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.
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@Nathanson Can you try this one too? – Jessie Mar 20 '15 at 19:44
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Here is some useful info http://math.stackexchange.com/questions/265690/continued-fraction-of-a-square-root – turkeyhundt Mar 20 '15 at 19:45
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1I would suggest giving it a try yourself, working systematically and carefully - otherwise you'll get the answer, but you won't really understand how to do it yourself. – Mark Bennet Mar 20 '15 at 19:45
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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.
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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 $$x=1+\frac{4}{1+x}$$
$$x=1\frac{4}{2\frac{4}{2\frac{4}{2\frac{4}{2.....}}}}$$ $$\sqrt{5}=1\frac{4}{2\frac{4}{2\frac{4}{2\frac{4}{2.....}}}}$$
E.H.E
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