Multiplication of continued fraction

I know the continued fractions for $\sqrt2$ and $\sqrt3$:

$\sqrt2=1+\cfrac1{2+\cfrac1{2+\cfrac1{2+\cfrac1{...}}}}$

$\sqrt3=1+\cfrac1{1+\cfrac1{2+\cfrac1{1+\cfrac1{2+...}}}}$

Since $\sqrt6=\sqrt2{\sqrt3}$, the continued fractions should be able to be multiplied to form $\sqrt6$.

How do I accomplish this?

Solution 1:

I didn't really get an answer, so I spent a solid three hours working it out, and I think I found it.

Let $\sqrt2=1+a$ and $\sqrt3=1+b$ meaning that given the continued fractions of both $\sqrt2$ and $\sqrt3$, we have:

$a={1\over{2+a}}$ and $b={1\over{1+{1\over2+b}}}$

Therefore, given that $$\sqrt6=\sqrt2\sqrt3,$$ we obtain $$\sqrt6=(1+a)(1+b)=1+ab+a+b;$$

also given that that $ab={{2+b}\over{6+3a+2b+ab}}$

You can find that:

$\sqrt6=1+{{2+b}\over{6+3a+2b+ab}}+a+b$

This can be further compacted down, but it is presently 12:45AM where I am and I need some rest.