Find the infinite simple continued fractions for ... Find the infinite simple continued fractions for $\sqrt{2};\sqrt{5};\sqrt{6};\sqrt{7};\sqrt{8}$.


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*I have solved similar equations for continued fractions but only using a fraction, if someone could please demonstrate how to do this to ANY of these values I will be good from there just need an example to work off of. I realize I can just get these values off of Wolfram Alpha but I need to know how to actually work through them. Thank you in advance!

 A: Just an example: $\sqrt{7}$. Since $4<7<9$, $\left\lfloor \sqrt{7}\right\rfloor = 2$, so:
$$ \sqrt{7} = 2+(\sqrt{7}-2) = \color{blue}{2}+\frac{1}{\frac{\sqrt{7}+2}{3}}\tag{1}.$$
Since $2+\sqrt{7}\in (4,5)$, $\left\lfloor\frac{\sqrt{7}+2}{3}\right\rfloor =1$, so:
$$ \frac{\sqrt{7}+2}{3} = 1+\frac{\sqrt{7}-1}{3} = 1+\frac{1}{\frac{\sqrt{7}+1}{2}}$$
and by plugging this identity back into $(1)$ we get:
$$\sqrt{7}=\color{blue}{2}+\frac{1}{\color{blue}{1}+\frac{1}{\frac{\sqrt{7}+1}{2}}}.\tag{2}$$
Now $\frac{1+\sqrt{7}}{2}\in(1,2)$, hence:
$$ \frac{\sqrt{7}+1}{2}=1+\frac{\sqrt{7}-1}{2} = 1+\frac{1}{\frac{\sqrt{7}+1}{3}}$$
and by plugging it back into $(2)$ we get:
$$\sqrt{7}=\color{blue}{2}+\frac{1}{\color{blue}{1}+\frac{1}{\color{blue}{1}+\frac{1}{\frac{\sqrt{7}+1}{3}}}}.\tag{3}$$
Continuing that way, we have: 
$$\frac{\sqrt{7}+1}{3} = 1+\frac{\sqrt{7}-2}{3} = 1+\frac{1}{\sqrt{7}+2}$$
hence:

$$ \color{red}{\sqrt{7}}=[2;1,1,1,2+\sqrt{7}]=\color{red}{[2;1,1,1,\overline{4,1,1,1}]}.\tag{4}$$

Now you just have to check that the same algorithm leads to:
$$ \sqrt{2} = [1;\overline{2}],\quad \sqrt{5}=[2;\overline{4}],\quad \sqrt{6}=[2;\overline{2,4}],\quad \sqrt{8}=[2;\overline{1,4}]$$
(yes, I took the most complex example on purpose).
