How $\sqrt{2}=1+\frac{1}{\sqrt{2}+1}$?

Question

I have found it in the chapter about chain fractionals. I am unable to transform it to such state.

$$\sqrt{2}=1+\sqrt{2}-1=?=1+\frac{1}{\sqrt{2}+1}$$

Answer 1

Hint :

Multiply the term $\sqrt2-1$ by $\dfrac{\sqrt2+1}{\sqrt2+1}$.

Answer 2

$\frac{1}{\sqrt 2+1}$

$=\frac{\sqrt 2-1}{(\sqrt 2+1)(\sqrt 2-1)}$

$=\sqrt 2-1$

Answer 3

We have $\sqrt{2}-1=\frac{1}{\sqrt{2}+1}$ because of $(\sqrt{2}-1)(\sqrt{2}+1)=1$. The claim follows now.

Answer 4

Starting from,

$$1+\frac{1}{\sqrt{2}+1}$$

$$\implies \frac{\sqrt2+1+1}{\sqrt{2}+1}$$ Now we rationalize it

$$\implies\frac{\sqrt2+2}{\sqrt{2}+1}\times{\frac{\sqrt2-1}{\sqrt2-1}}$$

$$\implies\frac{2-\sqrt2+2\sqrt2-2}{2-1}$$

$$\implies \sqrt2$$

Answer 5

$$\implies1+\frac{1}{\sqrt{2}+1}$$

$$\implies \frac{\sqrt2+1+1}{\sqrt{2}+1}$$

$$\implies \frac{\sqrt2+2}{\sqrt{2}+1}$$

now take $\sqrt2$ common from numerator

$$\implies \frac{\sqrt2({1+\sqrt{2}})}{(\sqrt{2}+1)}$$

cancel common term from both denominator and numerator and you will get the answer

$$\implies \sqrt2$$