The euclidean algorithm and continued fractions

2 Continued Fractions for Numbers During the 1500 years following Euclid, it was real-ized by mathematicians of the Indian and Ara-bic schools that the application of the Euclidean algor

Princeton Companion to Mathematics Proof 1

The Euclidean Algorithm and

Continued Fractions

By Keith Ball

1 The Euclidean Algorithm

The fundamental theorem of arithmetic (see

p ??), which states that every integer can be

factored into primes in a unique way, has been

known since antiquity The usual proof depends

upon what is known as the Euclidean algorithm,

which constructs the highest common factor (h,

say) of two numbers m and n In doing so, it shows

that h can be written in the form am + bn for some

pair of integers a, b (not necessarily positive) For

example, the highest common factor of 17 and 7

is 1 and sure enough we can express 1 as the

com-bination 1 = 5× 17 − 12 × 7.

The algorithm works as follows Assume that m

is larger than n and start by dividing m by n to

yield a quotient q1and a remainder r1 that is less

than n Then we have

m = q1n + r1. (1.1)

Now since r1< n we may divide n by r1 to obtain

a second quotient and remainder:

n = q2r1+ r2. (1.2)

Continue in this way, dividing r1 by r2, r2 by r3,

and so on The remainders get smaller each time

but cannot go below zero So the process must

stop at some point with a remainder of 0—with

a division that comes out exactly For instance, if

m = 165 and n = 70, the algorithm generates the

sequence of divisions

165 = 2× 70 + 25, (1.3)

70 = 2× 25 + 20, (1.4)

25 = 1× 20 + 5, (1.5)

20 = 4× 5 + 0. (1.6) The process guarantees that the last non-zero

remainder, 5 in this case, is the highest common

factor of m and n On the one hand, the last line

shows that 5 is a factor of the previous

remain-der 20 Now the last but one line shows that 5 is

also a factor of the remainder 25 that occurred one

step earlier, because 25 is expressed as a combina-tion of 20 and 5 Working back up the algorithm

we conclude that 5 is a factor of both m = 165 and n = 70 So 5 is certainly a common factor of

m and n.

On the other hand, the last-but-one line shows that 5 can be written as a combination of 25 and

20 with integer coefficients Since the previous line shows that 20 can be written as a combination of

70 and 25 we can write 5 in terms of 70 and 25:

5 = 25− 20 = 25 − (70 − 2 × 25) = 3 × 25 − 70.

Continuing back up the algorithm we can express

25 in terms of 165 and 70 and conclude that

5 = 3× (165 − 2 × 70) − 70 = 3 × 165 − 7 × 70.

This shows that 5 is the highest common factor

of 165 and 70 because any factor of 165 and 70 would automatically be a factor of 3×165−7×70:

that is, a factor of 5 Along the way we have shown that the highest common factor can be expressed

as a combination of the two original numbers m and n.

2 Continued Fractions for Numbers

During the 1500 years following Euclid, it was real-ized by mathematicians of the Indian and Ara-bic schools that the application of the Euclidean

algorithm to a pair of integers m and n could be encoded in a formula for the ratio m/n The

equa-tion (1.1) can be written

m

n = q1+

r1

n = q1+

1

F ,

where F = n/r1 Now equation (1.2) expresses F

as

F = q2+r2

r1

.

The next step of the algorithm will produce an

expression for r1/r2 and so on If the algorithm

stops after k steps, then we can put these expres-sions together to get what is called the continued

fraction for m/n:

m

n = q1+

1

q2+ 1

q3+ . + 1

.

2 Princeton Companion to Mathematics Proof

For example,

165

70 = 2 +

1

2 +1+11

.

The continued fraction can be constructed

directly from the ratio 165/70 = 2.357 14

with-out reference to the integers 165 and 70 We start

by subtracting from 2.357 14 the largest whole

number we can: namely 2 Now we take the

recip-rocal of what is left: 1/0.357 14 = 2.8 Again

we subtract off the largest integer we can, 2, which

tells us that q2= 2 The reciprocal of 0.8 is 1.25 so

q3= 1 and then, finally, 1/0.25 = 4, so q4= 4 and

the continued fraction stops

The mathematician John Wallis, who worked

in the seventeenth century, seems to have been

the first to give a systematic account of continued

fractions and to recognize that continued-fraction

expansions exist for all numbers (not only rational

numbers), provided that we allow the continued

fraction to have infinitely many levels If we start

with any positive number, we can build its

con-tinued fraction in the same way as for the ratio

2.357 14 For example, if the number is π =

3.141 592 65 , we start by subtracting 3, then

take the reciprocal of what is left: 1/0.141 59 =

7.062 51 So for π we get that the second

quo-tient is 7 Continuing the process we build the

con-tinued fraction

π = 3 + 1

7 +15+ 11

292+ 1 1+ .

. (2.1)

The numbers 3, 7, 15, and so on, that appear in

the fraction are called the partial quotients of π.

The continued fraction for a real number can be

used to approximate it by rational numbers If we

truncate the continued fraction after several steps,

we are left with a finite continued fraction which is

a rational number: for example, by truncating the

fraction (2.1), one level down we get the familiar

approximation π ≈ 3 + 1/7 = 22/7; at the second

level we get the approximation 3 + 1/(7 + 1/15) =

333/106 The truncations at different levels thus

generate a sequence of rational approximations:

the sequence for π begins

3, 22/7, 333/106, 355/113,

Whatever positive number x we start with, the

sequence of continued-fraction approximations will

approach x as we move further down the

frac-tion Indeed, the formal interpretation of the equa-tion (2.1) is precisely that the successive

trunca-tions of the fraction approach π.

Naturally, in order to get better approximations

to a number x we need to take more

“compli-cated” fractions—fractions with larger numerator and denominator The continued-fraction

approx-imations to x are best approxapprox-imations to x in the following sense: if p/q is one of these fractions, then

it is impossible to find any fraction r/s that is closer than p/q to x, but which has denominator s smaller than q.

Moreover, if p/q is one of the approximations coming from the continued fraction for x, then the error x − p/q cannot be too large relative to the

size of the denominator q; specifically, it is always

true that

x − p q

1

This error estimate shows just how special the continued-fraction approximations are: if you pick

a denominator q without thinking, and then select the numerator p that makes p/q closest to x, the only thing you can guarantee is that x lies between (p −1/2)/q and (p+1/2)/q So the error could be as

large as 1/(2q), which is much bigger than 1/(q2)

if q is a large integer.

Sometimes a continued-fraction approximation

to x can have even smaller error than is

guar-anteed by (2.2) For example, the approximation

π ≈ 355/113 that we get by truncating (2.1) at

the third level is exceptionally accurate, the reason being that the next partial quotient, 292, is rather large So we are not changing the fraction much

by ignoring the tail 1/(292 + 1/(1 + ..)) In this sense, the most difficult number to approximate

by fractions is the one with the smallest possible partial quotients, i.e the one with all its partial quotients equal to 1 This number,

1 + 1+1 ..

can be easily calculated because the sequence of partial quotients is periodic: it repeats itself If we

call the number φ, then φ −1 is 1/(1 + 1/(1 + )).

The reciprocal of this number is exactly the

con-tinued fraction (2.3) for φ Hence

1

φ − 1 = φ,

Princeton Companion to Mathematics Proof 3

which in turn implies that φ2−φ = 1 The roots of

this quadratic equation are (1 +√

5)/2 = 1.618

and (1− √ 5)/2 = −0.618 Since the number we

are trying to find is positive, it is the first of these

roots: the so-called golden ratio.

It is quite easy to show that, just as (2.3)

represents the positive solution of the equation

x2 − x − 1 = 0, any other periodic continued

fraction represents a root of a quadratic equation

This fact seems to have been understood already

in the sixteenth century It is quite a lot trickier

to prove the converse: that the continued fraction

of any quadratic surd is periodic This was

estab-lished by Lagrange during the eighteenth century

and is closely related to the existence of units in

quadratic number fields (see Algebraic

Num-berson p ??).

3 Continued Fractions for

Functions

Several of the most important functions in

math-ematics are most easily described using infinite

sums For example, the exponential function has

the infinite series

ex = 1 + x + x

2

2 +· · · + x n

n! +· · ·

There are also a number of functions that have

sim-ple continued-fraction expansions: continued

frac-tions involving a variable like x These are

proba-bly the most important continued fractions

histor-ically

For example, the function x → tan x has the

continued fraction

1− x2

3− x2

5− ..

valid for any value of x other than the odd

mul-tiples of π/2, where the tangent has a vertical

asymptote

Whereas the infinite series of a function can

be truncated to provide polynomial

approxima-tions to the function, truncation of the

contin-ued fraction provides approximations by rational

functions: functions that are ratios of polynomials.

Thus if we truncate the fraction for the tangent

after one level, we get the approximation

tan x ≈ x

1− x2/3 =

3x

3− x2.

This continued fraction, and the rapidity with

which its truncations approach tan x, played the central role in the proof that π is irrational: that π

is not the ratio of two whole numbers The proof was found by Johann Lambert in the 1760s He

used the continued fraction to show that if x is a rational number (other than 0), then tan x is not But tan π/4 = 1 (which certainly is rational), so

π/4 cannot be.