Prove that the number e is irrational. Rational and irrational numbers

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.

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.

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

But…. Let's look at this proof of the ancient Greeks somewhat critically. And if you are more careful in simple mathematics, then you can see the following in it:

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.

The ancient mathematicians already knew about a segment of unit length: they knew, for example, the incommensurability of the diagonal and the side of the square, which is equivalent to the irrationality of the number.

Irrational are:

Examples of proof of irrationality

Root of 2

Let us assume the opposite: it is rational, that is, it is represented in the form of an irreducible fraction, where and are integers. Let's square the supposed equality:

.

It follows that even is even and . Let it be where the whole is. Then

Therefore, even means even and . We found that and are even, which contradicts the irreducibility of the fraction . This means that the original assumption was incorrect, and it is an irrational number.

Binary logarithm of the number 3

Let us assume the opposite: it is rational, that is, it is represented as a fraction, where and are integers. Since , and can be chosen to be positive. Then

But even and odd. We get a contradiction.

e

Story

The concept of irrational numbers was implicitly adopted by Indian mathematicians in the 7th century BC, when Manava (c. 750 BC - c. 690 BC) figured out that the square roots of some natural numbers, such as 2 and 61 cannot be expressed explicitly.

The first proof of the existence of irrational numbers is usually attributed to Hippasus of Metapontus (c. 500 BC), a Pythagorean who found this proof by studying the lengths of the sides of the pentagram. At the time of the Pythagoreans, it was believed that there was a single unit of length, sufficiently small and indivisible, which entered any segment an integer number of times. However, Hippasus argued that there is no single unit of length, since the assumption of its existence leads to a contradiction. He showed that if the hypotenuse of an isosceles right triangle contains an integer number of unit segments, then this number must be both even and odd. The proof looked like this:

• The ratio of the length of the hypotenuse to the length of the leg of an isosceles right triangle can be expressed as a:b, Where a And b chosen as the smallest possible.
• According to the Pythagorean theorem: a² = 2 b².
• Because a- even, a must be even (since the square of an odd number would be odd).
• Because the a:b irreducible b must be odd.
• Because a even, we denote a = 2y.
• Then a² = 4 y² = 2 b².
• b² = 2 y², therefore b- even, then b even.
• However, it has been proven that b odd. Contradiction.

Greek mathematicians called this ratio of incommensurable quantities alogos(unspeakable), but according to the legends they did not pay due respect to Hippasus. There is a legend that Hippasus made the discovery while on a sea voyage and was thrown overboard by other Pythagoreans “for creating an element of the universe that denies the doctrine that all entities in the universe can be reduced to integers and their ratios.” The discovery of Hippasus posed a serious problem for Pythagorean mathematics, destroying the underlying assumption that numbers and geometric objects were one and inseparable.

see also

Notes

Numerical systems
Counting sets	Integers ()

The very concept of an irrational number is structured in such a way that it is defined through the negation of the property “to be rational,” therefore proof by contradiction is the most natural here. It is possible, however, to offer the following reasoning.

How do rational numbers fundamentally differ from irrational numbers? Both of them can be approximated by rational numbers with any given accuracy, but for rational numbers there is an approximation with “zero” accuracy (by this number itself), but for irrational numbers this is no longer the case. Let's try to "play" on this.

First of all, let's note this simple fact. Let $%\alpha$%, $%\beta$% be two positive numbers that approximate each other with an accuracy of $%\varepsilon$%, that is, $%|\alpha-\beta|=\varepsilon$%. What happens if we replace the numbers with their inverses? How will the accuracy change? It is easy to see that $$\left|\frac1\alpha-\frac1\beta\right|=\frac(|\alpha-\beta|)(\alpha\beta)=\frac(\varepsilon)(\alpha\ beta),$$ which will be strictly less than $%\varepsilon$% for $%\alpha\beta>1$%. This statement can be considered as an independent lemma.

Now let's set $%x=\sqrt(2)$%, and let $%q\in(\mathbb Q)$% be a rational approximation of the number $%x$% with an accuracy of $%\varepsilon$%. We know that $%x>1$%, and regarding the approximation $%q$% we require the inequality $%q\ge1$%. All numbers smaller than $%1$% will have worse approximation accuracy than $%1$% itself, and therefore we will not consider them.

To each of the numbers $%x$%, $%q$% we add $%1$%. Obviously, the approximation accuracy will remain the same. Now we have the numbers $%\alpha=x+1$% and $%\beta=q+1$%. Moving on to the reciprocal numbers and applying the “lemma”, we will come to the conclusion that our approximation accuracy has improved, becoming strictly less than $%\varepsilon$%. We have met the required condition $%\alpha\beta>1$% even with a margin: in fact, we know that $%\alpha>2$% and $%\beta\ge2$%, from which we can conclude that accuracy improves at least $%4$% times, that is, it does not exceed $%\varepsilon/4$%.

And here is the main point: according to the condition, $%x^2=2$%, that is, $%x^2-1=1$%, which means that $%(x+1)(x- 1)=1$%, that is, the numbers $%x+1$% and $%x-1$% are inverse to each other. This means that $%\alpha^(-1)=x-1$% will be an approximation to the (rational) number $%\beta^(-1)=1/(q+1)$% with an accuracy strictly less $%\varepsilon$%. It remains to add $%1$% to these numbers, and it turns out that the number $%x$%, that is, $%\sqrt(2)$%, has a new rational approximation equal to $%\beta^(- 1)+1$%, that is, $%(q+2)/(q+1)$%, with “improved” accuracy. This completes the proof, since for rational numbers, as we noted above, there is an “absolutely accurate” rational approximation with an accuracy of $%\varepsilon=0$%, where the accuracy cannot, in principle, be increased. But we managed to do this, which speaks to the irrationality of our numbers.

In fact, this reasoning shows how to construct specific rational approximations for $%\sqrt(2)$% with ever-improving accuracy. We must first take the approximation $%q=1$%, and then apply the same replacement formula: $%q\mapsto(q+2)/(q+1)$%. This process produces the following: $$1,\frac32,\frac75,\frac(17)(12),\frac(41)(29),\frac(99)(70)$$ and so on.