# CBSE Class 10 Maths Notes & Important Questions Chapter 1 Real Numbers ## 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.

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

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

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

1. 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.
2. 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.
3. 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.
4. 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.
5. 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).
1. 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.
2. 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.
3. 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)
4. 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.

1. 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

The sum of an irrational number and a rational number is always:
a) Irrational
b) Rational
c) Integer
d) Prime

Which of the following is a rational number?
a) √3
b) π
c) -2/3
d) e

Which of the following is an irrational number?
a) 0.25
b) 1/5
c) √7
d) -4

Which of the following is not a real number?
a) 3/0
b) √(-9)
c) 5/7
d) π

Which of the following is not a rational number?
a) 0.5
b) 2.25
c) 2/3
d) √2

The square root of which of the following numbers is not a real number?
a) 16
b) -16
c) 0
d) 25

Which of the following numbers is not a prime number?
a) 3
b) 7
c) 11
d) 15

Which of the following is not a perfect square?
a) 36
b) 64
c) 49
d) 51

Which of the following is a prime number?
a) 27
b) 37
c) 50
d) 64

## 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 = 24 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:

1. To locate HCF and LCM of two or more positive integers.
2. To prove irrationality of numbers.
3. 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 2m5n, 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 2m5n, ‘m’ and ‘n’ being non-negative integers, then x has a non-terminating repeating decimal expansion.