Transcendental number in simple words. Liouville transcendental numbers

which for a \u003d 1 served us to determine the sum of a geometric progression. Assuming the Gauss theorem to be proved, we assume that a \u003d a 1 is a root of equation (17), so that

	) \u003d a n + a		a n − 1			a n − 2		a 1 + a

Subtracting this expression from f (x) and rearranging the terms, we get the identity

f (x) \u003d f (x) - f (a1) \u003d (xn - a n 1) + an − 1 (xn − 1 - a n 1 −1) +. ... ... + a1 (x - a1).

(21) Using now formula (20), we can select the factor x - a 1 from each term and then take it out of the bracket, and the degree of the polynomial remaining in the brackets will be less by one. Rearranging the terms again, we get the identity

f (x) \u003d (x - a1) g (x),

where g (x) is a polynomial of degree n - 1:

g (x) \u003d xn − 1 + bn − 2 xn − 2 +. ... ... + b1 x + b0.

(We are not interested in the calculation of the coefficients denoted by b.) Let us apply the same argument to the polynomial g (x). By Gauss's theorem, there exists a root a2 of the equation g (x) \u003d 0, so that

g (x) \u003d (x - a2) h (x),

where h (x) is a new polynomial of degree n - 2. Repeating these arguments n - 1 times (of course, the application of the principle of mathematical induction is implied), we finally arrive at the expansion

f (x) \u003d (x - a1) (x - a2). ... ... (x - an).

From identity (22) it follows not only that the complex numbers a1, a2,

An are the roots of equation (17), but also the fact that equation (17) has no other roots. Indeed, if the number y were a root of equation (17), then from (22) it would follow

f (y) \u003d (y - a1) (y - a2). ... ... (y - an) \u003d 0.

But we have seen (p. 115) that the product of complex numbers is zero if and only if one of the factors is zero. So, one of the factors y - ar is equal to 0, that is, y \u003d ar, as required.

§ 6.

1. Definition and questions of existence. An algebraic number is any number x, real or imaginary, satisfying some algebraic equation kind

an xn + an − 1 xn − 1 +. ... ... + a1 x + a0 \u003d 0 (n\u003e 1, an 6 \u003d 0),

130 MATHEMATICAL NUMERICAL SYSTEM chap. II

where the numbers ai are integers. So, for example, the number 2 is algebraic, since it satisfies the equation

x2 - 2 \u003d 0.

In the same way, an algebraic number is any root of any equation with integer coefficients of the third, fourth, fifth, whatever degree, and regardless of whether or not it is expressed in radicals. The concept of an algebraic number is a natural generalization of the concept of a rational number, which corresponds to the special case n \u003d 1.

Not every real number is algebraic. This follows from the following theorem stated by Cantor: the set of all algebraic numbers is countable. Since the set of all real numbers is uncountable, there must necessarily be real numbers that are not algebraic.

Let us indicate one of the methods for recalculating the set of algebraic numbers. To each equation of the form (1) we associate a positive integer

h \u003d | an | + | an − 1 | +. ... ... + | a1 | + | a0 | + n,

which we will call for the sake of brevity the "height" of the equation. For each fixed value of n, there are only a finite number of equations of the form (1) with height h. Each of these equations has at most n roots. Therefore, there can exist only a finite number of algebraic numbers generated by equations with height h; therefore, all algebraic numbers can be arranged in the form of a sequence, first listing those of them that are generated by the equations of height 1, then - of height 2, etc.

This proof of the countability of the set of algebraic numbers establishes the existence of real numbers that are not algebraic. Such numbers are called transcendental (from the Latin transcendere - to go over, to exceed); Euler gave them this name because they "surpass the power of algebraic methods."

Cantor's proof of the existence of transcendental numbers is not constructive. Arguing theoretically, one could construct a transcendental number using a diagonal procedure performed on an imaginary list of decimal expansions of all algebraic numbers; but such a procedure is devoid of any practical significance and would not lead to a number whose expansion into a decimal (or some other) fraction could actually be written. The most interesting problems with transcendental numbers are in proving that certain, concrete numbers (this includes the numbers p and e, for which see pages 319–322) are transcendental.

ALGEBRAIC AND TRANSCENDENT NUMBERS

** 2. Liouville's theorem and the construction of transcendental numbers. The proof of the existence of transcendental numbers even before Cantor was given by J. Liouville (1809–1862). It makes it possible to actually construct examples of such numbers. Liouville's proof is more difficult than Cantor's, and this is not surprising, since constructing an example is generally more difficult than proving its existence. In presenting Liouville's proof below, we mean only a trained reader, although knowledge of elementary mathematics is quite enough to understand the proof.

As Liouville discovered, irrational algebraic numbers have the property that they cannot be approximated by rational numbers with very to a large extent accuracy, unless the denominators of the approximating fractions are extremely large.

Suppose that the number z satisfies an algebraic equation with integer coefficients

f (x) \u003d a0 + a1 x + a2 x2 +. ... ... + an xn \u003d 0 (an 6 \u003d 0),

but does not satisfy the same lower degree equation. Then

x itself is said to be an algebraic number of degree n. For example,

the number z \u003d 2 is an algebraic number of degree 2, since it satisfies the equation x2 - 2 \u003d 0√ of degree 2, but does not satisfy the equation of the first degree; the number z \u003d 3 2 is of degree 3, since it satisfies the equation x3 - 2 \u003d 0, but does not satisfy (as we will show in Chapter III) an equation of a lower degree. Algebraic number of degree n\u003e 1

cannot be rational, since the rational number z \u003d p q satisfies

satisfies the equation qx - p \u003d 0 of degree 1. Each irrational number z can be approximated with any degree of accuracy using a rational number; this means that you can always specify a sequence of rational numbers

p 1, p 2,. ... ...

q 1 q 2

with infinitely growing denominators, which has the property

that

p r → z. qr

Liouville's theorem states: whatever the algebraic number z of degree n\u003e 1, it cannot be approximated by a rational

sufficiently large denominators, the inequality

z - p q			\u003e q n1 +1.

MATHEMATICAL NUMERICAL SYSTEM

We are going to give a proof of this theorem, but earlier we will show how you can use it to construct transcendental numbers. Consider the number

z \u003d a1 10−1! + a2 10−2! + a3 10−3! +. ... ... + am 10 − m! +. ... ... \u003d \u003d 0, a1 a2 000a3 00000000000000000a4 000. ... ... ,

where ai denote arbitrary digits from 1 to 9 (the easiest way would be to put all ai equal to 1), and the symbol n !, as usual (see page 36), denotes 1 · 2 ·. ... ... · N. A characteristic property of the decimal expansion of such a number is that groups of zeros that are rapidly increasing in length alternate in it with individual digits other than zero. Let us denote by zm the final decimal fraction obtained when we take in the expansion all terms up to am · 10 − m! inclusive. Then we obtain the inequality

Suppose z was an algebraic number of degree n. Then, setting in Liouville's inequality (3) p q \u003d zm \u003d 10 p m! , we must have

| z - zm | \u003e 10 (n + 1) m!

for sufficiently large values \u200b\u200bof m. Comparing the last inequality with inequality (4) gives

10 (n + 1) m!		10 (m + 1)!			10 (m + 1)! - 1

whence (n + 1) m! \u003e (m + 1)! - 1 for sufficiently large m. But this is not true for values \u200b\u200bof m greater than n (let the reader take the trouble to give a detailed proof of this statement). We have come to a contradiction. So the number z is transcendental.

It remains to prove Liouville's theorem. Suppose that z is an algebraic number of degree n\u003e 1 satisfying equation (1), so that

f (zm) \u003d f (zm) - f (z) \u003d a1 (zm - z) + a2 (zm 2 - z2) +. ... ... + an (zm n - zn).

Dividing both sides by zm - z and using the algebraic formula

u n - v n \u003d un − 1 + un − 2 v + un − 3 v2 +. ... ... + uvn − 2 + vn − 1, u - v

we get:

	f (zm)	A1 + a2 (zm + z) + a3 (zm 2 + zm z + z2) +. ... ...
	zm - z
		An (zm n − 1 +.. + Zn − 1). (6)

ALGEBRAIC AND TRANSCENDENT NUMBERS

Since zm tends to z, then for sufficiently large m the rational number zm will differ from z by less than one. Therefore, for sufficiently large m, the following rough estimate can be made:

	f (zm)		< |a1 | + 2|a2 |(|z| + 1) + 3|a3 |(|z| + 1)2
zm - z

N | an | (| z | + 1) n − 1 \u003d M, (7)

moreover, the number M on the right is constant, since z does not change during the proof. We now choose m so large that

fraction z m \u003d p m denominator q m was larger than M; thenqm

| z - zm | \u003e	| f (zm) |			| f (zm) |

| f (zm) | \u003d							−q n
							1 p +. ... ... + a
The rational number zm \u003d			cannot be the root of the equation

since then it would be possible to extract the factor (x - zm) from the polynomial f (x), and, therefore, z would satisfy an equation of degree lower than n. So, f (zm) 6 \u003d 0. But the numerator on the right-hand side of equality (9) is an integer and, therefore, in absolute value, it is at least is equal to one... Thus, a comparison of relations (8) and (9) implies the inequality

| z - zm | \u003e
					q n + 1

is precisely the constituent content of the indicated theorem.

Over the past several decades, research on the possibility of approximating algebraic numbers by rational ones has advanced much further. For example, the Norwegian mathematician A. Thue (1863–1922) established that in Liouville's inequality (3) the exponent n + 1 can be replaced by a smaller exponent n 2 + 1.

K.L.Siegel showed that one can take an even smaller (even smaller

for large n) exponent 2 n.

Transcendental numbers have always been a topic that attracts the attention of mathematicians. But until relatively recently, among the numbers that are interesting in themselves, very few were known whose transcendental character was established. (From the transcendence of the number p, which will be discussed in Chapter III, it follows that the circle cannot be squared with a ruler and compass.) In his speech at the Paris International Congress of Mathematics in 1900, David Hilbert proposed thirty mathematical

ALGEBRA OF SETS

problems admitting a simple formulation, some - even quite elementary and popular, of which not only was not solved, but did not even seem capable of being solved by the means of mathematics of that era. These "Hilbert problems" had a strong stimulating influence throughout the subsequent period in the development of mathematics. Almost all of them were gradually resolved, and in many cases their solution was associated with clearly expressed successes in the sense of developing more general and deeper methods. One problem that seemed rather hopeless was

proof that the number

is transcendental (or at least irrational). For three decades, there was not even a hint of such an approach to the issue from anyone's side that would open up hope for success. Finally, Siegel and, independently of him, the young Russian mathematician A. Gelfond discovered new methods for proving the transcendence of many

numbers that matter in mathematics. In particular, it was established

transcendence not only of the Hilbert number 2 2, but also of an entire rather extensive class of numbers of the form ab, where a is an algebraic number other than 0 and 1, and b is an irrational algebraic number.

SUPPLEMENT TO CHAPTER II

Algebra of sets

1. General theory. The concept of a class, or collection, or set of objects is one of the most fundamental in mathematics. The set is determined by some property ("attribute") A, which each considered object must either have or not have; those objects that have property A form a set A. So, if we consider integers and property A is to "be simple", then the corresponding set A consists of all primes 2, 3, 5, 7,. ... ...

Mathematical set theory assumes that new sets can be formed from sets using certain operations (just as new numbers are obtained from numbers by means of addition and multiplication). The study of operations on sets is the subject of "set algebra", which has much in common with ordinary numerical algebra, although in some ways it differs from it. The fact that algebraic methods can be applied to the study of non-numerical objects, which are sets, is illus-

ALGEBRA OF SETS

shows a great commonality of ideas in modern mathematics. IN recent times it turned out that the algebra of sets throws new World on many areas of mathematics, for example, measure theory and probability theory; it is also useful for organizing mathematical concepts and clarification of their logical connections.

In what follows, I will denote some constant set of objects, the nature of which is indifferent, and which we can call the universal set (or the universe of reasoning), and

A, B, C,. ... ... there will be some subsets of I. If I is the collection of all natural numbers, then A, say, can denote the set of all even numbers, B - the set of all odd numbers, C - the set of all primes, and so on. If I denotes the collection of all points on the plane, then A can be a set of points inside some circle, B - a set of points inside another circle, etc. It is convenient for us to include I itself in the number of "subsets", as well as an "empty" set that does not contain any elements. The goal pursued by such an artificial extension is to preserve the position that each property A corresponds to a certain set of elements from I possessing this property. If A is a universally fulfilled property, an example of which is (if we are talking about numbers) the property to satisfy the trivial equality x \u003d x, then the corresponding subset of I will be I itself, since each element has such a property; on the other hand, if A is some internally contradictory property (like x 6 \u003d x), then the corresponding subset contains no elements at all, it is “empty” and is denoted by a symbol.

They say that the set A is a subset of the set B, in short, "A is included in B", or "B contains A" if there is no such element in the set A that would not also be in the set B. This relation corresponds to the notation

A B, or B A.

For example, the set A of all integers divisible by 10 is a subset of the set B of all integers divisible by 5, since every number divisible by 10 is also divisible by 5. The relation AB does not exclude the relation B A. If and both, then

This means that each element of A is at the same time an element of B, and vice versa, so that the sets A and B contain exactly the same elements.

The relationship A B between sets is in many ways similar to the relationship a 6 b between numbers. In particular, note the following

ALGEBRA OF SETS

the following properties of this ratio:

1) A A.

2) If A B and B A, then A \u003d B.

3) If A B and B C then A C.

For this reason, the relationship A B is sometimes called the "order relationship." The main difference between the ratio under consideration and the ratio a 6 b between numbers is that between any two given (real) numbers a and b, at least one of the ratios a 6 b or b 6 a is certainly carried out, while for the ratio AB between the sets a similar statement is not true. For example, if A is a set consisting of numbers 1, 2, 3,

and B is a set consisting of numbers 2, 3, 4,

then neither the relation A B nor the relation B A. holds. For this reason, the subsets A, B, C,. ... ... the sets I are "partially ordered", while the real numbers a, b, c,. ... ...

form a "well-ordered" set.

Note, by the way, that from the definition of the relation A B it follows that, whatever the subset A of the set I,

Property 4) may seem somewhat paradoxical, but if you think about it, it logically strictly corresponds to the exact meaning of the definition of the sign. Indeed, the relation A would be violated only

in if the empty set contained an element that would not be contained in A; but since the empty set contains no elements at all, this cannot be, whatever the A.

We now define two operations on sets that formally possess many algebraic properties of addition and multiplication of numbers, although in their inner content they are completely different from these arithmetic operations. Let A and B be some two sets. By union, or "logical sum", A and B means a set consisting of those elements that are contained either in A, or

in B (including those elements that are contained in both A and B). This set is denoted A + B.1 By "intersection", or "logical product", A and B means a set consisting of those elements that are contained in both A and B. This set is denoted by AB.2

Among the important algebraic properties of the operations A + B and AB, we list the following. The reader will be able to verify their validity based on the definition of the operations themselves:

	A + (B + C) \u003d (A + B) + C. 9)
	A (B + C) \u003d AB + AC.		A + (BC) \u003d (A + B) (A + C).
	The ratio A B is equivalent to each of the two relations

Verification of all these laws is a matter of the most elementary logic. For example, rule 10) states that the set of elements contained in either A or A is just the set A; rule 12) states that the set of those elements that are contained in A and at the same time are contained either in B or C, coincides with the set of elements that are either contained simultaneously in A and B, or are simultaneously contained in A and C The logical reasoning used in the proofs of this kind of rules is conveniently illustrated if we agree to represent the sets A, B, C,. ... ... in the form of some figures on a plane and we will be very careful not to miss any of the logical possibilities that arise when it comes to the presence of common elements of two sets or, on the contrary, the presence in one set of elements that are not contained in the other.

ALGEBRA OF SETS

The reader undoubtedly drew attention to the fact that laws 6), 7), 8), 9) and 12) are outwardly identical with the well-known commutative, associative and distributive laws of ordinary algebra. It follows from this that all the rules of ordinary algebra that follow from these laws are also valid in the algebra of sets. On the contrary, laws 10), 11) and 13) have no analogues in ordinary algebra, and they give the algebra of sets a simpler structure. For example, the binomial formula in the algebra of sets reduces to the simplest equality

(A + B) n \u003d (A + B) (A + B). ... ... (A + B) \u003d A + B,

which follows from Law 11). Laws 14), 15) and 17) say that the properties of the sets and I with respect to the operations of union and intersection of sets are very similar to the properties of the numbers 0 and 1 with respect to the operations of numerical actions of addition and multiplication. But Law 16) has no analogue in numerical algebra.

It remains to define one more operation in the algebra of sets. Let A be some subset of the universal set I. Then by the complement A in I we mean the set of all elements of I that are not contained in A. For this set, we introduce the notation A0. So, if I is the set of all natural numbers, and A is the set of all prime numbers, then A0 is the set consisting of all composite numbers and the number 1. The operation of passing from A to A0, for which there is no analog in ordinary algebra, has the following properties :

	A + A0 \u003d I.		AA0 \u003d.
	0 \u003d I.		I0 \u003d.

23) A 00 \u003d A.

24) A B is equivalent to B0 A0.

25) (A + B) 0 \u003d A0 B0. 26) (AB) 0 \u003d A0 + B0.

We again leave the verification of these properties to the reader.

Laws 1) –26) underlie the algebra of sets. They have the remarkable property of "duality" in the following sense:

If in one of the laws 1) -26) replace the corresponding

(in each of their occurrences), the result is again one of the same laws. For example, law 6) turns into law 7), 12) - in 13), 17) - in 16), etc. It follows that each theorem that can be derived from laws 1) –26) corresponds to another , A theorem "dual" to it, which is obtained from the first one by means of the indicated permutations of symbols. Indeed, since the proof

ch. II ALGEBRA OF SETS 139

the first theorem consists of a consistent application (on different stages the reasoning) of some of laws 1–26), then the application of the “dual” laws at the appropriate stages will constitute a proof of the “dual” theorem. (For a similar "duality" in geometry, see Chapter IV.)

2. Application to mathematical logic. Verification of the laws of set algebra was based on the analysis of the logical meaning of the relation A B and the operations A + B, AB and A0. We can now reverse this process and consider laws 1) –26) as the basis for the "algebra of logic". Let's say more precisely: that part of logic that concerns sets, or, which is essentially the same, the properties of the objects under consideration, can be reduced to a formal algebraic system based on the laws 1) –26). The logical "conditional universe" defines the set I; each property A defines a set A consisting of those objects in I that have this property. The rules for translating ordinary logical terminology into the language of sets are clear from

the following examples:
"Neither A, nor B"			(A + B) 0, or, which is the same, A0 B0
"It is not true that both A and B"			(AB) 0, or, which is the same, A0 + B0
is B ", or
"If A, then B",
"From A follows B"
"Some A is B"
"No A is B"			AB \u003d
"Some A is not B"			AB0 6 \u003d
"There is no A"

In terms of algebra of sets, the syllogism "Barbara" denoting that "if every A is B and every B is C, then every A is C", takes a simple form:

3) If A B and B C, then A C.

Similarly, the "law of contradiction", which states that "an object cannot simultaneously possess and not possess some property," is written in the form:

20) AA 0 \u003d,

a “The law of the excluded third”, which says that “an object must either have or not have some property,” is written:

19) A + A 0 \u003d I.

ALGEBRA OF SETS

Thus, that part of logic, which is expressible in terms of symbols, +, · and 0, can be interpreted as a formal algebraic system, subject to the laws 1) –26). On the basis of the fusion of the logical analysis of mathematics and the mathematical analysis of logic, a new discipline was created - mathematical logic, which is currently in the process of rapid development.

From an axiomatic point of view, noteworthy is the remarkable fact that assertions 1) –26), together with all other theorems of set algebra, can be logically deduced from the following three equalities:

27) A + B \u003d B + A,

(A + B) + C \u003d A + (B + C),

(A0 + B0) 0 + (A0 + B) 0 \u003d A.

Hence it follows that the algebra of sets can be constructed as a purely deductive theory, like Euclidean geometry, on the basis of these three propositions taken as axioms. If these axioms are accepted, then the operation AB and the relation A B are defined in terms of A + B and A0:

	denotes the set (A0 + B0) 0,
	B means A + B \u003d B.

A completely different example of a mathematical system in which all the formal laws of the algebra of sets are satisfied is given by the system of eight numbers 1, 2, 3, 5, 6, 10, 15, 30: here a + b denotes, by

definition, the common least multiple of a and b, ab is the common greatest divisor of a and b, a b is the statement “b is divisible by a” and a0 is the number 30 a. Su-

the development of such examples has led to the study of general algebraic systems satisfying the laws (27). Such systems are called "Boolean algebras" after George Boole (1815–1864), an English mathematician and logician whose book An investigation of the laws of thought appeared in 1854.

3. One of the applications to the theory of probability. Set algebra is closely related to probability theory and allows you to look at it in a new light. Consider the simplest example: imagine an experiment with a finite number of possible outcomes, all of which are thought of as "equally possible." An experiment might, for example, be that we draw a card at random from a well-reshuffled full deck. If we denote the set of all outcomes of the experiment by I, and A denotes some subset of I, then the probability that the outcome of the experiment turns out to belong to the subset A is defined as the ratio

p (A) \u003d number of elements of A. number of elements I

ALGEBRA OF SETS

If we agree to denote the number of elements in some set A by n (A), then the last equality can be given the form

In our example, assuming that A is a subset of clubs, we get
n (A) \u003d 13, n (I) \u003d 52 and p (A) \u003d

The ideas of algebra of sets are discovered when calculating probabilities when it is necessary, knowing the probabilities of some sets, to calculate the probabilities of others. For example, knowing the probabilities p (A), p (B), and p (AB), you can calculate the probability p (A + B):

p (A + B) \u003d p (A) + p (B) - p (AB).

It will not be difficult to prove this. We have

n (A + B) \u003d n (A) + n (B) - n (AB),

since the elements contained simultaneously in A and B, that is, the elements of AB, are counted twice when calculating the sum n (A) + n (B), and, therefore, it is necessary to subtract n (AB) from this sum in order to calculate n (A + B) was produced correctly. Then dividing both sides of the equality by n (I), we obtain relation (2).

A more interesting formula is obtained when we are talking about three sets A, B, C from I. Using relation (2), we have

p (A + B + C) \u003d p [(A + B) + C] \u003d p (A + B) + p (C) - p [(A + B) C].

Law (12) from the previous paragraph gives us (A + B) C \u003d AC + BC. This implies:

p [(A + B) C)] \u003d p (AC + BC) \u003d p (AC) + p (BC) - p (ABC).

Substituting the p [(A + B) C] value and the p (A + B) value taken from (2) into the relation obtained earlier, we arrive at the formula we need:

p (A + B + C) \u003d p (A) + p (B) + p (C) - p (AB) - p (AC) - p (BC) + p (ABC). (3)

Consider the following experiment as an example. Three numbers 1, 2, 3 are written in any random order. What is the likelihood that at least one of the numbers will be in the proper (in terms of numbering) place? Let A be the set of permutations in which the number 1 is in the first place, B is the set of permutations in which the number 2 is in the second place, and C is the set of permutations in which the number 3 is in the third place. We need to calculate p (A + B + C). It's clear that

p (A) \u003d p (B) \u003d p (C) \u003d 2 6 \u003d 1 3;

indeed, if any digit is in its proper place, then there are two possibilities to permute the remaining two digits from the total number of 3 · 2 · 1 \u003d 6 possible permutations of three digits. Further,

Exercise. Derive the corresponding formula for p (A + B + C + D) and apply it to an experiment with 4 digits. The corresponding probability is 5 8 \u003d 0.6250.

The general formula for the union of n sets is

p (A1 + A2 +.. + An) \u003d
	p (Ai) -
			p (Ai Aj) + p (Ai Aj Ak) -. ... ... ± p (A1 A2... An), (4)
where symbols				denote the summation over all possible

combinations containing one, two, three,. ... ... , (n - 1) letters from the number A1, A2,. ... ...

An. This formula can be established by means of mathematical induction - just like formula (3) was derived from formula (2).

From formula (4), we can conclude that if n digits are 1, 2, 3,. ... ... , n are written in any order, then the probability that at least one of the digits will be in the proper place is

pn \u003d 1 -

where the last term is preceded by a + or - sign, depending on whether n is even or odd. In particular, for n \u003d 5 this probability is

p5 \u003d 1 - 2! + 3! - 4! + 5! \u003d 30 \u003d 0.6333. ... ...

In Chapter VIII we will see that as n tends to infinity, the expression

1 1 1 1 Sn \u003d 2! - 3! + 4! -. ... ... ± n!

tends to the limit 1 e, the value of which, with five decimal places,

is equal to 0.36788. Since it is clear from formula (5) that pn \u003d 1 - Sn, it follows that as n → ∞

pn → 1 - e ≈ 0.63212.

Ilya Shchurov

Mathematician Ilya Shchurov on decimal fractions, transcendence and irrationality of Pi.

How did one help build the first cities and great empires? How did you inspire the outstanding minds of humanity? What role did it play in making money? How "one" merged with zero to rule modern world? The history of the unit is inextricably linked with the history of European civilization. Terry Jones embarks on a humorous journey to piece together the amazing story of our simplest number. With the help of computer graphics in this program, the unit comes to life in a variety of ways. From the history of unity, it becomes clear where modern numbers came from, and how the invention of zero saved us from the need to use Roman numerals today.

Jacques Sesiano

We know little about Diophantus. It seems he lived in Alexandria. No Greek mathematician mentions him until the 4th century, so he probably lived in the middle of the 3rd century. The most main job Diophantus, "Arithmetic" (Ἀριθμητικά), took place at the beginning of 13 "books" (βιβλία), that is, chapters. Today we have 10 of them, namely: 6 in the Greek text and 4 others in the medieval Arabic translation, which place in the middle of the Greek books: books I-III in Greek, IV-VII in Arabic, VIII-X in Greek ... "Arithmetic" by Diophantus is primarily a collection of problems, about 260 in total. To tell the truth, there is no theory; there are only general instructions in the introduction of the book, and special notes in some problems, when necessary. "Arithmetic" already has the features of an algebraic treatise. First, Diophantus uses different signs to express the unknown and its degrees, also some calculations; like all algebraic symbolism of the Middle Ages, its symbolism comes from mathematical words. Then, Diophantus explains how to solve the problem in an algebraic way. But Diophantus' problems are not algebraic in the usual sense, because almost everything is reduced to solving an indefinite equation or systems of such equations.

George Shabbat

Course program: History. First estimates. The problem of commensurability of the circumference of a circle with its diameter. Infinite series, products and other expressions for π. Convergence and its quality. Expressions containing π. Sequences rapidly converging to π. Modern methods computing π, using computers. On the irrationality and transcendence of π and some other numbers. No prior knowledge is required to understand the course.

Scientists at Oxford University stated that the earliest known use of the number 0 to indicate the absence of a digit value (as in the number 101) should be considered the text of the Indian manuscript Bakhshali.

Vasily Pispanen

Who hasn't played the name-the-biggest-number game as a kid? Millions, trillions and other "-ons" are already difficult to imagine in the mind, but we will try to make out the "mastodon" in mathematics - Graham's number.

Victor Kleptsyn

The real number can be approximated arbitrarily accurately by rational ones. How good can such an approximation be compared to its complexity? For example, breaking off the decimal notation of the number x by kth digit after the decimal point, we get an approximation x≈a / 10 ^ k with an error of the order of 1/10 ^ k. And in general, having fixed the denominator q of the approximating fraction, we can definitely get an approximation with an error of the order of 1 / q. Can you do better? The familiar π≈22 / 7 approximation gives an error of the order of 1/1000 - that is, clearly much better than one would expect. And why? Are we lucky that π has such an approximation? It turns out that for any irrational number there are infinitely many fractions p / q that approximate it better than 1 / q ^ 2. This is stated by Dirichlet's theorem - and we will begin the course with its slightly non-standard proof.

In 1980, the Guinness Book of Records repeated Gardner's claims, further fueling public interest in this number. Graham's number is an unimaginable number of times larger than other well-known large numbers such as googol, googolplex, and even more than Skuse's and Moser's numbers. In fact, the entire observable universe is too small to contain the usual decimal notation of the Graham number.

Dmitry Anosov

Lectures are given by Dmitry Viktorovich Anosov, Doctor of Physics and Mathematics, Professor, Academician of the Russian Academy of Sciences. Summer School "Contemporary Mathematics", Dubna. July 16-18, 2002

It is impossible to answer this question correctly, since the number series has no upper limit. So, to any number it is enough just to add one to get an even larger number. Although the numbers themselves are infinite, they do not have very many names of their own, since most of them are content with names made up of smaller numbers. It is clear that in the finite set of numbers that humanity has awarded with its own name, there must be some largest number. But what is it called and what is it equal to? Let's try to figure it out and at the same time find out how big numbers mathematicians have invented.

Every transcendental real number is irrational, but the converse is not true. For example, the number $\backslash \backslash \; sqrt\; 2$ - irrational, but not transcendental: it is the root of the polynomial $x\; ^\; 2-2$ (and therefore algebraic).

The order on the set of real transcendental numbers is isomorphic to the order on the set of irrational numbers.

The measure of irrationality of almost any transcendental number is 2.

Examples of

History

For the first time the concept of a transcendental number was introduced by J. Liouville in 1844, when he proved the theorem that an algebraic number cannot be approximated too well by a rational fraction.

| title3 \u003d Extension tools
number systems | title4 \u003d Hierarchy of numbers | list4 \u003d

$-1,\; \backslash \backslash ;\; 0,\; \backslash \backslash ;\; 1,\; \backslash \backslash ;\; \backslash \backslash \; ldots$ Whole numbers $-1,\; \backslash \backslash ;\; 1,\; \backslash \backslash ;\; \backslash \backslash \; frac\; (1)\; (2),\; \backslash \backslash ;\; \backslash \backslash ;\; 0\; (,)\; 12,\; \backslash \backslash \; frac\; (2)\; (3),\; \backslash \backslash ;\; \backslash \backslash \; ldots$ Rational numbers $-1,\; \backslash \backslash ;\; 1,\; \backslash \backslash ;\; \backslash \backslash ;\; 0\; (,)\; 12,\; \backslash \backslash \; frac\; (1)\; (2),\; \backslash \backslash ;\; \backslash \backslash \; pi,\; \backslash \backslash ;\; \backslash \backslash \; sqrt\; (2),\; \backslash \backslash ;\; \backslash \backslash \; ldots$ Real numbers

$-1,\; \backslash \backslash ;\; \backslash \backslash \; frac\; (1)\; (2),\; \backslash \backslash ;\; 0\; (,)\; 12,\; \backslash \backslash ;\; \backslash \backslash \; pi,\; \backslash \backslash ;\; 3i\; +\; 2,\; \backslash \backslash ;\; e\; ^\; (i\; \backslash \backslash \; pi\; /\; 3),\; \backslash \backslash ;\; \backslash \backslash \; ldots$

Complex numbers

$1,\; \backslash \backslash ;\; i,\; \backslash \backslash ;\; j,\; \backslash \backslash ;\; k,\; \backslash \backslash ;\; 2i\; +\; \backslash \backslash \; pi\; j-\; \backslash \backslash \; frac\; (1)\; (2)\; k,\; \backslash \backslash ;\; \backslash \backslash \; dots$ Quaternions $1,\; \backslash \backslash ;\; i,\; \backslash \backslash ;\; j,\; \backslash \backslash ;\; k,\; \backslash \backslash ;\; l,\; \backslash \backslash ;\; m,\; \backslash \backslash ;\; n,\; \backslash \backslash ;\; o,\; \backslash \backslash ;\; 2\; -\; 5l\; +\; \backslash \backslash \; frac\; (\backslash \backslash \; pi)\; (3)\; m,\; \backslash \backslash ;\; \backslash \backslash \; Transcendental\; numbers$ Number Ray Biquaternion Excerpt from the Transcendental Number

- How can you be healthy ... when you suffer morally? Is it possible to remain calm in our time when a person has a feeling? - said Anna Pavlovna. - You all evening with me, I hope?

- And the holiday of the English envoy? Today is Wednesday. I have to show myself there, ”said the prince. - My daughter will pick me up and take me.
- I thought this holiday was canceled. Je vous avoue que toutes ces fetes et tous ces feux d "artifice commencent a devenir insipides. [I confess, all these celebrations and fireworks are becoming unbearable.]
“If you knew that you wanted it, the holiday would have been canceled,” said the prince, out of habit, like a clock, saying things that he didn’t want to be believed.
- Ne me tourmentez pas. Eh bien, qu "a t on decide par rapport a la depenche de Novosiizoff? Vous savez tout. [Don't torture me. Well, what did you decide on the occasion of Novosiltsov's dispatch? You all know.]
- How can I tell you? - said the prince in a cold, bored tone. - Qu "at on decide? On a decide que Buonaparte a brule ses vaisseaux, et je crois que nous sommes en train de bruler les notres. [What did you decide? We decided that Bonaparte burned his ships; and we, too, seem ready to burn ours.] - Prince Vasily always spoke lazily, as an actor speaks the role of an old play.Anna Pavlovna Sherer, on the contrary, despite her forty years, was filled with animation and impulses.
To be an enthusiast became her social position, and sometimes, when she did not even want to, she, in order not to deceive the expectations of people who knew her, became an enthusiast. The restrained smile that played constantly on Anna Pavlovna's face, although it did not go to her obsolete features, expressed, as with spoiled children, the constant consciousness of her sweet flaw, from which she does not want, cannot and does not find it necessary to correct.
In the middle of a conversation about political actions, Anna Pavlovna flared up.
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- Oh, don't tell me about Austria! I don’t understand anything, maybe, but Austria never wanted and does not want war. She betrays us. Russia alone should be the savior of Europe. Our benefactor knows his high calling and will be faithful to it. This is one thing I believe in. Our kind and wonderful sovereign will greatest role in the world, and he is so virtuous and good that God will not leave him, and he will fulfill his calling to crush the hydra of the revolution, which is now even more terrible in the person of this murderer and villain. We alone must redeem the blood of the righteous ... Who can we hope for, I ask you? ... England, with its commercial spirit, will not understand and cannot understand the full height of the soul of Emperor Alexander. She refused to clear Malta. She wants to see, is looking for an afterthought of our actions. What did they say to Novosiltsov? ... Nothing. They did not understand, they cannot understand the selflessness of our emperor, who wants nothing for himself and wants everything for the good of the world. And what did they promise? Nothing. And what they promised, and that will not happen! Prussia has already declared that Bonaparte is invincible and that the whole of Europe can do nothing against him ... And I do not believe in a single word either to Hardenberg or Gaugwitz. Cette fameuse neutralite prussienne, ce n "est qu" un piege. [This notorious neutrality of Prussia is only a trap.] I believe in one God and in the high destiny of our dear emperor. He will save Europe! ... ”She suddenly stopped with a smile of mockery at her fervor.

The number is called algebraicif it is a root of some polynomial with integer coefficients

a n x n + a n-1 x n-1 + ... + a 1 x + a 0 (i.e., the root of the equation a n x n + a n-1 x n-1 + ... + a 1 x + a 0 \u003d 0where a n, a n-1, ..., a 1, a 0 --- whole numbers, n 1, a n 0).

The set of algebraic numbers is denoted by the letter .

It is easy to see that any rational number is algebraic. Indeed, is the root of the equation qx-p \u003d 0 with integer coefficients a 1 \u003d q and a 0 \u003d -p... So, .

However, not all algebraic numbers are rational: for example, the number is the root of the equation x 2 -2 \u003d 0, hence, --- algebraic number.

For a long time, an important question for mathematics remained unresolved: Do non-algebraic real numbers ? It was not until 1844 that Liouville first gave an example of a transcendental (i.e., non-algebraic) number.

The construction of this number and the proof of its transcendence are very difficult. It is much easier to prove the existence theorem for transcendental numbers, using considerations about the equivalence and non-equivalence of numerical sets.

Namely, let us prove that the set of algebraic numbers is countable. Then, since the set of all real numbers is uncountable, we will establish the existence of non-algebraic numbers.

Let's build a one-to-one correspondence between and some subset ... This will mean that - of course either countable. But since then is infinite, and therefore countable.

Let be some algebraic number. Consider all polynomials with integer coefficients, the root of which is, and choose among them the polynomial P of minimum degree (that is, it will not be a root of any polynomial with integer coefficients of less degree).

For example, for a rational number such a polynomial has degree 1, and for a number it has degree 2.

We divide all the coefficients of the polynomial P by their greatest common divisor. We obtain a polynomial whose coefficients are coprime in the aggregate (their greatest common divisor is 1). Finally, if the leading coefficient a n is negative, we multiply all the coefficients of the polynomial by -1 .

The resulting polynomial (that is, a polynomial with integer coefficients, the root of which is a number having the minimum possible degree, coprime coefficients and a positive leading coefficient) is called the minimum polynomial of the number.

It can be proved that such a polynomial is uniquely determined: each algebraic number has exactly one minimal polynomial.

The number of real roots of a polynomial is not more than its degree. Hence, it is possible to number (for example, in ascending order) all the roots of such a polynomial.

Now any algebraic number is completely determined by its minimum polynomial (i.e., the set of its coefficients) and the number that distinguishes this polynomial from other roots: (a 0, a 1, ..., a n-1, a n, k).

So, to each algebraic number we put in correspondence a finite set of integers, and from this set it is uniquely reconstructed (that is, different sets correspond to different numbers).

Let us number in ascending order all the primes (it is easy to show that there are infinitely many of them). We get an infinite sequence (p k): p 1 \u003d 2,p 2 \u003d 3, p 3 \u003d 5, p 4 \u003d 7, ... Now the set of integers (a 0, a 1, ..., a n-1, a n, k) you can match the work

(this number is positive and rational, but not always natural, because among the numbers a 0, a 1, ..., a n-1, may be negative). Note that this number is an irreducible fraction, since the prime factors included in the expansion of the numerator and denominator are different. Note also that two irreducible fractions with positive numerators and denominators are equal if and only if their numerators are equal and their denominators are equal.

Consider now the through display:

(a 0, a 1, ..., a n-1, a n, k) \u003d

Since we assigned different sets of integers to different algebraic numbers, and to different sets --- various rational numbers, then we have thus established a one-to-one correspondence between the set and some subset ... Therefore, the set of algebraic numbers is countable.

Since the set of real numbers is uncountable, we have proved the existence of non-algebraic numbers.

However, the existence theorem does not indicate how to determine whether given number algebraic. And this question is sometimes very important for mathematics.

Transcendental number— complex numberthat is not algebraic, that is, it is not a root of any nonzero polynomial with rational coefficients.

The existence of transcendental numbers was first established by J. Liouville in 1844; he also constructed the first examples of such numbers. Liouville noted that algebraic numbers cannot approximate "too well" by rational numbers. Namely, Liouville's theorem says that if an algebraic number is a root of a polynomial of degree with rational coefficients, then for any rational number the inequality

where the constant depends only on. This statement implies a sufficient criterion for transcendence: if the number is such that for any constant there is an infinite set of rational numbers satisfying the inequalities

then it is transcendental. Subsequently, such numbers were called Liouville numbers. An example of such a number is

Another proof of the existence of transcendental numbers was obtained by G. Cantor in 1874 on the basis of the set theory he created. Cantor proved that the set of algebraic numbers is countable and the set of real numbers is uncountable, which implies that the set of transcendental numbers is uncountable. However, unlike Liouville's proof, this reasoning does not allow us to give an example of even one such number.

Liouville's work gave rise to a whole section of the theory of transcendental numbers - the theory of approximation of algebraic numbers by rational or, more generally, algebraic numbers. Liouville's theorem was strengthened and generalized in the works of many mathematicians. This made it possible to build new examples of transcendental numbers. So, K. Mahler showed that if is a non-constant polynomial that takes non-negative integer values \u200b\u200bfor all natural numbers, then for any natural number, where is the notation of a number in the base numeral system, is transcendental, but is not a Liouville number. For example, for and we get the following elegant result: the number

transcendental, but not a Liouville number.

In 1873 S. Hermit, using other ideas, proved the transcendence of the Neper number (the base of the natural logarithm):

Developing the ideas of Hermite, F. Lindemann in 1882 proved the transcendence of number, thereby putting an end to the ancient problem of squaring a circle: using a compass and a ruler, it is impossible to construct a square equal in size (that is, having the same area) to a given circle. More generally, Lindemann showed that for any algebraic number the number is transcendental. Equivalent formulation: for any algebraic number other than and, its natural logarithm is a transcendental number.

In 1900, at a congress of mathematicians in Paris, D. Hilbert, among 23 unsolved problems in mathematics, pointed to the following, formulated in a particular form by L. Euler:

Let be and - algebraic numbers, and transcendental? In particular, are the numbers transcendental and?

This problem can be reformulated in the following form, close to Euler's original formulation:

Let be and - algebraic numbers other than and, moreover, the ratio of their natural logarithms irrational. Will there be a number transcendental?

The first partial solution of the problem was obtained in 1929 by A.O. Gel'fond, who, in particular, proved the transcendence of number. In 1930, RO Kuz'min improved Gel'fond's method, in particular, he was able to prove the transcendence of a number. The complete solution of the Euler-Hilbert problem (in the affirmative sense) was obtained in 1934 independently by A.O. Gel'fond and T. Schneider.

A. Baker in 1966 generalized the theorems of Lindemann and Gelfond-Schneider, proving, in particular, the transcendence of the product of an arbitrary finite number of numbers of the form and with algebraic ones under natural restrictions.

In 1996. Yu.V. Nesterenko proved the algebraic independence of the values \u200b\u200bof the Eisenstein series and, in particular, of the numbers and. This means the transcendence of any number of the form, where a nonzero rational function with algebraic coefficients. For example, the sum of the series will be transcendental

In 1929-1930. K. Mahler in a series of works suggested new method proofs of the transcendence of the values \u200b\u200bof analytic functions satisfying functional equations of a certain form (later such functions were called Mahler functions).

The methods of the theory of transcendental numbers have found application in other branches of mathematics, in particular, in the theory of Diophantine equations.