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 us note this simple fact. Let $%\alpha$%, $%\beta$% be two positive numbers, which 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. 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 $%\varepsilon$%. 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.
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 don't rush 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 of this, from k√2 = n get the number √2 – irrational (like that " wished"the ancient Greeks!), then it is necessary to have in them, at least , number " k» in the form not whole (!!!) numbers. And this is precisely 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.
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.
The set of irrational numbers is usually denoted by a capital Latin letter I (\displaystyle \mathbb (I) ) in bold style without shading. Thus: I = R ∖ Q (\displaystyle \mathbb (I) =\mathbb (R) \backslash \mathbb (Q) ), that is, the set of irrational numbers is the difference between the sets of real and rational numbers.
The existence of irrational numbers, more precisely, segments incommensurable with a segment of unit length, was already known to ancient mathematicians: they knew, for example, the incommensurability of the diagonal and the side of a square, which is equivalent to the irrationality of the number.
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Irrational are:
Examples of proof of irrationality
Root of 2
Let's assume the opposite: 2 (\displaystyle (\sqrt (2))) rational, that is, represented as a fraction m n (\displaystyle (\frac (m)(n))), Where m (\displaystyle m) is an integer, and n (\displaystyle n)- natural number.
Let's square the supposed equality:
2 = m n ⇒ 2 = m 2 n 2 ⇒ m 2 = 2 n 2 (\displaystyle (\sqrt (2))=(\frac (m)(n))\Rightarrow 2=(\frac (m^(2 ))(n^(2)))\Rightarrow m^(2)=2n^(2)).Story
Antiquity
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 square roots 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. At the time of the Pythagoreans, it was believed that there was a single unit of length, sufficiently small and indivisible, which included an integer number of times in any segment [ ] .
There is no exact data on which number was proven irrational by Hippasus. According to legend, he found it by studying the lengths of the sides of the pentagram. Therefore, it is reasonable to assume that this was the golden ratio [ ] .
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 challenged Pythagorean mathematics serious problem, destroying the assumption underlying the entire theory that numbers and geometric objects united and inseparable.
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. The 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 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.
