Fraction m/n we will consider irreducible (after all, a reducible fraction can always be reduced to an irreducible form). Squaring both sides of the equation, we get m^2=2n^2. From this we conclude that m^2, and then 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, which cannot be, since the fraction m/n irreducible. There is a contradiction. It remains to be concluded that our assumption is wrong and rational number m/n equal to √2 does not exist.”

That's all their proof.

Critical evaluation 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, one must precisely determine their purpose;

2) When the ancients decided that the number m is even, then in their accepted equality m = 2k they (intentionally or out of ignorance!) did not quite “correctly” characterize the number “ k ". But here is the number k- this is whole(Whole!) and completely famous a number that clearly defines the found even number m. And don't be it found numbers " k» the ancients could not further « use» and number m ;

3) And when from equality 2 k^2 = n^2 the ancients got the number n^2 is even, and at the same time n- even, they should have do not rush with a conclusion about emerging controversy", but it is better to make sure of the limit accuracy accepted by them choice» numbers « n ».

And how could they do it? Yes, simple!
See: from their equation 2 k^2 = n^2 one could easily obtain the following equality k√2 = n. And here there is nothing reprehensible in any way - after all, they received from equality m/n=√2 another adequate equality m^2=2n^2 ! And no one crossed them!

But in the new equality k√2 = n with obvious INTEGER numbers k and n it is clear that from always get the number √2 - rational . Is always! Because it contains numbers k and n- famous WHOLE!

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"ancient Greeks!"), then they must have, least , number " k" as non-integer (!!!) numbers. And the ancient Greeks just didn’t have this!

Hence the CONCLUSION: the above proof of the irrationality of the number √2, made by the ancient Greeks 2400 years ago, frankly wrong and mathematically incorrect, to say the least - it's just false .

In the small F-6 brochure shown above (see photo above), issued in Krasnodar (Russia) in 2015 with a total circulation of 15,000 copies. (obviously, with a sponsorship) a new, extremely correct from the point of view of mathematics and extremely true] proof of the irrationality of the number √2, which could have taken place long ago, were it not for rigid " prepo n" to the study of the antiquities of History.

The very concept of an irrational number is so arranged 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 fundamentally rational numbers differ from irrational ones? Both of them can be approximated by rational numbers with any given precision, but for rational numbers there is an approximation with "zero" precision (the number itself), but for irrational numbers this is no longer the case. Let's try to play with it.

First of all, we note such a simple fact. Let $%\alpha$%, $%\beta$% be two positive numbers that approximate each other with an accuracy of $%\varepsilon$%, i.e. $%|\alpha-\beta|=\varepsilon$%. What happens if we reverse the numbers? How does this change the accuracy? 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 assertion can be regarded as an independent lemma.

Now let's put $%x=\sqrt(2)$%, and let $%q\in(\mathbb Q)$% be a rational approximation of $%x$% with precision $%\varepsilon$%. We know that $%x>1$%, and as for the $%q$% approximation, we require that the inequality $%q\ge1$% be satisfied. For all numbers less than $%1$%, the approximation accuracy will be worse than that of $%1$% itself, and therefore we will not consider them.

Let's add $%1$% to each of the numbers $%x$%, $%q$%. Obviously, the approximation accuracy will remain the same. Now we have the numbers $%\alpha=x+1$% and $%\beta=q+1$%. Passing to reciprocals and applying the "lemma", we will come to the conclusion that our approximation accuracy has improved, becoming strictly less than $%\varepsilon$%. The required condition $%\alpha\beta>1$% is met even with a margin: in fact, we know that $%\alpha>2$% and $%\beta\ge2$%, from which we can conclude that the accuracy is improved by at least $%4$% times, i.e. it does not exceed $%\varepsilon/4$%.

And here is the main point: by 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. And 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 than $%\varepsilon$%. 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 $%\beta^(- 1)+1$%, i.e. $%(q+2)/(q+1)$%, with "improved" accuracy. This completes the proof, since rational numbers, as we noted above, have an "absolutely exact" rational approximation with an accuracy of $%\varepsilon=0$%, where the accuracy cannot be increased in principle. And we managed to do it, which speaks of the irrationality of our number.

In fact, this argument shows how to construct concrete 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.

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 as \(\frac(157304)(10000000)\) ;
\(0.333(3)…\) - 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.

Impossible because it 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 (Task from the OGE). The value 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) \(\sqrt(18)\cdot \sqrt(7)=\sqrt(9\cdot 2\cdot 7)=3\sqrt(14)\) 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, \(\frac(-5)(1)\) , so 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.

With a segment of unit length, ancient mathematicians already knew: 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:

Irrationality Proof Examples

Root of 2

Assume the contrary: it is rational, that is, it is represented as an irreducible fraction, where and are integers. Let's square the supposed equality:

From this it follows that even, therefore, even and . Let where the whole. Then

Therefore, even, therefore, even and . We have obtained that and are even, which contradicts the irreducibility of the fraction . Hence, the original assumption was wrong, and is an irrational number.

Binary logarithm of the number 3

Assume the contrary: it is rational, that is, it is represented as a fraction, where and are integers. Since , and can be taken positive. Then

But it's clear, it's odd. We get a contradiction.

e

Story

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

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 a pentagram. In the time of the Pythagoreans, it was believed that there is a single unit of length, sufficiently small and indivisible, which is an integer number of times included in any segment. 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 at the same time. 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 selected as the smallest possible.
• According to the Pythagorean theorem: a² = 2 b².
• Because a² 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, denote a = 2y.
• Then a² = 4 y² = 2 b².
• b² = 2 y², therefore b is even, then b even.
• However, it has been proven that b odd. Contradiction.

Greek mathematicians called this ratio of incommensurable quantities alogos(inexpressible), but according to the legends, Hippasus was not paid due respect. 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, which denies the doctrine that all entities in the universe can be reduced to whole numbers and their ratios." The discovery of Hippasus posed a serious problem for Pythagorean mathematics, destroying the underlying assumption that numbers and geometric objects are one and inseparable.

see also

Notes

What numbers are irrational? irrational number is not a rational real number, i.e. it cannot be represented as a fraction (as a ratio of two integers), where m is an integer, n- natural number . irrational number can be represented as an infinite non-periodic decimal fraction.

irrational number cannot be exact. Only in the format 3.333333…. For example, Square root of two - is an irrational number.

What is the irrational number? Irrational number(unlike rational ones) is called an infinite decimal non-periodic fraction.

Many irrational numbers often denoted by a capital Latin letter in bold without shading. That.:

Those. the set of irrational numbers is the difference between the sets of real and rational numbers.

Properties of irrational numbers.

• The sum of 2 non-negative irrational numbers can be a rational number.
• Irrational numbers define Dedekind sections in the set of rational numbers, in the lower class which do not have the a large number, and there is no smaller one in the upper one.
• Every real transcendental number is an irrational number.
• All irrational numbers are either algebraic or transcendent.
• The set of irrational numbers is everywhere dense on the number line: between each pair of numbers there is an irrational number.
• The order on the set of irrational numbers is isomorphic to the order on the set of real transcendental numbers.
• The set of irrational numbers is infinite, is a set of the 2nd category.
• The result of every arithmetic operation on rational numbers (except division by 0) is a rational number. The result of arithmetic operations on irrational numbers can be either a rational or an irrational number.
• The sum of a rational and an irrational number will always be an irrational number.
• The sum of irrational numbers can be a rational number. For example, let x irrational, then y=x*(-1) also irrational; x+y=0, and the number 0 rational (if, for example, we add the root of any degree of 7 and minus the root of the same degree of seven, we get a rational number 0).

Irrational numbers, examples.

γ — ζ (3) — ρ — √2 — √3 — √5 — φ — δs — α — e — π — δ