Overview of Class 10 Maths Chapter 1: Real Numbers
The transition of students to the upper grades might feel like a nightmare to a Class 10 student trying to handle different subjects mixed with numerous assignments. The first chapter of the NCERT Maths textbook for Class 10, which deals with real numbers, is already covered with ToppersSky complete solutions crafted according to the latest CBSE syllabus, for instance. A very cool method of presenting the easy to understand solutions has been adopted wherein, the difficult concepts will be taught step by step using animated demonstrations. Following to ToppersSky with the NCERT Class 10 Maths Chapter 1 solutions, students are going to not only acquire a solid base on real numbers but also get their fundamental understanding up to the level where they will be self-confident enough to tackle even difficult problems alone.
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers – Quick View | ToppersSky
A prime number p divides a², then p also divides a, this is illustrated very simply with animated step-by-step logic. The sum or difference of a rational and an irrational number always results in an irrational number, this is shown through very simple visual examples. The product or quotient of a non-zero rational number and an irrational number is irrational, this is made very easy to understand with concept animations. For any prime number p, √p is irrational, this is explained with very clear visual reasoning.
Important Topics Covered in Class 10 Maths Chapter 1 – Real Numbers
The very first chapter of the Class 10 Maths syllabus is Real Numbers, which is among the most important foundational chapters. The chapter consists of four major sections and is integral in reinforcing basic mathematical concepts. For that reason, the ToppersSky’s animated learning method facilitates the students’ comprehension and application of the concepts much easier. It is absolutely necessary that the students take the time to meticulously study every single topic so as to create the right understanding, conquer the problem-solving methods, and thus, confidently answer questions involving Real Numbers.
The most important aspects dealt with in the Real Numbers discussion include:
- The meaning of Real Numbers
- The Fundamental Theorem of Arithmetic (HCF and LCM)
- Irrational Numbers re-examined
1.1 Introduction
In the introduction of Class 10 Maths Chapter 1, students revise the concepts of real numbers and irrational numbers studied in Class 9. With the help of animated pictures, the writer has mastered the art of disclosing the future main issues, such as Euclid’s Division Algorithm and the Fundamental Theorem of Arithmetic, in a very nice and easy-to-follow way. The Fundamental Theorem of Arithmetic reveals the unique representation of every composite number as a product of prime factors. By means of visuals and gradual animation, the students are going to have no problem understanding this very crucial concept, which has extensive and significant uses in math.
1.2 The Fundamental Theorem of Arithmetic
To put it simply, by the application of the aforementioned theorem, all composite numbers can be written as a multiplication of prime numbers. The uniqueness of this prime factorisation is such that every natural number has this kind of factorisation as the order of prime factors does not matter. With the help of ToppersSky’s animated explanations, students can grasp this concept effortlessly through simple examples that are based on the following fundamentals:
HCF (Highest Common Factor):
The HCF of two or more numbers the biggest number that can divide all the numbers without any remainder is defined as such.
Example: The HCF of numbers 60 and 75 is 15.
LCM (Least Common Multiple):
The LCM of two or more numbers the smallest number that every one of the numbers can divide without any remainder is defined as such.
Example: The LCM of 2, 4, and 5 is 20.
For any two integers a and b that are positive,
HCF(a, b) × LCM(a, b) = a × b
Thus, using this theorem, a natural number can be factorized into prime numbers like 253 = 11 × 23 and 4 = 2 × 2, which is made visually clear by the step-by-step animations on ToppersSky.
1.3 Revisiting Irrational Numbers
In this part of ToppersSky’s NCERT Solutions for Class 10 Maths Chapter 1, the learner reviews the idea of irrational numbers familiar from earlier classes. With the help of animated explanations step by step, the learner gets to know and prove that √p is an irrational number where p is a prime number.
A number n is an irrational number if it is not possible to express it in the form x/y where x and y are integers and y is not equal to 0. Learners through very basic visual examples come to the conclusion that numbers of the kind √2, √3, etc. are indeed irrational.
The Real Numbers Chapter Discusses:
- Euclid’s Division Algorithm: This method, which was presented by the ancient mathematician Euclid, demonstrates how numbers are divided to determine the quotient and remainder. With a series of animated step-by-step explanations, students will have no difficulty grasping the process of this algorithm and its application in problem-solving.
- Fundamental Theorem of Arithmetic: This part teaches the concept of rational and irrational numbers through simple and clear reasoning. Rational numbers can be expressed as the fractions 1/2 or 3/4, whereas irrational numbers can’t be represented as fractions, for instance, the square root of two. With the help of visual animations, these ideas are conveyed in a very comprehensible manner.
- Decimal Expansions of Real Numbers: In this part of the curriculum, learners are able to grasp how digits are put together to form a number in decimal system, e.g. expressing 1/2 as 0.5. It also clarifies the concept of irrational numbers expressed through decimals that are non-terminating and non-repeating, using simple-to-follow animated illustrations that make the idea more comprehensible.
Conclusion
ToppersSky’s NCERT Solutions for Class 10 Maths Chapter 1 – Real Numbers are a great resource for students as the clear explanations are provided through step-by-step animated solutions. Students get to learn the basics of real numbers very easily. The students need to concentrate on understanding the main points such as rational numbers, irrational numbers, and their properties since they are the basis for the more advanced mathematical topics. It is a fact that in the board exams, three questions on average are taken out from this chapter, which makes it an area of an important and easy score. Through ToppersSky’s animated solutions, students are able to create a strong conceptual base, which is a must for higher-level Mathematics. The regular practice done with these solutions increases the confidence level and helps the students to perform well in the exams.
Class 10 Maths Chapter 1 Real Numbers: Practice & Exercises
Exercise 1.1
1. Use Euclid’s division algorithm to find the HCF of:
(i) Find the HCF of 135 and 225
Ans: We have to find the HCF of 135 and 225 by using Euclid’s Division Algorithm.
According to Euclid’s division algorithm, the HCF of any two positive integers a and b, where a > b, is found as follows:
First find the values of q and r, where a = bq + r, 0 ≤ r < b.
If r = 0, the HCF is b. If r ≠ 0, apply Euclid’s lemma to b and r.
Continue the steps till the remainder becomes zero. When the remainder is zero, the divisor is the HCF.
Let a = 225 and b = 135.
Since a > b, using the division algorithm, we get:
a = bq + r
⇒ 225 = 135 × 1 + 90
Here:
b = 135
q = 1
r = 90
Since r ≠ 0, we apply Euclid’s lemma to b (new divisor) and r (new remainder). We get:
⇒ 135 = 90 × 1 + 45
Here:
b = 90
q = 1
r = 45
Since r ≠ 0, we again apply Euclid’s lemma to b and r. We get:
⇒ 90 = 45 × 2 + 0
Now, we get r = 0, thus we can stop at this stage.
When the remainder is zero, the divisor is the HCF.
Therefore, the HCF of 135 and 225 is 45.
(ii) Find the HCF of 196 and 38220
Ans: We have to find the HCF of 196 and 38220 by using Euclid’s Division Algorithm.
According to Euclid’s division algorithm, the HCF of any two positive integers a and b, where a > b, is found as follows:
First find the values of q and r, where a = bq + r, 0 ≤ r < b.
If r = 0, the HCF is b. If r ≠ 0, apply Euclid’s lemma to b and r.
Continue the steps till the remainder becomes zero. When the remainder is zero, the divisor is the HCF.
Let a = 38220 and b = 196.
Since a > b, using the division algorithm, we get:
a = bq + r
⇒ 38220 = 196 × 195 + 0
Here:
b = 196
q = 195
r = 0
Since we get r = 0, we can stop at this stage.
When the remainder is zero, the divisor is the HCF.
Therefore, the HCF of 196 and 38220 is 196.
(iii) Find the HCF of 867 and 255
Ans: We have to find the HCF of 867 and 255 by using Euclid’s Division Algorithm.
According to Euclid’s division algorithm, the HCF of any two positive integers a and b, where a > b, is found as follows:
First find the values of q and r, where a = bq + r, 0 ≤ r < b.
If r = 0, the HCF is b. If r ≠ 0, apply Euclid’s lemma to b and r.
Continue the steps till the remainder becomes zero. When the remainder is zero, the divisor is the HCF.
Let a = 867 and b = 255.
Since a > b, using the division algorithm, we get:
a = bq + r
⇒ 867 = 255 × 3 + 102
Here:
b = 255
q = 3
r = 102
Since r ≠ 0, we apply Euclid’s lemma to b (new divisor) and r (new remainder). We get:
⇒ 255 = 102 × 2 + 51
Here:
b = 102
q = 2
r = 51
Since r ≠ 0, we again apply Euclid’s lemma to b and r. We get:
⇒ 102 = 51 × 2 + 0
Now, we get r = 0, thus we can stop at this stage.
When the remainder is zero, the divisor is the HCF.
Therefore, the HCF of 867 and 255 is 51.
2. Show that any positive odd integer is of the form 6q + 1, 6q + 3, or 6q + 5, where q is some integer.
Ans: Let a be any positive integer and let b = 6. By Euclid’s Division Algorithm, we can write
a = bq + r, where 0 ≤ r < b
Here, 0 ≤ r < 6. Substituting the value of b, we get
a = 6q + r
Now consider the possible values of r:
If r = 0, then a = 6q
If r = 1, then a = 6q + 1
If r = 2, then a = 6q + 2
If r = 3, then a = 6q + 3
If r = 4, then a = 6q + 4
If r = 5, then a = 6q + 5
Thus, a can be written as 6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4, or 6q + 5.
Now observe:
6q + 1 = 2(3q) + 1 = 2k1 + 1
6q + 3 = 2(3q + 1) + 1 = 2k2 + 1
6q + 5 = 2(3q + 2) + 1 = 2k3 + 1
where k1, k2, and k3 are integers. Hence, 6q + 1, 6q + 3, and 6q + 5 are all of the form 2k + 1 and are therefore odd.
Thus, any positive odd integer can be expressed in the form 6q + 1, 6q + 3, or 6q + 5, where q is an integer.
3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Ans: Given that an army contingent of 616 members is to march behind an army band of 32 members in a parade, and both groups must march in the same number of columns.
We need to find the maximum number of columns in which they can march. This is given by the Highest Common Factor (HCF) of 616 and 32.
We use Euclid’s Division Algorithm to find the HCF.
Let a = 616 and b = 32. Since a > b, applying the division algorithm:
a = bq + r
616 = 32 × 19 + 8
Here,
b = 32
q = 19
r = 8
Since r ≠ 0, we now apply Euclid’s lemma to b (new divisor) and r (new remainder):
32 = 8 × 4 + 0
Since the remainder is now 0, we stop the process. When the remainder becomes zero, the divisor at that stage is the HCF.
Therefore, the HCF of 616 and 32 is 8.
Hence, the maximum number of columns in which both groups can march is 8 columns.
4. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
(Hint: Let x be any positive integer and show that it is of the form 3q, 3q + 1, or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.)
Ans: Let a be any positive integer and let b = 3. By Euclid’s division algorithm, we can write:
a = bq + r, where 0 ≤ r < b
Here, 0 ≤ r < 3. Substituting the value of b, we get:
a = 3q + r
Now consider the possible values of r:
If r = 0, then a = 3q
If r = 1, then a = 3q + 1
If r = 2, then a = 3q + 2
Therefore, any positive integer a is of the form 3q, 3q + 1, or 3q + 2.
Now, squaring each of these forms:
a2 = (3q)2, (3q + 1)2, or (3q + 2)2
Applying the identity (a + b)2 = a2 + 2ab + b2, we get:
(3q)2 = 9q2 = 3(3q2) = 3m, where m = 3q2
(3q + 1)2 = 9q2 + 6q + 1 = 3(3q2 + 2q) + 1
⇒ a2 = 3m + 1, where m = 3q2 + 2q
(3q + 2)2 = 9q2 + 12q + 4
= 3(3q2 + 4q + 1) + 1
⇒ a2 = 3m + 1, where m = 3q2 + 4q + 1
Hence, the square of any positive integer is either of the form 3m or 3m + 1, where m is an integer.
5. Use Euclid’s Division Lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Ans: Let a be any positive integer and b = 3.
By Euclid’s division algorithm, we get:
a = bq + r, where 0 ≤ r < b
Here, 0 ≤ r < 3
Substituting the values, we get:
⇒ a = 3q + r
If r = 0, then
⇒ a = 3q + 0
⇒ a = 3q
If r = 1, then
⇒ a = 3q + 1
If r = 2, then
⇒ a = 3q + 2
Therefore, a = 3q or 3q + 1 or 3q + 2.
Case 1: When a = 3q
⇒ a3 = (3q)3
⇒ a3 = 27q3
⇒ a3 = 9(3q3)
⇒ a3 = 9m, where m = 3q3
Case 2: When a = 3q + 1
⇒ a3 = (3q + 1)3
Using the identity (a + b)3 = a3 + 3a2b + 3ab2 + b3, we get:
⇒ a3 = 27q3 + 27q2 + 9q + 1
⇒ a3 = 9(3q3 + 3q2 + q) + 1
⇒ a3 = 9m + 1, where m = 3q3 + 3q2 + q
Case 3: When a = 3q + 2
⇒ a3 = (3q + 2)3
Using the identity (a + b)3, we get:
⇒ a3 = 27q3 + 54q2 + 36q + 8
⇒ a3 = 9(3q3 + 6q2 + 4q) + 8
⇒ a3 = 9m + 8, where m = 3q3 + 6q2 + 4q
Therefore, the cube of any positive integer is always of the form 9m, 9m + 1, or 9m + 8.
Exercise 1.2 (Page No. 11)
1. Express each number as product of its prime factors:
(i) 140
Ans: We know that the procedure of writing a number as the product of prime numbers is known as the prime factorization. Prime numbers that can be multiplied to obtain the original number are known as prime factors.
⇒ 140 = 2 × 2 × 5 × 7
∴ 140 = 22 × 5 × 7
Therefore, the prime factors of 140 are 2, 5, and 7.
(ii) 156
Ans: We know that the procedure of writing a number as the product of prime numbers is known as the prime factorization. Prime numbers that can be multiplied to obtain the original number are known as prime factors.
⇒ 156 = 2 × 2 × 3 × 13
∴ 156 = 22 × 3 × 13
Therefore, the prime factors of 156 are 2, 3, and 13.
(iii) 3825
Ans: We know that the procedure of writing a number as the product of prime numbers is known as the prime factorization. Prime numbers that can be multiplied to obtain the original number are known as prime factors.
⇒ 3825 = 3 × 3 × 5 × 5 × 17
∴ 3825 = 32 × 52 × 17
Therefore, the prime factors of 3825 are 3, 5, and 17.
(iv) 5005
Ans: We know that the procedure of writing a number as the product of prime numbers is known as the prime factorization. Prime numbers that can be multiplied to obtain the original number are known as prime factors.
⇒ 5005 = 5 × 7 × 11 × 13
∴ 5005 = 5 × 7 × 11 × 13
Therefore, the prime factors of 5005 are 5, 7, 11, and 13.
(v) 7429
Ans: We know that the procedure of writing a number as the product of prime numbers is known as the prime factorization. Prime numbers that can be multiplied to obtain the original number are known as prime factors.
⇒ 7429 = 17 × 19 × 23
∴ 7429 = 17 × 19 × 23
Therefore, the prime factors of 7429 are 17, 19, and 23.
2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = Product of two numbers
(i) 26 and 91
Answer: First, we write the prime factors of 26 and 91.
⇒ 26 = 2 × 13
⇒ 91 = 7 × 13
Now, we know that HCF is the highest factor among the common factors of two numbers.
Therefore, the HCF of 26 and 91 is 13.
Now, we know that LCM is the least common multiple. To find the LCM, we multiply each factor the maximum number of times it occurs in any of the numbers.
Then, the LCM of 26 and 91 will be:
⇒ 2 × 7 × 13 = 182
Therefore, the LCM of 26 and 91 is 182.
Now, the product of the two numbers is:
⇒ 26 × 91 = 2366
Product of LCM and HCF is:
⇒ 13 × 182 = 2366
We get LCM × HCF = Product of two numbers.
Hence, the required result is verified.
(ii) 510 and 92
Answer: First, we write the prime factors of 510 and 92. We get:
⇒ 510 = 2 × 3 × 5 × 17
⇒ 92 = 2 × 2 × 23
Now, we know that HCF is the highest factor among the common factors of two numbers.
Therefore, the HCF of 510 and 92 is 2.
Now, we know that LCM is the least common multiple. To find the LCM, we multiply each factor the maximum number of times it occurs in any number.
Then, the LCM of 510 and 92 will be:
⇒ 2 × 2 × 3 × 5 × 17 × 23 = 23460
Therefore, the LCM of 510 and 92 is 23460.
Now, the product of the two numbers is:
⇒ 510 × 92 = 46920
Product of LCM and HCF is:
⇒ 2 × 23460 = 46920
We get LCM × HCF = Product of two numbers.
The desired result has been verified.
(iii) 336 and 54
Answer: First, we write the prime factors of 336 and 54. We get:
⇒ 336 = 2 × 2 × 2 × 2 × 3 × 7
and
⇒ 54 = 2 × 3 × 3 × 3
Now, we know that HCF is the highest factor among the common factors of two numbers.
Therefore, the HCF of 336 and 54 is 2 × 3 = 6.
Now, we know that LCM is the least common multiple. To find the LCM, we multiply each factor the maximum number of times it occurs in any number.
Then, the LCM of 336 and 54 will be:
⇒ 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 = 3024
Therefore, the LCM of 336 and 54 is 3024.
Now, the product of the two numbers is:
⇒ 336 × 54 = 18144
Product of LCM and HCF is:
⇒ 6 × 3024 = 18144
We get LCM × HCF = Product of two numbers.
The desired result has been verified.
3. Find the LCM and HCF of the following integers by applying the prime factorization method.
(i) 12, 15 and 21
Ans: The procedure of writing a number as the product of prime numbers is known as the prime factorization.
The prime factors of 12, 15 and 21 are as follows:
12 = 2 × 2 × 3
15 = 3 × 5
21 = 3 × 7
Now, we know that HCF is the highest factor among the common factors of the given numbers.
Therefore, the HCF of 12, 15 and 21 is 3.
Now, we know that LCM is the least common multiple. To find the LCM, we take each prime factor the maximum number of times it occurs in any of the numbers.
LCM of 12, 15 and 21 = 2 × 2 × 3 × 5 × 7 = 420
Therefore, the LCM of 12, 15 and 21 is 420.
(ii) 17, 23 and 29
Ans: The procedure of writing a number as the product of prime numbers is known as the prime factorization.
The prime factors of 17, 23 and 29 are as follows:
17 = 17 × 1
23 = 23 × 1
29 = 29 × 1
Now, we know that HCF is the highest factor among the common factors of the given numbers.
Therefore, the HCF of 17, 23 and 29 is 1.
Now, we know that LCM is the least common multiple. To find the LCM, we multiply each prime factor the maximum number of times it occurs in any of the numbers.
LCM of 17, 23 and 29 = 17 × 23 × 29 = 11339
Therefore, the LCM of 17, 23 and 29 is 11339.
(iii) 8, 9 and 25
Ans: The procedure of writing a number as the product of prime numbers is known as the prime factorization.
The prime factors of 8, 9 and 25 are as follows:
8 = 2 × 2 × 2
9 = 3 × 3
25 = 5 × 5
Now, we know that HCF is the highest factor among the common factors of the given numbers. Since there is no common factor other than 1:
Therefore, the HCF of 8, 9 and 25 is 1.
Now, we know that LCM is the least common multiple. To find the LCM, we multiply each prime factor the maximum number of times it occurs in any of the numbers.
LCM of 8, 9 and 25 = 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800
Therefore, the LCM of 8, 9 and 25 is 1800.
4. Given that HCF (306, 657) = 9, find LCM (306, 657).
Ans: We have been given the HCF of two numbers (306, 657) = 9. We have to find the LCM of (306, 657).
Now, we know that:
LCM × HCF = Product of the two numbers
Substituting the values, we get:
LCM × 9 = 306 × 657
⇒ LCM = (306 × 657) / 9
∴ LCM = 22338
Therefore, the LCM of (306, 657) is 22338.
5. Check whether 6n can end with the digit 0 for any natural number n.
Ans: We have to check whether 6n can end with the digit 0 for any natural number n.
By the divisibility rule, if any number ends with the digit 0, it must be divisible by both 2 and 5.
Thus, the prime factorization of 6n is:
6n = (2 × 3)n
Now, we observe that for any value of n, 6n is not divisible by 5.
Therefore, 6n cannot end with the digit 0 for any natural number n.
Exercise 1.2 (Page No. 14)
1. Prove that √5 is irrational.
Ans: We have to prove that √5 is irrational. We will use the method of contradiction to prove it.
Let √5 be a rational number of the form a / b, where b ≠ 0 and a and b are co-prime, i.e., they have only 1 as a common factor.
Let √5 = a⁄b
Now, squaring both sides, we get:
(√5)2 = (a⁄b)2
⇒ 5 = a2⁄b2
⇒ a2 = 5b2 ……(1)
If a2 is divisible by 5, then a is also divisible by 5.
Let a = 5k, where k is any integer.
Again, squaring both sides, we get:
⇒ a2 = (5k)2
Substituting the value in equation (1), we get:
⇒ (5k)2 = 5b2
⇒ b2 = 5k2 ……(2)
If b2 is divisible by 5, then b is also divisible by 5.
From equations (1) and (2), we conclude that a and b have 5 as a common factor.
This contradicts our assumption that a and b are co-prime.
Therefore, √5 is irrational.
2. Prove that 3 + 2√5 is irrational.
Ans: We have to prove that 3 + 2√5 is irrational. We will use the method of contradiction to prove it.
Let 3 + 2√5 be a rational number of the form a⁄b, where b ≠ 0 and a and b are co-prime, i.e., they have only 1 as a common factor.
Let 3 + 2√5 = a⁄b
⇒ 2√5 = a⁄b − 3
⇒ √5 = 1/2 ( a⁄b − 3 ) ……(1)
From equation (1), we can say that 1/2 ( a⁄b − 3 ) is rational, so √5 must be rational.
But this contradicts the fact that √5 is irrational. Hence, the assumption is false.
Therefore, 3 + 2√5 is irrational.
Hence proved.
Exercise 1.3 (Page No. 17)
1. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational and of the form p⁄q, what can you say about the prime factors of q?
(i) 43.123456789
Ans: Given a decimal expansion 43.123456789.
The given number has a terminating decimal expansion. Hence, we can write the number as 43123456789⁄1000000000, which is of the form p⁄q.
Therefore, the number 43.123456789 is a rational number.
Since the number has a terminating decimal expansion, the prime factors of q must be of the form 2n5m.
(ii) 0.120120012000120000……
Ans: Given a decimal expansion 0.120120012000120000……
On observing the given expansion, we see that the number has a non-terminating and non-repeating decimal expansion.
Hence, we cannot express it in the form p⁄q.
Therefore, the given number is irrational.
(iii) 43. 123456789
Ans: Given the decimal expansion 43. 123456789.
The given number has a non-terminating but repeating decimal expansion. Hence, the number can be expressed in the form p⁄q.
Therefore, the number 43. 123456789 is a rational number.
But the factors of the denominator are not of the form 2n5m. The denominator also has prime factors other than 2 and 5.
FAQs
1. Is π exactly equal to 22/7?
No, π is an irrational number, while 22/7 is only a rational approximation of π. The decimal expansion of π is non-terminating and non-repeating, whereas the decimal form of 22/7 is repeating. This difference is an important concept in Class 10 Real Numbers.
2. Are NCERT Solutions for Real Numbers just for copying homework?
No, NCERT Only are meant for learning, not just copying answers. They help you check your method, not only the final result. By using Class 10 Real Numbers NCERT Solutions, students can understand the correct steps and logic needed to solve problems according to board guidelines.
3. Do I only need to learn the final formulas in Class 10 Maths Chapter 1?
No, understanding the logic and derivation once NCERT and reasoning behind concepts like Euclid’s Division Lemma and the Fundamental Theorem of Arithmetic, not just the final formulas. Many students think that memorising the formula
HCF × LCM = Product of the numbers is enough. However, exams often test your conceptual understanding by asking you to apply the prime factorisation method or to write formal proofs, such as proving the and.
4. Is Euclid’s Division Lemma only used to find HCF?
No, although finding the HCF is its most common use in this chapter, Euclid’s Division Lemma is a more fundamental concept. It is also used to prove general properties of integers, such as showing that any positive odd integer can be written in the form 4 q+1 or 4 q+3.
5. Does the Fundamental Theorem of Arithmetic just say we can find prime factors?
No, the theorem states something much more important. It says that every composite number can be written as a product of prime numbers,prime factorisation is unique, except for the order of the factors. The in part is the most critical aspect of the theorem and is often tested in exams.
