) \to x_1 = x_2$

Plug in $x_1$ and $x_2$ to your function, and then use some creative algebra to arrive at the necessary conclusion !