Try the Free Math Solver or Scroll down to Tutorials!












Please use this form if you would like
to have this math solver on your website,
free of charge.

Approximation of irrational numbers

Let α be an irrational number. Its decimal representation is then nonterminating and nonrepeating. One can write

where A is the integral part of α. For instance,

To approximate sqrt(2), we may take its truncated decimal expansions as follows:

All numbers in the preceding p sequence are rational and they converge to
sqrt(2) from below. In fact, the difference between sqrt(2)and its n-th rational
approximation is smaller than 101-n. To achieve a good approximation,
one needs to choose n sufficiently large:

Could a good approximation of sqrt(2) have a not so large denominator?
Indeed, if possible, we would like to operate with smaller integers. Take a
look at the following example.

EXAMPLE. Among all rational fractions with denominator 12, the one
closest to sqrt(2) is 17/12,

At the same time,

Obviously, 17/12 is a better approximation than 1.41, and its denominator
is about 8 times smaller

Let’s introduce a measure of our approximation of α by a rational number
p/q, the approximation coefficient:

If is small and q is not so large, then the coefficient is small.
Conversely, if is small, then is even smaller. Thus the size of k tells
us about the quality of our approximation.

EXAMPLE. Out of three approximations, 3/2, 7/5, and 1.41, to sqrt(2), which
one has the smallest k? A quick computation gives the answer:

Given the size of its denominator, 7/5 is a very good choice. But 17/12 is,
of course, better than all three.

How to find these good approximations? Do they always exist? The true
answer has to do with continued fractions. We’ll only look at a partial
explanation here.

THEOREM. Let α be a given irrational number, and let N be any positive
integer. Then there exists a rational fraction p/q such that

In particular,

The preceding fact has an interesting illustration. Consider the family of
horizontal lines y = q, where q = 0, 1, 2, 3, . . . , and the family of parallel
lines x = αy − p, where p = 0,±1,±2,±3, . . . . The two families intersect
in a lattice of points with components (αq − p, q). For instance, the line
y = 3 meets the line x = αy − 5 at (3α − 5, 3).

Thus to each rational fraction p/q we put in correspondence a point
(αq − p, q), whose x-component has absolute value k and whose
y-component is the denominator q. Now draw a rectangle with vertices

The theorem says that for any
choice of N, the rectangle must contain a lattice point in its interior. Even
if N is very very large, i.e. the rectangle is very very thin.

Plot and experiment to get a better understanding. If you carefully
examine this geometric construction, you may see a way to prove the