Any whole number or simple fraction can be written as a continued fraction. Continued fractions appear a little strange and look like this:

1
a0 + ----------------------
1
a1 + ---------------
1
a2 + --------
a3 + ...

To make it easier to write down, the above fraction is written as:

[a0 ; a1, a2, a3 ...]

For example, the continued fraction for 40/33 is:

1
1 + ----------------------
1
4 + ---------------
1
1 + ---------
1
2 + ---
2

= [1; 4, 1, 2, 2]

You can check this on a calculator by starting at the bottom of the fraction and working upwards. First calculate 1/2, then add 2, then do 1/x , then add 1, then do 1/x again, then add 4, do 1/x again and finally add one to get 1.21212... = 40/33.

To work out the numbers that go in a continued fraction you go in the opposite direction. Start with 40/33. = 1.21212.

1.2121212

Subtract the whole number part, in this case 1

value remaining | continued fraction so far |

0.2121212 | 1 + ... |

Take 1/x

value remaining | continued fraction so far |

4.7142857 | 1 + 1/ ... |

Subtract the whole number part, in this case 4

value remaining | continued fraction so far |

0.7142857 | 1 + 1/( 4 + ...) |

Take 1/x

value remaining | continued fraction so far |

1.4000000 | 1 + 1/( 4 + 1/ ...) |

Subtract the whole number part, in this case 1

value remaining | continued fraction so far |

0.4000000 | 1 + 1/( 4 + 1/ (1 +...)) |

Take 1/x

value remaining | continued fraction so far |

2.5000000 | 1 + 1/( 4 + 1/ (1 + 1/ ...)) |

Subtract the whole number part, in this case 2

value remaining | continued fraction so far |

0.5000000 | 1 + 1/( 4 + 1/ (1 + 1/ (2 + ...)) |

Take 1/x

value remaining | continued fraction so far |

2.0000000 | 1 + 1/( 4 + 1/ (1 + 1/ (2 + 1/...)) |

We are left with a whole number so put this number in our continued fraction and we have reached the end.

40/33 = 1 + 1/( 4 + 1/ (1 + 1/ (2 + 1/2))

You can use the same method to work out continued fractions for square roots. Then the continued fraction goes on forever (unless the square root is a whole number eg √4 = 2):

1
√2 = 1 + ---------------------
1
2 + --------------
1
2 + --------
2 + ...

= [1; 2, 2, 2, 2, ...]

For square roots, the continued fraction soon forms a repeating pattern.

√2 = [1; 2, 2, ...]

√3 = [1; 1, 2, 1, 2, ...]

√5 = [2; 4, 4, ...]

√6 = [2; 2, 4, 2, 4, ...]

√7 = [2; 1, 1, 1, 4, 1, 1, 1, 4, ...]

For other numbers, such as π, the continued fraction does not form a pattern:

π = [3; 7, 15, 1, 292, 1, 1, 1, 2. ...]

By using the start of a continued fraction we can determine a fraction that is close to the irrational number. For the case of π:

[3;] | = | 3/1 | = | 3 | = | π - 0.14 |

[3; 7] | = | 22/7 | = | 3.1428 | = | π + 0.0012 |

[3; 7, 15] | = | 333/106 | = | 3.141509 | = | π - 0.000083 |

[3; 7, 15, 1] | = | 355/113 | = | 3.14159292 | = | π + 0.00000027 |

[3; 7, 15, 1, 292] | = | 103993/33102 | = | 3.14159265301 | = | π - 0.00000000058 |

The 292 in the continued fraction for π is quite large. This indicates that we will have to use much larger numbers in the faction to estimate π, before we find a better approximation than

355/113. So 355/113 will be quite a good approximation for π.