4 irrational numbers with examples. What are rational and irrational numbers

The set of irrational numbers is usually denoted by capital Latin letter I (\displaystyle \mathbb (I) ) in bold with no fill. In this way: 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 that are 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 the square, which is equivalent to the irrationality of the number.

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Irrational are:

Irrationality Proof Examples

Root of 2

Let's say 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 Manawa (c. 750 BC - c. 690 BC) found 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 is a single unit of length, sufficiently small and indivisible, which is an integer number of times included in any segment [ ] .
There is no exact data on the irrationality of which number was proved 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(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 Hippas put before Pythagorean mathematics serious problem, destroying the assumption underlying the whole theory that numbers and geometric objects are one and inseparable.

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 . So the original assumption was wrong, and - ir rational 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 assumption underlying the whole theory that numbers and geometric objects are one and inseparable.

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Notes

Numerical systems
Counting
sets Natural numbers () Integers ()

rational number is a number represented by an ordinary fraction m/n, where the numerator m is an integer and the denominator n is a natural number. Any rational number can be represented as a periodic infinite decimal fraction. The set of rational numbers is denoted by Q.

If a real number is not rational, then it is irrational number . Decimal fractions expressing irrational numbers are infinite and not periodic. The set of irrational numbers is usually denoted by the capital Latin letter I.

The real number is called algebraic, if it is a root of some polynomial (nonzero degree) with rational coefficients. Any non-algebraic number is called transcendent.

Some properties:
When solving problems, it is convenient, together with the irrational number a + b√ c (where a, b are rational numbers, c is an integer that is not a square of a natural number), to consider the number “conjugate” with it a - b√ c: its sum and product with the original - rational numbers. So a + b√ c and a – b√ c are the roots of a quadratic equation with integer coefficients.

Problems with solutions

1. Prove that

a) number √ 7;

b) number lg 80;

c) number √ 2 + 3 √ 3;

is irrational.

a) Assume that the number √ 7 is rational. Then, there are coprime p and q such that √ 7 = p/q, whence we obtain p 2 = 7q 2 . Since p and q are coprime, then p 2, and hence p is divisible by 7. Then р = 7k, where k is some natural number. Hence q 2 = 7k 2 = pk, which contradicts the fact that p and q are coprime.

So, the assumption is false, so the number √ 7 is irrational.

b) Assume that the number lg 80 is rational. Then there are natural p and q such that lg 80 = p/q, or 10 p = 80 q , whence we get 2 p–4q = 5 q–p . Considering that the numbers 2 and 5 are coprime, we get that the last equality is possible only for p–4q = 0 and q–p = 0. Whence p = q = 0, which is impossible, since p and q are chosen to be natural.

So, the assumption is false, so the number lg 80 is irrational.

c) Let's denote this number by x.

Then (x - √ 2) 3 \u003d 3, or x 3 + 6x - 3 \u003d √ 2 (3x 2 + 2). After squaring this equation, we get that x must satisfy the equation

x 6 - 6x 4 - 6x 3 + 12x 2 - 36x + 1 = 0.

Its rational roots can only be the numbers 1 and -1. The check shows that 1 and -1 are not roots.

So, the given number √ 2 + 3 √ 3 is irrational.

2. It is known that the numbers a, b, √ a –√ b ,- rational. Prove that √ a and √ b are also rational numbers.

Consider the product

(√ a - √ b) (√ a + √ b) = a - b.

Number √ a + √ b , which is equal to the ratio of numbers a – b and √ a –√ b , is rational because the quotient of two rational numbers is a rational number. Sum of two rational numbers

½ (√ a + √ b) + ½ (√ a - √ b) = √ a

is a rational number, their difference,

½ (√ a + √ b) - ½ (√ a - √ b) = √ b,

is also a rational number, which was to be proved.

3. Prove that there are positive irrational numbers a and b for which the number a b is natural.

4. Are there rational numbers a, b, c, d satisfying the equality

(a+b √ 2 ) 2n + (c + d√ 2 ) 2n = 5 + 4√ 2 ,

where n is a natural number?

If the equality given in the condition is satisfied, and the numbers a, b, c, d are rational, then the equality is also satisfied:

(a-b √ 2 ) 2n + (c – d√ 2 ) 2n = 5 – 4√ 2.

But 5 – 4√ 2 (a – b√ 2 ) 2n + (c – d√ 2 ) 2n > 0. The resulting contradiction proves that the original equality is impossible.

Answer: they don't exist.

5. If segments with lengths a, b, c form a triangle, then for all n = 2, 3, 4, . . . segments with lengths n √ a , n √ b , n √ c also form a triangle. Prove it.

If segments with lengths a, b, c form a triangle, then the triangle inequality gives

Therefore we have

( n √ a + n √ b ) n > a + b > c = ( n √ c ) n ,

N √ a + n √ b > n √ c .

The remaining cases of checking the triangle inequality are considered similarly, from which the conclusion follows.

6. Prove that the infinite decimal fraction 0.1234567891011121314... integers in order) is an irrational number.

As you know, rational numbers are expressed as decimal fractions, which have a period starting from a certain sign. Therefore, it suffices to prove that this fraction is not periodic with any sign. Suppose that this is not the case, and some sequence T, consisting of n digits, is the period of a fraction, starting from the mth decimal place. It is clear that there are non-zero digits after the mth digit, so there is a non-zero digit in the sequence of digits T. This means that starting from the m-th digit after the decimal point, among any n digits in a row there is a non-zero digit. However, in the decimal notation of this fraction, there must be a decimal notation for the number 100...0 = 10 k , where k > m and k > n. It is clear that this entry will occur to the right of the m-th digit and contain more than n zeros in a row. Thus, we obtain a contradiction, which completes the proof.

7. Given an infinite decimal fraction 0,a 1 a 2 ... . Prove that the digits in its decimal notation can be rearranged so that the resulting fraction expresses a rational number.

Recall that a fraction expresses a rational number if and only if it is periodic, starting from some sign. We divide the numbers from 0 to 9 into two classes: in the first class we include those numbers that occur in the original fraction a finite number of times, in the second class - those that occur in the original fraction an infinite number of times. Let's start writing out a periodic fraction, which can be obtained from the original permutation of the digits. First, after zero and a comma, we write in random order all the numbers from the first class - each as many times as it occurs in the entry of the original fraction. The first class digits written will precede the period in the fractional part of the decimal. Next, we write down the numbers from the second class in some order once. We will declare this combination a period and repeat it an infinite number of times. Thus, we have written out the required periodic fraction expressing some rational number.

8. Prove that in each infinite decimal fraction there is a sequence of decimal digits of arbitrary length, which occurs infinitely many times in the expansion of the fraction.

Let m be an arbitrarily given natural number. Let's break this infinite decimal fraction into segments, each with m digits. There will be infinitely many such segments. On the other hand, various systems, consisting of m digits, there are only 10 m , i.e., a finite number. Consequently, at least one of these systems must be repeated here infinitely many times.

Comment. For irrational numbers √ 2 , π or e we don't even know which digit is repeated infinitely many times in the infinite decimals that represent them, although each of these numbers can easily be shown to contain at least two distinct such digits.

9. Prove in an elementary way that the positive root of the equation

is irrational.

For x > 0, the left side of the equation increases with x, and it is easy to see that at x = 1.5 it is less than 10, and at x = 1.6 it is greater than 10. Therefore, the only positive root of the equation lies inside the interval (1.5 ; 1.6).

We write the root as an irreducible fraction p/q, where p and q are some coprime natural numbers. Then, for x = p/q, the equation will take the following form:

p 5 + pq 4 \u003d 10q 5,

whence it follows that p is a divisor of 10, therefore, p is equal to one of the numbers 1, 2, 5, 10. However, writing out fractions with numerators 1, 2, 5, 10, we immediately notice that none of them falls inside the interval (1.5; 1.6).

So, the positive root of the original equation cannot be represented as common fraction, which means it is an irrational number.

10. a) Are there three points A, B and C on the plane such that for any point X the length of at least one of the segments XA, XB and XC is irrational?

b) The coordinates of the vertices of the triangle are rational. Prove that the coordinates of the center of its circumscribed circle are also rational.

c) Does there exist a sphere on which there is exactly one rational point? (A rational point is a point for which all three Cartesian coordinates are rational numbers.)

a) Yes, there are. Let C be the midpoint of segment AB. Then XC 2 = (2XA 2 + 2XB 2 – AB 2)/2. If the number AB 2 is irrational, then the numbers XA, XB and XC cannot be rational at the same time.

b) Let (a 1 ; b 1), (a 2 ; b 2) and (a 3 ; b 3) be the coordinates of the vertices of the triangle. The coordinates of the center of its circumscribed circle are given by the system of equations:

(x - a 1) 2 + (y - b 1) 2 \u003d (x - a 2) 2 + (y - b 2) 2,

(x - a 1) 2 + (y - b 1) 2 \u003d (x - a 3) 2 + (y - b 3) 2.

It is easy to check that these equations are linear, which means that the solution of the considered system of equations is rational.

c) Such a sphere exists. For example, a sphere with the equation

(x - √ 2 ) 2 + y 2 + z 2 = 2.

Point O with coordinates (0; 0; 0) is a rational point lying on this sphere. The remaining points of the sphere are irrational. Let's prove it.

Assume the opposite: let (x; y; z) be a rational point of the sphere, different from the point O. It is clear that x is different from 0, since for x = 0 there is a unique solution (0; 0; 0), which we cannot now interested. Let's expand the brackets and express √ 2 :

x 2 - 2√ 2 x + 2 + y 2 + z 2 = 2

√ 2 = (x 2 + y 2 + z 2)/(2x),

which cannot be for rational x, y, z and irrational √ 2 . So, O(0; 0; 0) is the only rational point on the sphere under consideration.

Problems without solutions

1. Prove that the number

\[ \sqrt(10+\sqrt(24)+\sqrt(40)+\sqrt(60)) \]

is irrational.

2. For what integers m and n does the equality (5 + 3√ 2 ) m = (3 + 5√ 2 ) n hold?

3. Is there a number a such that the numbers a - √ 3 and 1/a + √ 3 are integers?

4. Can the numbers 1, √ 2, 4 be members (not necessarily adjacent) of an arithmetic progression?

5. Prove that for any positive integer n the equation (x + y √ 3 ) 2n = 1 + √ 3 has no solutions in rational numbers (x; y).

A rational number is a number that can be represented as a fraction, where . Q is the set of all rational numbers.

Rational numbers are divided into: positive, negative and zero.

Each rational number can be associated with a single point on the coordinate line. The relation "to the left" for points corresponds to the relation "less than" for the coordinates of these points. It can be seen that every negative number is less than zero and every positive number; of two negative numbers, the one whose modulus is greater is less. So, -5.3<-4.1, т.к. |5.3|>|4.1|.

Any rational number can be represented as a decimal periodic fraction. For example, .

Algorithms for operations on rational numbers follow from the rules of signs for the corresponding operations on zero and positive fractions. Q performs division other than division by zero.

Any linear equation, i.e. equation of the form ax+b=0, where , is solvable on the set Q, but not any quadratic equation kind , is solvable in rational numbers. Not every point on a coordinate line has a rational point. Even at the end of the 6th century BC. n. e in the school of Pythagoras, it was proved that the diagonal of a square is not commensurate with its height, which is tantamount to the statement: "The equation has no rational roots." All of the above led to the need to expand the set Q, the concept of an irrational number was introduced. Denote the set of irrational numbers by the letter J .

On a coordinate line, all points that do not have rational coordinates have irrational coordinates. , where r– sets real numbers. in a universal way assignments of real numbers are decimals. Periodic decimals define rational numbers, and non-periodic decimals define irrational numbers. So, 2.03 (52) is a rational number, 2.03003000300003 ... (the period of each following digit “3” is written one zero more) is an irrational number.

The sets Q and R have the properties of positivity: between any two rational numbers there is a rational number, for example, ecoi a
For every irrational number α one can specify a rational approximation both with a deficiency and with an excess with any accuracy: a< α
The operation of extracting a root from some rational numbers leads to irrational numbers. Extracting the root of a natural degree is an algebraic operation, i.e. its introduction is connected with the solution of an algebraic equation of the form . If n is odd, i.e. n=2k+1, where , then the equation has a single root. If n is even, n=2k, where , then for a=0 the equation has a single root x=0, for a<0 корней нет, при a>0 has two roots that are opposite to each other. Extracting a root is the reverse operation of raising to a natural power.

The arithmetic root (for brevity, the root) of the nth degree of a non-negative number a is a non-negative number b, which is the root of the equation. The root of the nth degree from the number a is denoted by the symbol. For n=2, the degree of the root 2 is not indicated: .

For example, , because 2 2 =4 and 2>0; , because 3 3 =27 and 3>0; does not exist because -four<0.

For n=2k and a>0, the roots of equation (1) are written as and . For example, the roots of the equation x 2 \u003d 4 are 2 and -2.

For n odd, equation (1) has a single root for any . If a≥0, then - the root of this equation. If a<0, то –а>0 and - the root of the equation. So, the equation x 3 \u003d 27 has a root.

All rational numbers can be represented as a common fraction. This applies to whole numbers (for example, 12, -6, 0), and final decimal fractions (for example, 0.5; -3.8921), and infinite periodic decimal fractions (for example, 0.11(23); -3 ,(87)).

However infinite non-recurring decimals cannot be represented as ordinary fractions. That's what they are irrational numbers(i.e. irrational). An example of such a number is π, which is approximately equal to 3.14. However, what it exactly equals cannot be determined, since after the number 4 there is an endless series of other numbers in which repeating periods cannot be distinguished. At the same time, although the number π cannot be expressed exactly, it has a specific geometric meaning. The number π is the ratio of the length of any circle to the length of its diameter. Thus irrational numbers do exist in nature, as do rational numbers.

Another example of irrational numbers is the square roots of positive numbers. Extracting roots from some numbers gives rational values, from others - irrational. For example, √4 = 2, i.e. the root of 4 is a rational number. But √2, √5, √7 and many others result in irrational numbers, that is, they can be extracted only with an approximation, rounded to a certain decimal place. In this case, the fraction is obtained non-periodic. That is, it is impossible to say exactly and definitely what the root of these numbers is.

So √5 is a number between 2 and 3, since √4 = 2, and √9 = 3. We can also conclude that √5 is closer to 2 than to 3, since √4 is closer to √5 than √9 to √5. Indeed, √5 ≈ 2.23 or √5 ≈ 2.24.

Irrational numbers are also obtained in other calculations (and not only when extracting roots), they are negative.

In relation to irrational numbers, we can say that no matter what unit segment we take to measure the length expressed by such a number, we cannot definitely measure it.

In arithmetic operations, irrational numbers can participate along with rational ones. At the same time, there are a number of regularities. For example, if only rational numbers are involved in an arithmetic operation, then the result is always a rational number. If only irrational ones participate in the operation, then it is impossible to say unequivocally whether a rational or irrational number will turn out.

For example, if you multiply two irrational numbers √2 * √2, you get 2 - this is a rational number. On the other hand, √2 * √3 = √6 is an irrational number.

If an arithmetic operation involves a rational and an irrational number, then an irrational result will be obtained. For example, 1 + 3.14... = 4.14... ; √17 - 4.

Why is √17 - 4 an irrational number? Imagine that you get a rational number x. Then √17 = x + 4. But x + 4 is a rational number, since we assumed that x is rational. The number 4 is also rational, so x + 4 is rational. However, a rational number cannot be equal to the irrational √17. Therefore, the assumption that √17 - 4 gives a rational result is incorrect. The result of an arithmetic operation will be irrational.

However, there is an exception to this rule. If we multiply an irrational number by 0, we get a rational number 0.
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