## CBSE Class 10 Maths Notes Chapter 1 Real Numbers

here are some brief notes on CBSE Class 10 Maths Chapter 1: Real Numbers:

Real Numbers:

- A real number is a number that can be represented on a number line.
- It can be rational (i.e. a fraction or terminating decimal) or irrational (i.e. a non-terminating, non-repeating decimal).
- Every real number can be written as a decimal, either terminating or non-terminating, repeating or non-repeating.

Euclid’s Division Algorithm:

- Euclid’s division algorithm is a method of finding the HCF (highest common factor) of two positive integers.
- According to the algorithm, we keep dividing the larger number by the smaller number until we get a remainder of 0.
- The HCF of the two numbers is the last non-zero remainder obtained in this process.

Fundamental Theorem of Arithmetic:

- The fundamental theorem of arithmetic states that every composite number can be expressed as a product of prime numbers, uniquely up to the order of the factors.
- This theorem can be used to find the prime factorisation of a given number.

Rational and Irrational Numbers:

- Rational numbers are those that can be expressed in the form of a/b, where a and b are integers and b is not equal to 0.
- Irrational numbers are those that cannot be expressed in the form of a/b, where a and b are integers.

Decimal Expansion of Rational Numbers:

- The decimal expansion of a rational number is either terminating (i.e. has a finite number of decimal places) or non-terminating repeating (i.e. has a repeating pattern of decimal places).
- The decimal expansion of a rational number can be found by dividing the numerator by the denominator.

Square Roots:

- The square root of a number is a value that, when multiplied by itself, gives the original number.
- The square root of a non-perfect square number is an irrational number.
- The square root of a perfect square number is a rational number.

I hope these notes help you!

## CBSE Class 10 Maths Important Questions Chapter 1 Real Numbers

- Prove that √2 is irrational.
- Prove that the square of any odd integer is of the form 4q + 1.
- Show that there is no rational number whose square is 10.
- Express 0.66666… in the form of p/q where p and q are integers.
- If a and b are any two positive integers, prove that (a + b) and (a – b) have the same HCF as (a) and (b).
- Prove that there is no rational number whose square is 27.
- Prove that the product of two irrational numbers can be rational.
- If x = √3 + √2, find the value of x² + 1/x².
- Find the HCF of 4052 and 12576 using Euclid’s division algorithm.
- Show that √5 + √3 is irrational.

**CBSE Class 10 Maths Important Questions Answers Chapter 1 Real Numbers**

- To prove that √2 is irrational, we assume the opposite i.e. √2 is rational. Then, we can write √2 = p/q, where p and q are coprime integers. Squaring both sides, we get 2 = p²/q². This implies that p² is even, which in turn implies that p is even (as the square of an odd number is odd). Let p = 2k for some integer k. Substituting this in √2 = p/q, we get √2 = 2k/q. Squaring both sides, we get 2 = 4k²/q², which simplifies to q² = 2k². This implies that q² is even, which in turn implies that q is even. But this contradicts the assumption that p and q are coprime. Hence, our assumption that √2 is rational is false, and therefore √2 is irrational.
- Let n be any odd integer. Then, n can be written as 2k + 1 for some integer k. The square of n is (2k + 1)² = 4k² + 4k + 1 = 4(k² + k) + 1. Let q = k² + k. Then, the square of n is of the form 4q + 1.
- Let’s assume that there exists a rational number p/q whose square is 10. Then, we have (p/q)² = 10. This implies that p² = 10q². Since 10 is not a perfect square, it must have a prime factorisation that contains a prime factor raised to an odd power. Let p be such a prime factor. Then, p divides 10 and hence p divides p² and 10q². This implies that p divides p² – 10q², which is a contradiction as p is a prime factor of p² but not of 10q². Therefore, there is no rational number whose square is 10.
- Let x = 0.66666…. Multiplying both sides by 10, we get 10x = 6.6666…. Subtracting x from 10x, we get 9x = 6. Simplifying, we get x = 2/3.
- Let d be the HCF of a and b. Then, we can write a = dx and b = dy, where x and y are coprime integers. Now, (a + b) = d(x + y) and (a – b) = d(x – y). Let h be the HCF of (a + b) and (a – b). Then, h divides (a + b) – (a – b) = 2b and h divides (a + b) + (a – b) = 2a. Hence, h divides 2d. Now, since x and y are coprime, x + y and x – y are coprime. Therefore, h = 1 or 2. But h cannot be 2 as otherwise d would be an even factor of a and b, which contradicts the fact that d is the HCF. Hence, h = 1, and therefore (a + b) and (a – b) have the same HCF as (a) and (b).

- Let’s assume that there exists a rational number p/q whose square is 27. Then, we have (p/q)² = 27. This implies that p² = 27q². But 27 is not a perfect square, and hence its prime factorization contains a prime factor raised to an odd power. Let p be such a prime factor. Then, p divides 27 and hence p divides p² and 27q². This implies that p divides p² – 27q², which is a contradiction as p is a prime factor of p² but not of 27q². Therefore, there is no rational number whose square is 27.
- The product of two irrational numbers can be rational. For example, let’s consider √2 and -√2. Both are irrational numbers as they cannot be expressed as a ratio of two integers. However, their product is (-√2)(√2) = -2, which is a rational number.
- Given x = √3 + √2, we can find the value of x² + 1/x² as follows: x² + 1/x² = (√3 + √2)² + 1/((√3 + √2)²) = 3 + 2√3√2 + 2 + 1/(3 + 2√3√2 + 2) = 5 + 2√6 + 1/(5 + 2√6)
- To find the HCF of 4052 and 12576 using Euclid’s division algorithm, we perform the following steps:

Step 1: Divide the larger number by the smaller number and find the remainder. 12576 ÷ 4052 = 3 with a remainder of 1412

Step 2: Replace the larger number with the smaller number and the smaller number with the remainder obtained in the previous step. 4052 ÷ 1412 = 2 with a remainder of 1228

Step 3: Repeat step 1 and step 2 until the remainder becomes 0. 1412 ÷ 1228 = 1 with a remainder of 184 1228 ÷ 184 = 6 with a remainder of 4 184 ÷ 4 = 46 with a remainder of 0

Since the remainder becomes 0, the HCF of 4052 and 12576 is the last non-zero remainder obtained, which is 4.

- To show that √5 + √3 is irrational, we can follow a similar proof as we did for √2 being irrational. Let’s assume that √5 + √3 is rational, i.e., it can be expressed as a ratio of two integers. Then, we can write √5 + √3 = p/q, where p and q are coprime integers. Squaring both sides, we get 5 + 2√15 + 3 = p²/q². This implies that 2√15 = p²/q² – 8. Since the left-hand side is irrational (2√15), the right-hand side must also be irrational. But the right-hand side is a difference of two rational numbers, which contradicts the fact that it is irrational. Hence, our assumption that √5 + √3 is rational is false, and therefore √5 + √3 is irrational.

**CBSE Class 10 Maths Important Questions Answers MCQs Chapter 1 Real Numbers**

The product of two irrational numbers can be:

a) Rational

b) Irrational

c) Integer

d) Prime

Answer: b) Irrational

The sum of an irrational number and a rational number is always:

a) Irrational

b) Rational

c) Integer

d) Prime

Answer: a) Irrational

Which of the following is a rational number?

a) √3

b) π

c) -2/3

d) e

Answer: c) -2/3

Which of the following is an irrational number?

a) 0.25

b) 1/5

c) √7

d) -4

Answer: c) √7

Which of the following is not a real number?

a) 3/0

b) √(-9)

c) 5/7

d) π

Answer: b) √(-9)

Which of the following is not a rational number?

a) 0.5

b) 2.25

c) 2/3

d) √2

Answer: d) √2

The square root of which of the following numbers is not a real number?

a) 16

b) -16

c) 0

d) 25

Answer: b) -16

Which of the following numbers is not a prime number?

a) 3

b) 7

c) 11

d) 15

Answer: d) 15

Which of the following is not a perfect square?

a) 36

b) 64

c) 49

d) 51

Answer: d) 51

Which of the following is a prime number?

a) 27

b) 37

c) 50

d) 64

Answer: b) 37

## CBSE Class 10 Maths Notes Chapter 1 Real Numbers

**R = Real Numbers:**

All rational and irrational numbers are called real numbers.

**I = Integers:**

All numbers from (…-3, -2, -1, 0, 1, 2, 3…) are called integers.

**Q = Rational Numbers:**

Real numbers of the form pq, q ≠ 0, p, q ∈ I are rational numbers.

- All integers can be expressed as rational, for example, 5 = 51
- Decimal expansion of rational numbers terminating or non-terminating recurring.

**Q’ = Irrational Numbers:**

Real numbers which cannot be expressed in the form pq and whose decimal expansions are non-terminating and non-recurring.

- Roots of primes like √2, √3, √5 etc. are irrational

**N = Natural Numbers:**

Counting numbers are called natural numbers. N = {1, 2, 3, …}

**W = Whole Numbers:**

Zero along with all natural numbers are together called whole numbers. {0, 1, 2, 3,…}

**Even Numbers:**

Natural numbers of the form 2n are called even numbers. (2, 4, 6, …}

**Odd Numbers:**

Natural numbers of the form 2n -1 are called odd numbers. {1, 3, 5, …}

- Why can’t we write the form as 2n+1?

**Remember this!**

- All Natural Numbers are whole numbers.
- All Whole Numbers are Integers.
- All Integers are Rational Numbers.
- All Rational Numbers are Real Numbers.

**Prime Numbers:**

The natural numbers greater than 1 which are divisible by 1 and the number itself are called prime numbers, Prime numbers have two factors i.e., 1 and the number itself For example, 2, 3, 5, 7 & 11 etc.

- 1 is not a prime number as it has only one factor.

**Composite Numbers:**

The natural numbers which are divisible by 1, itself and any other number or numbers are called composite numbers. For example, 4, 6, 8, 9, 10 etc.

Note: 1 is neither prime nor a composite number.

**I. Euclid’s Division lemma**

Given two positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0 ≤ r ≤ b.

Notice this. Each time ‘r’ is less than b. Each ‘q’ and ‘r’ is unique.

**II. Application of lemma**

Euclid’s Division lemma is used to find HCF of two positive integers. Example: Find HCF of 56 and 72 ?

Steps:

- Apply lemma to 56 and 72.
- Take bigger number and locate ‘b’ and ‘r’. 72 = 56 × 1 + 16
- Since 16 ≠ 0, consider 56 as the new dividend and 16 as the new divisor. 56 = 16 × 3 + 8
- Again, 8 ≠ 0, consider 16 as new dividend and 8 as new divisor. 16 = 8 × 2 + 0

**Since remainder is zero, divisor (8) is HCF.**

Although Euclid’s Division lemma is stated for only positive integers, it can be extended for all integers except zero, i.e., b ≠ 0.

**III. Constructing a factor tree**

Steps

- Write the number as a product of prime number and a composite number

Example:

Factorize 48 - Repeat the process till all the primes are obtained

∴ Prime factorization of 48 = 2^{4}x 3

**IV. Fundamental theorem of Arithmetic**

Every composite number can be expressed as a product of primes, and this expression is unique, apart from the order in which they appear.**Applications:**

- To locate HCF and LCM of two or more positive integers.
- To prove irrationality of numbers.
- To determine the nature of the decimal expansion of rational numbers.

**1. Algorithm to locate HCF and LCM of two or more positive integers:**

**Step I:**

Factorize each of the given positive integers and express them as a product of powers of primes in ascending order of magnitude of primes.**Step II:**

To find HCF, identify common prime factor and find the least powers and multiply them to get HCF.**Step III:**

To find LCM, find the greatest exponent and then multiply them to get the LCM.

**2. To prove Irrationality of numbers:**

- The sum or difference of a rational and an irrational number is irrational.
- The product or quotient of a non-zero rational number and an irrational number is irrational.

**3. To determine the nature of the decimal expansion of rational numbers:**

- Let x = p/q, p and q are co-primes, be a rational number whose decimal expansion terminates. Then the prime factorization of’q’ is of the form 2
^{m}5^{n}, m and n are non-negative integers. - Let x = p/q be a rational number such that the prime factorization of ‘q’ is not of the form 2
^{m}5^{n}, ‘m’ and ‘n’ being non-negative integers, then x has a non-terminating repeating decimal expansion.

**Alert!**

- 2
^{3}can be written as: 2^{3}= 2^{3}5^{0} - 5
^{2}can be written as: 5^{2}= 2^{0}5^{2}

- CBSE Class 10 Maths Notes & Important Questions Chapter 15 Probability
- CBSE Class 10 Maths Notes & Important Questions Chapter 14 Statistics
- CBSE Class 10 Maths Notes & Important Questions Chapter 13 Surface Areas and Volumes
- CBSE Class 10 Maths Notes & Important Questions Chapter 11 Constructions
- CBSE Class 10 Maths Notes & Important Questions Chapter 10 Circles
- CBSE Class 10 Maths Notes & Important Questions Chapter 9 Some Applications of Trigonometry
- CBSE Class 10 Maths Notes & Important Questions Chapter 8 Introduction to Trigonometry
- CBSE Class 10 Maths Notes & Important Questions Chapter 7 Coordinate Geometry
- CBSE Class 10 Maths Notes & Important Questions Chapter 6 Triangles
- CBSE Class 10 Maths Notes & Important Questions Chapter 5 Arithmetic Progressions
- CBSE Class 10 Maths Notes & Important Questions Chapter 4 Quadratic Equations
- CBSE Class 10 Maths Notes & Important Questions Chapter 3 Pair of Linear equations in Two Variables
- CBSE Class 10 Maths Notes & Important Questions Chapter 2 Polynomials
- CBSE Class 10 Maths Notes & Important Questions Chapter 1 Real Numbers