Unsolved mathematical. Millennium goals. What did Grigory Perelman prove

  1. 1 Murad :

    We considered the equality Zn = Xn + Yn to be the Diophantus equation or Fermat's Great Theorem, and this is the solution of the equation (Zn- Xn) Xn = (Zn - Yn) Yn. Then Zn =-(Xn + Yn) is a solution to the equation (Zn + Xn) Xn = (Zn + Yn) Yn. These equations and solutions are related to the properties of integers and operations on them. So we don't know the properties of integers?! With such limited knowledge, we will not reveal the truth.
    Consider the solutions Zn = +(Xn + Yn) and Zn =-(Xn + Yn) when n = 1. Integers + Z are formed using 10 digits: 0, 1, 2, 3, 4, 5, 6, 7 , 8, 9. They are divisible by 2 whole numbers+X - even, last right digits: 0, 2, 4, 6, 8 and +Y - odd, last right digits: 1, 3, 5, 7, 9, i.e. + X = + Y. The number of Y = 5 - odd and X = 5 - even numbers is: Z = 10. Satisfies the equation: (Z - X) X = (Z - Y) Y, and the solution + Z = + X + Y= +(X + Y).
    Integers -Z consist of the union of -X for even and -Y for odd, and satisfies the equation:
    (Z + X) X = (Z + Y) Y, and the solution -Z = - X - Y = - (X + Y).
    If Z/X = Y or Z / Y = X, then Z = XY; Z / -X = -Y or Z / -Y = -X, then Z = (-X)(-Y). Division is checked by multiplication.
    Single-digit positive and negative numbers consist of 5 odd and 5 odd numbers.
    Consider the case n = 2. Then Z2 = X2 + Y2 is a solution to the equation (Z2 – X2) X2 = (Z2 – Y2) Y2 and Z2 = -(X2 + Y2) is a solution to the equation (Z2 + X2) X2 = (Z2 + Y2) Y2. We considered Z2 = X2 + Y2 to be the Pythagorean theorem, and then the solution Z2 = -(X2 + Y2) is the same theorem. We know that the diagonal of a square divides it into 2 parts, where the diagonal is the hypotenuse. Then the equalities are valid: Z2 = X2 + Y2, and Z2 = -(X2 + Y2) where X and Y are legs. And more solutions R2 = X2 + Y2 and R2 =- (X2 + Y2) are circles, centers are the origin of the square coordinate system and with radius R. They can be written as (5n)2 = (3n)2 + (4n)2 , where n are positive and negative integers, and are 3 consecutive numbers. Also solutions are 2-bit XY numbers that starts at 00 and ends at 99 and is 102 = 10x10 and count 1 century = 100 years.
    Consider solutions when n = 3. Then Z3 = X3 + Y3 are solutions of the equation (Z3 – X3) X3 = (Z3 – Y3) Y3.
    3-bit numbers XYZ starts at 000 and ends at 999 and is 103 = 10x10x10 = 1000 years = 10 centuries
    From 1000 cubes of the same size and color, you can make a rubik of about 10. Consider a rubik of the order +103=+1000 - red and -103=-1000 - blue. They consist of 103 = 1000 cubes. If we decompose and put the cubes in one row or on top of each other, without gaps, we get a horizontal or vertical segment of length 2000. Rubik is a large cube, covered with small cubes, starting from the size 1butto = 10st.-21, and you cannot add to it or subtract one cube.
    - (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9+10); + (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9+10);
    - (12 + 22 + 32 + 42 + 52 + 62 + 72 + 82 + 92+102); + (12 + 22 + 32 + 42 + 52 + 62 + 72 + 82 + 92+102);
    - (13 + 23 + 33 + 43 + 53 + 63 + 73 + 83 + 93+103); + (13 + 23 + 33 + 43 + 53 + 63 + 73 + 83 + 93+103).
    Each integer is 1. Add 1(ones) 9 + 9 =18, 10 + 9 =19, 10 +10 =20, 11 +10 =21, and the products:
    111111111 x 111111111 = 12345678987654321; 1111111111 x 111111111 = 123456789987654321.
    0111111111x1111111110= 0123456789876543210; 01111111111x1111111110= 01234567899876543210.
    These operations can be performed on 20-bit calculators.
    It is known that +(n3 - n) is always divisible by +6, and - (n3 - n) is divisible by -6. We know that n3 - n = (n-1)n(n+1). This is 3 consecutive numbers (n-1)n(n+1), where n is even, then divisible by 2, (n-1) and (n+1) odd, divisible by 3. Then (n-1) n(n+1) is always divisible by 6. If n=0, then (n-1)n(n+1)=(-1)0(+1), n=20, then(n-1)n (n+1)=(19)(20)(21).
    We know that 19 x 19 = 361. This means that one square is surrounded by 360 squares, and then one cube is surrounded by 360 cubes. The equality is fulfilled: 6 n - 1 + 6n. If n=60, then 360 - 1 + 360, and n=61, then 366 - 1 + 366.
    The following generalizations follow from the above statements:
    n5 - 4n = (n2-4) n (n2+4); n7 - 9n = (n3-9) n (n3+9); n9 –16 n= (n4-16) n (n4+16);
    0… (n-9) (n-8) (n-7) (n-6) (n-5) (n-4) (n-3) (n-2) (n-1)n(n +1) (n+2) (n+3) (n+4) (n+5) (n+6) (n+7) (n+8) (n+9)…2n
    (n+1) x (n+1) = 0123… (n-3) (n-2) (n-1) n (n+1) n (n-1) (n-2) (n-3 )…3210
    n! = 0123… (n-3) (n-2) (n-1) n; n! = n (n-1) (n-2) (n-3)…3210; (n+1)! =n! (n+1).
    0 +1 +2+3+…+ (n-3) + (n-2) + (n-1) +n=n (n+1)/2; n + (n-1) + (n-2) + (n-3) +…+3+2+1+0=n (n+1)/2;
    n (n+1)/2 + (n+1) + n (n+1)/2 = n (n+1) + (n+1) = (n+1) (n+1) = (n +1)2.
    If 0123… (n-3) (n-2) (n-1) n (n+1) n (n-1) (n-2) (n-3)…3210 x 11=
    = 013… (2n-5) (2n-3) (2n-1) (2n+1) (2n+1) (2n-1) (2n-3) (2n-5)…310.
    Any integer n is a power of 10, has: – n and +n, +1/ n and -1/ n, odd and even:
    - (n + n +…+ n) = -n2; – (n x n x…x n) = -nn; – (1/n + 1/n +…+ 1/n) = – 1; – (1/n x 1/n x…x1/n) = -n-n;
    + (n + n +…+ n) =+n2; + (n x n x…x n) = + nn; + (1/n +…+1/n) = + 1; + (1/n x 1/n x…x1/n) = + n-n.
    It is clear that if any integer is added to itself, then it will increase by 2 times, and the product will be a square: X = a, Y = a, X + Y = a + a = 2a; XY = a x a = a2. This was considered Vieta's theorem - a mistake!
    If we add and subtract the number b to the given number, then the sum does not change, but the product changes, for example:
    X \u003d a + b, Y \u003d a - b, X + Y \u003d a + b + a - b \u003d 2a; XY \u003d (a + b) x (a -b) \u003d a2-b2.
    X = a +√b, Y = a -√b, X+Y = a +√b + a – √b = 2a; XY \u003d (a + √b) x (a - √b) \u003d a2- b.
    X = a + bi, Y = a - bi, X + Y = a + bi + a - bi = 2a; XY \u003d (a + bi) x (a -bi) \u003d a2 + b2.
    X = a + √b i, Y = a - √bi, X+Y = a + √bi+ a - √bi =2a, XY = (a -√bi) x (a -√bi) = a2+b.
    If we put integer numbers instead of letters a and b, then we get paradoxes, absurdities, and mistrust of mathematics.

Pierre de Fermat, reading the "Arithmetic" of Diophantus of Alexandria and reflecting on its problems, had the habit of writing down the results of his reflections in the form of brief remarks in the margins of the book. Against the eighth problem of Diophantus in the margins of the book, Fermat wrote: " On the contrary, it is impossible to decompose neither a cube into two cubes, nor a bi-square into two bi-squares, and, in general, no power greater than a square into two powers with the same exponent. I have discovered a truly marvelous proof of this, but these margins are too narrow for it.» / E.T.Bell "Creators of Mathematics". M., 1979, p.69/. I bring to your attention an elementary proof of the farm theorem, which can be understood by any high school student who is fond of mathematics.

Let us compare Fermat's commentary on the Diophantine problem with the modern formulation of Fermat's great theorem, which has the form of an equation.
« The equation

x n + y n = z n(where n is an integer greater than two)

has no solutions in positive integers»

The comment is in a logical connection with the task, similar to the logical connection of the predicate with the subject. What is affirmed by the problem of Diophantus, on the contrary, is affirmed by Fermat's commentary.

Fermat's comment can be interpreted as follows: if quadratic equation with three unknowns has an infinite number of solutions on the set of all triples of Pythagorean numbers, then, conversely, an equation with three unknowns in a degree greater than the square

There is not even a hint of its connection with the Diophantine problem in the equation. His assertion requires proof, but it does not have a condition from which it follows that it has no solutions in positive integers.

The variants of the proof of the equation known to me are reduced to the following algorithm.

  1. The equation of Fermat's theorem is taken as its conclusion, the validity of which is verified with the help of proof.
  2. The same equation is called original the equation from which its proof must proceed.

The result is a tautology: If an equation has no solutions in positive integers, then it has no solutions in positive integers.". The proof of the tautology is obviously wrong and devoid of any meaning. But it is proved by contradiction.

  • An assumption is made that is the opposite of that stated by the equation to be proved. It should not contradict the original equation, but it does. To prove what is accepted without proof, and to accept without proof what is required to be proved, does not make sense.
  • Based on the accepted assumption, absolutely correct mathematical operations and actions are performed to prove that it contradicts the original equation and is false.

Therefore, for 370 years now, the proof of the equation of Fermat's Last Theorem has remained an impossible dream of specialists and lovers of mathematics.

I took the equation as the conclusion of the theorem, and the eighth problem of Diophantus and its equation as the condition of the theorem.


"If the equation x 2 + y 2 = z 2 (1) has an infinite set of solutions on the set of all triples of Pythagorean numbers, then, conversely, the equation x n + y n = z n , where n > 2 (2) has no solutions on the set of positive integers."

Proof.

BUT) Everyone knows that equation (1) has an infinite number of solutions on the set of all triples of Pythagorean numbers. Let us prove that no triple of Pythagorean numbers, which is a solution to equation (1), is a solution to equation (2).

Based on the law of reversibility of equality, the sides of equation (1) are interchanged. Pythagorean numbers (z, x, y) can be interpreted as the lengths of the sides right triangle, and the squares (x2, y2, z2) can be interpreted as the areas of squares built on its hypotenuse and legs.

We multiply the squares of equation (1) by an arbitrary height h :

z 2 h = x 2 h + y 2 h (3)

Equation (3) can be interpreted as the equality of the volume of a parallelepiped to the sum of the volumes of two parallelepipeds.

Let the height of three parallelepipeds h = z :

z 3 = x 2 z + y 2 z (4)

The volume of the cube is decomposed into two volumes of two parallelepipeds. We leave the volume of the cube unchanged, and reduce the height of the first parallelepiped to x and the height of the second parallelepiped will be reduced to y . The volume of a cube is greater than the sum of the volumes of two cubes:

z 3 > x 3 + y 3 (5)

On the set of triples of Pythagorean numbers ( x, y, z ) at n=3 there can be no solution to equation (2). Consequently, on the set of all triples of Pythagorean numbers, it is impossible to decompose a cube into two cubes.

Let in equation (3) the height of three parallelepipeds h = z2 :

z 2 z 2 = x 2 z 2 + y 2 z 2 (6)

The volume of a parallelepiped is decomposed into the sum of the volumes of two parallelepipeds.
We leave the left side of equation (6) unchanged. On its right side the height z2 reduce to X in the first term and up to at 2 in the second term.

Equation (6) turned into the inequality:

The volume of a parallelepiped is decomposed into two volumes of two parallelepipeds.

We leave the left side of equation (8) unchanged.
On the right side of the height zn-2 reduce to xn-2 in the first term and reduce to y n-2 in the second term. Equation (8) turns into the inequality:

z n > x n + y n (9)

On the set of triples of Pythagorean numbers, there cannot be a single solution of equation (2).

Consequently, on the set of all triples of Pythagorean numbers for all n > 2 equation (2) has no solutions.

Obtained "post miraculous proof", but only for triplets Pythagorean numbers. This is lack of evidence and the reason for the refusal of P. Fermat from him.

b) Let us prove that equation (2) has no solutions on the set of triples of non-Pythagorean numbers, which is the family of an arbitrarily taken triple of Pythagorean numbers z=13, x=12, y=5 and the family of an arbitrary triple of positive integers z=21, x=19, y=16

Both triplets of numbers are members of their families:

(13, 12, 12); (13, 12,11);…; (13, 12, 5) ;…; (13,7, 1);…; (13,1, 1) (10)
(21, 20, 20); (21, 20, 19);…;(21, 19, 16);…;(21, 1, 1) (11)

The number of members of the family (10) and (11) is equal to half the product of 13 by 12 and 21 by 20, i.e. 78 and 210.

Each member of the family (10) contains z = 13 and variables X and at 13 > x > 0 , 13 > y > 0 1

Each member of the family (11) contains z = 21 and variables X and at , which take integer values 21 > x >0 , 21 > y > 0 . The variables decrease sequentially by 1 .

The triples of numbers of the sequence (10) and (11) can be represented as a sequence of inequalities of the third degree:

13 3 < 12 3 + 12 3 ;13 3 < 12 3 + 11 3 ;…; 13 3 < 12 3 + 8 3 ; 13 3 > 12 3 + 7 3 ;…; 13 3 > 1 3 + 1 3
21 3 < 20 3 + 20 3 ; 21 3 < 20 3 + 19 3 ; …; 21 3 < 19 3 + 14 3 ; 21 3 > 19 3 + 13 3 ;…; 21 3 > 1 3 + 1 3

and in the form of inequalities of the fourth degree:

13 4 < 12 4 + 12 4 ;…; 13 4 < 12 4 + 10 4 ; 13 4 > 12 4 + 9 4 ;…; 13 4 > 1 4 + 1 4
21 4 < 20 4 + 20 4 ; 21 4 < 20 4 + 19 4 ; …; 21 4 < 19 4 + 16 4 ;…; 21 4 > 1 4 + 1 4

The correctness of each inequality is verified by raising the numbers to the third and fourth powers.

The cube of a larger number cannot be decomposed into two cubes of smaller numbers. It is either less than or greater than the sum of the cubes of the two smaller numbers.

The bi-square of a larger number cannot be decomposed into two bi-squares of smaller numbers. It is either less than or greater than the sum of the bi-squares of smaller numbers.

As the exponent increases, all inequalities, except for the leftmost inequality, have the same meaning:

Inequalities, they all have the same meaning: the degree of the larger number is greater than the sum of the degrees of the smaller two numbers with the same exponent:

13n > 12n + 12n ; 13n > 12n + 11n ;…; 13n > 7n + 4n ;…; 13n > 1n + 1n (12)
21n > 20n + 20n ; 21n > 20n + 19n ;…; ;…; 21n > 1n + 1n (13)

The leftmost term of sequences (12) (13) is the weakest inequality. Its correctness determines the correctness of all subsequent inequalities of the sequence (12) for n > 8 and sequence (13) for n > 14 .

There can be no equality among them. An arbitrary triple of positive integers (21,19,16) is not a solution to equation (2) of Fermat's Last Theorem. If an arbitrary triple of positive integers is not a solution to the equation, then the equation has no solutions on the set of positive integers, which was to be proved.

WITH) Fermat's commentary on the Diophantus problem states that it is impossible to decompose " in general, no power greater than the square, two powers with the same exponent».

Kisses a power greater than a square cannot really be decomposed into two powers with the same exponent. I don't kiss a power greater than the square can be decomposed into two powers with the same exponent.

Any randomly chosen triple of positive integers (z, x, y) may belong to a family, each member of which consists of a constant number z and two numbers less than z . Each member of the family can be represented in the form of an inequality, and all the resulting inequalities can be represented as a sequence of inequalities:

z n< (z — 1) n + (z — 1) n ; z n < (z — 1) n + (z — 2) n ; …; z n >1n + 1n (14)

The sequence of inequalities (14) begins with inequalities whose left side is less than the right side and ends with inequalities whose right side is less than the left side. With increasing exponent n > 2 the number of inequalities on the right side of sequence (14) increases. With an exponent n=k all the inequalities of the left side of the sequence change their meaning and take on the meaning of the inequalities of the right side of the inequalities of the sequence (14). As a result of the increase in the exponent of all inequalities, the left side is greater than the right side:

z k > (z-1) k + (z-1) k ; z k > (z-1) k + (z-2) k ;…; zk > 2k + 1k ; zk > 1k + 1k (15)

With a further increase in the exponent n>k none of the inequalities changes its meaning and does not turn into equality. On this basis, it can be argued that any arbitrarily taken triple of positive integers (z, x, y) at n > 2 , z > x , z > y

In an arbitrary triple of positive integers z can be an arbitrarily large natural number. For all natural numbers not greater than z , Fermat's Last Theorem is proved.

D) No matter how big the number z , in the natural series of numbers before it there is a large but finite set of integers, and after it there is an infinite set of integers.

Let us prove that the entire infinite set of natural numbers greater than z , form triples of numbers that are not solutions to the equation of Fermat's Last Theorem, for example, an arbitrary triple of positive integers (z+1,x,y) , wherein z + 1 > x and z + 1 > y for all values ​​of the exponent n > 2 is not a solution to the equation of Fermat's Last Theorem.

A randomly chosen triple of positive integers (z + 1, x, y) may belong to a family of triples of numbers, each member of which consists of a constant number z + 1 and two numbers X and at , taking different values, smaller z + 1 . Family members can be represented as inequalities whose constant left side is less than, or greater than, the right side. The inequalities can be arranged in order as a sequence of inequalities:

With a further increase in the exponent n>k to infinity, none of the inequalities in the sequence (17) changes its meaning and does not become an equality. In sequence (16), the inequality formed from an arbitrarily taken triple of positive integers (z + 1, x, y) , can be in its right side in the form (z + 1) n > x n + y n or be on its left side in the form (z+1)n< x n + y n .

In any case, the triple of positive integers (z + 1, x, y) at n > 2 , z + 1 > x , z + 1 > y in sequence (16) is an inequality and cannot be an equality, i.e., cannot be a solution to the equation of Fermat's Last Theorem.

It is easy and simple to understand the origin of the sequence of power inequalities (16), in which the last inequality of the left side and the first inequality of the right side are inequalities of the opposite sense. On the contrary, it is not easy and difficult for schoolchildren, high school students and high school students to understand how a sequence of inequalities (17) is formed from a sequence of inequalities (16), in which all inequalities have the same meaning.

In sequence (16), increasing the integer degree of inequalities by 1 turns the last inequality on the left side into the first inequality of the opposite meaning on the right side. Thus, the number of inequalities on the ninth side of the sequence decreases, while the number of inequalities on the right side increases. Between the last and first power inequalities of the opposite meaning, there is a power equality without fail. Its degree cannot be an integer, since there are only non-integer numbers between two consecutive natural numbers. The power equality of a non-integer degree, according to the condition of the theorem, cannot be considered a solution to equation (1).

If in the sequence (16) we continue to increase the degree by 1 unit, then the last inequality of its left side will turn into the first inequality of the opposite meaning of the right side. As a result, there will be no inequalities on the left side and only inequalities on the right side, which will be a sequence of increasing power inequalities (17). A further increase in their integer degree by 1 unit only strengthens its power inequalities and categorically excludes the possibility of the appearance of equality in an integer degree.

Therefore, in general, no integer power of a natural number (z+1) of the sequence of power inequalities (17) can be decomposed into two integer powers with the same exponent. Therefore, equation (1) has no solutions on an infinite set of natural numbers, which was to be proved.

Therefore, Fermat's Last Theorem is proved in all generality:

  • in section A) for all triplets (z, x, y) Pythagorean numbers (Fermat's discovery is a truly miraculous proof),
  • in section C) for all members of the family of any triple (z, x, y) pythagorean numbers,
  • in section C) for all triplets of numbers (z, x, y) , not large numbers z
  • in section D) for all triples of numbers (z, x, y) natural series of numbers.

Changes were made on 05.09.2010

Which theorems can and which cannot be proven by contradiction

The Explanatory Dictionary of Mathematical Terms defines proof by contradiction of a theorem opposite to the inverse theorem.

“Proof by contradiction is a method of proving a theorem (sentence), which consists in proving not the theorem itself, but its equivalent (equivalent), opposite inverse (reverse to opposite) theorem. Proof by contradiction is used whenever the direct theorem is difficult to prove, but the opposite inverse is easier. When proving by contradiction, the conclusion of the theorem is replaced by its negation, and by reasoning one arrives at the negation of the condition, i.e. to a contradiction, to the opposite (the opposite of what is given; this reduction to absurdity proves the theorem.

Proof by contradiction is very often used in mathematics. The proof by contradiction is based on the law of the excluded middle, which consists in the fact that of the two statements (statements) A and A (negation of A), one of them is true and the other is false./ Explanatory dictionary of mathematical terms: A guide for teachers / O. V. Manturov [and others]; ed. V. A. Ditkina.- M.: Enlightenment, 1965.- 539 p.: ill.-C.112/.

It would not be better to declare openly that the method of proof by contradiction is not a mathematical method, although it is used in mathematics, that it is a logical method and belongs to logic. Is it valid to say that proof by contradiction is "used whenever a direct theorem is difficult to prove", when in fact it is used if, and only if, there is no substitute for it.

The characteristic of the relationship between the direct and inverse theorems also deserves special attention. “An inverse theorem for a given theorem (or to a given theorem) is a theorem in which the condition is the conclusion, and the conclusion is the condition of the given theorem. This theorem in relation to the converse theorem is called the direct theorem (initial). At the same time, the converse theorem to the converse theorem will be the given theorem; therefore, the direct and inverse theorems are called mutually inverse. If the direct (given) theorem is true, then the converse theorem is not always true. For example, if a quadrilateral is a rhombus, then its diagonals are mutually perpendicular (direct theorem). If the diagonals in a quadrilateral are mutually perpendicular, then the quadrilateral is a rhombus - this is not true, i.e., the converse theorem is not true./ Explanatory dictionary of mathematical terms: A guide for teachers / O. V. Manturov [and others]; ed. V. A. Ditkina.- M.: Enlightenment, 1965.- 539 p.: ill.-C.261 /.

This characteristic The relation of direct and inverse theorems does not take into account the fact that the condition of the direct theorem is taken as given, without proof, so that its correctness is not guaranteed. The condition of the inverse theorem is not taken as given, since it is the conclusion of the proven direct theorem. Its correctness is confirmed by the proof of the direct theorem. This essential logical difference between the conditions of the direct and inverse theorems turns out to be decisive in the question of which theorems can and which cannot be proved by the logical method from the contrary.

Let's assume that there is a direct theorem in mind, which can be proved by the usual mathematical method, but it is difficult. Let us formulate it in general form in short form So: from BUT should E . Symbol BUT has the meaning given condition theorem accepted without proof. Symbol E is the conclusion of the theorem to be proved.

We will prove the direct theorem by contradiction, logical method. The logical method proves a theorem that has not mathematical condition, and logical condition. It can be obtained if the mathematical condition of the theorem from BUT should E , supplement with the opposite condition from BUT do not do it E .

As a result, a logical contradictory condition of the new theorem was obtained, which includes two parts: from BUT should E and from BUT do not do it E . The resulting condition of the new theorem corresponds to the logical law of the excluded middle and corresponds to the proof of the theorem by contradiction.

According to the law, one part of the contradictory condition is false, another part is true, and the third is excluded. The proof by contradiction has its own task and goal to establish exactly which part of the two parts of the condition of the theorem is false. As soon as the false part of the condition is determined, it will be established that the other part is the true part, and the third is excluded.

According to the explanatory dictionary of mathematical terms, “proof is reasoning, during which the truth or falsity of any statement (judgment, statement, theorem) is established”. Proof contrary there is a discussion in the course of which it is established falsity(absurdity) of the conclusion that follows from false conditions of the theorem being proved.

Given: from BUT should E and from BUT do not do it E .

Prove: from BUT should E .

Proof: The logical condition of the theorem contains a contradiction that requires its resolution. The contradiction of the condition must find its resolution in the proof and its result. The result turns out to be false if the reasoning is flawless and infallible. The reason for a false conclusion with logically correct reasoning can only be a contradictory condition: from BUT should E and from BUT do not do it E .

There is no shadow of a doubt that one part of the condition is false, and the other in this case is true. Both parts of the condition have the same origin, are accepted as given, assumed, equally possible, equally admissible, etc. In the course of logical reasoning, not a single logical feature has been found that would distinguish one part of the condition from the other. Therefore, to the same extent, from BUT should E and maybe from BUT do not do it E . Statement from BUT should E may be false, then the statement from BUT do not do it E will be true. Statement from BUT do not do it E may be false, then the statement from BUT should E will be true.

Therefore, it is impossible to prove the direct theorem by contradiction method.

Now we will prove the same direct theorem by the usual mathematical method.

Given: BUT .

Prove: from BUT should E .

Proof.

1. From BUT should B

2. From B should AT (according to the previously proved theorem)).

3. From AT should G (according to the previously proved theorem).

4. From G should D (according to the previously proved theorem).

5. From D should E (according to the previously proved theorem).

Based on the law of transitivity, from BUT should E . The direct theorem is proved by the usual method.

Let the proven direct theorem have a correct converse theorem: from E should BUT .

Let's prove it by ordinary mathematical method. The proof of the inverse theorem can be expressed in symbolic form as an algorithm of mathematical operations.

Given: E

Prove: from E should BUT .

Proof.

1. From E should D

2. From D should G (by the previously proved inverse theorem).

3. From G should AT (by the previously proved inverse theorem).

4. From AT do not do it B (the converse is not true). That's why from B do not do it BUT .

In this situation, it makes no sense to continue the mathematical proof of the inverse theorem. The reason for the situation is logical. It is impossible to replace an incorrect inverse theorem with anything. Therefore, this inverse theorem cannot be proved by the usual mathematical method. All hope is to prove this inverse theorem by contradiction.

In order to prove it by contradiction, it is required to replace its mathematical condition with a logical contradictory condition, which in its meaning contains two parts - false and true.

Inverse theorem claims: from E do not do it BUT . Her condition E , from which follows the conclusion BUT , is the result of proving the direct theorem by the usual mathematical method. This condition must be retained and supplemented with the statement from E should BUT . As a result of the addition, a contradictory condition of the new inverse theorem is obtained: from E should BUT and from E do not do it BUT . Based on this logically contradictory condition, the converse theorem can be proved by the correct logical reasoning only, and only, logical opposite method. In a proof by contradiction, any mathematical actions and operations are subordinate to logical ones and therefore do not count.

In the first part of the contradictory statement from E should BUT condition E was proved by the proof of the direct theorem. In the second part from E do not do it BUT condition E was assumed and accepted without proof. One of them is false and the other is true. It is required to prove which of them is false.

We prove with the correct logical reasoning and find that its result is a false, absurd conclusion. The reason for a false logical conclusion is the contradictory logical condition of the theorem, which contains two parts - false and true. The false part can only be a statement from E do not do it BUT , wherein E accepted without proof. This is what distinguishes it from E statements from E should BUT , which is proved by the proof of the direct theorem.

Therefore, the statement is true: from E should BUT , which was to be proved.

Conclusion: only that converse theorem is proved by the logical method from the contrary, which has a direct theorem proved by the mathematical method and which cannot be proved by the mathematical method.

The conclusion obtained acquires an exceptional importance in relation to the method of proof by contradiction of Fermat's great theorem. The overwhelming majority of attempts to prove it are based not on the usual mathematical method, but on the logical method of proving by contradiction. The proof of Fermat Wiles' Great Theorem is no exception.

Dmitry Abrarov in his article "Fermat's Theorem: the Phenomenon of Wiles' Proofs" published a commentary on the proof of Fermat's Last Theorem by Wiles. According to Abrarov, Wiles proves Fermat's Last Theorem with the help of a remarkable finding by the German mathematician Gerhard Frey (b. 1944) relating a potential solution to Fermat's equation x n + y n = z n , where n > 2 , with another completely different equation. This new equation is given by a special curve (called the Frey elliptic curve). The Frey curve is given by a very simple equation:
.

“It was precisely Frey who compared to every solution (a, b, c) Fermat's equation, that is, numbers satisfying the relation a n + b n = c n the above curve. In this case, Fermat's Last Theorem would follow."(Quote from: Abrarov D. "Fermat's Theorem: the phenomenon of Wiles proof")

In other words, Gerhard Frey suggested that the equation of Fermat's Last Theorem x n + y n = z n , where n > 2 , has solutions in positive integers. The same solutions are, by Frey's assumption, the solutions of his equation
y 2 + x (x - a n) (y + b n) = 0 , which is given by its elliptic curve.

Andrew Wiles accepted this remarkable discovery of Frey and, with its help, through mathematical method proved that this finding, that is, Frey's elliptic curve, does not exist. Therefore, there is no equation and its solutions that are given by a non-existent elliptic curve. Therefore, Wiles should have concluded that there is no equation of Fermat's Last Theorem and Fermat's Theorem itself. However, he takes the more modest conclusion that the equation of Fermat's Last Theorem has no solutions in positive integers.

It may be an undeniable fact that Wiles accepted an assumption that is directly opposite in meaning to what is stated by Fermat's Last Theorem. It obliges Wiles to prove Fermat's Last Theorem by contradiction. Let's follow his example and see what happens from this example.

Fermat's Last Theorem states that the equation x n + y n = z n , where n > 2 , has no solutions in positive integers.

According to the logical method of proof by contradiction, this statement is preserved, accepted as given without proof, and then supplemented with a statement opposite in meaning: the equation x n + y n = z n , where n > 2 , has solutions in positive integers.

The hypothesized statement is also accepted as given, without proof. Both statements, considered from the point of view of the basic laws of logic, are equally admissible, equal in rights and equally possible. By correct reasoning, it is required to establish which of them is false, in order to then establish that the other statement is true.

Correct reasoning ends with a false, absurd conclusion, the logical cause of which can only be a contradictory condition of the theorem being proved, which contains two parts of a directly opposite meaning. They were the logical cause of the absurd conclusion, the result of proof by contradiction.

However, in the course of logically correct reasoning, not a single sign was found by which it would be possible to establish which particular statement is false. It can be a statement: the equation x n + y n = z n , where n > 2 , has solutions in positive integers. On the same basis, it can be the statement: the equation x n + y n = z n , where n > 2 , has no solutions in positive integers.

As a result of the reasoning, there can be only one conclusion: Fermat's Last Theorem cannot be proven by contradiction.

It would be a very different matter if Fermat's Last Theorem were inverse theorem, which has a direct theorem proved by the usual mathematical method. In this case, it could be proven by contradiction. And since it is a direct theorem, its proof must be based not on the logical method of proof by contradiction, but on the usual mathematical method.

According to D. Abrarov, Academician V. I. Arnold, the most famous contemporary Russian mathematician, reacted to Wiles's proof "actively skeptical". The academician stated: “this is not real mathematics – real mathematics is geometric and has strong links with physics.”

By contradiction, it is impossible to prove either that the equation of Fermat's Last Theorem has no solutions, or that it has solutions. Wiles' mistake is not mathematical, but logical - the use of proof by contradiction where its use does not make sense and does not prove Fermat's Last Theorem.

Fermat's Last Theorem is not proved even with the help of the usual mathematical method, if it is given: the equation x n + y n = z n , where n > 2 , has no solutions in positive integers, and if it is required to prove in it: the equation x n + y n = z n , where n > 2 , has no solutions in positive integers. In this form, there is not a theorem, but a tautology devoid of meaning.

Note. My BTF proof was discussed on one of the forums. One of the participants in Trotil, a specialist in number theory, made the following authoritative statement entitled: "A brief retelling of what Mirgorodsky did." I quote it verbatim:

« BUT. He proved that if z 2 \u003d x 2 + y , then z n > x n + y n . This is a well-known and quite obvious fact.

AT. He took two triples - Pythagorean and non-Pythagorean and showed by simple enumeration that for a specific, specific family of triples (78 and 210 pieces) BTF is performed (and only for it).

WITH. And then the author omitted the fact that from < in a subsequent degree may be = , not only > . A simple counterexample is the transition n=1 in n=2 in a Pythagorean triple.

D. This point does not contribute anything essential to the BTF proof. Conclusion: BTF has not been proven.”

I will consider his conclusion point by point.

BUT. In it, the BTF is proved for the entire infinite set of triples of Pythagorean numbers. Proven by a geometric method, which, as I believe, was not discovered by me, but rediscovered. And it was opened, as I believe, by P. Fermat himself. Fermat might have had this in mind when he wrote:

"I have discovered a truly marvelous proof of this, but these margins are too narrow for it." This assumption of mine is based on the fact that in the Diophantine problem, against which, in the margins of the book, Fermat wrote, we are talking about solutions to the Diophantine equation, which are triples of Pythagorean numbers.

An infinite set of triples of Pythagorean numbers are solutions to the Diophatian equation, and in Fermat's theorem, on the contrary, none of the solutions can be a solution to the equation of Fermat's theorem. And Fermat's truly miraculous proof has a direct bearing on this fact. Later, Fermat could extend his theorem to the set of all natural numbers. On the set of all natural numbers, BTF does not belong to the "set of exceptionally beautiful theorems". This is my assumption, which can neither be proved nor disproved. It can be both accepted and rejected.

AT. In this paragraph, I prove that both the family of an arbitrarily taken Pythagorean triple of numbers and the family of an arbitrarily taken non-Pythagorean triple of numbers BTF is satisfied. This is a necessary, but insufficient and intermediate link in my proof of the BTF. The examples I have taken of the family of a triple of Pythagorean numbers and the family of a triple of non-Pythagorean numbers have the meaning of specific examples that presuppose and do not exclude the existence of similar other examples.

Trotil's statement that I "showed by simple enumeration that for a specific, specific family of triples (78 and 210 pieces) BTF is fulfilled (and only for it) is without foundation. He cannot refute the fact that I could just as well take other examples of Pythagorean and non-Pythagorean triples to get a specific family of one and the other triple.

Whatever pair of triples I take, checking their suitability for solving the problem can be carried out, in my opinion, only by the method of "simple enumeration". Any other method is not known to me and is not required. If he did not like Trotil, then he should have suggested another method, which he does not. Without offering anything in return, to condemn the "simple search", which in this case irreplaceable, incorrect.

WITH. I omitted = between< и < на основании того, что в доказательстве БТФ рассматривается уравнение z 2 \u003d x 2 + y (1), in which the degree n > 2 — whole positive number. From the equality between the inequalities it follows obligatory consideration of equation (1) with a non-integer value of the degree n > 2 . Trotil counting compulsory consideration of equality between inequalities, actually considers necessary in the BTF proof, consideration of equation (1) with non-integer degree value n > 2 . I did this for myself and found that equation (1) with non-integer degree value n > 2 has a solution of three numbers: z, (z-1), (z-1) with a non-integer exponent.

Fermat's interest in mathematics was somehow unexpected and rather adulthood. In 1629, a Latin translation of Pappus's work, containing a brief summary of Apollonius' results on the properties of conic sections, fell into his hands. Fermat, a polyglot, an expert in law and ancient philology, suddenly sets out to completely restore the course of reasoning of the famous scientist. With the same success, a modern lawyer can try to independently reproduce all the proofs from a monograph from problems, say, of algebraic topology. However, the unthinkable enterprise is crowned with success. Moreover, by delving into geometric constructions of the ancients, he makes an amazing discovery: in order to find the maxima and minima of the areas of figures, ingenious drawings are not needed. It is always possible to compose and solve some simple algebraic equation, the roots of which determine the extremum. He came up with an algorithm that would become the basis of differential calculus.

He quickly moved on. He found sufficient conditions the existence of maxima, learned to determine the inflection points, drew tangents to all known curves of the second and third order. A few more years, and he finds a new purely algebraic method for finding quadratures for parabolas and hyperbolas of arbitrary order (that is, integrals of functions of the form y p = Cx q and y p x q \u003d C), calculates areas, volumes, moments of inertia of bodies of revolution. It was a real breakthrough. Feeling this, Fermat begins to seek communication with the mathematical authorities of the time. He is confident and longs for recognition.

In 1636 he wrote the first letter to His Reverend Marin Mersenne: “Holy Father! I am extremely grateful to you for the honor you have done me by giving me the hope that we will be able to talk in writing; ...I will be very glad to hear from you about all the new treatises and books on Mathematics that have appeared in the last five or six years. ... I also found many analytical methods for various problems, both numerical and geometric, for which Vieta's analysis is insufficient. All this I will share with you whenever you want, and, moreover, without any arrogance, from which I am freer and more distant than any other person in the world.

Who is Father Mersenne? This is a Franciscan monk, a scientist of modest talents and a wonderful organizer, who for 30 years headed the Parisian mathematical circle, which became the true center of French science. Subsequently, the Mersenne circle, by decree of Louis XIV, will be transformed into the Paris Academy of Sciences. Mersenne tirelessly carried on a huge correspondence, and his cell in the monastery of the Order of the Minims on the Royal Square was a kind of "post office for all the scientists of Europe, from Galileo to Hobbes." Correspondence then replaced scientific journals, which appeared much later. Meetings at Mersenne took place weekly. The core of the circle was made up of the most brilliant natural scientists of that time: Robertville, Pascal Father, Desargues, Midorge, Hardy and, of course, the famous and universally recognized Descartes. Rene du Perron Descartes (Cartesius), a mantle of nobility, two family estates, the founder of Cartesianism, the “father” of analytic geometry, one of the founders of new mathematics, as well as Mersenne’s friend and comrade at the Jesuit College. This wonderful man will be Fermat's nightmare.

Mersenne found Fermat's results interesting enough to bring the provincial into his elite club. The farm immediately strikes up a correspondence with many members of the circle and literally falls asleep with letters from Mersenne himself. In addition, he sends completed manuscripts to the court of pundits: “Introduction to flat and solid places”, and a year later - “The method of finding maxima and minima” and “Answers to B. Cavalieri's questions”. What Fermat expounded was absolutely new, but the sensation did not take place. Contemporaries did not flinch. They didn’t understand much, but they found unambiguous indications that Fermat borrowed the idea of ​​the maximization algorithm from Johannes Kepler’s treatise with the funny title “The New Stereometry of Wine Barrels”. Indeed, in Kepler's reasoning there are phrases like “The volume of the figure is greatest if on both sides of the place the greatest value the decrease is at first insensitive.” But the idea of ​​a small increment of a function near an extremum was not at all in the air. The best analytical minds of that time were not ready for manipulations with small quantities. The fact is that at that time algebra was considered a kind of arithmetic, that is, mathematics of the second grade, a primitive improvised tool developed for the needs of base practice (“only merchants count well”). Tradition prescribed to adhere to purely geometric methods of proofs, dating back to ancient mathematics. Fermat was the first to understand that infinitesimal quantities can be added and reduced, but it is rather difficult to represent them as segments.

It took almost a century for Jean d'Alembert to admit in his famous Encyclopedia: Fermat was the inventor of the new calculus. It is with him that we meet the first application of differentials for finding tangents.” At the end of the 18th century, Joseph Louis Comte de Lagrange spoke out even more clearly: “But the geometers - Fermat's contemporaries - did not understand this new kind of calculus. They saw only special cases. And this invention, which appeared shortly before Descartes' Geometry, remained fruitless for forty years. Lagrange is referring to 1674, when Isaac Barrow's "Lectures" were published, covering Fermat's method in detail.

Among other things, it quickly became clear that Fermat was more inclined to formulate new problems than to humbly solve the problems proposed by the meters. In the era of duels, the exchange of tasks between pundits was generally accepted as a form of clarifying issues related to chain of command. However, the Farm clearly does not know the measure. Each of his letters is a challenge containing dozens of complex unsolved problems, and on the most unexpected topics. Here is an example of his style (addressed to Frenicle de Bessy): “Item, what least square, which, when reduced by 109 and added to one, will give a square? If you don't send me common solution, then send the quotient for these two numbers, which I chose small, so as not to make it very difficult for you. After I get your answer, I will suggest some other things to you. It is clear, without special reservations, that in my proposal it is required to find integers, since in the case fractional numbers the most insignificant arithmetician could reach the goal.” Fermat often repeated himself, formulating the same questions several times, and openly bluffed, claiming that he had an unusually elegant solution to the proposed problem. There were no direct errors. Some of them were noticed by contemporaries, and some of the insidious statements misled readers for centuries.

Mersenne's circle reacted adequately. Only Robertville, the only member of the circle who had problems with the origin, maintains a friendly tone of letters. The good shepherd Father Mersenne tried to reason with the "Toulouse impudent". But Farm does not intend to make excuses: “Reverend Father! You write to me that the posing of my impossible problems angered and cooled Messrs. Saint-Martin and Frenicle, and that this was the reason for the termination of their letters. However, I want to object to them that what seems impossible at first is actually not, and that there are many problems that, as Archimedes said...” etc.

However, Farm is disingenuous. It was to Frenicle that he sent the problem of finding a right-angled triangle with integer sides whose area is equal to the square of an integer. He sent it, although he knew that the problem obviously had no solution.

The most hostile position towards Fermat was taken by Descartes. In his letter to Mersenne dated 1938 we read: “because I found out that this is the same person who had previously tried to refute my “Dioptric”, and since you informed me that he sent it after he had read my “Geometry ” and in surprise that I did not find the same thing, i.e. (as I have reason to interpret it) sent it with the aim of entering into rivalry and showing that he knows more about it than I do, and since more of your letters, I learned that he had a reputation as a very knowledgeable geometer, then I consider myself obliged to answer him. Descartes will later solemnly designate his answer as “the small trial of Mathematics against Mr. Fermat”.

It is easy to understand what infuriated the eminent scientist. First, in Fermat's reasoning, coordinate axes and the representation of numbers by segments constantly appear - a device that Descartes comprehensively develops in his just published "Geometry". Fermat comes to the idea of ​​replacing the drawing with calculations on his own, in some ways even more consistent than Descartes. Secondly, Fermat brilliantly demonstrates the effectiveness of his method of finding minima on the example of the problem of the shortest path of a light beam, refining and supplementing Descartes with his "Dioptric".

The merits of Descartes as a thinker and innovator are enormous, but let's open the modern "Mathematical Encyclopedia" and look at the list of terms associated with his name: "Cartesian coordinates" (Leibniz, 1692), "Cartesian sheet", "Descartes ovals". None of his arguments went down in history as Descartes' Theorem. Descartes is primarily an ideologist: he is the founder of a philosophical school, he forms concepts, improves the system of letter designations, but there are few new ones in his creative heritage. specific techniques. In contrast, Pierre Fermat writes little, but on any occasion he can come up with a lot of witty mathematical tricks (see ibid. "Fermat's Theorem", "Fermat's Principle", "Fermat's method of infinite descent"). They probably quite rightly envied each other. The collision was inevitable. With the Jesuit mediation of Mersenne, a war broke out that lasted two years. However, Mersenne turned out to be right before history here too: the fierce battle between the two titans, their tense, to put it mildly, polemic contributed to the understanding of the key concepts of mathematical analysis.

Fermat is the first to lose interest in the discussion. Apparently, he spoke directly with Descartes and never again offended his opponent. In one of his last works, "Synthesis for refraction", the manuscript of which he sent to de la Chaumbra, Fermat mentions "the most learned Descartes" through the word and in every possible way emphasizes his priority in matters of optics. Meanwhile, it was this manuscript that contained the description of the famous "Fermat's principle", which provides an exhaustive explanation of the laws of reflection and refraction of light. Curtseys to Descartes in a work of this level were completely unnecessary.

What happened? Why did Fermat, putting aside pride, went to reconciliation? Reading Fermat's letters of those years (1638 - 1640), one can assume the simplest: during this period, his scientific interests changed drastically. He abandons the fashionable cycloid, ceases to be interested in tangents and areas, and for a long 20 years forgets about his method of finding the maximum. Having great merits in the mathematics of the continuous, Fermat completely immerses himself in the mathematics of the discrete, leaving the hateful geometric drawings to his opponents. Numbers are his new passion. As a matter of fact, the entire "Theory of Numbers", as an independent mathematical discipline, owes its birth entirely to the life and work of Fermat.

<…>After Fermat's death, his son Samuel published in 1670 a copy of Arithmetic belonging to his father under the title "Six books of arithmetic by the Alexandrian Diophantus with comments by L. G. Basche and remarks by P. de Fermat, Senator of Toulouse." The book also included some letters from Descartes and full text works by Jacques de Bigly "A New Discovery in the Art of Analysis", written on the basis of Fermat's letters. The publication was an incredible success. An unprecedented bright world opened up before the astonished specialists. The unexpectedness, and most importantly, the accessibility, democratic nature of Fermat's number-theoretic results gave rise to a lot of imitations. At that time, few people understood how the area of ​​a parabola was calculated, but every student could understand the formulation of Fermat's Last Theorem. A real hunt began for the unknown and lost letters of the scientist. Until the end of the XVII century. Every word of his that was found was published and republished. But the turbulent history of the development of Fermat's ideas was just beginning.

Unsolvable problems are 7 most interesting mathematical problems. Each of them was proposed at one time by well-known scientists, as a rule, in the form of hypotheses. For many decades, mathematicians all over the world have been racking their brains over their solution. Those who succeed will be rewarded with a million US dollars offered by the Clay Institute.

Clay Institute

This name is a private non-profit organization headquartered in Cambridge, Massachusetts. It was founded in 1998 by Harvard mathematician A. Jeffey and businessman L. Clay. The aim of the Institute is to popularize and develop mathematical knowledge. To achieve this, the organization gives awards to scientists and sponsors promising research.

At the beginning of the 21st century, the Clay Mathematical Institute offered a prize to those who solve problems that are known as the most difficult unsolvable problems, calling their list Millennium Prize Problems. From the "Hilbert List" it included only the Riemann hypothesis.

Millennium Challenges

The Clay Institute list originally included:

  • the Hodge cycle hypothesis;
  • equations of quantum theory Yang-Mills;
  • the Poincaré hypothesis;
  • the problem of equality of classes P and NP;
  • the Riemann hypothesis;
  • on the existence and smoothness of its solutions;
  • Birch-Swinnerton-Dyer problem.

These open mathematical problems are of great interest because they can have many practical implementations.

What did Grigory Perelman prove

In 1900, the famous philosopher Henri Poincaré suggested that any simply connected compact 3-manifold without boundary is homeomorphic to a 3-sphere. Its proof in the general case was not found for a century. Only in 2002-2003, the St. Petersburg mathematician G. Perelman published a number of articles with a solution to the Poincaré problem. They had the effect of an exploding bomb. In 2010, the Poincaré hypothesis was excluded from the list of "Unsolved Problems" of the Clay Institute, and Perelman himself was offered to receive a considerable remuneration due to him, which the latter refused without explaining the reasons for his decision.

The most understandable explanation of what the Russian mathematician managed to prove can be given by imagining that a rubber disk is pulled onto a donut (torus), and then they try to pull the edges of its circumference into one point. Obviously this is not possible. Another thing, if you make this experiment with a ball. In this case, a seemingly three-dimensional sphere, resulting from a disk, the circumference of which was pulled to a point by a hypothetical cord, will be three-dimensional in the understanding ordinary person, but mathematically two-dimensional.

Poincaré suggested that a three-dimensional sphere is the only three-dimensional "object" whose surface can be contracted to a single point, and Perelman was able to prove this. Thus, the list of "Unsolvable problems" today consists of 6 problems.

Yang-Mills theory

This mathematical problem was proposed by its authors in 1954. The scientific formulation of the theory is as follows: for any simple compact gauge group, the quantum spatial theory created by Yang and Mills exists, and at the same time has a zero mass defect.

Speaking in a language understandable to an ordinary person, the interactions between natural objects (particles, bodies, waves, etc.) are divided into 4 types: electromagnetic, gravitational, weak and strong. For many years, physicists have been trying to create a general field theory. It should become a tool for explaining all these interactions. Yang-Mills theory is a mathematical language with which it became possible to describe 3 of the 4 main forces of nature. It does not apply to gravity. Therefore, it cannot be considered that Yang and Mills succeeded in creating a field theory.

In addition, the nonlinearity of the proposed equations makes them extremely difficult to solve. For small coupling constants, they can be approximately solved in the form of a series of perturbation theory. However, it is not yet clear how these equations can be solved with strong coupling.

Navier-Stokes equations

These expressions describe processes such as air flows, fluid flow, and turbulence. For some special cases, analytical solutions of the Navier-Stokes equation have already been found, but so far no one has succeeded in doing this for the general one. At the same time, numerical simulation for specific values speed, density, pressure, time and so on allows you to achieve excellent results. It remains to be hoped that someone will be able to apply the Navier-Stokes equations in the opposite direction, that is, calculate the parameters with their help, or prove that there is no solution method.

Birch-Swinnerton-Dyer problem

The category of "Unsolved Problems" also includes the hypothesis proposed by English scientists from the University of Cambridge. Even 2300 years ago, the ancient Greek scientist Euclid gave a complete description of the solutions to the equation x2 + y2 = z2.

If for each of prime numbers count the number of points on the curve modulo it, you get an infinite set of integers. If you specifically “glue” it into 1 function of a complex variable, then you get the Hasse-Weyl zeta function for a third-order curve, denoted by the letter L. It contains information about the behavior modulo all prime numbers at once.

Brian Burch and Peter Swinnerton-Dyer conjectured about elliptic curves. According to it, the structure and number of the set of its rational solutions are related to the behavior of the L-function at the identity. Unproven on this moment the Birch-Swinnerton-Dyer conjecture depends on the description of algebraic equations of the 3rd degree and is the only relatively simple general way to calculate the rank of elliptic curves.

To understand the practical importance of this problem, it is enough to say that in modern cryptography a whole class of asymmetric systems is based on elliptic curves, and domestic standards digital signature.

Equality of classes p and np

If the rest of the Millennium Challenges are purely mathematical, then this one is related to the actual theory of algorithms. The problem concerning the equality of the classes p and np, also known as the Cooke-Levin problem, can be formulated in understandable language as follows. Suppose that a positive answer to a certain question can be checked quickly enough, i.e., in polynomial time (PT). Then is the statement correct that the answer to it can be found fairly quickly? Even simpler it sounds like this: is it really not more difficult to check the solution of the problem than to find it? If the equality of the classes p and np is ever proved, then all selection problems can be solved for PV. At the moment, many experts doubt the truth of this statement, although they cannot prove the opposite.

Riemann hypothesis

Until 1859, no pattern was identified that would describe how prime numbers are distributed among natural numbers. Perhaps this was due to the fact that science dealt with other issues. However, by the middle of the 19th century, the situation had changed, and they became one of the most relevant that mathematics began to deal with.

The Riemann Hypothesis, which appeared during this period, is the assumption that there is a certain pattern in the distribution of prime numbers.

Today, many modern scientists believe that if it is proven, then many of the fundamental principles of modern cryptography, which form the basis of a significant part of the mechanisms of electronic commerce, will have to be revised.

According to the Riemann hypothesis, the nature of the distribution of prime numbers may differ significantly from what is currently assumed. The fact is that so far no system has been discovered in the distribution of prime numbers. For example, there is the problem of "twins", the difference between which is 2. These numbers are 11 and 13, 29. Other prime numbers form clusters. These are 101, 103, 107, etc. Scientists have long suspected that such clusters exist among very large prime numbers. If they are found, then the stability of modern crypto keys will be in question.

Hodge Cycle Hypothesis

This hitherto unsolved problem was formulated in 1941. Hodge's hypothesis suggests the possibility of approximating the shape of any object by "gluing" together simple bodies of higher dimensions. This method has been known and successfully used for a long time. However, it is not known to what extent the simplification can be made.

Now you know what unsolvable problems exist at the moment. They are the subject of research by thousands of scientists around the world. It remains to be hoped that in the near future they will be resolved, and their practical application will help humanity enter a new round of technological development.

Hello!

There is an opinion that it is not profitable to engage in science today - one cannot become rich! But I hope that today's post will show you that this is far from the case. Today I will tell you how, by doing fundamental research, you can earn a tidy sum.

At any stage of development, any of the sciences has always faced a number of unresolved problems and tasks that haunted scientists. Physics - cold thermonuclear fusion, mathematics - Goldbach's hypothesis, medicine - a cure for cancer, and so on. Some of them are so important (for one reason or another) that a reward is due for their solution. And sometimes this reward is very, very decent.

In a number of sciences, this reward can be Nobel Prize. But they don’t give it for mathematical discoveries, and today I would like to talk about mathematics.

Mathematics - the queen of sciences, offers you a sea of ​​unsolved problems and interesting tasks, but today we will talk about only seven. They are also called the Millennium Targets.

It would seem, tasks, and tasks? What is special about them? The fact is that their solution has not been found for many years, and for the solution of each of them, the Clay Institute promised a reward of $ 1 million! Agree, not a lot. Of course, not the Nobel Prize, the size of which is approximately 1.5 million, but it will also do.

Here is their list:

  • Equality of classes P and NP
  • Hodge hypothesis
  • Poincare conjecture (solved)
  • Riemann hypothesis
  • Quantum Yang-Mills theory
  • Existence and smoothness of solutions of the Navier-Stokes equations
  • Birch-Swinnerton-Dyer hypothesis

So let's take a closer look at each of them.

1.Equality of classes P and NP

This problem is one of the most important problems in the theory of algorithms, and, I bet, many of you, at least indirectly, have heard about it. What is this problem and what is its essence? Imagine that there is a certain class of problems to which we can quickly give an answer, that is, quickly find a solution for them. This class of problems in the theory of algorithms is called the P class. And there is a class of problems for which we can quickly check the correctness of their solution - this is the NP class. And so far, it is not known whether these classes are equal or not. That is, it is not known whether it is possible, at least in theory, to find such an algorithm by which we can quickly find a solution to the problem, as well as check its correctness.

Classic example. Let a set of numbers be given, for example: 50, 2, 47, 5, 21, 4, 78, 1. Problem: is it possible to choose among these numbers such that their sum will give 100? Answer: you can, for example, 50 + 47 + 2 + 1 = 100. It is easy to check the correctness of the solution. We apply the addition operation four times and that's it. It's just a matter of picking up those numbers. At first glance, this is much more difficult to do. That is, finding a solution to a problem is more difficult than checking it. From the point of view of banal erudition, this is true, but mathematically this has not been proven, and there is hope that this is not so.

And so what? What if it turns out that the classes P and NP are equal? Everything is simple. Class equality means that there are algorithms for solving many problems that work much faster than currently known (as mentioned above).

Naturally, far from one attempt was made to prove or disprove this hypothesis, but none was successful. The latest attempt was made by the Indian mathematician Vinay Deolalikar. According to the author of the problem statement, Stephen Cook, this solution was "a relatively serious attempt to solve the problem of P vs NP". But, unfortunately, a number of errors were found in the presented proof, which the author promised to correct.

2. Hodge hypothesis

The complex is the sum of the simple parts. As a result of studying complex objects, mathematicians have developed methods for their approximation by gluing objects of increasing dimension. But it has not yet been clarified to what extent this kind of approximation can be carried out, and the geometric nature of some objects that are used in the approximation remains unclear.

3.Poincaré's hypothesis

The Poincaré hypothesis is currently the only one of the seven Millennium Challenges that has been solved. It is gratifying to note that our compatriot Grigory Yakovlevich Perelman, part-time reclusive genius, became the author of the decision. You can talk about it a lot and interestingly, but let's focus on the hypothesis itself.

Formulation:

Every simply connected compact 3-manifold without boundary is homeomorphic to a 3-sphere.

Or the generalized Poincare conjecture:

For any natural number n, any manifold of dimension n is homotopy equivalent to a sphere of dimension n if and only if it is homeomorphic to it.

In a simple way, the essence of the problem is as follows. If we take an apple and cover it with a rubber film, then with the help of deformations, without tearing the film, we can turn the apple into a dot or a cube, but in no way can we turn it into a donut. Cube, 3D sphere and even three-dimensional space identical to each other, up to deformation.

Despite such a simple formulation, the hypothesis remained unproven for hundreds of years. Although in mathematics, sometimes, the simpler the formulation, the more difficult the proof (we all remember Fermat's Last Theorem).

Let's return to Comrade Perelman. This gentleman is also famous for the fact that he refused the million put to him, stating the following: “Why do I need your money, if I have the whole Universe in my hands?” I wouldn't be able to do that. As a result of the refusal, the allocated million was granted to young French and American mathematicians.

Finally, I would like to note that the Poincaré hypothesis has absolutely no practical application (!!!).

4. The Riemann hypothesis.

The Riemann Hypothesis is probably the most famous (along with the Poincaré Hypothesis) of the seven Millennium Problems. One of the reasons for its popularity among non-professional mathematicians is that it has a very simple formulation.

All non-trivial zeros of the Riemann zeta function have a real part equal to ?.

Agree, it's quite simple. And the apparent simplicity was the reason for many attempts to prove this hypothesis. Unfortunately, so far to no avail.

A large number of unsuccessful attempts to prove the Riemann hypothesis gave rise to doubts about its validity among some mathematicians. Among them is John Littlewood. But the ranks of skeptics are not so numerous and most of the mathematical community tends to believe that the Riemann hypothesis is, nevertheless, correct. Indirect confirmation of this is the validity of a number of similar statements and hypotheses.

Many algorithms and statements in number theory have been formulated with the assumption that the above conjecture is true. Thus, the proof of the validity of the Riemann hypothesis will confirm the foundation of number theory, and its refutation of number theory will “shake” the very foundation.

And, finally, one rather famous, but very interesting fact. Once David Gilbert was asked: “What will be your first actions if you sleep for 500 years and wake up?” "I'll ask if the Riemann Hypothesis has been proven."

5. Yang-Mills theory

One of the gauge theories quantum physics with a non-Abelian gauge group. This theory was proposed in the middle of the last century, but for a long time it was considered as a purely mathematical technique that has nothing to do with the real nature of things. But later, on the basis of the Yang-Mills theory, the main theories were built standard model- quantum chromodynamics and the theory of weak interactions.

Problem formulation:

For any simple compact gauge group quantum theory Yang - Mills for space exists and has a non-zero mass defect.

The theory is perfectly confirmed by the results of experiments and the results of computer simulations, but has not received theoretical proof.

6. Existence and smoothness of solutions of the Navier-Stokes equations

One of the most important problems in hydrodynamics, and the last of the unsolved problems of classical mechanics.

The Navier-Stokes equation, supplemented by Maxwell's equations, heat transfer equations, etc., is used in solving many problems of electrohydrodynamics, magnetohydrodynamics, fluid and gas convection, thermal diffusion, etc.

The equations themselves are a system of partial differential equations. The equations have two parts:

  • equations of motion
  • continuity equations

Finding a complete analytical solution of the Navier-Stokes equations is greatly complicated by their nonlinearity and strong dependence on boundary and initial conditions.

7. Birch-Swinnerton-Dyer hypothesis

The last of the millennium problems is the Birch-Swinnerton-Dyer hypothesis.

The hypothesis states that

the rank of an elliptic curve r over Q is equal to the order of zero of the Hasse-Weil zeta function

E(L,s) at the point s = 1.

This conjecture is the only relatively simple way to determine the rank of elliptic curves, which, in turn, are the main objects of study in modern number theory and cryptography.

That's all the problems of the millennium. I apologize for the fact that some issues are covered much less than others. This is due to the lack of information on these problems and the impossibility of quite simply (without involving cumbersome and complex mathematics) to state their essence. The Clay Institute offered a $1 million reward for solving each of the problems. Dare! There is a chance to earn good money by moving forward fundamental science, because six out of seven problems have not yet been solved.

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