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# 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 10^{1-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

theorem.