Good Approximations
Solve the quadratic equation $x^2 = 7x + 1$. This equation is equivalent to $x = 7 + \frac{1}{x}$ which has solutions given by the infinite continued fraction
$$ x = 7 + {1\over\displaystyle 7+ { 1\over \displaystyle 7+ { 1\over \displaystyle 7+ \cdots }}}. $$
This is because, if we think of this last equation as being $x = 7 + {1\over y}$, then clearly $y = x$. Show that the sequence of numbers $$7 + {1\over\displaystyle 7}, \quad 7 + {1\over\displaystyle 7+ { 1\over \displaystyle 7}}, \quad 7 + {1\over\displaystyle 7+ { 1\over \displaystyle 7+ { 1\over \displaystyle 7 }}}, \quad \cdots$$ gives better and better approximations to one of the solutions of the original quadratic equation. [Note that to find these approximations you can simply repeat the steps: 'take reciprocal, add 7', over and over again.]
Find integers $a$ and $b$, with $b$ less than 400, such that ${a\over b}$ , is equal to $\sqrt 53$ correct to six significant figures.
Now consider $x^2 = 5x + 1$, ...
[See the articles Continued Fractions I and Continued Fractions II. ]
In this example we see continued fractions used to give rational approximations to irrational numbers.
The following solution was done by Ling Xiang Ning, Raffles Institution, Singapore.
Using the quadratic formula to solve the equation
$ x^2 = 7x + 1$
$x^2 - 7x - 1 = 0$
I find that $x$ is $(7 \pm \sqrt{53} )/2$. The positive solution is approximately $7.140054945$
This equation is equivalent to $x = 7 + 1/x$ and hence to the sequence of continued fractions mentioned in the problem. These continued fractions give better and better approximations to the positive root of the quadratic equation and I shall do them one by one.
$ 7+\frac{1}{7} = \frac{50}{7} = 7.142857142 $
$ 7+\frac{1}{7+\frac{1}{7}} = \frac{357}{50} = 7.14 $
$ 7+\frac{1}{7+\frac{1}{7+\frac{1}{7}}} = \frac{2549}{357} = 7.140056022 $
$7+\frac{1}{7+\frac{1}{7+\frac{1}{7 + \frac{1}{7}}}} = \frac{18200}{2549} = 7.140054923 $
To find a rational approximation to $\sqrt{53}$ we take, as above, \[ \frac{7 + \sqrt{53}}{2} \approx \frac{2549}{357} \]
which gives \[ \sqrt{53} \approx 2 (\frac{2549}{357}) - 7 \approx \frac{2599}{357}. \]
$ 5+\frac{1}{5} = \frac{26}{5} = 5.2 $
$5+\frac{1}{5+\frac{1}{5}} = \frac{135}{26} = 5.192307692 $
$5+\frac{1}{5+\frac{1}{5+\frac{1}{5}}} = \frac{701}{135} = 5.192592592 $
$5+\frac{1}{5+\frac{1}{5+\frac{1}{5+\frac{1}{5}}}} = \frac{3640}{701} = 5.192582025 $
Similarly, using the equation, $x^2 = 5x + 1$, which has solutions $\frac{5\pm \sqrt{29} }{2}$ , we can find a rational approximation to $\sqrt{29}$. The positive root is approximately $5.192582404$. The sequence of continued fractions is:
Why do this problem?
For a better understanding of rational and irrational numbers.
Possible approach
Use this problem as part of a lesson series on number to include some or all of:
- proof root 2 is irrational
- converting periodic decimals to rational numbers
- proof that every rational number has a periodic decimal expansion
- the rational numbers are countable (see Route to Infinity )
- the irrational numbers are uncountable (see the article Infinity is not a number ).