189-NCERT New Syllabus Grade 10 Real Numbers Ex-1.2

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NCERT New Syllabus Mathematics

Class: 10

Exercise 1.2

Topic: Real Numbers

Understanding Real Numbers: Key to Mathematical Mastery.

Welcome to the path of discovering one of the most fundamental topics in mathematics: real numbers. This chapter from the NCERT class 10 syllabus offers the framework for comprehending advanced mathematical concepts by concentrating on properties, theorems, and real-number operations. In this blog, we'll look at exercise 1.2, breaking down each difficulty carefully and addressing it. Whether you're studying for an exam or creating a solid foundation, this post will easily guide you through each answer. Let's discover the power of real numbers together!

EXERCISE 1.2

Q1. Prove that √5 is irrational.

Explanation:

1) We can prove this using indirect proof. It is also known as proof by contradiction.

Solution:

1) Let us assume, for the sake of contradiction, that √5 is a rational number.

2) Therefore, we can express √5 = p/q where p and q are coprime integers 

(having no common factor other than 1) and q ≠ 0.

(√5)² = (p/q)²

5 = (p)²/(q)²

p² = 5q²  ---------- equation 1

3) Equation (1) shows that 5 divides p², meaning that 5 must also divide p (since

if a prime divides the square of a number, it must divide the number itself).

4) Let $p=5r$ for some integer r. Substituting this value into equation (1):

(5r)² = 5(q)²

25r² = 5q²  ---------- equation 2

5) Dividing equation (2) by 5, we get:

q² = 5r²

6) Thus, we have shown that 5 divides both $p$ and $q$, which contradicts our initial

assumption that $p$ and $q$ are coprime.

7) Therefore, our assumption that √5 is a rational number must be incorrect.

8) Hence, √5 is an irrational number.

Q2. Prove that 3 + 2√5 is irrational.

Explanation:

1) We can prove this using indirect proof. It is also known as proof by contradiction.

Solution:

1) Let us assume, for contradiction, that 3 + 2√5 is a rational number.

2) Therefore, we can express 3 + 2√5 = p/q where p and q are coprime integers

(with no common factor other than 1) and q ≠ 0.

3 + 2√5 = p/q 

2√5 = (p/q) – 3

2√5 = [(p – 3q)/q]

√5 = (p – 3q)/2q

3) Since p and q are co-primes, (p – 3q)/2q is rational, so √5 is also rational.

However, this contradicts the well-known fact that √5 is irrational.

4) Therefore, our assumption was wrong, and 3 + 2√5 is an irrational number.

Q3. Prove that the following are irrationals :

(i) 1/√2    (ii) 7√5    (iii) 6 + √2

Explanation:

1) We can prove this using indirect proof. It is also known as proof by contradiction.

Solution:

(i) 1/√2

1) Assume, for contradiction, that 1/√2 is a rational number.

2) Thus, 1/√2 = p/q where p and q are co-primes and q ≠ 0. (i.e., no common

factors other than 1).

1/√2 = p/q

√2 = q/p

3) Since p and q are co-primes, q/p is a rational number, which implies √2 is 

rational. But this contradicts the known fact that which contradicts the fact that √2 is irrational.

4) Therefore, 1/√2 is an irrational number.

(ii) 7√5

1) Let us assume, for contradiction, that 7√5 is a rational number.

2) So, 7√5 = p/q where p and q are co-primes and q ≠ 0. (i.e., no common

factors other than 1).

7√5 = p/q

√5 = p/7q

3) Since p and q are co-primes, p/7q is a rational number, which implies √5 is 

rational. However, this contradicts the fact that √5 is irrational.

4) Thus, 7√5 is an irrational number.

(iii) 6 + √2

1) Assume that 6 + √2 is a rational number.

2) Therefore, 6 + √2 = p/q where p and q are co-primes integers and q ≠ 0. That is, there is no common factor other than 1.

6 + √2 = p/q 

√2 = (p/q) – 6

√2 = (p – 6q)/q

3) Since p and q are co-primes, (p – 6q)/q is rational, which implies √2 is rational. But this contradicts the fact that √2 is irrational.

4) Therefore, 6 + √2 is irrational number.

Conclusion: Unveiling the Power of Real Numbers

In this chapter on Real Numbers, we've explored the foundational blocks of mathematics, laying the groundwork for a deeper understanding of algebra and beyond. From the beauty of irrational numbers to the precision of the Euclidean algorithm, real numbers are at the core of every mathematical operation. As you continue to delve into the world of numbers, remember that these principles stretch far beyond the classroom, influencing technology, science, and everyday calculations. Keep exploring, keep calculating, and unlock the endless possibilities of real numbers!

#RealNumbersUnveiled #Class10Math #NCERTSyllabus #NumberTheory #AlgebraEssentials #MathInLife #MathIsBeautiful #NCERTClass10 #Mathematics #NCERTMaths #Grade10Maths #MathSyllabus #NCERTSolutions #MathTips #LearnMath #MathConcepts #MathMadeEasy #RealNumberSystem #MathHelp #PrimeNumbers #MathForStudents #CBSEMath #MathEducation #MathLearning #simple method

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NCERT New Syllabus Class 10 - Polynomials Exercise 2.1

Anil Satpute

188-NCERT New Syllabus Grade 10 Real Numbers Ex-1.1

NCERT New Syllabus Mathematics

Class: 10

Exercise 1.1

Topic: Real Numbers

Understanding Real Numbers: Key to Mathematical Mastery.

Welcome to the path of discovering one of the most fundamental topics in mathematics: real numbers. This chapter from the NCERT class 10 syllabus offers the framework for comprehending advanced mathematical concepts by concentrating on properties, theorems, and real-number operations. In this blog, we'll look at exercise 1.1, breaking down each difficulty carefully and addressing it. Whether you're studying for an exam or creating a solid foundation, this post will easily guide you through each answer. Let's discover the power of real numbers together!

1) Theorem 1.1 (Fundamental Theorem of Arithmetic):

Every composite number may be written as a unique product of prime numbers, regardless of how the prime elements are organized. This means that each composite number has a unique prime factorisation except the prime number sequence.

2) For any two positive integers a and b, the relationship between their HCF and

LCM is given by: HCF (a, b) × LCM (a, b) = a × b.

3) This formula allows us to easily calculate the LCM of two positive integers once

we know their HCF. 

EXERCISE 1.1

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

(i) 140     (ii) 156     (iii) 3825     (iv) 5005     (v) 7429

Explanation:

Every composite number can be represented as a product of prime numbers. 

Solution:

(i) 140

1) Prime factor of 140:            Note: Apply divisibility test of 2.

140 = 2 x 70                    Note: Apply divisibility test of 2.

140 = 2 x 2 x 35              Note: Apply divisibility test of 5.

140 = 2 x 2 x 5 x 7

140 = 2² x 5 x 7

2) The prime factorization of 140 is 2² x 5 x 7.

(ii) 156

1) Prime factor of 156:        Note: Apply divisibility test of 2.

156 = 2 x 78                Note: Apply divisibility test of 2.

156 = 2 x 2 x 39          Note: Apply divisibility test of 3.

156 = 2 x 2 x 3 x 13

156 = 2² x 3 x 13

2) The prime factorization of 156 is 2² x 3 x 13.

(iii) 3825

1) Prime factor of 3825:        Note: Apply divisibility test of 3.

3825 = 3 x 1275            Note: Apply divisibility test of 3.

3825 = 3 x 3 x 425        Note: Apply divisibility test of 5.

3825 = 3 x 3 x 5 x 85    Note: Apply divisibility test of 5.

3825 = 3 x 3 x 5 x 5 x 17

3825 = 3² x 5² x 17

2) The prime factorization of 3825 is 3² x 5² x 17.

(iv) 5005

1) Prime factor of 5005:         Note: Apply divisibility test of 5.

5005 = 5 x 1001             Note: Apply divisibility test of 7.

5005 = 5 x 7 x 143         Note: Apply divisibility test of 11.

5005 = 5 x 7 x 11 x 13

2) The prime factorization of 5005 is 5 x 7 x 11 x 13.

Note: Click here to know the details of the divisibility test of 7.

(v) 7429

Note:

a) 2 is not a factor of 7429 as it is not an even number. So 4, 8, 16, and so on

are also not the factors of 7429.

b) 3 is not a factor of 7429 as the sum of the digits is 22, which is not divisible

by 3. So 6, 9, 12, 15, and so on are also not factors 7429.

c) 5 is not a factor of 7429 as the unit placed digit is not 0 or 5. So 5, 10, 15,

and so on are also not factors 7429.

d) 7 is not a factor of 7429. So 14, 21, and so on are also not factors 7429.

 

Note: Click here to know the details of the divisibility test of 7.

 

1) Prime factor of 7429:           Note: Try directly dividing by 17.

7429 = 17 x 437               Note: Try directly dividing by 19.

7429 = 17 x 19 x 23

2) The prime factorization of 7429 is 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

Explanation:

1) To determine the LCM and HCF of given numbers, we first need to identify their

prime factors. 

2) The HCF (Highest Common Factor) is found by multiplying the smallest powers

of the common prime factors between the numbers.

3) The LCM (Least Common Multiple) is calculated by multiplying the greatest

powers of each prime factor present in the numbers.

4) The relationship between HCF and LCM is given by the formula:

HCF (p, q) × LCM (p, q) = p × q, where p and q are any two numbers that are positive integers.

Solution:

(i) 26 and 91

1) First, find the prime factors of 26:

26 = 2 x 13

2) Next, find the prime factors of 91:

91 = 7 x 13

3) The common factor between 26 and 91 is 13.

4) Therefore, the HCF of 26 and 91 is 13.

5) To get the LCM, multiply the individual prime factors:

LCM = 2 x 7 x 13

LCM = 14 x 13

LCM = 182.

6) Thus, the LCM of 26 and 91 is 182.

7) Therefore, LCM of 26 and 91 is 182, and HCF of 26 and 91 is 13.

8) We know 26 x 91 = 2366 -------- (equation 1).

9) Also, HCF x LCM = 13 x 182 = 2366 -------- (equation 2).

10) From Equation 1 and Equation 2, we confirm that:

HCF(26,91) x LCM(26,91) = 26 x 91. 

(ii) 510 and 92

1) First, find the prime factors of 510:

510 = 2 x 255

510 = 2 x 3 x 85

510 = 2 x 3 x 5 x 17

2) Next, find the prime factors of 92:

92 = 2 x 46

92 = 2 x 2 x 23

3) The common factor between 510 and 92 is 2.

4) Therefore, the HCF of 510 and 92 is 2.

5) To get the LCM, multiply the individual prime factors:

LCM = 2 x 2 x 3 x 5 x 17 x 23.

LCM = 4 x 15 x 17 x 23.

LCM = 23460.

6) Thus, the LCM of 510 and 92 is 23460.

7) Therefore, the LCM of 510 and 92 is 23460, and the HCF of 510 and 92 is 2.

8) We know 510 x 92 = 46920 -------- (equation 1)

9) Also, HCF x LCM = 2 x 23460 = 46920 -------- (equation 2).

10) From Equation 1 and Equation 2, we confirm that:

HCF(510,92) x LCM(510,92) = 510 x 92.

(iii) 336 and 54

1) First, find the prime factors of 336.

336 = 2 x 168

336 = 2 x 2 x 84

336 = 2 x 2 x 2 x 42

336 = 2 x 2 x 2 x 2 x 21

336 = 2 x 2 x 2 x 2 x 3 x 7

2) Next, find the prime factors of  54.

54 = 2 x 27

54 = 2 x 3 x 9

54 = 2 x 3 x 3 x 3

3) The common factor between 336 and 54 is 2 x 3 = 6.

4) Therefore, the HCF of 336 and 54 is 6.

5) To get the LCM, multiply the individual prime factors:

LCM = 2 x 2 x 2 x 2 x 3 x 3 x 3 x 7

LCM = 16 x 27 x 7
LCM = 3024.

6) Thus, the LCM of 336 and 54 is 3024

7) Therefore, LCM of 336 and 54 is 3024, and HCF of 336 and 54 is 6.

8) We know 336 x 54 = 18144 -------- (equation 1).

9) Also, HCF x LCM = 6 x 3024 = 18144 -------- (equation 2).

10) From Equation 1 and Equation 2, we confirm that: 

HCF(336, 54) x LCM(336, 54) = 336 x 54.

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

Solution:

(i) 12, 15 and 21

1) Now we will find the prime factors of 12.

12 = 2 x 6

12 = 2 x 2 x 3

2) Now we will find the prime factors of 15.

15 = 3 x 5

3) Now we will find the prime factors of 21.

21 = 3 x 7

4) Here, the common factor is 3.

5) So HCF of 12, 15, and 21 is 3.

6) Simmilarly, LCM = 2 x 2 x 3 x 5 x 7 = 420.

7) Thus, LCM is 420, and HFC is 3.

(ii) 17, 23 and 29

1) Now we will find the prime factors of 17.

17 = 1 x 17

2) Now we will find the prime factors of 23.

23 = 1 x 23

3) Now we will find the prime factors of 29.

29 = 1 x 29

4) Here common factor is 1.

5) The HCF of 17, 23, and 29 is 1.

6) Simmilarly, LCM = 1 x 17 x 23 x 29 = 11339.

7) Thus, LCM is 11339, and HFC is 1.

(iii) 8, 9 and 25

1) Now we will find the prime factors of 8.

8 = 1 x 2 x 4

8 = 1 x 2 x 2 x 2

2) Now we will find the prime factors of 9.

9 = 1 x 3 x 3

3) Now we will find the prime factors of 25.

25 = 1 x 5 x 5

4) Here common factor is 1.

5) So HCF of 8, 9, and 25 is 1.

6) Simmilarly, LCM = 1 x 2 x 2 x 2 x 3 x 3 x 5 x 5 = 1800.

7) Thus, LCM is 1800, and HFC is 1.

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

Explanation:

1) We know that HCF (p, q) × LCM (p, q) = p × q, where p and q are any two numbers that are positive integers.

2) Out of the above 3 values, if 2 values are given, we can find the 3rd value.

Solution:

1) Here HCF (306, 657) = 9.

2) So, HCF of 306 and 657 = 9.

3) Products of 306 and 657 are 306 x 657

4) We know that HCF (p, q) × LCM (p, q) = p × q, so, 

[LCM (306, 657)] = [306 x 657] / HCF (306, 657)

[LCM (306, 657)] = [306 x 657] / 9

[LCM (306, 657)] = [34 x 657]

[LCM (306, 657)] = [22338]

5) Thus, LCM (306, 657) = 22338.

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

Explanation:

1) If any number ends with the digit 0, then that number must be divisible by 2 and 5 simultaneously.

2) To check whether 6ⁿ ends with 0, we must find whether 6ⁿ has 2 and 5 as prime factors or not.

Solution:

1) Factors of 6ⁿ = (2 x 3)ⁿ  

2) Here 2 is the factor of 6ⁿ, but 5 is not a factor of 6ⁿ.

3) So 6ⁿ can't end with 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.

Explanation:

1) Prime numbers have only 2 factors. 1 and the number itself. e.g. the prime number 5 has two factors. 1 and 5 itself.

2) The positive Number, which has factors other than 1, is known as a composite number.

Solution:

1) The first expression is (7 × 11 × 13) + 13. We will simplify it.

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

= 13 (77 + 1) 

= 13 (78)

= 13 (2 x 39)

= 13 (2 x 3 x 13)

2) The first expression has factors as (2 x 3 x 13 x 13)

3) The number obtained from (7 × 6 × 5 × 4 × 3 × 2 × 1) + 5 is a composite number.

4) The second expression is (7 × 11 × 13) + 13. We will simplify it.

(7 × 6 × 5 × 4 × 3 × 2 × 1) + 5 = 5 ((7 x 6 x 4 x 3 x 2 x 1) + 1)

= 5 (1008 + 1) 

= 5 (1009)

5) The second expression has factors as (5 x 1009)

6) The number obtained from (7 × 6 × 5 × 4 × 3 × 2 × 1) + 5 is a composite

number.

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?

Explanation:

1) Sonia takes 18 minutes and Ravi takes 12 minutes to complete one round. 

2) Now, we need to find out how many minutes will they meet again at the same point. 

3) Here we will find LCM of 18 and 12 to get the time when both meet again at the starting point.

Solution:

1) To find the LCM of 18 and 12, we first need to determine their prime factors.

2) Prime factorization of 18:

18 = 2 x 9

18 = 2 x 3 x 3

3) Prime factorization of 12:

12 = 2 x 6

12 = 2 x 2 x 3

4) The common prime factor is 2 x 3 = 6.

5) The LCM (Least Common Multiple) is the product of all the highest powers of the

prime factors:

LCM = 2 x 2 x 3 x 3 = 36.

6) Therefore, the LCM is 36.

7) After 36 minutes Sonia and Ravi will meet again at the starting point.

Conclusion: Unveiling the Power of Real Numbers

In this chapter on Real Numbers, we've explored the foundational blocks of mathematics, laying the groundwork for a deeper understanding of algebra and beyond. From the beauty of irrational numbers to the precision of the Euclidean algorithm, real numbers are at the core of every mathematical operation. As you continue to delve into the world of numbers, remember that these principles stretch far beyond the classroom, influencing technology, science, and everyday calculations. Keep exploring, keep calculating, and unlock the endless possibilities of real numbers!

#RealNumbersUnveiled #Class10Math #NCERTSyllabus #NumberTheory #AlgebraEssentials #MathInLife #MathIsBeautiful #NCERTClass10 #Mathematics #NCERTMaths #Grade10Maths #MathSyllabus #NCERTSolutions #MathTips #LearnMath #MathConcepts #MathMadeEasy #RealNumberSystem #MathHelp #PrimeNumbers #MathForStudents #CBSEMath #MathEducation #MathLearning #simple method

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Anil Satpute

187 🚀 Announcing the Launch of Step-by-Step Solutions for Class 10 Maths (New NCERT Syllabus) - October 7, 2024! 🎉

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Anil Satpute