Solution

Find a simple continued fraction expansion for $\sqrt{3}$.

To avoid the possibility of round-off error in a calculator, let us keep exact values throughout our calculation.

First, we separate the integer part of $\sqrt{3}$ from the fractional part:

$$\sqrt{3} = 1 + (\sqrt{3} - 1)$$

Reciprocating the fractional part twice (which doesn't change it's value), and then multiplying by a well-chosen value of one that involves the conjugate of $\sqrt{3}-1$ (again, not changing the overall value, we have

$$= 1 + \frac{1}{\displaystyle{\frac{1}{\sqrt{3}-1} \cdot \frac{\sqrt{3}+1}{\sqrt{3}+1}}}$$

Simplifying the bottom-most product, we arrive at

$$= 1 + \frac{1}{\displaystyle{\frac{\sqrt{3}+1}{2}}}$$

Now we start the process over. We separate the integer part of $\frac{\sqrt{3}+1}{2}$ from its fractional part:

$$= 1 + \frac{1}{\displaystyle{1+\frac{\sqrt{3}-1}{2}}}$$

Then, we reciprocate the fractional part twice and multiply by a well-chosen value of one that involves a conjugate

$$= 1 + \frac{1}{\displaystyle{1+\displaystyle{\frac{1}{\displaystyle{\frac{2}{\sqrt{3}-1} \cdot \frac{\sqrt{3}+1}{\sqrt{3}+1}}}}}}$$

and, like before, we simplify this bottom-most product

$$= 1 + \frac{1}{\displaystyle{1+\displaystyle{\frac{1}{\displaystyle{\sqrt{3}+1}}}}}$$

As we start to do it again, we notice something when we separate the integer part of $\sqrt{3}+1$ from its fractional part...

$$= 1 + \frac{1}{\displaystyle{1+\displaystyle{\frac{1}{\displaystyle{2 + (\sqrt{3}-1)}}}}}$$

The fractional part found is something we have seen already -- it is the same fractional part we discovered for $\sqrt{3}$ at the beginning of this argument! As such, this continued fraction will simply repeat from this point forward. Thus, the simple continued fraction representation for $\sqrt{3}$ is given by

$$\sqrt{3} = [1;\overline{1,2}]$$