Proof of the irrationality of the root of two using an example. Numbers. Irrational numbers. Rational numbers $\mathbb(Q)$

Understanding numbers, especially natural numbers, is one of the oldest math "skills." Many civilizations, even modern ones, have attributed certain mystical properties to numbers due to their enormous importance in describing nature. Although modern science and mathematics do not confirm these “magical” properties, the importance of number theory is undeniable.

Historically, a variety of natural numbers appeared first, then fairly quickly fractions and positive irrational numbers were added to them. Zero and negative numbers were introduced after these subsets of the set of real numbers. The last set, the set of complex numbers, appeared only with the development of modern science.

In modern mathematics, numbers are not introduced in historical order, although quite close to it.

Natural numbers $\mathbb(N)$

The set of natural numbers is often denoted as $\mathbb(N)=\lbrace 1,2,3,4... \rbrace $, and is often padded with zero to denote $\mathbb(N)_0$.

$\mathbb(N)$ defines the operations of addition (+) and multiplication ($\cdot$) with the following properties for any $a,b,c\in \mathbb(N)$:

1. $a+b\in \mathbb(N)$, $a\cdot b \in \mathbb(N)$ the set $\mathbb(N)$ is closed under the operations of addition and multiplication
2. $a+b=b+a$, $a\cdot b=b\cdot a$ commutativity
3. $(a+b)+c=a+(b+c)$, $(a\cdot b)\cdot c=a\cdot (b\cdot c)$ associativity
4. $a\cdot (b+c)=a\cdot b+a\cdot c$ distributivity
5. $a\cdot 1=a$ is a neutral element for multiplication

Since the set $\mathbb(N)$ contains a neutral element for multiplication but not for addition, adding a zero to this set ensures that it includes a neutral element for addition.

In addition to these two operations, the “less than” relations ($

1. $a b$ trichotomy
2. if $a\leq b$ and $b\leq a$, then $a=b$ antisymmetry
3. if $a\leq b$ and $b\leq c$, then $a\leq c$ is transitive
4. if $a\leq b$ then $a+c\leq b+c$
5. if $a\leq b$ then $a\cdot c\leq b\cdot c$

Integers $\mathbb(Z)$

Examples of integers:
$1, -20, -100, 30, -40, 120...$

Solving the equation $a+x=b$, where $a$ and $b$ are known natural numbers, and $x$ is an unknown natural number, requires the introduction of a new operation - subtraction(-). If there is a natural number $x$ satisfying this equation, then $x=b-a$. However, this particular equation does not necessarily have a solution on the set $\mathbb(N)$, so practical considerations require expanding the set of natural numbers to include solutions to such an equation. This leads to the introduction of a set of integers: $\mathbb(Z)=\lbrace 0,1,-1,2,-2,3,-3...\rbrace$.

Since $\mathbb(N)\subset \mathbb(Z)$, it is logical to assume that the previously introduced operations $+$ and $\cdot$ and the relations $ 1. $0+a=a+0=a$ there is a neutral element for addition
2. $a+(-a)=(-a)+a=0$ there is an opposite number $-a$ for $a$

Property 5.:
5. if $0\leq a$ and $0\leq b$, then $0\leq a\cdot b$

The set $\mathbb(Z)$ is also closed under the subtraction operation, that is, $(\forall a,b\in \mathbb(Z))(a-b\in \mathbb(Z))$.

Rational numbers $\mathbb(Q)$

Examples of rational numbers:
$\frac(1)(2), \frac(4)(7), -\frac(5)(8), \frac(10)(20)...$

Now consider equations of the form $a\cdot x=b$, where $a$ and $b$ are known integers, and $x$ is an unknown. For the solution to be possible, it is necessary to introduce the division operation ($:$), and the solution takes the form $x=b:a$, that is, $x=\frac(b)(a)$. Again the problem arises that $x$ does not always belong to $\mathbb(Z)$, so the set of integers needs to be expanded. This introduces the set of rational numbers $\mathbb(Q)$ with elements $\frac(p)(q)$, where $p\in \mathbb(Z)$ and $q\in \mathbb(N)$. The set $\mathbb(Z)$ is a subset in which each element $q=1$, therefore $\mathbb(Z)\subset \mathbb(Q)$ and the operations of addition and multiplication extend to this set according to the following rules, which preserve all the above properties on the set $\mathbb(Q)$:
$\frac(p_1)(q_1)+\frac(p_2)(q_2)=\frac(p_1\cdot q_2+p_2\cdot q_1)(q_1\cdot q_2)$
$\frac(p-1)(q_1)\cdot \frac(p_2)(q_2)=\frac(p_1\cdot p_2)(q_1\cdot q_2)$

The division is introduced as follows:
$\frac(p_1)(q_1):\frac(p_2)(q_2)=\frac(p_1)(q_1)\cdot \frac(q_2)(p_2)$

On the set $\mathbb(Q)$, the equation $a\cdot x=b$ has a unique solution for each $a\neq 0$ (division by zero is undefined). This means that there is an inverse element $\frac(1)(a)$ or $a^(-1)$:
$(\forall a\in \mathbb(Q)\setminus\lbrace 0\rbrace)(\exists \frac(1)(a))(a\cdot \frac(1)(a)=\frac(1) (a)\cdot a=a)$

The order of the set $\mathbb(Q)$ can be expanded as follows:
$\frac(p_1)(q_1)

The set $\mathbb(Q)$ has one important property: between any two rational numbers there are infinitely many other rational numbers, therefore, there are no two adjacent rational numbers, unlike the sets of natural numbers and integers.

Irrational numbers $\mathbb(I)$

Examples of irrational numbers:
$\sqrt(2) \approx 1.41422135...$
$\pi\approx 3.1415926535...$

Since between any two rational numbers there are infinitely many other rational numbers, it is easy to erroneously conclude that the set of rational numbers is so dense that there is no need to expand it further. Even Pythagoras made such a mistake in his time. However, his contemporaries already refuted this conclusion when studying solutions to the equation $x\cdot x=2$ ($x^2=2$) on the set of rational numbers. To solve such an equation, it is necessary to introduce the concept of a square root, and then the solution to this equation has the form $x=\sqrt(2)$. An equation like $x^2=a$, where $a$ is a known rational number and $x$ is an unknown one, does not always have a solution on the set of rational numbers, and again the need arises to expand the set. A set of irrational numbers arises, and numbers such as $\sqrt(2)$, $\sqrt(3)$, $\pi$... belong to this set.

Real numbers $\mathbb(R)$

The union of the sets of rational and irrational numbers is the set of real numbers. Since $\mathbb(Q)\subset \mathbb(R)$, it is again logical to assume that the introduced arithmetic operations and relations retain their properties on the new set. The formal proof of this is very difficult, so the above-mentioned properties of arithmetic operations and relations on the set of real numbers are introduced as axioms. In algebra, such an object is called a field, so the set of real numbers is said to be an ordered field.

In order for the definition of the set of real numbers to be complete, it is necessary to introduce an additional axiom that distinguishes the sets $\mathbb(Q)$ and $\mathbb(R)$. Suppose that $S$ is a non-empty subset of the set of real numbers. An element $b\in \mathbb(R)$ is called the upper bound of a set $S$ if $\forall x\in S$ holds $x\leq b$. Then we say that the set $S$ is bounded above. The smallest upper bound of the set $S$ is called the supremum and is denoted $\sup S$. The concepts of lower bound, set bounded below, and infinum $\inf S$ are introduced similarly. Now the missing axiom is formulated as follows:

Any non-empty and upper-bounded subset of the set of real numbers has a supremum.
It can also be proven that the field of real numbers defined in the above way is unique.

Complex numbers$\mathbb(C)$

Examples of complex numbers:
$(1, 2), (4, 5), (-9, 7), (-3, -20), (5, 19),...$
$1 + 5i, 2 - 4i, -7 + 6i...$ where $i = \sqrt(-1)$ or $i^2 = -1$

The set of complex numbers represents all ordered pairs of real numbers, that is, $\mathbb(C)=\mathbb(R)^2=\mathbb(R)\times \mathbb(R)$, on which the operations of addition and multiplication are defined as follows way:
$(a,b)+(c,d)=(a+b,c+d)$
$(a,b)\cdot (c,d)=(ac-bd,ad+bc)$

There are several forms of writing complex numbers, of which the most common is $z=a+ib$, where $(a,b)$ is a pair of real numbers, and the number $i=(0,1)$ is called the imaginary unit.

It is easy to show that $i^2=-1$. Extending the set $\mathbb(R)$ to the set $\mathbb(C)$ allows us to determine the square root of negative numbers, which was the reason for introducing the set of complex numbers. It is also easy to show that a subset of the set $\mathbb(C)$, given by $\mathbb(C)_0=\lbrace (a,0)|a\in \mathbb(R)\rbrace$, satisfies all the axioms for real numbers, therefore $\mathbb(C)_0=\mathbb(R)$, or $R\subset\mathbb(C)$.

The algebraic structure of the set $\mathbb(C)$ with respect to the operations of addition and multiplication has the following properties:
1. commutativity of addition and multiplication
2. associativity of addition and multiplication
3. $0+i0$ - neutral element for addition
4. $1+i0$ - neutral element for multiplication
5. Multiplication is distributive with respect to addition
6. There is a single inverse for both addition and multiplication.

Definition of an irrational number

Irrational numbers are those numbers that in decimal notation represent endless non-periodic decimal fractions.

So, for example, numbers obtained by taking the square root of natural numbers are irrational and are not squares of natural numbers. But not all irrational numbers are obtained by taking square roots, because the number pi obtained by division is also irrational, and you are unlikely to get it by trying to extract the square root of a natural number.

Properties of irrational numbers

Unlike numbers written as infinite decimals, only irrational numbers are written as non-periodic infinite decimals.
The sum of two non-negative irrational numbers can end up being a rational number.
Irrational numbers define Dedekind cuts in the set of rational numbers, in the lower class of which there is no largest number, and in the upper class there is no smaller one.
Any real transcendental number is irrational.
All irrational numbers are either algebraic or transcendental.
The set of irrational numbers on a line is densely located, and between any two of its numbers there is sure to be an irrational number.
The set of irrational numbers is infinite, uncountable and is a set of the 2nd category.
When performing any arithmetic operation on rational numbers, except division by 0, the result will be a rational number.
When adding a rational number to an irrational number, the result is always an irrational number.
When adding irrational numbers, we can end up with a rational number.
The set of irrational numbers is not even.

Numbers are not irrational

Sometimes it is quite difficult to answer the question of whether a number is irrational, especially in cases where the number is in the form of a decimal fraction or in the form of a numerical expression, root or logarithm.

Therefore, it will not be superfluous to know which numbers are not irrational. If we follow the definition of irrational numbers, then we already know that rational numbers cannot be irrational.

Irrational numbers are not:

First, all natural numbers;
Secondly, integers;
Third, ordinary fractions;
Fourthly, various mixed numbers;
Fifthly, these are infinite periodic decimal fractions.

In addition to all of the above, an irrational number cannot be any combination of rational numbers that is performed by the signs of arithmetic operations, such as +, -, , :, since in this case the result of two rational numbers will also be a rational number.

Now let's see which numbers are irrational:

Do you know about the existence of a fan club where fans of this mysterious mathematical phenomenon are looking for more and more information about Pi, trying to unravel its mystery? Any person who knows by heart a certain number of Pi numbers after the decimal point can become a member of this club;

Did you know that in Germany, under the protection of UNESCO, there is the Castadel Monte palace, thanks to the proportions of which you can calculate Pi. King Frederick II dedicated the entire palace to this number.

It turns out that they tried to use the number Pi in the construction of the Tower of Babel. But unfortunately, this led to the collapse of the project, since at that time the exact calculation of the value of Pi was not sufficiently studied.

Singer Kate Bush in her new disc recorded a song called “Pi”, in which one hundred and twenty-four numbers from the famous number series 3, 141… were heard.

Example:
\(4\) is a rational number, because it can be written as \(\frac(4)(1)\) ;
\(0.0157304\) is also rational, because it can be written in the form \(\frac(157304)(10000000)\) ;
\(0.333(3)...\) - and this is a rational number: can be represented as \(\frac(1)(3)\) ;
\(\sqrt(\frac(3)(12))\) is rational, since it can be represented as \(\frac(1)(2)\) . Indeed, we can carry out a chain of transformations \(\sqrt(\frac(3)(12))\) \(=\)\(\sqrt(\frac(1)(4))\) \(=\) \ (\frac(1)(2)\)

Irrational number is a number that cannot be written as a fraction with an integer numerator and denominator.

It's impossible because it's endless fractions, and even non-periodic ones. Therefore, there are no integers that, when divided by each other, would give an irrational number.

Example:
\(\sqrt(2)≈1.414213562…\) is an irrational number;
\(π≈3.1415926… \) is an irrational number;
\(\log_(2)(5)≈2.321928…\) is an irrational number.

Example (Assignment from the OGE). The meaning of which of the expressions is a rational number?
1) \(\sqrt(18)\cdot\sqrt(7)\);
2)\((\sqrt(9)-\sqrt(14))(\sqrt(9)+\sqrt(14))\);
3) \(\frac(\sqrt(22))(\sqrt(2))\);
4) \(\sqrt(54)+3\sqrt(6)\).

Solution:

1) \(\sqrt(18)\cdot \sqrt(7)=\sqrt(9\cdot 2\cdot 7)=3\sqrt(14)\) – the root of \(14\) cannot be taken, which means It is also impossible to represent a number as a fraction with integers, therefore the number is irrational.

2) \((\sqrt(9)-\sqrt(14))(\sqrt(9)+\sqrt(14))= (\sqrt(9)^2-\sqrt(14)^2)=9 -14=-5\) – there are no roots left, the number can be easily represented as a fraction, for example \(\frac(-5)(1)\), which means it is rational.

3) \(\frac(\sqrt(22))(\sqrt(2))=\sqrt(\frac(22)(2))=\sqrt(\frac(11)(1))=\sqrt( 11)\) – the root cannot be extracted - the number is irrational.

4) \(\sqrt(54)+3\sqrt(6)=\sqrt(9\cdot 6)+3\sqrt(6)=3\sqrt(6)+3\sqrt(6)=6\sqrt (6)\) is also irrational.

Fraction m/n we will consider it irreducible (after all, a reducible fraction can always be reduced to an irreducible form). By squaring both sides of the equality, we get m^2=2n^2. From here we conclude that m^2, and after this the number m- even. those. m = 2k. That's why m^2 = 4k^2 and therefore 4 k^2 =2n^2, or 2 k^2 = n^2. But then it turns out that n is also an even number, but this cannot be, since the fraction m/n irreducible. A contradiction arises. It remains to conclude: our assumption is incorrect and the rational number m/n, equal to √2, does not exist.”

That's all their proof.

A critical assessment of the evidence of the ancient Greeks

1) In the rational number adopted by the Greeks m/n numbers m And n- whole, but unknown(whether they even, whether they odd). And so it is! And in order to somehow establish any dependence between them, it is necessary to accurately determine their purpose;

2) When the ancients decided that the number m– even, then in the equality they accepted m = 2k they (intentionally or out of ignorance!) did not quite “correctly” characterize the number “ k " But here is the number k- This whole(WHOLE!) and quite famous a number that quite clearly defines what was found even number m. And don't be this way found numbers " k"the ancients could not in the future" use" and number m ;

3) And when from equality 2 k^2 = n^2 the ancients received the number n^2 is even, and at the same time n– even, then they would have to do not hurry with the conclusion about " the contradiction that has arisen", but it is better to make sure of the maximum accuracy accepted by them " choice» numbers « n ».

How could they do this? Yes, simple!
Look: from the equality they obtained 2 k^2 = n^2 one could easily obtain the following equality k√2 = n. And there is nothing reprehensible here - after all, they got from equality m/n=√2 is another equality adequate to it m^2=2n^2 ! And no one contradicted them!

But in the new equality k√2 = n for obvious INTEGERS k And n it is clear that from it Always get the number √2 - rational . Always! Because it contains numbers k And n- famous WHOLE ones!

But so that from their equality 2 k^2 = n^2 and, as a consequence, from k√2 = n get the number √2 – irrational (like that " wished"the ancient Greeks!), then it is necessary to have in them, least , number " k" as not whole (!!!) numbers. And this is exactly what the ancient Greeks did NOT have!

Hence the CONCLUSION: the above proof of the irrationality of the number √2, made by the ancient Greeks 2400 years ago, is frankly incorrect and mathematically incorrect, not to say rudely - it is simply fake .

In the small brochure F-6 shown above (see photo above), released in Krasnodar (Russia) in 2015 with a total circulation of 15,000 copies. (obviously with sponsorship investment) a new, extremely correct from the point of view of mathematics and extremely correct ] proof of the irrationality of the number √2 is given, which could have happened long ago if there were no hard " teacher n" to the study of the antiquities of History.

1.Proofs are examples of deductive reasoning and are different from inductive or empirical arguments. A proof must demonstrate that the statement being proven is always true, sometimes by listing all possible cases and showing that the statement holds in each of them. A proof may rely on obvious or generally accepted phenomena or cases known as axioms. Contrary to this, the irrationality of the “square root of two” is proven.
2. The intervention of topology here is explained by the very nature of things, which means that there is no purely algebraic way to prove irrationality, in particular based on rational numbers. Here is an example, the choice is yours: 1 + 1/2 + 1/4 + 1/8 ….= 2 or 1+1/2 + 1/4 + 1/8 …≠ 2 ???
If you accept 1+1/2 + 1/4 + 1/8 +…= 2, which is considered the “algebraic” approach, then it is not at all difficult to show that there exists n/m ∈ ℚ, which on an infinite sequence is irrational and finite number. This suggests that the irrational numbers are the closure of the field ℚ, but this refers to a topological singularity.
So for Fibonacci numbers, F(k): 1,1,2,3,5,8,13,21,34,55,89,144,233,377, … lim(F(k+1)/F(k)) = φ
This only shows that there is a continuous homomorphism ℚ → I, and it can be shown rigorously that the existence of such an isomorphism is not a logical consequence of the algebraic axioms.