In this article we see how Euclid's algorithm can be used to produce continued fractions and how these continued fractions can be truncated after a very few steps to give close rational approximations to irrational numbers.
In earlier articles we have seen a geometric interpretation of Euclid's algorithm in which we divide a rectangle up into squares plus a smaller rectangle and then divide the smaller rectangle into squares plus an even smaller rectangle and so on until we finally get to a rectangle that is completely made up of squares.
We have also discussed continued fractions. If you have not read the earlier articles, it might be a good idea to look at them before you read this one.
We start by evaluating the following continued fraction:
3+12+11+14.
The value of this is
3+12+45=3+514=4714.
Now draw (on graph paper) a rectangle of base 14 units and height 47 units; we shall call this a
14×47 rectangle.
 |
It is easy to see that this rectangle can be divided into THREE 14×14 squares and one 5×14 rectangle. Draw these squares inside the 14×47 rectangle, starting from the bottom. Now take the 5×14 rectangle and divide this into TWO 5×5 squares, and one 4×5 rectangle. Again, you should draw these squares in your diagram. Finally, the 4×5 rectangle can be divided into ONE 4×4 square and one 1×4 rectangle which can itself be divided into FOUR 1×1 squares. Notice that the number of squares found at each stage (written in capital letters) are exactly the numbers appearing in the continued fraction (which has 1 at the top of each fraction).
Let us try another example:
2+11+12+13=2+11+37=2+710=2710.
If we follow the instructions given above, we start with a 10×27 rectangle which we divide into TWO 10×10 squares and a 7×10 rectangle. This last rectangle is divided into ONE 7×7 square and a 3×7 rectangle. The 3×7rectangle divides into TWO 3×3squares and a 1×3 rectangle which then divides into THREE 1×1 squares. |
We can now try and reverse this process, and this will give us a way of finding the continued fraction for a given ratio. For example, starting with the number 43/30 we construct a 30×43rectangle (again, you should use graph paper here). Now divide this into a number of 30×30squares (how many?) and a rectangle (of what size?) If you continue in this way you should arrive at the continued fraction expansion for 43/30, and you can then check your answer by working out the continued fraction as we have done above. Of course, if your work is correct, the value of the continued fraction you get will be 43/30.
Here are two more examples to try. First start with the fraction 43/20, draw the squares and rectangles as above; then find the continued fraction for 43/20 , and check that it really does have the value 43/20. Now do the same with 43/10.
Euclid's algorithm is best known for finding the highest common factor of two whole numbers, which is then used in solving Diophantine equations such as ax+by=c where a,b and c are integers. Note that when the lengths of the edges of the original rectangle are whole numbers, because the rectangles produced by Euclid's algorithm get smaller and smaller and still have edges whose lengths are whole numbers, finally the process must terminate with a rectangle with an edge of 1 unit.
However Euclid's algorithm can also be used in the same way starting with any two numbers, not necessarily whole numbers or even rational numbers, in which case the associated continued fraction is usually an infinite fraction and the process does not terminate. This is useful for quickly finding good rational approximations to irrational numbers.
Suppose for example you want to find a rational approximation to π. We know that 227 is used as an approximation but where did that come from, and can we find a better rational approximation? To 7 decimal places π≈3.1415927 which gives 3141592710000000 as a rational approximation but we can do much better than that.
Taking the reciprocal of 0.1415927 we have
π≈3+17.0625133
which gives the rational approximation
π≈3+17=227 giving an approximation accurate to 3 significant figures.
For a better approximation, taking the reciprocal of 0.0625133 we have
π≈3+17+115.9966.
Repeating this algorithm we have
π≈3+17+115+11+1293+110.320556
and so on ... This gives the approximation (which is accurate to 7 significant figures):
π≈3+17+115+1=3+16113=355113.
