Haryana State Board HBSE 10th Class Maths Notes Chapter 1 Real Numbers Notes.

## Haryana Board 10th Class Maths Notes Chapter 1 Real Numbers

**Introduction**

We have studied about natural numbers, whole numbers, rational numbers, irrational numbers, and real numbers in earlier classes. We have also studied their properties and basic fundamental operations upon them. Recall that real number is the collection of rational and irrational numbers. It is denoted by R.

In this chapter, we shall learn about prove of irrationality of \(\sqrt{2}\), \(\sqrt{3}\)……, decimal expansion of rational numbers as terminating or non-terminating repeating decimal expansions, Euclid’s division lemma and fundamental theorem of arithmetic Euclid was the first Greek mathematician who suggests, the divisibility of two positive integers. He says that any positive integer a can be divided by another positive integer bin such a way that it leaves a remainder r such that 0 â‰¤ r â‰¤ b, and we compute HCF of two positive integers by the technique of Euclid’s division algorithm.

Fundamental theorem of arithmetic tells us that every composite number can be expressed as a product of primes in a unique way. Both of these results have wide and significant applications in the field of mathematics However, we shall discuss their use only in a few areas.

1. Real numbers: A collection of all rational and irrational numbers is called real numbers.

Example:- 1, 0, 1\(\frac{1}{2}\), 2.25, 4.010010001, \(\sqrt{3}\), Ï€ ….., etc.

2. Rational numbers: The numbers which can be written in the form \(\frac{p}{q}\), where p and q are integers and q â‰ 0, are called rational numbers.

Example : –\(\frac{4}{5}\), \(\frac{3}{7}\), \(\frac{9}{11}\) etc.

3. Irrational numbers: The numbers which cannot be written in the form \(\frac{p}{q}\), where p and q are integers and q â‰ 0, are called irrational numbers.

Example: \(\sqrt{3}\), \(\sqrt{19}\), 0.04004000400004…., etc.

4. Integers: The collection of whole numbers and their negatives is called integers.

Integers are of three type:

(a) Positive integers [1, 2, 3, …];

(b) Zero and

(c) Negative integers [- 1, -2, -3, …]

5. Positive integers/Natural numbers: The numbers which are used for counting objects are called positive integers or natural numbers.

Esrample: 1, 2, 3, 4, … etc.

6. Non-positive integers: Zero and the negative integers are collectively called non-positive integers.

Example : 0, -1, -2, -3, … etc.

7. Prime numbers: Numbers having exactly two factors (1 and the numbers itself) are called prime numbers.

Example : 2, 3, 5, 7, 11, 13, 17, ….etc.

8. Composite numbers: Numbers having more than two factors are called composite numbers. Composite numbers can be expressed as the product of primes.

Example : 4, 6, 12, 35, …. etc.

9. Co-primes: Two numbers are said to be coprimes, if they do not have a common factor other than 1. HCF of co-primes is 1.

Example : 2 and 3, 7 and 11.

10. Lemma : A lemma is a proven statement used for proving another statement.

11. Algorithm: An algorithm is a series of well defined steps which provides a procedure for solving a type of problem.

**Euclid’s Division Lemma**

Theorem 1.1.

For any two given positive integers a and b there exist unique whole numbers q and r such that

a = bq + r

where 0 â‰¤ r < b.

here a = dividend, b = divisor, q = quotient and r = remainder.

Dividend = divisor Ã— quotient + remainder

**Euclid’s Division Algorithm:**

To find the HCF of two positive integers, say a and b, with a > b, the following steps:

Step I: Apply Euclid’s division lemma to a and b. So, we find whole numbers q and r, such that a = bq + r, where 0 â‰¤ r < b.

Step II: If r = 0, b is the HCF of a and b. If r â‰ 0, apply the division lemma to b and r.

Step III: Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.

**Fundamental Theorem of Arithmetic**

Theorem 1.2

Every composite number can be expressed (factorized) as a product of primes, and this factorization is unique, apart from the order in which the prime factor recur. Let be a composite number. Then factorization of can be written as x = P_{1} Ã— P_{2} Ã— P_{3} Ã— …..P_{x}, where P_{1}, P_{2}, …. P_{x} are primes and written in ascending order.

For example:

25410 = 2 Ã— 3 Ã— 6 Ã— 7 Ã— 11 Ã— 11

= 2 Ã— 3 Ã— 6 Ã— 7 Ã— 112

Note: This method is also called the prime factorination method.

1. For any two positive integers a and b.

HCP (a, b) Ã— LCM (a, b) = a Ã— b

We can find other relations with the help of this formula.

(I) HCF (a, b) = \(\frac{a \times b}{\mathrm{LCM}(a, b)}\)

(II) LCM (a, b) = \(\frac{a \times b}{\mathrm{HCF}(a, b)}\)

(III) a = \(\frac{\mathrm{HCF}(a, b) \times \mathrm{LCM}(a, b)}{b}\)

(IV) \(b=\frac{\mathrm{HCF}(a, b) \times \mathrm{LCM}(a, b)}{a}\)

2. For any three positive integens x, y and z.

(a) HCF (x, y, z) Ã— LCM (x, y, z) â‰ x Ã— y Ã— z

(b) LCM (x, y, z)

= \(\frac{x \times y \times z \times \mathrm{HCF}(x, y, z)}{\mathrm{HCF}(x, y) \times \mathrm{HCF}(y, z) \times \mathrm{HCF}(z, x)}\)

(c) HCF (x, y, z)

= \(\frac{x \times y \times z \times \mathrm{LCM}(x, y, z)}{\mathrm{LCM}(x, y) \times \mathrm{LCM}(y, z) \times \mathrm{LCM}(z, x)}\)

**Revisiting Irrational Numbers**

Irrational Numbers : An irrational number is a non-terminating and non-recurring decimal, that is, it cannot be in the form \(\frac{a}{b}\), where p and q are both integers and q â‰ 0.

Example: (i) 0.12119111211112….

(ii) \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{5}\)……….. 3\(\sqrt{2}\), 5\(\sqrt{7}\), ……

(iii) 3\(\sqrt{2}\), 3\(\sqrt{5}\)…. and Ï€.

Theorem 1.3

Let p be a prime number. If p divides a^{2}, then p divides a, where a is a positive integer.

Theorem 1.4

Prove that \(\sqrt{2}\) is an irrational number.

Solution:

Let us assume that \(\sqrt{2}\) is a rational number. It can be expressed in the form of \(\frac{a}{b}\), where a, b are coprime positive integers and b â‰ 0.

âˆ´ \(\sqrt{2}\) = \(\frac{a}{b}\) [HCF of a and b is 1]

[because a and b are coprime.]

â‡’ 2 = \(\frac{a^2}{b^2}\) (Squaring both sides)

â‡’ 2b^{2} = a^{2} ……(i)

Therefore, 2 divides a^{2}, it follows that 2 divides a [By Theorem 1.3]

Let a = 2c [Where c is any integer]

Put a = 2c in equation (i), we get

â‡’ \(b^2=\frac{4 c^2}{2}=2 c^2\) ……(ii)

It means bis divided by 2 and so b is divided by 2 [By Theorem 1.3]

From (i) and (ii) we say that 2 is a common factor of both a and b. But this contradict the fact that a and b are coprime, so they have no common factor. So our assumption that \(\sqrt{2}\) is a rational number is wrong.

Therefore, \(\sqrt{2}\) is an irrational number. Proved

**Revisiting Rational Numbers And Their Decimal Expansions**

1. Rational Numbers: Rational numbers are of the form of \(\frac{p}{q}\), where p and q are integers

and q â‰ 0.

2. Rational numbers in decimal form: A rational number has either a terminating or nonterminating but recurring decimal.

For example: (i) \(\frac{2}{5}\) = 0.4 [Terminating decimal expansion].

(ii) \(\frac{1}{3}\) = 0.333 …… [Non-terminating but recurring decimal expansion]

3. Theorem 1.5 : Let x be a rational number whose decimal expension terminates. Then x can be expressed in the form \(\frac{p}{q}\), where p and q are coprime, and the prime factorisation of q in of the form 2^{n}5^{m}, where n, m are non negative integers.

For example:

(i) 1.04 = \(\frac{104}{100}=\frac{104}{10^2}=\frac{104}{2^2 \times 5^2}\)

(ii) 0.125 = \(\frac{125}{1000}=\frac{125}{10^3}=\frac{125}{2^3 \times 5^3}\)

4. Theorem 1.6 : Let x = \(\frac{p}{q}\) be a rational number, such that p and q are co-prime and the prime factorisation of q is of the form 2^{m}5^{n} or 2^{n}5^{m}, where m, n, are non-negative integers. Then has terminating decimal expansion.

For Example : \(\frac{23}{40}=\frac{23}{2^3 \times 5^1}\)

Hence \(\frac{23}{40}\) has a terminating decimal.

5. Theorem 1.7: Let x = \(\frac{p}{q}\) be a rational number, such that p and q are coprime and the prime factorisation of q is not of the form 2^{m}5^{n} where m, n are non-negative integers. Then x has a non terminating repeating (recurring) decimal expansion.

For example : \(\frac{53}{343}\)

âˆµ 343 = 73 [Not of the form 2^{m}5^{n}, where m and n are non-negative integers]

âˆ´ \(\frac{53}{343}\) has non terminating repeating decimal.