How to find irrational approximates

Question

Say I have a rational number, $n$, that approximates an irrational number of the form:

$$n \approx {a+\sqrt b \over c}$$

in terms of being irrational.

What is a good way of finding the unknown integer constants $a$, $b$, and $c$?

Ex: $n = {33385282\over 20633239} \approx 1.618034 \approx \phi \approx {1 + \sqrt 5 \over 2}$ (note that $\phi$ (the golden ratio) is merely for example)

Thus $a = 1$, $b = 5$, and $c = 2$ for this example.

I know that since $n^2 \approx {a^2 + 2a\sqrt b + b \over c^2}$,you could find $a$ and $b$ by multiplying $n^2$ by $c^2$, and then subtracting $xnc$, finding the value of $x$ that makes the difference an integer. Then $x = 2a$, and $b$ can be solved using substitution algebra. But, all that implies that $c$ is known, and that's where I'm stuck.

Ex: $c$ is known, thus $c = 2$

$n^2c^2 \approx 10.472135$

$n^2c^2 - nc \approx 7.2360680$

$n^2c^2 - 2nc \approx 4$ ($x$ is now $2$ because $4$ is an interger)

$a = x/2 = {2/2} = 1$, $ b = \text algebra = 5$

So how do I find $c$ first, or what is a better way to go about solving $a$, $b$, and $c$ other than by trial?

Answer 1

You can do this with simple continued fractions. First, you are guaranteed to get an irrational by finding the complete scf for your rational number, then simply repeat that again and again forever. Such "purely periodic" scf's correspond to "reduced" quadratic irrationals.

Meanwhile, if you get good enough with scf's you may be able to spot when repetition seems to begin of its own accord. If so, you fill in the pattern, and get a simpler quadratic irrational, probably not reduced.

Answer 2

Of course, ther are many irrationaly that are approximately your given rational and what you are looking for is somewhat the "simplest" among all close enough quadratic irrationals.

I suggest, you transform your given fraction into a continued fraction. Just doing a few steps, you should find $$ \frac{33385282}{20633239}=1+\frac1{1+\frac1{1+\frac1{1+\frac{3010349}{4870847}}}}$$ and in fact you will get a few more $1$s upon proceeding, but the pattern is already apparent. Therefore we make the educated guess that our irrational number is the one with this pattern repeating indefinitely as continued fraction, i.e. $$ \alpha=1+\frac1{1+\frac1{1+\frac1{1+\frac1{1+\frac1{1+\frac1{1+\frac1{1+\frac1{\ddots}}}}}}}}$$ As this means $\alpha=1+\frac 1\alpha$, we arrive at $\alpha=\frac{1+\sqrt 5}{2}$.