Continued Fraction for Root 5 How can I find the continued fraction expansion for the square root of 5. Do this without the use of a calculator and show all the steps.
 A: It is the same idea as $\sqrt{7}$ and $\sqrt{3}$, or any $a+b\sqrt{c}$.
First determine the integer part of $\sqrt{5}$. We know that $2<\sqrt{5}<3$.


*

*Then it is only about extracting the integer part of improper fractions,


*taking reciprocals of proper fractions


*and multiplying numerator and denominator by the conjugate number when there is an irrational number in a denominator.

The case of $\sqrt{5}$ is short:
We obtain $\sqrt{5}=2+(\sqrt{5}-2)=2+\frac{1}{\frac{1}{\sqrt{5}-2}}$.
With the fraction $\frac{1}{\sqrt{5}-2}=\frac{\sqrt{5}+2}{1}=4+(\sqrt{5}-2)=4+\frac{1}{\frac{1}{\sqrt{5}-2}}$.
Therefore $$\sqrt{5}=2+\frac{1}{\frac{1}{\sqrt{5}-2}}=2+\frac{1}{\frac{\sqrt{5}+2}{1}}=2+\frac{1}{4+(\sqrt{5}-2)}=2+\frac{1}{4+\frac{1}{\frac{1}{\sqrt{5}-2}}}.$$ Since the $\frac{1}{\sqrt{5}-2}$ has appeared before the continued fraction will continue to spit $4$'s. The result is that
$$\sqrt{5}=[2| 4,4,4,...]$$
For the case of quadratic irrational numbers (solutions of quadratic equations) there is always going to appear a periodicity (and conversely periodicity only appears in this cases [and rational numbers]). So, eventually you get a fraction that you have already treated before.
A: We can use the 
$$x^2-5=0$$
$$x^2=5$$
$$x^2+x=5+x$$
$$x(x+1)=5+x$$
$$x=\frac{x+5}{x+1}$$
or
$$x=1+\frac{4}{1+x}$$
$$x=1\frac{4}{2\frac{4}{2\frac{4}{2\frac{4}{2.....}}}}$$
$$\sqrt{5}=1\frac{4}{2\frac{4}{2\frac{4}{2\frac{4}{2.....}}}}$$
