Real numbers are all the numbers that can be found on the number line. This includes both rational numbers (like 7, -3, 0.5, and 4/3) and irrational numbers (like √2) . They encompass integers, fractions, and decimals, representing a continuous, unbroken set of values.

Rational numbers such as integers (-2, 0, 1), fractions (1/2, 2.5) and irrational numbers such as √3, π(22/7), etc., are all real numbers.

Classification of Real Numbers

Natural Numbers: Natural Numbers are a set of counting numbers. They are denoted by N.
N = {1, 2, 3, 4..........∞}
Whole Numbers: Whole numbers are a set of natural numbers plus zero.
W = {0, 1, 2, 3 ......... ∞}
Integers: Integers is a set of whole numbers and negative of all natural numbers.
Z = { -3, -2, -1, 0, 1, 2, 3}
Rational Numbers: All the numbers that can be written in the p/q form where p and q are integers and q ≠ 0 are called rational numbers.
E.g. 8/11, -3/17
Irrational Numbers: All the numbers that cannot be written in the p/q form are called irrational numbers. All the non-terminating and non-repeating decimal numbers are irrational numbers.
E.g. √5, √3, √5 + √3, π

Fundamental Theorem of Arithmetic

To understand the fundamental theorem of Arithmetic, first, it is important to know what are composite numbers and prime numbers.

Composite Number

Composite Numbers are those numbers that have at least one factor other than one and the number itself.

Consider a number, 10. Now, the factors of 10 are 1, 2, 5 and 10. So it is a composite number.

Prime Number

Prime Numbers are those numbers that have exactly two factors: 1 and the number itself.

Let us take one more number 23. Now, the factors of 23 are 1 and 23. That means it has two factors 1 and the number itself, which is called a prime number.

Prime and Composite Numbers

Theorem: Fundamental Theorem of Arithmetic states that every composite number can be expressed as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.

Any composite number can be written as a product of primes in one way only as long as we are not particular about the order in which the primes occur.

Let us consider an example here: we will use a tree diagram to show the factors of 270.

270 = 2 x 33 x 5

Here, in the prime factorization of 270, the prime numbers appearing in both cases are the same, only the order in which they appear is different.

Therefore, the prime factorization of 270 is unique except for the order in which the primes occur.

Example 1: Check whether 15n can end with the digit zero for any natural number n.

Sol:

Example 2: Explain, why (7 × 6 × 5 × 4 × 3 × 2 × 1) + 5 and (3 × 5 × 13 × 46) + 23 is a composite number?

Sol:

HCF and LCM by Prime Factorisation Method

In this method, we first express the given numbers as a product of prime factors separately. Then, HCF is the product of the smaller power of each common prime factor in the numbers, and LCM is the product of the greatest power of each prime factor involved in the numbers.

For any two positive integers a and b,

HCF (a, b) × LCM (a, b) = a × b

Example 3: Find the LCM and HCF of 120 and 144 by the fundamental arithmetic theorem.

Sol:

120 = 23 × 3 × 5

144 = 24 × 32

Now, HCF is the product of the smallest power of each common prime factor in the numbers.

HCF (120, 144) = 23 × 3 = 8 × 3 = 24

LCM is the product of the greatest power of each prime factor involved in the numbers.

LCM (120,144) = 24 × 32 × 5 = 16 × 9 × 5 = 720

Example 4: If two positive integers p and q can be expressed as p = ab2 and q = a3b, where a, b are prime numbers, find the LCM (p, q).

Sol:

Given: p = ab2 and q = a3b

LCM is the product of the greatest power of each prime factor involved in the numbers.

LCM (p, q) = a3 × b2 = a3b2

Example 5: Write the HCF and LCM of the smallest odd composite number and the smallest odd prime number.

Sol:

The smallest odd composite number is 9, and the smallest odd prime number is 3.

9 = 32

3 = 31

Now, the smallest power of the common prime factor is 31.

HCF (9, 3) = 3

The greatest power of the common prime factor is 32.

LCM (9, 3) = 32 = 9

Example 6: If HCF (253,440) = 11 and LCM (253,440)= 253 × R. Find the value of R.

Sol:

We know that,

HCF (a, b) × LCM (a, b) = a × b

∴ HCF (253, 440) × LCM (253, 440) = 253 × 440

11 × 253 × R = 253 × 440

R = 253 x 440 / 253 x 11

R = 40

Example 7: Ravi and Shikha drive around a circular sports field. Ravi takes 16 min to complete one round, while Shikha completes the round in 20 min. If both start at the same point, at the same time, and go in the same direction, then how much time will they meet at the starting point?

Sol:

Time taken by Ravi to drive one round of the circular field = 16 min. Time taken by Shikha to drive one round of the circular field =20 min.

The time after which they will again meet at the starting point will be equal to the LCM of 16 min and 20 min.

16 = 24

20 = 22 × 5

LCM(16, 20) = 24 × 5 = 16 × 5 = 80

Therefore, Ravi and Shikha will meet again at the starting point after 80 min.

MULTIPLE CHOICE QUESTION

Try yourself: According to the Fundamental Theorem of Arithmetic, which of the following statements is true?

Every prime number can be expressed as a product of composites.

CORRECT ANSWER

Every composite number can be expressed as a product of primes, and this factorization is unique.

Every natural number can be expressed as a product of two prime numbers.

Every prime number can be expressed as a product of other prime numbers.

Correct Answer: B

The correct answer is:

b) Every composite number can be expressed as a product of primes, and this factorization is unique.

Reason:

The Fundamental Theorem of Arithmetic states that every integer greater than 1 is either a prime itself or can be written as a product of prime numbers, and this factorization is unique (apart from the order of the primes).

Revisiting Irrational Numbers

Irrational numbers are those numbers that cannot be written in the form p/q, where p and q are integers and q ≠0. E.g.,√2, √3, √15

The square roots of all the numbers do not give an irrational number.

For example, √2 is an irrational number, but √4 = 2, which is rational.

Therefore, the square roots of all prime numbers are irrational.

If p is a prime number, then √p is an irrational number.

Theorem : If a prime number p divides a2, then p divides a, where a is a positive integer.

Theorem: Prove that √2 is an irrational number.

Sol:

Example 8: Show that 3√2 is an irrational number.

Sol:

Example 9: Show that 5 - √3 is irrational.

Sol:

Important Definitions

1. Real Numbers

Real numbers constitute the union of all rational and irrational numbers.
In general, all the arithmetic operations can be performed on these numbers and they can be represented on the number line, also.

2. Fundamental Theorem of Arithmetic

Every composite number can be factorized as a product of primes and this factorization is unique, apart from the order in which the prime factors occur.
Example: 54=2×3×3×3
Therefore, 54 is represented as a product of prime factors (One 2 and three 3s) ignoring the arrangement of the factors.
Also, the number 120 can be factorized as 2 × 2 × 2 × 3 × 5. The prime factors are 2, 2, 2, 3, and 5. The order of the factors doesn't matter, as the Fundamental Theorem of Arithmetic guarantees a unique factorization regardless of the arrangement of the prime factors.

Example 1: What is the HCF of 36 and 48 using prime factorization?
Solution: To find the HCF using prime factorization, follow these steps:
1. Find the prime factors of each number.
Prime factors of 36: 2 × 2 × 3 × 3 = (22 × 32)
Prime factors of 48: 2 × 2 × 2 × 2 × 3 = (24 × 3)
2. Find the common prime factors.
Common prime factors: 2 × 2 × 3 = (22 × 3)
3. Multiply the common prime factors to get the HCF.
HCF = 2 × 2 × 3 = 12
So, the HCF of 36 and 48 using prime factorization is 12.

Example 2 : Consider the number 4n, where n is a natural number. Is there any value of n for which 4n ends with the digit zero? Justify your answer.
Solution:If 4n, for any n, were to end with the digit zero, it would need to be divisible by 10. For a number to be divisible by 10, its prime factorization must include both the primes 2 and 5.
The prime factorization of 4n is:
4n=(2)2n
This shows that the only prime factor in 4n is 2. Since there is no 5 in the factorization, 4n cannot be divisible by 10.
Thus, by the Fundamental Theorem of Arithmetic, which ensures the uniqueness of prime factorization, there are no natural numbers n for which 4n ends with the digit zero.

3. Rational Numbers

The decimal expansion of every rational number is either terminating or non-terminating repeating.

Terminating Decimal Representation: Let x be a rational number with a terminating decimal representation. Then we can express x as p/q where p and q are coprime, and the prime factorization of q is of the form 2n 5m, where n, m are some non-negative integers.
Non-terminating Decimal Representation: The rational number p/q will have a non-terminating repeating(recurring) decimal representation if, in standard form, the prime factorization of q is not of form 2n 5m, where n, m is some non-negative integers.

4. Irrational Number

A number is irrational if and only if, its decimal representation is non-terminating and non-repeating (non-recurring).
OR

A number that cannot be expressed in the form of p/q, q ≠ 0 and p, q ∈ I, will be an irrational number. The set of irrational numbers is generally denoted by S.

Example 3: Prove that 3 + 2√5 is irrational.
Solution: Let 3 + 2√5 be a rational number.
Then the co-primes x and y of the given rational number where (y ≠ 0) is such that:
3 + 2√5 = x/y
Rearranging, we get,
2√5 = (x/y) - 3
√5 = 1/2[(x/y) - 3]
Since x and y are integers, thus, 1/2[(x/y) - 3] is a rational number.
Therefore, √5 is also a rational number. But this confronts the fact that √5 is irrational.
Thus, our assumption that 3 + 2√5 is a rational number is wrong.
Hence, 3 + 2√5 is irrational.

MULTIPLE CHOICE QUESTION

Try yourself: Find out the decimal expansion of 13/125.

CORRECT ANSWER

Terminating decimal expansion

Non-terminating repeating decimal expansion

Non-terminating non-repeating decimal expansion

None of these

Correct Answer: A

We are asked to determine whether 13/125 has a terminating decimal expansion without actually dividing.

A rational number p/q will have a terminating decimal expansion if:

p and q are co-prime, and
the prime factorisation of q has only 2s and/or 5s (i.e., q is of the form 2ⁿ × 5ᵐ)

Step 1: Check if 13 and 125 are co-prime

They have no common factors other than 1, so yes, they are co-prime.

Step 2: Prime factorisation of the denominator (125)

125 = 5 × 5 × 5 = 5³

This is of the form 2⁰ × 5³.

Since the denominator has only the prime factor 5, it satisfies the condition.

So, 13/125 has a terminating decimal expansion, and

13 ÷ 125 = 0.104

5. Prime Number

A prime number is a natural number greater than 1 that has exactly two distinct positive divisors: 1 and the number itself.

In other words, a prime number cannot be formed by multiplying two smaller natural numbers. The first few prime numbers are 2, 3, 5, 7, 11, 13, 17, and so on.

Example 4: Express each number as a product of its prime factors:
(i) 140
(ii) 156
Solution: (i) 140
Using the division of a number by prime numbers method, we can get the product of prime factors of 140.
Therefore, 140 = 2 × 2 × 5 × 7 × 1 = 22 × 5 × 7
(ii) 156
Using the division of a number by prime numbers method, we can get the product of prime factors of 156.
Hence, 156 = 2 × 2 × 13 × 3 = 22 × 13 × 3

Important Formulas with Examples

LCM & HCF of Two Numbers

For any two positive integers p and q, we have:

HCF (p, q) x LCM [p, q] = p x q

Example 5: Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution: As we know that,
HCF × LCM = Product of the two given numbers
So, 9 × LCM = 306 × 657
LCM = (306 × 657)/9
LCM = 22,338
Therefore, LCM (306,657) = 22,338

Worksheet:

Rational + Irrational Numbers = Real Numbers

True and False

Q1: State whether the given statement is true or false :

(i) The product of two rationals is always rational

(ii) The product of two irrationals is an irrational

(iii) The product of a rational and an irrational is irrational

(iv) The sum of two rationals is always rational

(v) The sum of two irrationals is an irrational.

Very Short Questions

Q.2. Classify the following numbers as rational or irrational:

(i) 3.1416

(ii) 3.142857

(iii) 2.040040004......

(iv) 3.121221222...

(v) 3√3

Q.3. Find the prime factorization of 1152

Q4: Show that the product of two numbers 60 and 84 is equal to the product of their HCF and LCM

Q5: If p and q are two coprime numbers, then find the HCF and LCM of p and q.

Short Questions

Q6: Given that H.C.F. (306, 657) = 9, find L.C.M. (306, 657)

Q7: The H.C.F. and L.C.M. of two numbers are 12 and 240 respectively. If one of these numbers is 48; find the other numbers.

Long Questions

Q8: In a school, the duration of a period in junior section is 40 minutes and in senior section is 1 hour: If the first bell for each section ring at 9:00 a.m., when will the two bells ring together again?

Q9: The HCF of 408 and 1032 is expressible in the form 1032 m - 2040. Find the value of m. Also, find the LCM of 408 and 1032.

Q10: Prove that

(i) √2 is irrational number

(ii) √3 is irrational numberSimilarly √5, √7, √11...... are irrational numbers.

Key Questions

Q1: Use Euclid's division lemma to show that the square of any positive integer is of the form 3m or 3m + 1.

Solution:

Let a be any positive integer. By Euclid's lemma, a = 3q, 3q + 1 or 3q + 2. Squaring each case gives a² = 9q², 9q² + 6q + 1 or 9q² + 12q + 4, which can be written in the form 3m or 3m + 1.

Q2: Prove that √5 is irrational.

Solution:

Assume √5 is rational, so it can be written as p/q in lowest form. Squaring gives 5q² = p², implying p is divisible by 5. Let p = 5k, then substituting gives q also divisible by 5, which contradicts lowest form. Hence √5 is irrational.

Q3: State and prove the Fundamental Theorem of Arithmetic.

Solution:

The Fundamental Theorem of Arithmetic states that every composite number can be expressed as a product of prime numbers uniquely, apart from the order. If a number is written in two different ways as products of primes, then by comparing factors, both forms must be identical, proving uniqueness.

Q4: Find the HCF and LCM of 120 and 144 using prime factorisation.

Solution:

120 = 2³ × 3 × 5 and 144 = 2⁴ × 3². HCF is the product of lowest powers of common primes = 2³ × 3 = 24. LCM is the product of highest powers of primes = 2⁴ × 3² × 5 = 720.

Q5: Show that the product of two consecutive positive integers is divisible by 2.

Solution:

Let the integers be n and n + 1. Their product is n(n + 1). One of the two consecutive integers must be even, so their product is always divisible by 2.

Q6: Find a rational number between √2 and √3.

Solution:

√2 ≈ 1.41 and √3 ≈ 1.73. A number between them is 1.5, which is rational and lies between √2 and √3.

Q7: Show that 6n cannot end with digit 0 for any natural number n.

Solution:

For a number to end with 0, it must contain factors 2 and 5. The prime factorisation of 6ⁿ contains only 2 and 3. Since 5 is not present, 6ⁿ cannot end with 0.

Q8: Determine whether the decimal expansion of 43/(2⁴ × 5³) is terminating.

Solution:

The denominator has only prime factors 2 and 5. Therefore, the decimal expansion will be terminating.

Q9: Find HCF of 52 and 117 using Euclid's algorithm.

Solution:

117 = 52 × 2 + 13

52 = 13 × 4 + 0

Hence HCF = 13.

Q10: If HCF of two numbers is 12 and their product is 1800, find their LCM.

Solution:

Product of numbers = HCF × LCM

1800 = 12 × LCM

LCM = 150.

Q11: Show that √2 + √3 is irrational.

Solution:

Assume √2 + √3 is rational. Then √3 = rational - √2, which would be irrational, contradicting assumption. Hence √2 + √3 is irrational.

Q12: Prove that there are infinitely many prime numbers.

Solution:

Assume finite primes p₁, p₂, ..., pₙ. Consider N = p₁p₂...pₙ + 1. N is not divisible by any of these primes, so either N is prime or has a new prime factor, contradicting the assumption. Hence infinitely many primes exist.

Q13: Express 0.375 as a rational number.

Solution:

0.375 = 375/1000 = 3/8.

Q14: Find the LCM and HCF of 98 and 28.

Solution:

98 = 2 × 7² and 28 = 2² × 7. HCF = 2 × 7 = 14 and LCM = 2² × 7² = 196.

Q15: Check whether 7ⁿ can end with digit 0 for any natural number n.

Solution:

7ⁿ contains only factor 7. Since factor 5 is required to end with 0 and it is absent, 7ⁿ cannot end with digit 0.

Short Answer Questions

Q1: Prove that √3 is irrational.

Let √3 be rational in the simplest form of P/q.
i.e., p and q are integers having no common factor other than 1 and q ≠ 0.
Now, √3 = p/q
Squaring both sides, we have
⇒ (√3)2 = (p/q)2
⇒ 3 = p2/q2
⇒ 3q2 = p2 ............(1)
Since 3q2 is divisible by 3
∴ p2 is also divisible by 3
⇒ p is divisible by 3 ..........(2)
Let p = 3c for some integer 'c'.
Substituting p = 3c in (1), we have:
⇒ 3q2  = (3c)2
⇒ 3q2 = 9c2
⇒ q2 = 3c2
3c2 is divisible by 3
∴ q2 is divisible by 3
⇒ q is divisible by 3 ...(3)
From (2) and (3)
3 is a common factor of 'p' and 'q'. But this contradicts our assumption that p and q are having no common factor other than 1.
∴ Our assumption that √3 is rational is wrong.
Thus, √3 is an irrational.

Q2: If 'p' is prime, prove that √p is irrational.

Let √p be rational in the simplest form a/b, where p is prime.
∴ a and b are integers having no common factor other than 1 and b ≠ 0.
⇒ Now, √p = a/b
⇒ Squaring both sides, we have
⇒ pb2 = a2 ...(1)
⇒ Since pb2 is divisible by p, a2 is also divisible by p.
⇒ a is also divisible by p ...(2)
Let a = pc for some integer c.
⇒ Substituting a = pc in (1), we have
pb2 = (pc)2
⇒ pb2 = p2c2
⇒ b2 = pc2
pc2 is divisible by p,
∴ b2 is divisible by p
⇒ b is divisible by p ...(3)
From (2) and (3),
⇒ p is a common factor of 'a' and 'b'. But this contradicts our assumption that a and b are co-prime.
∴ Our assumption that √p is rational is wrong. Thus, √p is irrational if p is prime.

Q3: Find the HCF of 18 and 24 using prime factorisation.

Using factor tree method, we have:
∴ 18 = 2 × 3 × 3 = 2 × 32
24 = 2 × 2 × 2 × 3 = 23 × 3
HCF = Product of common prime factors with lowest powers.
⇒ HCF (18, 24) = 3 × 2 = 6

Q4: Find the LCM of 10, 30 and 120.

∴ 10 = 2 × 5 = 21 × 51
30 = 2 × 3 × 5 = 21 × 31 × 51
120 = 2 × 2 × 2 × 3 × 5 = 23 × 31 × 51
LCM = Product of each prime factor with highest powers
⇒ LCM of 10, 30 and 120 = 23 × 3 × 5 = 120.

Q5: Find the LCM and HCF of 1296 and 5040 by prime factorisation method.

and
∴ 5040 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 7
= 24 × 32 × 5 × 7 and
⇒ 1296 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3
= 24 × 34
∴ LCM = Product of each prime factor with highest powers
= 24 × 34 × 5 × 7
= 16 × 81 × 5 × 7 = 45360
HCF = Product of common prime factors with lowest powers
= 24 × 32
= 16 × 9 = 144

Q6: Show that 3√2 is irrational.

Let 3√2 be a rational number
∴ p/q = 3√2 where p and q are prime to each other and q ≠ 0.
∴ p/3q = √2 ...(1)
Since, p is integer and 3q is also integer (3q≠ 0).
∴ p/3q is a rational number.
From (1), √2 is a rational number.
But this contradicts the fact that √2 is irrational. Therefore, our assumption that 3√2 is rational is incorrect.
Hence, 3√2 is irrational.

Q7: Show that 2 - √3 is an irrational number.

Let 2 - √3 is rational.
∴ It can be expressed as p/q where p and q are integers (prime to each other) such that q ≠ 0.
∴ 2 - √3 = p/q
⇒   ...(1)
∵ p is an integer}
∴ q is an integer}
⇒ p/q is a rational number.
∴  is a rational number. ...(2)
From (1) and (2), √3 is a rational number. This contradicts the fact that √3 is an irrational number.
∴ Our assumption that (2 -√3) is a rational number is not correct. Thus, (2 - √3) is irrational.

Q8: State whether is a rational number or not.

= 1.23333..... is a non-terminating repeating decimal.
∴ is a rational number.
3/4 is in the form of p/q, where q ≠ 0 [Here 4 ≠ 0]
∴ 3/4 is a rational number.
Since the sum of two rational numbers is a rational number.
Therefore, is a rational number.

Q9: The LCM of two numbers is 45 times their HCF. If one of the numbers is 225 and sum of their LCM and HCF is 1150, find the other number.

One of the numbers = 225
Let the other number = x
Also LCM = 45 (HCF) ...(1)
And LCM + HCF =1150
⇒ (45 HCF) + HCF = 1150
⇒ 46 HCF = 1150
⇒ HCF = 1150/46 = 25
From (1),
LCM = 45 × 25
∴ LCM × HCF = (45 × 25) × 25
Now, LCM × HCF = Product of the numbers
∴ x × 225 = (45 × 25) × 25
⇒
= 125
Thus, the required number is 125.

Q10: Three different containers contain 496 litres, 403 litres and 713 litres of a mixture. What is the capacity of the biggest container that can measure all the different quantities exactly?

For the capacity of the biggest container, we have to find the HCF.
HCF: By Long Division method
First find the HCF of two numbers, 496 and 403
The HCF of 496 and 403 = 31
Now find the HCF of 31 and 713
HCF of 713 and 31 is 31
So, the maximum capacity is 31 liters.

Q11: Prove that (5+3√2) is an irrational number.

Sol:

Let (5+3√2) is a rational number.
∴ (5+3√2) = a/b [where 'a' and 'b' are co-prime integers and b ≠ 0
⇒
⇒
⇒
'a' and 'b' are integers,
∴ is a rational number.
⇒ √2 is a rational number.
But this contradicts the fact that √2 is an irrational number.
∴ Our assumption that (5+3√2) is a rational is incorrect.
⇒ (5+3√2) is an irrational number.

Q12: Prove that 3-√5 is an irrational number.

Sol:

Let (3-√5) is a rational number.
∴ 3-√5 = p/q , such that p and q are co-prime integers and q ≠ 0.
⇒
⇒
Since, p and q are integers,
∴ is a rational number.
⇒ √5 is a rational number.
But this contradicts the fact that √5 is an irrational number.
∴ Our assumption that (3-√5) is a rational number' is incorrect.
⇒ (3-√5) is an irrational number.

Q13: Show that there is no positive integer 'p' for which is rational.

Sol:

If possible let there be a positive integer p for which = a/b is equal to a rational i.e. where a and b are positive integers.
Now
Also,
Since a, b are integer
are rationals
⇒ (p + 1) and (p - 1) are perfect squares of positive integers, which is not possible (because any two perfect squares differ at least by 3). Hence, there is no positive integer p for which is rational.

Q14: Prove that is irrational, where p and q are primes.

Sol:

Let be rational
Let it be equal to 'r'
i.e.
Squaring both sides, we have
⇒
⇒ ...(i)
Since, p, q are both rationals
Also, r2 is rational (∵ r is rational)
∴ RHS of (i) is a rational number
⇒ LHS of (i) should be rational i.e.√q should be rational.
But √q is irrational (∵ p is prime).
∴ We have arrived at a contradiction.
Thus, our supposition is wrong.
Hence, √p+√q is irrational.

Unit Test

Time: 1 hour

M.M. 30

Attempt all questions.

Question numbers 1 to 5 carry 1 mark each.
Question numbers 6 to 8 carry 2 marks each.
Question numbers 9 to 11 carry 3 marks each.
Question number 12 & 13 carry 5 marks each.

Q1: Which of the following numbers is irrational? (1 Mark)

(a) √25

(b) 3.14

(d) -7

Q2: What is the value of (5² + 12²)? (1 Mark)

(a) 169

(b) 144

(d) 169√2

Q3: Which one is not a prime number? (1 Mark)

(a) 1

(b) 2

(d) 5

Q4: State whether "√16" is a rational number or not. (1 Mark)

Q5:What is the prime factorization of 96? (1 Mark)

Q6: Prove that the square of any positive integer of the form (5k + 1) is one more than a multiple of 8, where "k" is an integer. (2 Marks)

Q7: Prove that √2 is irrational. (2 Marks)

Q8: Find the LCM (Least Common Multiple) of 15 and 20. (2 Marks)

Q9: Sonia takes 18 minutes to complete one round of a circular track, while Ravi takes 12 minutes for the same. If they start together from the same point, after how many minutes will they meet again at the starting point? (3 Marks)

Q10: Find the HCF (Highest Common Factor) of 72 and 96 using the prime factorization method. (3 Marks)

Q11: Check whether 6n can end with the digit 0 for any natural number n(3 Marks)

Q12: Given that p is a rational number and q is an irrational number, prove that their sum (p + q) is an irrational number. (5 Marks)

Q13: Prove that 5√3 - 3√75 is an irrational number. (5 Marks)

NCERT Solutions: Real Numbers (Exercise 1.1)

Q1. Express each number as a product of its prime factors:

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

Solutions:

(i) 140

Prime factors of 140 = 2, 2, 5, 7

= 22 x 5 x 7

(ii) 156

Prime factors of 156 = 2 x 2 x 3 x 13

= 22 x 3 x 13

(iii) 3825

Prime factors of 3825 = 3 x 3 x 5 x 5 x 17

= 32 x 52 x 17

(iv) 5005

Prime factors of 5005 = 5 x 7 x 11 x 13

(v) 7429

Prime factors of 7429 = 17 x 19 x 23

Q2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.

(i) 26 and 91

(ii) 510 and 92

(iii) 336 and 54

Solutions:

(i) 26 and 91

Prime factors of 26 = 2 x 13

Prime factors of 91 = 7 x 13

HCF of 26 and 91 = 13

LCM of 26 and 91 = 2 x 7 x 13

= 14 x 13

= 182

Product of two numbers = 26 x 91

= 2,366

LCM x HCF = 182 x13 = 2,366

So, product of two numbers = LCM x HCF

(ii) 510 and 92

Prime factors of 510 = 2 x 3 x 5 x17

Prime factors of 92 = 2x 2 x 23

HCF of two numbers = 2

LCM of two numbers = 2 x 2 x 3 x 5 x 17 x 23

= 23460

Product of two numbers = 510x92

= 46920

LCM x HCF = 2 x 23460

=46920

Product of two numbers = LCM x HCF

(iii) 336 and 54

Prime factors of 336 = 2 x 2 x 2 x 2 x 3 x 7

Prime factors of 54 = 2 x 3 x 3 x 3

HCF of two numbers = 6

LCM of two numbers = 2 x 2 x 2 x 2 x 3 x 3 x 3 x 7

= 24 x 33 x 7 = 3024

Product of two numbers = 336 x 54

=18144

LCM x HCF = 3024 x 6 =18144

Product duct of two numbers = LCM x HCF

Q3. Find the LCM and HCF of the following integers by applying the prime factorisation method.

(i) 12, 15 and 21

(ii) 17, 23 and 29

(iii) 8, 9 and 25

Solutions:

(i) 12, 15 and 21

Prime factors of 12 = 2 x 2 x 3

= 22 x 3

Prime factors of 15 = 3 x 5

Prime factors of 21 = 3 x 7

HCF of 12,15 and 21 = 3

LCM of 12,15 and 21 = 22 x 3 x 5 x 7

= 420

(ii) 17, 23 and 29

Prime factors of 17 = 17 x 1

Prime factors of 23 = 23 x 1

Prime factors of 29 = 29 x 1

HCF of 17, 23 and 29 = 1

LCM of 17, 23 and 29 = 17 x 23 x 29

= 11339

(iii) 8, 9 and 25

Prime factors of 8 = 2 x 2 x 2 x1

= 23 x 1

Prime factors of 9 = 3 x 3 x1

= 32 x 1

Prime factors of 25 = 5 x 5 x 1

= 52 x 1

HCF of 8, 9 and 25 = 1

LCM of 8,9 and 25 = 2 x 2 x 2 x 3 x 3 x 5 x 5

= 1800

Q4. Given that HCF (306, 657) = 9, find LCM (306, 657).

Solution:

Given, HCF (306, 657) = 9.

We have to find,

LCM (306, 657) = ?

We know that

LCM x HCF = Product of two numbers

LCM x 9 = 306x 657

LCM = 34 x 657

LCM = 22338

Q5. Check whether 6n can end with the digit 0 for any natural number n.

Solution:

Prime factors of 6n = (2 x 3)n = (2)n (3)n

You can observe clearly, 5 is not in the prime factors of 6n.

That means 6n will not be divisible by 5.

Therefore, 6n cannot end with the digit 0 for any natural number n.

Q6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Solution:

Checking, 7 × 11 × 13 + 13

It can be observed that,

7 x 11 x 13 + 13 = 13 (7 x11 +1)

= 13( 77 + 1)

= 13 x 78

= 13 x13 x 6 x1

= 13 x13 x 2 x 3 x1

The given number has 2, 3, 13 and, 1 as its factors. Therefore, it is a composite number.

Checking, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

Q7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

Solution:

LCM of 18 and 12,

18 = 2 x 3 x 3

12 = 2 x 2 x 3

LCM of 12 and 18 = 2 x 2 x 3 x 3 = 36

Therefore, Ravi and Sonia will meet together at starting point after 36 minutes.

HOTS

Q1. Show that 6n cannot end with the digit 0.

Ans:

∵ All composite numbers ending with 0 have 10 as a factor.

∴10 is a factor of 6n

∴ We have: 6n = 10 × p, where p is a natural number

⇒ (2 × 3)n = 2 × 5 × p

⇒ 2n × 3n = 2 × 5 × p

i.e. 5 is a prime factor of 2n × 3n which is not possible.

Thus, our assumption was wrong and 6n cannot end with the digit 0.

Q2. Show that there is no integer for which

is rational.

Ans:

If possible, let 'n' be a positive integer such that

is rational.

Let

(using componendo and dividendo)

Since

are rational numbers.

∴

are also rational.

i.e. are perfect squares of +ve integers. It is not possible.

[∵ two perfect squares differ by 3] ⇒ Our assumption is not possible.

Q3. If

is an irrational number, then show that

is also irrational.

Ans:

If possible, let

be a rational number.

But it contradicts our assumption thatis irrational. Thus, is irrational.

Q4. How many prime numbers are there of the form 10n + 3, where n is a whole number 1 < n < 10?

Ans:

Substituting 1, 2, 3, .................... 10 in 10 n + 3, we have:

Thus, the required numbers are 13, 23, 43, 53, 73 and 83.

Value-Based Questions

Q5. To raise funds for an orphanage various sports activities are planned in a school. In a cycling activity around a circular field Sonia takes 24 minutes to complete one round of the field while Ram takes 18 minutes for the same, suppose they both start at the same point and at the same time and go in the same direction

(i) After how many minutes will they meet again at the starting point?

(ii) Which concept of mathematics is used in the above problem?

(iii) By organising sports activities for raising funds for an orphanage which value is depicted by the school?

Ans:

(i) Time taken by Sonia to complete a round = 24 minutes

Time taken by Ram to complete a round = 18 minutes

The LCM of 24 and 18 gives the exact-number of minutes after which they meet at the starting point again.

24 = 2 × 2 × 2 × 3

18 = 2 × 3 × 3

∴ LCM of 24 and 18 = 2 × 2 × 2 × 3 × 3 = 72

⇒ They will meet again at the starting point-after 72 minutes.

(ii) Real numbers

(iii) Helping needy persons.

Q6. A dairy owner decided to supply pure milk to his customers. He collected 120 litres of pure cow-milk and 180 litres of pure buffalo-milk. He fills the two kinds of milk in containers of equal capacity.

(i) What is the greatest capacity of such container?

(ii) Which mathematical concept is used in the above problem?

(iii) By supplying pure-milk which value is depicted by the dairy-owner?

Ans:

(i) The maximum capacity of a pack of milk is equal to the HCF of 120 and 180.

We have

∴ HCF of 120 and 180 = 60

⇒ Capacity of each container = 60 litre

(ii) Real Numbers

(iii) Positive contribution to social-health.

Q7. In a food safety drive a team of food inspectors collected samples of adulterated 420 pieces of burfis and 148 pieces of ladoos. They want to pack these pieces in such a way that each pack has the same number of pieces of sweets.

(i) What is the maximum number of sweets pieces that can be placed in each pack?

(ii) Which mathematical concept is used in the above problem?

(iii) By collected the samples of adulterated sweets which value is depicted by the food inspectors?

Ans:

(i) For packing maximum number of sweets pieces in each pack, we find the HCF of 420 and 148.

∴ We have

⇒ HCF of 420 and 148 is 4.

⇒ Required number of sweets pieces

= 4

(ii) Real numbers

(iii) Check on public health.

Q8. A school has a book bank for the help of students from the economically weaker section. Students of class X donated the following number of books to the book bank.

English : 96

Hindi : 240

Mathematics : 336

These books have to be stacked in such a way that all the books are stored topic-wise and number of copies in each stack is the same.

(i) What is the maximum number of books in each stack?

(ii) Which mathematical concept is used in the above problem?

(iii) By donating books through the school book bank which value is depicted by the students of class X?

Ans:

(i) Maximum number of books in each pack is the HCF of 96, 240 and 336.

⇒ HCF of 96, 240 and 336 is 48.

⇒ The required maximum number of copies in each pack

= 48

(ii) Real numbers

(iii) Charity

Q9. Raghunath has to distribute bananas to the inhabitants of old-age-homes in a town. There are two old-age homes in the town-home-A and home-B. There are 32 old persons in home-A and 36 old persons in home B.

(i) Determine the minimum number of bananas required for the two old-age-homes that can be distributed equally among the inhabitants of home-A or home-B.

(ii) Which mathematical concept is used in the above problem?

(iii) By donating bananas to inhabitants of old-age homes which value is depicted by Raghunath?

Ans:

(i) The bananas are to be distributed equally among old persons of home-A and home-B.

The required number of bananas is the least multiple of 32 as well as 36.

Since, 32 = 2 × 2 × 2 × 2 × 2

= 25

36 = 2 × 2 × 3 × 3

= 22 × 32

∴ LCM of 32 and 36 = 25 × 32

= 288

Required number of bananas

= 288

(ii) Real numbers

(iii) Positive attitude towards senior citizens.

Very Short Answer Type Questions

Q1: For any two integers, the product of the integers = the product of their HCF and LCM. Is this relation true for three or more integers?

Ans: No

Q2: The decimal expansion of the rational number 43/2453 will terminate after how many places of decimals?

Ans: We have

= 0.0215

So, it will terminate after 4 places of decimals.

Q3: Write one rational and one irrational number lying between 0.25 and 0.32.

Ans: Given numbers are 0.25 and 0.32.

Clearly,

0.30 = 30100 = 310

Thus 0.30 is a rational number lying between 0.25 and 0.32. Also 0.280280028000.....has non-terminating non-repeating decimal expansion. It is an irrational number lying between 0.25 and 0.32

Q4: Express 98 as a product of its primes.

Ans: 98 = 2 × 7 × 7 = 2 × 7²

Q5: HCF and LCM of two numbers is 9 and 459 respectively. If one of the numbers is 27, find the other number.

Ans: We know,

1st number × 2nd number = HCF × LCM

⇒ 27 × 2nd number = 9 × 459

⇒ 2nd number = 9×45927 = 153

Q6: If HCF(336, 54) is 6 = , find LCM(336, 54).

Ans: HCF × LCM = Product of number

6×LCM = 336×54

LCM = (336 x 54)/6

= 56 × 54 = 3024

Thus LCM of 336 and 54 is 3024.

Q7: a and b are two positive integers such that the least prime factor of a is 3 and the least prime factor of b is 5. Then calculate the least prime factor of (a+b).

Ans: Given that a is a positive integer and 3 is least prime factor of a. Also, b is a positive integer and 5 is a least prime factor of b.

Since, least prime factor of a is 3, it implies that a is an odd number.

Similarly, b is also an odd number and we know that

odd + odd = even

So, a + b = even

The least prime factor of (a + b) is 2.

Q8: Calculate the HCF of 33×5 and 32×52.

Ans: We have 33×5 = 32 × 5 × 3

32×52 = 32 × 5 × 5

HCF (33×5, 32×52) = 32 ×5

9 × 5 = 45

Q9: Calculate 3/8 in the decimal form.

Ans:

38 = 323 = 2 × 5323 × 53

= 375103 = 3751,000

= 0.375

Q10: If HCF (a, b) = 12 and a × b = 1800, then find LCM (a,b).

Ans: HCF (a, b) × LCM (a, b) = a × b

Substituting the values we have

12× LCM (a, b) = 1800

LCM (a, b) = 1,80012 = 150

Q11: Find LCM of numbers whose prime factorisation are expressible as 3 × 52 and 32 × 72.

Ans: LCM (3×52, 32×72) = 32×52×72

= 9 × 25 × 49

= 11025

Q12: Find the least number that is divisible by all numbers between 1 and 10 (both inclusive).

Ans: The required number is the LCM of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,

LCM = 2×2×3×2×3×5×7 = 2520

Short Answer Type Questions- I

Q13: Find the LCM of 96 and 360 by using fundamental theorem of arithmetic.

Ans: 96 = 25 × 3

360 = 23 × 32 × 5

LCM = 25 × 32 × 5 = 32 × 9 × 5 = 1440

Q14: Find the largest number which divides 70 and 125 leaving remainder 5 and 8 respectively.

Ans: It is given that on dividing 70 by the required number, there is a remainder 5.

This means that 70 - 5 = 65 is exactly divisible by the required number.

Similarly,

125 - 8 = 117 is also exactly divisible by the required number.

65 = 5 × 13

117 = 32 × 13

HCF = 13

Required number = 13

Q15: Find HCF and LCM of 404 and 96 and verify that HCF × LCM = Product of the two given numbers.

Ans: 404 = 2×2×101

= 22 ×101

96 = 2×2×2×2×2×3

= 25×3

HCF (404, 96) = 22 = 4

LCM (404, 96) = 101×25×3 = 9696

HCF × LCM = 4×9696 = 38784

Also, 404×96 = 38784

Hence, HCF × LCM = Product of 404 and 96

Q16: Given that HCF (306,1314) = 18. Find LCM of 306,1314

Ans: HCF (306,1314) = 18

LCM (306,1314) = ?

Let a = 306 and b = 1314, then we have

LCM (a,b) × HCF (a,b) = a×b

Substituting values we have,

LCM (a,b)× 18 = 306 × 1314

LCM (a, b) = 306 x 131418

LCM (306, 1314) = 22,338

Q17: Complete the following factor tree and find the composite number x.

Ans: z = 371/7 = 53

y = 1855×3 = 5565

x = 2 × y = 2×5565 = 11130

Thus complete factor tree is as given below.

Q18: Check whether 4n can end with the digit 0 for any natural number n.

Ans: If the number 4n for any n, were to end with the digit zero, then it would be divisible by 5 and 2.

That is, the prime factorization of 4n would contain the prime 5 and 2. This is not possible because the only prime in the factorization of 4n = 22n is 2. So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorization of 4n. So, there is no natural number n for which 4n ends with the digit zero. Hence 4n cannot end with the digit zero.

Q19: Write the denominator of the rational number 257/500 in the form 2m × 5n, where m and n are non-negative integers. Hence write its decimal expansion without actual division.

Ans: 500 = 25 × 20

= 52 × 5 × 4

= 53 × 22

Here, denominator is 500 which can be written as 22 × 53.

Now decimal expansion,

257500 = 257 × 22 × 22 × 53 = 514103

= 0.514

Q20: Write a rational number between √2 and √3.

Ans: √2 = √200100 and √3 = √300100

We need to find a rational number x such that:

1/10√200 < x < 1/10√300

Choosing any perfect square such as 225 or 256 in between 200 and 300, we have:

x = √225100 = 1510 = 53

Similarly, if we choose 256, then we have:

x = √256100 = 1610 = 85

Q21: Can two numbers have 15 as their HCF and 175 as their LCM? Give reasons.

Ans: No,

LCM = Product of the highest power of each factor involved in the numbers.

HCF = Product of the smallest power of each common factor.

We can conclude that LCM is always a multiple of HCF, i.e., LCM = k × HCF

We are given that,

LCM = 175 and HCF = 15

175 = k × 15

⇒ 11.67 = k

But in this case, LCM ≠ k × HCF

Therefore, two numbers cannot have LCM as 175 and HCF as 15.

Q22: Explain why (17 × 5 × 11 × 3 × 2 + 2 × 11) is a composite number?

Ans: 17 × 5 × 11 × 3 × 2 + 2 × 11 ...(i)

= 2 × 11 × (17 × 5 × 3 + 1)

= 2 × 11 × (255 + 1)

= 2 × 11 × 256

Number (i) is divisible by 2, 11 and 256, it has more than 2 prime factors.

Therefore (17 × 5 × 11 × 3 × 2 + 2 × 11) is a composite number.

Short Answer Type Questions- II

Q23: Prove that 3 + 2√3 is an irrational number.

Ans: Let us assume to the contrary, that 3 + 2√3 is rational.

So that we can find integers a and b (b ≠ 0).

Such that 3 + 2√3 = ab, where a and b are coprime.

Rearranging the equations, we get

Here,

is a rational number, since

a and

b are integers.

Thus,

would be rational.

But this contradicts the known fact that is irrational.

Hence, our assumption was wrong.

Therefore, 3 + 2√3 is an irrational number.

Q24: Prove that 2+√3/5 is an irrational number, given that √3 is an irrational number.

Ans: Assume that 2+√3/5 is a rational number. Therefore, we can write it in the form of p/q where p and q are co-prime integers and q ≠ 0.

2 + √35 = pq

2 + √3 = 5pq

√3 = 5p - 2q

√3 = 5p - 2qq

Since, p and q are co-prime integers, then 5p-2q/q is a rational number. But this contradicts the fact that √3 is an irrational number. So, our assumption is wrong. Therefore 2+√3/5 is an irrational number.

Q25: Write the smallest number which is divisible by both 306 and 657.

Ans: The smallest number that is divisible by two numbers is obtained by finding the LCM of these numbers

Here, the given numbers are 306 and 657.

306 = 6 × 51 = 3×2×3×17

657 = 9 × 73 = 3×3×73

LCM (306, 657) = 2×3×3×17×73 = 22338

Hence, the smallest number which is divisible by 306 and 657 is 22338.

Q26: Three bells toll at intervals of 9, 12, 15 minutes respectively. If they start tolling together, after what time will they next toll together?

Ans: 9 = 32, 12 = 22 × 3, 15 = 3 × 5

LCM = 22 × 32 × 5 = 4 × 9 × 5 = 180 minutes or 3 hours

They will next toll together after 3 hours.

Q27: Two tankers contain 850 liters and 680 liters of petrol. Find the maximum capacity of a container which can measure the petrol of each tanker in the exact number of times.

Ans: To find the maximum capacity of a container which can measure the petrol of each tanker in the exact number of times, we find the HCF of 850 and 680.

850 = 2 × 52 × 17

680 = 23 × 5 × 17

HCF = 2 × 5 × 17 = 170

Maximum capacity of the container = 170 liters.

Q28: 144 cartons of Coke cans and 90 cartons of Pepsi cans are to be stacked in a canteen. If each stack is of the same height and if it equal contain cartons of the same drink, what would be the greatest number of cartons each stack would have?

Ans: The required answer will be HCF of 144 and 90.

144 = 24 × 32

90 = 2 × 32 × 5

HCF (144, 90) = 2 × 32 = 18

Thus each stack would have 18 cartons.

Q29: If p is prime number, then prove that √p is an irrational.

Ans: Let p be a prime number and if possible, let √p be rational

Thus,

√p = mn

where m and n are co-primes and n ≠0 .

Squaring on both sides, we get

p = m2n2

or, pn2 = m2 ...(1)

Here p divides pn2. Thus p divides m2 and in result p also divides m.

Let m = pq for some integer q and putting m = pq in eq. (1), we have

pn2 = p2q2

or, n2 = pq2

Here, p divides pq2. Thus p divides n2 and in result p also divides n.

[∵ p is prime and p divides n2 ⇒ p divides n]

Thus p is a common factor of m and n but this contradicts the fact that m and n are primes. The contradiction arises by assuming that √p is rational.

Hence, √p is irrational.

Q30: Amita, Sneha, and Raghav start preparing cards for all persons of an old age home. In order to complete one card, they take 10, 16 and 20 minutes respectively. If all of them started together, after what time will they start preparing a new card together?

Ans: To find the earliest (least) time, they will start preparing a new card together, we find the LCM of 10, 16 and 20.

10 = 2 × 5

16 = 24

20 = 22 × 5

LCM = 24 × 5 = 16 × 5 = 80 minutes

They will start preparing a new card together after 80 minutes.

Q31: If two positive integers x and y are expressible in terms of primes as x = p2q3 and y = p3q, what can you say about their LCM and HCF. Is LCM a multiple of HCF? Explain.

Ans: x = p2q3 and y = p3q

LCM = p3q3

HCF = p2q ...(i)

Now, LCM = p3q3

⇒ LCM = pq2 (p2q)

⇒ LCM = pq2 (HCF)

Yes, LCM is a multiple of HCF.

Explanation:

Let a = 12 = 22 × 3

b = 18 = 2 × 32

HCF = 2 × 3 = 6 ...(ii)

LCM = 22 × 32 = 36

LCM = 6 × 6

LCM = 6 (HCF) [From (ii)]

Here LCM is 6 times HCF.

Q32: Show that any positive odd integer is of the form 4q + 1 or 4q + 3 where q is a positive integer.

Ans: Let a be a positive odd integer

By Euclid's Division algorithm:

a = 4q + r ...[where q, r are positive integers and 0 ≤ r < 4]

a = 4q

or 4q + 1

or 4q + 2

or 4q + 3

But 4q and 4q + 2 are both even

a is of the form 4q + 1 or 4q + 3.

Long AnsweType Questions

Q33: Prove that √5 is an irrational number.

Ans: Assume that √5 be a rational number then we have

√5 = a/b

where a and b are co-primes and b≠0.

a = b√5

Squaring both the sides, we have

a2 = 5b2

Thus 5 is a factor of a2 and in result 5 is also a factor of a.

Let a = 5c where c is some integer, then we have

a2 = 25c2

Substituting a2 = 5b2 we have,

5b2 = 25c2

⇒ b2 = 5c2

Thus 5 is a factor of b2 and in result 5 is also a factor of b.

Thus √5 is a common factor of a and b. But this contradicts the fact that a and b are co-primes. Thus, our assumption that √5 is rational number is wrong.

Hence, √5 is irrational.

Q34: Prove that n2 - n is divisible by 2 for every positive integer n.

Ans: We have n2 - n = n(n-1)

Thus n2 - n is product of two consecutive positive integers.

Any positive integer is of the form 2q or 2q + 1, for some integer q.

Case 1: n = 2q

If n = 2q we have

n(n-1) = 2q(2q-1)

= 2m

where m = q(2q-1) which is divisible by 2.

Case 2: n = 2q+1

If n = 2q+1, we have

n(n-1) = (2q+1) (2q+1-1)

= 2q(2q+1)

= 2m

where m = q(2q+1) which is divisible by 2.

Hence, n2 - n is divisible by 2 for every positive integer n.

Case Based Type Questions

A. Read the following text and answer the following questions on the basis of the same:

To enhance the reading skills of grade X students, the school nominates you and two of your friends to set up a class library. There are two sections- section A and section B of grade X. There are 32 students in section A and 36 students in section B.

MULTIPLE CHOICE QUESTION

Try yourself: If p and q are positive integers such that p = ab2 and q = a2b, where a, b are prime numbers, then the LCM (p, q) is

a3b3

a3b2

CORRECT ANSWER

a2b2

Correct Answer: D

Given, p = ab2 = a × b × b

q = a2b = a × a × b

LCM of (p, q) = a2b2

MULTIPLE CHOICE QUESTION

Try yourself: If the product of two positive integers is equal to the product of their HCF and LCM is true then, the HCF (32 , 36) is

CORRECT ANSWER

Correct Answer: B

Prime factor 32 is 2×2×2×2×2

Prime factor 36 is 2×2×3×3

HCF is 2×2 = 4

HCF of 32 and 36 is 4

MULTIPLE CHOICE QUESTION

Try yourself: 7 × 11 × 13 × 15 + 15 is a

Prime number

CORRECT ANSWER

Composite number

Neither prime nor composite

None of the above

Correct Answer: B

7 × 11 × 13 × 15 + 15 is composite number.

Take 15 common

15 ( 7 × 11 × 13 + 1)

So, given number has factor other than 1 and itself so it is a composite number

MULTIPLE CHOICE QUESTION

Try yourself: In a school there are two sections – section A and section B of Class VI . There are 32 students in section A and 36 in section B. Then, the minimum number of books required for their class library so that they can be distributed equally among students of section A or section B is ____.

144

128

CORRECT ANSWER

288

272

Correct Answer: C

We have to find the LCM of 32 and 36.

LCM(32, 36) = 25 × 9 = 288

Hence, the minimum number of books required to distribute equally among students of section A and section B are 288.

B. Read the following text and answer the following questions on the basis of the same:

A garden consists of 135 rose plants planted in certain number of columns. There are another set of 225 marigold plants, which is to be planted in the same number of columns.

MULTIPLE CHOICE QUESTION

Try yourself:

A garden has 135 rose plants and 225 marigold plants.Find the sum of the exponents of the prime factors of the total number of plants.

CORRECT ANSWER

Correct Answer: D

Total number of plants = 135 + 225 = 360

The prime factors of 360 = 2 × 2 × 2 × 3 × 3 × 5

= 23 × 32 × 51

Exponent of 2 = 3
Exponent of 3 = 2
Exponent of 5 = 1

∴ Sum of exponents = 3 + 2 + 1 = 6.

MULTIPLE CHOICE QUESTION

Try yourself: A garden consists of 135 rose plants planted in certain number of columns. There are another set of 225 marigold plants, which is to be planted in the same number of columns.

CORRECT ANSWER

Correct Answer: C

HCF of (135,225) is 45

Number of rows of Rose plants = 135/45 = 3

Number of rows of marigold plants = 225/45 = 5

Total number of rows = 3 + 5 = 8

MULTIPLE CHOICE QUESTION

Try yourself: Find the sum of exponents of the prime factors of the maximum number of columns in which they can be planted.

CORRECT ANSWER

Correct Answer: B

We have proved that the maximum number of columns = 45

So, prime factors of 45

= 3 × 3 × 5 = 32 × 51

∴ Sum of exponents = 2 + 1 = 3.

NCERT Solutions: Real Numbers (Exercise 1.2)

Q1: Prove that √5 is irrational.

Sol: Let us assume, that √5 is rational number.

i.e. √5 = x/y (where, x and y are co-primes)

y√5= x

Squaring both the sides, we get,

(y√5)2 = x2

⇒ 5y2 = x2...................................... (1)

Thus, x2 is divisible by 5, so x is also divisible by 5.

Let us say, x = 5k, for some value of k and substituting the value of x in equation (1), we get,

5y2 = (5k)2

⇒ y2 = 5k2

is divisible by 5 it means y is divisible by 5.

Clearly, x and y are not co-primes. Thus, our assumption about √5 is rational is incorrect.

Hence, √5 is an irrational number.

Q2: Prove that 3 + 2√5 + is irrational.

Sol: Let us assume 3 + 2√5 is rational.

Then we can find co-prime a and b (b ≠ 0) such that 3 + 2√5 = a/b

Rearranging, we get,

Since a and b are integers, 1/2 (a/b -3) is a rational number.

Therefore, √5 is also a rational number. But this contradicts the fact that √5 is irrational.

So, we conclude that 3 + 2√5 is irrational.

Q3: Prove that the following are irrationals:

(i) 1/√2

(ii) 7√5

(iii) 6 + √2

Sol:

(i) 1/√2

Let us assume 1/√2 is rational.

Then we can find co-prime x and y (y ≠ 0) such that 1/√2 = x/y

Rearranging, we get,

√2 = y/x

Since, x and y are integers, thus, √2 is a rational number, but we know the fact that √2 is irrational.

This is a contradiction.

Hence, we can conclude that 1/√2 is irrational.

(ii) 7√5

Let us assume 7√5 is a rational number.

Then we can find co-prime x and y (y ≠ 0) such that 7√5 = x/y

Rearranging, we get,

√5 = x/7y

Since, x and y are integers, thus, √5 is a rational number, which contradicts the fact that √5 is irrational.

Hence, we can conclude that 7√5 is irrational.

(iii) 6 +√2

Let us assume 6 +√2 is a rational number.

Then we can find co-primes x and y (y ≠ 0) such that 6 +√2 = x/y⋅

Rearranging, we get,

√2 = (x/y) - 6

Since, x and y are integers, thus (x/y) - 6 is a rational number and therefore, √2 is rational. But we know the fact that √2 is an irrational number.

This is a contradiction.

Hence, we can conclude that 6 +√2 is irrational.

Very Short Answer Type Questions: Real Numbers

Q1: Given that HCF (150, 100) = 50. Find LCM (150, 100).

Sol: LCM × HCF = Product of the two numbers

∴ 150 × 100 = LCM × HCF

⇒ LCM × 50 = 150 × 100

⇒

Q2: Given that LCM (26, 91) = 182. Find their HCF.

Sol: ∵ HCF × LCM = Product of the two numbers

∴ HCF × 182 = 26 × 91

⇒

Q3: The LCM and HCF of the two numbers are 240 and 12 respectively. If one of the numbers is 60, then find the other number.

Sol: Let the required number be 'x'.

∵ LCM × HCF = Product of the two numbers

∴ 60 × x = 240 × 12

⇒

Q4: The decimal expansion of the rational number,

will terminate after how many places of decimal?

Sol:

Thus, will terminate after 4 places of decimal.

Q5: What is the exponent of 3 in the prime factorisation of 864.

Sol:

Making prime factors of 864. ,⇒ 864 = 3 × 3 × 3 × 2 × 2 × 2 × 2 × 2= 33 × 25

∴ Exponent of 3 in prime factorisation of 864 = 3.

Q6: State the fundamental theorem of arithmetic.

Sol: Fundamental Theorem of Arithmetic: Every composite number can be expressed as a product of prime numbers, and this factorisation is unique, except for the order of the prime factors.

A composite number can be broken down into prime factors.
This factorisation is unique; the same primes can be arranged in different orders.
For example, 2 × 3 × 5 is the same as 3 × 2 × 5.

This theorem is essential in number theory and has significant implications in various mathematical fields.

Q7: Define an irrational number.

Sol: Irrational numbers are those that cannot be expressed as a fraction of two integers. They have the following characteristics:

Their decimal expansion does not terminate or repeat.
They cannot be written in the form p/q, where q ≠ 0.

Examples of irrational numbers include: √2, √3, π

Q8: Write the condition for a rational number which can have a terminating decimal expansion.

Sol: A rational number x = p/q can have a terminating decimal expansion if the prime factorisation of q is of the form of 2n · 5m, where m and n are non-negative integers.

Q9: Write the condition for a rational number which has a non-terminating repeating decimal expansion.

Sol: A rational number x = p/q can have a non-terminating repeating decimal expansion if:

The prime factorisation of q is not of the form 2n x 5m,
where n and m are non-negative integers.

Q10: Can two numbers have 24 as their HCF and 7290 as their LCM? Give reasons.

Sol: No, because HCF always divides LCM but here 24 does not divide 7290.

Q11: If 6n is a number such that n is a natural number. Check whether there is any value of n ∈ N for which 6n is divisible by 7.

Sol: ∵ 6 = 2 × 3

∴ 6n = (2 × 3)n = 2n × 3n

i.e., the prime factorisation of 6n does not contain the prime number 7 thus the number 6n is not divisible by 7.

Q12: Write 98 as the product of its prime factors.

Sol: ∵

The prime factorisation of 98 = 2 × 7 × 7

⇒ 98 = 2 × 72

Q13: Without actually performing the long division, state whether

will have a terminating or non-terminating repeating decimal expansion.

Sol: Let =

∵ Prime factors of q are not of the for 2n · 5m.

If the prime factors of q are only 2 and 5, the decimal expansion is terminating.
If q has any prime factors other than 2 or 5, the decimal expansion is non-terminating repeating.

∴

will have a non-terminating repeating decimal expansion.

Q14: Without actually performing the long division, state whether 17/3125 will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

Sol: ∵ The denominator of 17/3125 is given by

3125 = 5 × 5 × 5 × 5 × 5

= 1 × 55

= 20 × 55 |∵ 20 = 1

∴

Since the prime factorisation includes only the primes 2 and 5, it indicates that:

The decimal expansion of 17/3125 is a terminating decimal.

Q15: Express 156 as a product of its prime factors.

Sol: ∵ 156 = 2 × 78

= 2 × 2 × 39

= 2 × 2 × 3 × 13

∴ 156 = 22 × 3 × 13

Q16: If the product of two numbers is 20736 and their LCM is 384, find their HCF.

Sol: ∵ LCM × HCF = Product of two numbers

∴ 384 × HCF = 20736

⇒ HCF = 20736 /384 = 54.

Q17: Find the LCM and HCF of 120 and 144 by the Fundamental Theorem of Arithmetic.

Sol: We have 120 = 2 × 2 × 2 × 3 × 5 = 23 × 3 × 5

144 = 2 × 2 × 2 × 2 × 3 × 3 = 24 × 32

∴ LCM = 24 × 32 × 5 = 720

HCF = 23 × 3 = 24

Q18: Find the HCF × LCM for the numbers 100 and 190.

Solution: HCF × LCM = 1st Number × 2nd Number

= 100 × 190 = 19000.

Q19: Find the (HCF × LCM) for the numbers 105 and 120.

Solution: HCF × LCM = 1st number × 2nd number

= 105 × 120 = 12600.

Q20: Write a rational number between √2 and √3.

Sol: ∵ √2 = 1.41 ..... and

√3 = 1.73 .....

∴ one rational number between 1.41 .....and 1.73 ..... is 1.5

i.e., one rational number between √2 and √3 is 1.5.

Long Answer Questions: Real Numbers

Q1: Kerosene, paraffin, or lamp oil is a combustible hydrocarbon liquid which is derivative from petroleum. Kerosen's uses vary from fuel for oil lamps to cleaning agents , jet fuel , heating oil or fuel for cooking Two oil tankers contain 825 litres and 675 litres of kerosene oil respectively.

(a) Find the maximum capacity of a container which can measure the Kerosene oil of both the tankers when used an exact number of times.

(b) How many times we have to use container for both the tanker to fill?

Sol:
(a) HCF of 825 and 625
825 = 3 x 5 x 5 x 11
675 = 3 x 3 x 3 x5 x 5
HCF = 3 x 5 x 5 = 75
Maximum capacity reqired is 75 litres
(b) The first tanker will require 875/75 = 11 times to fill
The second tanker will require 675/75 = 9 times to fill

Q2: The sum of LCM and HCF of two numbers is 7380.If the LCM of these numbers is 7340 more than their HCF. Find the product of the two numbers.

Sol: LCM + HCF = 7380
LCM - HCF = 7340
2LCM = 14720
LCM = 14720/2
LCM = 7360
LCM + HCF = 7380
7360+ HCF = 7380
HCF = 7380 - 7360
HCF = 20
HCF x LCM = product of numbers
20 x 7360= product of numbers
147200 = product of numbers

Q3: If two positive integers x and y are expressible in terms of primes as x = p2q3 and y = p3q, what can you say about their LCM and HCF. Is LCM a multiple of HCF? Explain.

Sol: x = p2q3 and y = p3q
LCM = p3q3
HCF = p2q .....(i)
Now, LCM = p3q3
⇒ LCM = pq2 (p2q)
⇒ LCM = pq2 (HCF)
Yes, LCM is a multiple of HCF.
Explanation:
Let a = 12 = 22 × 3
b = 18 = 2 × 32
HCF = 2 × 3 = 6 ...(ii)
LCM = 22 × 32 = 36
LCM = 6 × 6
LCM = 6 (HCF) ...[From (ii)]
Here LCM is 6 times HCF.

Q4: Use Euclid's division algorithm to find the HCF of :

(i) 135 and 225

Sol:
Apply Euclid's Division Algorithm:
Step 1: Divide the larger number by the smaller.
225 = 135 × 1 + 90
Step 2: Now divide 135 by 90.
135 = 90 × 1 + 45
Step 3: Now divide 90 by 45.
90 = 45 × 2 + 0
Since the remainder is now 0, we stop here.
HCF = 45 (the last non-zero remainder)

Q5: A charitable trust donates 28 different books of Maths,16 different books of science and 12 different books of Social Science to the poor students. Each student is given maximum number of books of only one subject of his interest and each student got equal number of books

(a) Find the number of books each student got.

(b) Find the total number of students who got books.

Sol:
(i) HCF of 28,16 and 12 is 4
Therefore maximum number of books each student get is 4
(ii) Number of maths books 28/4 = 7
Number of science books 16/4 = 4
Number of social science = 12/4 = 3
Total books = 7 + 4 + 3 =14

Q6: Amita, Sneha, and Raghav start preparing cards for all persons of an old age home. In order to complete one card, they take 10, 16 and 20 minutes respectively. If all of them started together, after what time will they start preparing a new card together?

Sol: To find the earliest (least) time, they will start preparing a new card together, we find the LCM of 10, 16 and 20.
10 = 2 × 5
16 = 24
20 = 22 × 5
LCM = 24 × 5 = 16 × 5 = 80 minutes
They will start preparing a new card together after 80 minutes.

Q7: Dudhnath has two vessels containing 720 ml and 405 ml of milk respectively. Milk from these containers is poured into glasses of equal capacity to their brim. Find the minimum number of glasses that can be filled.

Sol: 1st vessel = 720 ml; 2nd vessel = 405 ml
We find the HCF of 720 and 405 to find the maximum quantity of milk to be filled in one glass.
405 = 34 × 5
720 = 24 × 32 × 5
HCF = 32 × 5 = 45 ml = Capacity of glass
No. of glasses filled from 1st vessel = 720/45 = 16
No. of glasses filled from 2nd vessel = 405/45 = 9
Total number of glasses = 25

Q8: There are 104 students in class X and 96 students in class IX in a school. In a house examination, the students are to be evenly seated in parallel rows such that no two adjacent rows are of the same class.

(a) Find the maximum number of parallel rows of each class for the seating arrange¬ment.

(b) Also, find the number of students of class IX and also of class X in a row.

Sol:
104 = 23 × 13
96 = 25 × 3
HCF = 23 = 8
(a) Number of rows of students of class X = 104/8 = 13
Number maximum of rows class IX = 96/8 = 12
Total number of rows = 13 + 12 = 25
(b) No. of students of class IX in a row = 8
No. of students of class X in a row = 8
(c) The objective of school administration behind such an arrangement is fair and clean examination, so that no student can take help from any other student of his/her class.

Q9: Prove that 3 + 2√5 is irrational.

Sol:
Let us assume, to the contrary, that 3 + 2√5 is rational
So that we can find integers a and b (b ≠ 0), such that
3 + 2 √5 = ab, where a and b are coprime.
Rearranging this equation, we get
⇒ 2√5 = ab - 3 ⇒ √5 = a - 3b2b
⇒ √5 = a2b - 3b 2b ⇒ √5 = a2b - 3 2
Since a and b are integers, we get that a2b - 3 2 is rational and so √5 is rational. But this contradicts the fact that √5 is irrational.
So we conclude that 3 + 2√5 is irrational.

Q10: Prove that √5 is irrational.

Sol:
Consider that √5 is a rational number.
i.e. √5 = x/y (as well as, x and y are co-primes)
y√5= x
Squaring the both the sides, we observe,
(y√5)2 = x2
⇒5y2 = x2...................................... (1)
Therefore, x2 is divided by 5, so x is also divided by 5.
Consider, x = 5k, for some value of k and putting the value of x in equation (1), we observe,
5y2 = (5k)2
⇒y2 = 5k2
is divisible by 5 it means y is divisible by 5.
Clearly, x and y are not co-primes. Thus, our assumption about √5 is rational is incorrect.
Hence, √5 is an irrational number.

Case Based Questions: Real Numbers

Q1: Read the source below and answer the questions that follow:

Old age homes mean for senior citizens who are unable to stay with their families or destitute. These old age homes have special medical facilities for senior citizens such as mobile health care systems, ambulances, nurses and provision of well balanced meals.

Himanshu, Gaurav and Gagan start preparing greeting cards for each person of an old age home on new year. In order to complete one card, they take 10, 16 and 20 min respectively.

i. Co-prime numbers are those numbers which do not have any common factor other than 1. Is this statement true?

ii. Find the sum of the powers of all different prime factors of the numbers 10, 16 and 20.

iii. If all of them started together, then what time will they start preparing a new card together?

iv. What is the common time to make one card?

Ans:

i. True

ii. By prime factorisation, 10=2¹× 5¹ 16=2x2x2x2=24 20 2x2x5=22x5¹ .. Required sum = sum of the power of 2 + sum of the power of 5 = (1 + 4 + 2) + (1+1)=7+2=9

iii. The required number of minutes after which they start preparing a new card together is the LCM of 10, 16 and 20 min. Now, 10=2x5 16=2x2x2x2 20=2x2x5=22x5 .. LCM (10, 16, 20) = 24 x 51 = 16 x 5 = 80 min So, they will start preparing a new card together after 80 min i.e., 1 h 20 min.

iv. The common time to make one card = HCF of (10, 16, 20) = 2 min

Q2: Read the source below and answer the questions that follow:

In a morning walk, Naveeka, Arjun and Vedant step off together, their steps measuring 240 cm, 90 cm, 120 cm respectively. They want to go for a juice shop for a health issue, which is situated near by them.

i. Factor tree is a chain of factors, which is represented in the form of a tree. Is this statement true?

ii. Find the sum of the powers of all common prime factors of the numbers 240, 90 and 120.

iii. Find the minimum distance of shop from where they start to walk together, so that one can cover the distance in complete steps.

iv. Find the number of common steps covered by all of them to reach the juice shop.

Ans:

i. True

ii. By prime factorisation, 240=2x2x2x2x3x5=24 x 31 x 5¹ 90=2×3×3×5=2¹×32x5¹ 120=2×2×3×2×5=23×3¹×5¹ :. Required sum = sum of the power of 2+ sum of the power of 3+ sum of the power of 5=1+1+1=3.

iii. Minimum required distance to reach the juice shop = LCM (240, 90, 120) 240 = 2x2x2x2x3x5=24x3x5 90 = 2×3×3×5=2x32x5 and 120 = 2×2×2×3×5=23x3x5 Now, LCM=24x32x5=16x9 × 5=720 Hence, required minimum distance is 720 cm.

iv. The number of common steps covered by all of them HCF (240, 90, 120) = 2 x 3 x 5 =30

Q3: Read the source below and answer the questions that follow:

Teaching Mathematics through activities is a powerful approach that enhances students' understanding and engagement. Keeping this in mind, Ms. Mukta planned a prime number game for class 5 students. She announced the number 2 in her class and asked the first student to multiply it by a prime number and then pass it to the second student. The second student also multiplied it by a prime number and passed it to the third student. In this way by multiplying by a prime number, the last student got 173250.

Now, Mukta asked some questions as given below to the students:

i. What is the least prime number used by students?

ii. How many students are in the class?

iii. What is the highest prime number used by students?

iv. Which prime number has been used maximum times?

Ans:

So least prime no. used by students = 3(because 2 is announced by the teacher, so the least number used by the students is 3)

ii. As the last student got 173250 = 2 × 3 × 3 × 5 × 5 × 5 × 7 × 11

there are 7 factors other than 2, which is announced by teacher. So, Number of student = 7

iii. Highest prime number used by student = 11

iv. Prime number 5 is used maximum times i.e., 3 times.

Q4: Read the source below and answer the questions that follow:

Ms. Asha planned a prime factorization activity for her Class 10 students. She announced the number 3 in her class and asked the first student to multiply it by a prime number and pass it to the next student. Each student multiplied the result by a prime number and passed it further. The last student ended up with the number 231000.

Now, Ms. Asha asked the following questions:

i. What is the least prime number used by students?

ii. How many students are in the class?

iii. What is the highest prime number used by students?

iv. Which prime number has been used maximum times?

Ans:

i. The least prime number used is 2.

ii. The total number of students is 9 (since there are 9 prime factors).

iii. The highest prime number used is 11.

iv. The prime number used the maximum times is 2 and 5 (each used 3 times).

Assertions & Reason Type Questions: Real Numbers

Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:

Question 1:

Assertion : √x is an irrational number, where x is a prime number.

Reason : Square root of any prime number is an irrational number.

(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).

(d) Assertion (A) is false but reason (R) is true.

Correct Answer is Option (a)
As we know that square root of every prime number is an irrational number. So, both A and R are correct and R explains A.

Question 2:

Assertion : The HCF of two numbers is 18 and their product is 3072. Then their LCM = 169.

Reason : If a, b are two positive integers, then HCF x LCM = a x b.

(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).

(d) Assertion (A) is false but reason (R) is true.

Correct Answer is Option (d)
Here reason is true , Assertion is false.
We know that for any two numbers, Product of the two numbers = HCF x LCM = 18 x 169 = 3042 ≠ 3072

Question 3:

Assertion : 12n ends with the digit zero, where n is natural number.

Reason : Any number ends with digit zero, if its prime factor is of the form 2m x 5n, where m, n are natural numbers.

(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).

(d) Assertion (A) is false but reason (R) is true.

Correct Answer is Option (d)
12n = (2 x 2 x 3)n = 2n x 2n x 3n ,
Its prime factors do not contain 5n i.e., of the form 2m x 5n , where m, n are natural numbers.
Here assertion is incorrect but reason is correct.

Question 4: Assertion: The HCF of two numbers is 5 and their product is 150, then their LCM is 30

Reason: For any two positive integers a and b, HCF (a, b) + LCM (a, b) = a x b.

(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).

(d) Assertion (A) is false but reason (R) is true.

Correct Answer is Option (c)
We have,
LCM(a, b) x HCF(a, b) = a x b
LCM x 5 = 150
LCM 5/150 = 30

Question 5:

Assertion (A): If HCF (336, 54) = 6, then LCM (336,54) = 3000.

Reason (R): The sum of exponents of prime factors in the prime factorisation of 196 is 4.

(a) Both A and R are true and R is the correct explanation of A

(b) Both A and R are true but R is NOT the correct explanation of A

(d) A is false and R is True

Correct Answer is Option (d)
Let us consider the assertion,
∵ HCF × LCM = Product of numbers
∴ 6 × LCM = 336 × 54
⇒ LCM =
336 × 546

= 3024
Thus, the assertion is incorrect:
Now, let us consider the reason:
Prime factors of 196 = 22 × 72
∴ The sum of exponents of prime factors = 2 + 2 = 4.
So, the reason is correct:
Thus, assertion is incorrect but reason is correct.

Question 6:

Assertion (A): The product of two consecutive positive integers is divisible by 2.

Reason (R): 13233343563715 is a composite number.

(a) Both A and R are true and R is the correct explanation of A

(b) Both A and R are true but R is NOT the correct explanation of A

(d) A is false and R is True

Correct Answer is Option (b)
In case of assertion, Since, in the product of two consecutive positive integers, p = n(n + 1), one of n or (n + 1) is an even number.
Hence, the product of two consecutive positive integers is divisible by 2. So, it is correct.
Now, let us consider the reason:
Since, the given number ends in 5. It is a multiple of 5. Therefore, it is a composite number.
Thus, both assertion and reason are correct and reason is not the correct explanation for assertion.

Question 7:

Assertion (A): (7 × 13 × 11) + 11 and (7 × 6 × 5 × 4 × 3 × 2 × 1) + 3 have exactly composite numbers.

Reason (R): (3 × 12 × 101) + 4 is not a composite number.

(a) Both A and R are true and R is the correct explanation of A

(b) Both A and R are true but R is NOT the correct explanation of A

(d) A is false and R is True

Correct Answer is Option (c)
Firstly consider the assertion,
Since (7 × 13 × 11) + 11 = 11 × (7 × 13 + 1) = 11 × (91 + 1)
= 11 × 92 ⇒ 11 × 2 × 2 × 23
and (7 × 6 × 5 × 4 × 3 × 2 × 1) + 3 = 3 (7 × 6 × 5 × 4 × 2 × 1 + 1)
= 3 × (1681) ⇒ 3 × 41 × 41
Given numbers have more than two prime factors.
So, both the numbers are composite. Hence, assertion is correct.
Now let us consider the reason: 3 × 12 × 101 + 4 = 4(3 × 3 × 101 + 1)
= 4(909 + 1)
= 4(910)
= 2 × 2 × 2 × 5 × 7 × 13
= a composite number
[∵ Product of more than two prime factors]
Thus, reason is not correct.
Thus, assertion is correct but reason is incorrect.

Question 8:

Assertion (A): HCF of two or more numbers = Product of the smallest power of each common prime factor, involved in the numbers.

Reason (R): The HCF of 12, 21 and 15 is 3.

(a) Both A and R are true and R is the correct explanation of A

(b) Both A and R are true but R is NOT the correct explanation of A

(d) A is false and R is True

Correct Answer is Option (a)
Let a, a2 and a3 be three numbers, then we have the smallest power of a1, a2 and a3 is 1. So, HCF is a.
Now, let us consider the reason:
Prime factors of 12 = 22 × 3
Prime factors of 21 = 3 × 7
Prime factors of 15 = 3 × 5
∴ HCF of 12, 21 and 15 = 3, which is a common prime factor.
Thus both assertion and reason are correct and reason is the correct explanation for assertion.

Question 9:

Assertion (A): The decimal expansion of 15/1600 is 0.09375

Reason (R): The decimal expansion of 23/2352 is 0.115

(a) Both A and R are true and R is the correct explanation of A

(b) Both A and R are true but R is NOT the correct explanation of A

(d) A is false and R is True

Correct Answer is Option (d)
In case of assertion:
151600
=
152⁴ × 100
=
15 × 5⁴2⁴ × 5⁴ × 100
=
9375(2 × 5)⁴ × 100
=
93751000000
= 0.009375
So, assertion is incorrect.
Now, in case of Reason:
23
2³ × 5
²
=
23 × 52³ × 5² × 5
=
1152³ × 5³
=
115(2 × 5)³
=
1151000
= 0.115
So, reason is correct.
Thus, assertion is incorrect but reason is correct.

Visual Worksheet: Prime Factorisation (with Solutions)

Fundamental Theorem of Arithmetic

The Fundamental Theorem of Arithmetic states that every integer greater than 1 is either a prime number or can be expressed in the form of primes. In other words, all the natural numbers can be expressed in the form of the product of its prime factors. To recall, prime factors are the numbers that are divisible by 1 and itself only. For example, the number 35 can be written in the form of its prime factors as:
35 = 7 × 5
Here, 7 and 5 are the prime factors of 35
Similarly, another number 114560 can be represented as the product of its prime factors by using the prime factorization method,
114560 = 27 × 5 × 179
So, we have factorized 114560 as the product of the power of its primes.

Therefore, every natural number can be expressed in the form of the product of the power of its primes. This statement is known as the Fundamental Theorem of Arithmetic, unique factorization theorem or the unique-prime-factorization theorem.

Proof for Fundamental Theorem of Arithmetic

In number theory, a composite number is expressed in the form of the product of primes and this factorization is unique apart from the order in which the prime factor occurs.
From this theorem we can also see that not only a composite number can be factorized as the product of their primes but also for each composite number the factorization is unique, not taking into consideration the order of occurrence of the prime factors.
In simple words, there exists only a single way to represent a natural number by the product of prime factors. This fact can also be stated as:
The prime factorization of any natural number is said to be unique for except the order of their factors.
In general, a composite number "a" can be expressed as,
a = p1 p2 p3 ............ pn, where p1, p2, p3 ............ pn are the prime factors of a written in ascending order i.e. p1≤p2≤p3 ............ ≤pn.
Writing the primes in ascending order makes the factorization unique in nature.

Solved Examples

Question: In a formula racing competition the time taken by two racing cars A and B to complete 1 round of the track is 30 minutes and 45 minutes respectively. After how much time will the cars meet again at the starting point?
Solution:
As the time taken by car B is more compared to that of A to complete one round therefore it can be assumed that A will reach early and both the cars will meet again when A has already reached the starting point. This time can be calculated by finding the L.C.M of the time taken by each.
30 = 2 × 3 × 5
45 = 3 × 3 × 5
The L.C.M is 90.
Thus, both cars will meet at the starting point after 90 minutes.

What are Factors?

When a number is said to be a factor of any other second number, then the first number must divide the second number completely without leaving any remainder. In simple words, if a number (dividend) is exactly divisible by any number (divisor), then the divisor is a factor of that dividend.
Example: 4 is a factor of 24 i.e. 4 divides 24 exactly giving 6 as quotient and leaving zero as remainder. Alternatively, 6 is also a factor of 24 as it gives 4 as quotient on division. Therefore, 24 has 1, 24, 4, 6 as its factors in addition to 2, 3, 8 and 12 and all these numbers divide 24 exactly leaving no remainder.
If any natural number has only two factors i.e. 1 and the number itself as its factors, such numbers are called prime numbers.

Highest Common Factor (H.C.F)

The largest or greatest factor common to any two or more given natural numbers is termed as the HCF of given numbers. Also known as GCD (Greatest Common Divisor).
Example: Factors of the number 10 are 1, 2, 5 and 10
Factors of the number 15 are 1, 3, 5 and 15
We can see, the HCF (10, 15) = 5

Methods of Calculating Highest Common Factor

1. Listing all Factors
The simplest method of deriving the HCF is to simply list all the factors of the given number, identify all common factors, and choose the greatest of all the common factors, arriving at your Highest Common Factor(HCF).
Example:
16 = 1, 2, 4, 8, 16
40 = 1, 2, 4, 5, 8, 10, 20, 40
HCF (16,40) = 8 (Because 8 is the largest among all the common factors. Hence, it is the HCF.)
2. Prime Factorization
Given natural numbers are written as the product of prime factors. To obtain the highest common factor multiply all the common prime factors with the lowest degree (power).
Example:
20 = 2 × 2 × 5
12 = 2 × 2 × 3
HCF = 2 × 2 = 4
3. Continued Division
This method is used when we have to find the HCF of relatively larger numbers.
Step 1: Divide the larger number by the smaller number
Step 2: The remainder from Step 1 becomes the divisor, and divisor of Step 1 becomes the dividend.
Step 3: Continue this division process till remainder becomes zero.
Step 4: The divisor of the last division (when remainder is zero) is your HCF.

Finding the Highest Common Factors of Decimals

If the numbers given are decimal numbers, the method of finding HCF is very much similar to the Prime Factorization method, with a little modification.
Example: Say we have to find the HCF of 1.20 and 22.5
First, we convert both numbers to like decimals i.e 1.20 and 22.50
Now we express both numbers, without decimals, as a product of their prime numbers
120 = 2×2×2×3×5
2250 = 2×3×3×5×5×5
So, HCF (120, 2250) = 2×3×5 = 30
Therefore HCF (1.20, 22.5) = 0.30 (taking 2 decimal points)

Finding the Highest Common Factor of Fractions

So when finding the HCF of a fraction there is an easy formula to follow
HCF = HCF of numerators/ LCM of denominators
Example: Say we are to find the HCF of (4/5) and (3/7)
Here the HCF of the numerators 4 and 3 is 1 and the LCM of denominators 5 and 7 is 35.
Therefore the HCF of the two fractions is 1/35.

Solved Examples

Ques: Find the greatest number that divides 167 and 95, leaving 5 as the remainder.
Sol: Although the question sounds complicated, the solution is a simple application of HCF.
The said number divides 167 and leaves 5 as a remainder, So the number divides 162 (167-5) perfectly.
The said number also divides 95 and leaves a remainder of 5, So the number divides 90 (95-5) perfectly.
Therefore the solution is the HCF of 162 and 90, which is 18.

Multiples

A multiple of a number is a number that is the product of a given number and some other natural number. Multiples can be observed in a multiplication table.
Multiples of 2 are as follows:
2 × 1 = 2,
2 × 2 = 4,
2 × 3 = 6,
2 × 4 = 8,
2 × 5 = 10,
2 × 6 = 12,
2 × 7 = 14,
2 × 8 = 16,
and so on.

Least Common Multiple (L.C.M)

The least or smallest common multiple of any two or more given natural numbers are termed as LCM.
Example:
Multiples of 3 are 3, 6, 9, 12, 15, 18, 21....
Multiples of 5 are 5, 10, 15, 20, 25, 30....
LCM (3,5) = 15

Methods of calculating Lowest Common Multiple

1. Listing all the Multiples
This is the simplest method of finding the LCM of any given numbers. You write down several multiples of both the numbers and then identify the smallest common multiple among them.
Example:
Multiples of 5 = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50....
Multiples of 8 = 8, 16, 24, 32, 40, 48, 56, 64.....
LCM (5,8) = 40
2. Prime Factorization: Here given natural numbers are written as the product of prime factors. The lowest common multiple will be the product of all the prime factors with the highest degree (power).
Example:
10 = 2×5
12 = 2×2×3
Uncommon Prime Factors = 2,3,5
LCM(10,12) = 2×2×3×5 = 60
3. Long Division Method
Step 1: Write all the numbers in the first row divided by commas
Step 2: We divide the numbers by the lowest and most suitable prime number (should exactly divide at least 2 numbers)
Step 3: Write the quotient of the division in the next row, If the number is not exactly divisible by the prime number, bring it down as it is
Step 4: Continue to do the above steps, till only co-prime numbers are left in the last row.
Step 5: Multiply all prime numbers by which we have divided and all co-prime numbers left in the last row, This is your LCM

Finding the Least Common Multiple of Decimals

We follow the same method of prime factorization, with a few changes
Example:
Let's say we have to find the LCM of 2.5 and 0.35
First, we convert both numbers to like decimals i.e. 250 and 35
Now we express those two numbers as a product of their prime factors
250 = 2×5×5×5
35 = 5×7
LCM (250,3 5) = 2×5×5×7×5 = 1750
Therefore, LCM of (2.5, 0.35) = 17.50

Solved Examples

Ques: What is the smallest number that when divided by 20 and 48 separately gives the remainder of 7 every time?
Sol: The solution here is an application of the LCM principle. Here firstly we find the LCM of 20 and 48.
20 = 2×2×5
48 = 2×2×2×2×3
LCM (20 ,48) = 2×2×2×2×3×5 = 240
So the required number, that leaves a remainder of 7, is 247 (240+7)

Co-relation between HCF and LCM

So there is an interesting co-relation between H.C.F and L.C.M. of two numbers. The product of the H.C.F. and L.C.M. of any two numbers is always equal to the product of those two numbers. However the same is not applicable to three or more numbers.
LCM (a,b) × HCF (a,b) = (a × b)
Example:
Let the two numbers be 4 and 6.
HCF (4,6) = 2
LCM (4,6) = 12
HCF × LCM = 2 × 12 = 24
Product of the two numbers = 4 × 6 = 24

Applications of LCM and HCF

Now there are various real-life applications of LCM and HCF. The best way to understand these and grasp the concept of LCM and HCF is to learn via examples. So let us take a look at a few examples which will help you understand LCM and HCF.
Example 1: Find the greatest number that will divide 400, 435 and 541 leaving 9, 10 and 14 as remainders respectively.
Sol: The required number would be HCF of (400-9), (435-10) and (541-14)
So the HCF (391,425,527)
391 = 17 × 23
425 = 5 × 5 × 17
527 = 17 × 31
HCF = 17
Therefore the required number is 17.

Example 2: A, B, and C start to jog around a circular stadium. They complete their rounds in 36 seconds, 48 seconds and 42 seconds respectively. After how many seconds will they be together at the starting point?
Sol: The required time is the LCM of all their lap times. This is the earliest when all three will intersect at the same point.
The required time is the LCM (36,48,42)
LCM = 2 × 2 × 3 × 3 × 4 × 7
LCM = 1008
Therefore the required time is 1008 seconds

Example 3: Mr. Das has three classes. Each class has 24, 30 and 18 students respectively. Mr. Das wants to divide each class into groups so that every group in every class has the same number of students and there are no students left over. What is the maximum number of students he can put into each group?
Sol: We have to find the maximum number of students that can be put into each group.
This should give you an indication that here we have to calculate the HCF or GCD.
HFC (24,30,18) = 2 × 3 = 6
Therefore a maximum of 6 students can be put into each group.

Example 4: If the least prime factor of 'a' is 3 and the least prime factor of 'b' is 7, then find the least prime factor of (a+b)
Sol: When 3 will be the least prime factor of some a number, then 2 will not be a factor of such a number and thus it can be concluded that the number is odd. Thus, a = 2n+1 for some n.
Similarly when 7 will be the least prime factor of another b number, then again 2 will not be a factor of that number and it can be concluded that that number also is odd. Thus b = 2k - 1 for some k.
Adding a + b, we get
= a + b = 2n + 1 + 2k -1
= a + b = 2 (n+k)
a + b is clearly a even number as it is a multiple of 2. Thus, the least prime factor of (a+b) will be clearly 2.

Some Important Definitions

Real Numbers(R): All rational and irrational numbers are called real numbers.
Integers(I): All numbers from (...-3, -2, -1, 0, 1, 2, 3...) are called integers.
Rational Numbers(Q):Real numbers of the form p/q, q ≠ 0, p, q ∈ I are rational numbers.
All integers can be expressed as rational, for example, 5 = 5/1.
Decimal expansion of rational numbers terminating or non-terminating recurring.
Irrational Numbers(Q' ): Real numbers which cannot be expressed in the form p/q and whose decimal expansions are non-terminating and non-recurring.
Roots of primes like √2, √3, √5 etc. are irrational
Natural Numbers(N): Counting numbers are called natural numbers. N = {1, 2, 3, ...}
Whole Numbers(W): 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, ...}
Remember this!
All Natural Numbers are whole numbers.
All Whole Numbers are Integers.
All Integers are Rational Numbers.
All Rational Numbers are Real Numbers.
MULTIPLE CHOICE QUESTION
Try yourself: Which of the following statements is true based on the given information?
A
All natural numbers are irrational numbers.
B
All integers are irrational numbers.
C
All rational numbers are whole numbers.
CORRECT ANSWER
D
All rational numbers are real numbers.
Correct Answer: D
Natural numbers are a subset of whole numbers, which in turn are a subset of integers.
Integers are a subset of rational numbers.
Rational numbers can be expressed as fractions and are a subset of real numbers.
Therefore, it is correct to say that all rational numbers are real numbers. Options A, B, and C are incorrect based on the information provided.

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.
Note: 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.

Constructing a factor tree

Step 1: Write the number as a product of prime number and a composite number.
Step 2: Repeat the process till all the primes are obtained.
Example : Factorize 8190
So we have factorised 8190 as 2 × 3 × 3 × 5 × 7 × 13 as a product of primes, i.e., 8190 = 2 × 3² × 5 × 7 × 13 as a product of powers of primes

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.

Algorithm to locate HCF and LCM of two or more positive integers

Step 1: 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 2: To find HCF, identify common prime factor and find the least powers and multiply them to get HCF.
Step 3: To find LCM, find the greatest exponent and then multiply them to get the LCM.

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.

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.
MULTIPLE CHOICE QUESTION
Try yourself: Which of the following statements is true?
A
The sum or difference of a rational and irrational number is always rational.
B
The product or quotient of a non-zero rational number and an irrational number is always rational.
CORRECT ANSWER
C
If the prime factorization of 'q' in a rational number x = p/q is not of the form 2m5n, the decimal expansion of x is non-terminating but repeating.
D
The sum or difference of two irrational numbers is always irrational.

Correct Answer: C
Option A and B are incorrect because the sum, difference, product, or quotient of a rational and an irrational number is always irrational. Option D is incorrect because the sum or difference of two irrational numbers can be rational. For example, √2 - √2 = 0, which is a rational number. Option C is correct because, if 'q' in the rational number x = p/q is not of the form 2m5n, then x has a non-terminating repeating decimal expansion

Proof of Theorems : Real Numbers

Theorem 1.1 - Fundamental Theorem of Arithmetic

Statement: Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.
Given: Let N be a composite number. It can be expressed as a product of prime numbers.
To Prove: The prime factorization of N is unique, apart from the order of factors
Proof:
Consider any composite number N, which can be written as a product of primes.
For example, take 32760:
32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13
Written in exponent form: 23 × 32 × 5 × 7 × 13
Suppose there exists another way to factorize N into primes.
By the Fundamental Theorem of Arithmetic, the set of prime factors obtained will always be the same, only differing in order.
Thus, the prime factorization of a number is unique (except for the arrangement of factors).
Conclusion: The Fundamental Theorem of Arithmetic ensures that every composite number has a unique prime factorization, apart from the order of the factors

Theorem 1.2

Statement: If a prime number p divides a², then p also divides a, where a is a positive integer.
Given: Let p be a prime number and a be a positive integer such that p divides a².
To Prove: If p divides a², then p must also divide a.
Proof:
Let a be any positive integer, and let its prime factorization be:
a = p1 × p2 × ... × pn
Squaring both sides, we get:
a² = (p1 × p2 × ... × pn)² = p1² × p2² × ... × pn².
If a prime number p divides a², then p must be one of the prime factors of a².
Since each prime factor in a² appears as a square, p must already be a factor of itself.
This follows directly from the Fundamental Theorem of Arithmetic, which states that prime factorization is unique.
Therefore, if p divides a², then p must also divide a.

Conclusion:

The Theorem 1.2 proves that if a prime number divides the square of a number, then it must also divide the number itself. This theorem is widely used in number theory and proofs of irrationality, such as proving that √2 is irrational.

Theorem 1.3

Statement: √2 is an irrational number.
To Prove: √2 cannot be expressed as a fraction
pq
, where p and q are integers and q ≠ 0.
Proof :
Assume √2 is rational.
Then, we can write √2 as
pq
, where p and q are coprime integers (i.e., they have no common factor other than 1).
Squaring both sides:
2 =
p²q²
⇒ p² = 2q².
This means p² is divisible by 2, so p must also be divisible by 2 (from Theorem 1.2).
Let p = 2k for some integer k.
Substituting in p² = 2q², we get:
(2k)² = 2q² ⇒ 4k² = 2q² ⇒ q² = 2k².
This means q² is also divisible by 2, so q must also be divisible by 2.
But this contradicts our assumption that p and q are coprime (as both are divisible by 2).
Therefore, our assumption that √2 is rational must be false and√2 is rational.
Conclusion: Thus, √2 is irrational.

Real Numbers

Classification of Real Numbers

Fundamental Theorem of Arithmetic

Composite Number

Prime Number

HCF and LCM by Prime Factorisation Method

In this method, we first express the given numbers as a product of prime factors separately. Then, HCF is the product of the smaller power of each common prime factor in the numbers, and LCM is the product of the greatest power of each prime factor involved in the numbers.

Revisiting Irrational Numbers

Important Definitions

1. Real Numbers

2. Fundamental Theorem of Arithmetic

3. Rational Numbers

4. Irrational Number

5. Prime Number

Important Formulas with Examples

Worksheet:

True and False

Very Short Questions

Short Questions

Long Questions

Short Answer Questions

Unit Test

NCERT Solutions: Real Numbers (Exercise 1.1)

HOTS

Value-Based Questions

Very Short Answer Type Questions

Short Answer Type Questions- I

Q13: Find the LCM of 96 and 360 by using fundamental theorem of arithmetic.

Short Answer Type Questions- II

Q23: Prove that 3 + 2√3 is an irrational number.

Long AnsweType Questions

Case Based Type Questions

NCERT Solutions: Real Numbers (Exercise 1.2)

Very Short Answer Type Questions: Real Numbers

Long Answer Questions: Real Numbers

Case Based Questions: Real Numbers

Q1: Read the source below and answer the questions that follow:

Old age homes mean for senior citizens who are unable to stay with their families or destitute. These old age homes have special medical facilities for senior citizens such as mobile health care systems, ambulances, nurses and provision of well balanced meals.

Q2: Read the source below and answer the questions that follow:

In a morning walk, Naveeka, Arjun and Vedant step off together, their steps measuring 240 cm, 90 cm, 120 cm respectively. They want to go for a juice shop for a health issue, which is situated near by them.

Q3: Read the source below and answer the questions that follow:

Q4: Read the source below and answer the questions that follow:

Assertions & Reason Type Questions: Real Numbers

Visual Worksheet: Prime Factorisation (with Solutions)

Proof for Fundamental Theorem of Arithmetic

Solved Examples

What are Factors?

Highest Common Factor (H.C.F)

The largest or greatest factor common to any two or more given natural numbers is termed as the HCF of given numbers. Also known as GCD (Greatest Common Divisor).Example: Factors of the number 10 are 1, 2, 5 and 10Factors of the number 15 are 1, 3, 5 and 15We can see, the HCF (10, 15) = 5

Methods of Calculating Highest Common Factor

Finding the Highest Common Factors of Decimals

Finding the Highest Common Factor of Fractions

Solved Examples

Multiples

A multiple of a number is a number that is the product of a given number and some other natural number. Multiples can be observed in a multiplication table.Multiples of 2 are as follows:2 × 1 = 2,2 × 2 = 4,2 × 3 = 6,2 × 4 = 8,2 × 5 = 10,2 × 6 = 12,2 × 7 = 14,2 × 8 = 16,and so on.

Least Common Multiple (L.C.M)

The least or smallest common multiple of any two or more given natural numbers are termed as LCM.Example:Multiples of 3 are 3, 6, 9, 12, 15, 18, 21....Multiples of 5 are 5, 10, 15, 20, 25, 30....LCM (3,5) = 15

Methods of calculating Lowest Common Multiple

Finding the Least Common Multiple of Decimals

Solved Examples

Co-relation between HCF and LCM

Applications of LCM and HCF

Some Important Definitions

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.Note: 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.

Constructing a factor tree

Fundamental theorem of Arithmetic

Algorithm to locate HCF and LCM of two or more positive integers

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.

To determine the nature of the decimal expansion of rational numbers

Proof of Theorems : Real Numbers

Theorem 1.1 - Fundamental Theorem of Arithmetic

Theorem 1.2

Conclusion:

The Theorem 1.2 proves that if a prime number divides the square of a number, then it must also divide the number itself. This theorem is widely used in number theory and proofs of irrationality, such as proving that √2 is irrational.

Theorem 1.3

The largest or greatest factor common to any two or more given natural numbers is termed as the HCF of given numbers. Also known as GCD (Greatest Common Divisor).
Example: Factors of the number 10 are 1, 2, 5 and 10
Factors of the number 15 are 1, 3, 5 and 15
We can see, the HCF (10, 15) = 5

A multiple of a number is a number that is the product of a given number and some other natural number. Multiples can be observed in a multiplication table.
Multiples of 2 are as follows:
2 × 1 = 2,
2 × 2 = 4,
2 × 3 = 6,
2 × 4 = 8,
2 × 5 = 10,
2 × 6 = 12,
2 × 7 = 14,
2 × 8 = 16,
and so on.

The least or smallest common multiple of any two or more given natural numbers are termed as LCM.
Example:
Multiples of 3 are 3, 6, 9, 12, 15, 18, 21....
Multiples of 5 are 5, 10, 15, 20, 25, 30....
LCM (3,5) = 15

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

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.

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.