Chapter 1: Real Numbers (NCERT Solutions)

Exercise 1.1: Fundamental Theorem of Arithmetic

Q1. Express each number as a product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429

Solution:

(i) 140
140 = 2 × 70
= 2 × 2 × 35
= 2 × 2 × 5 × 7
= 2² × 5 × 7

(ii) 156
156 = 2 × 78
= 2 × 2 × 39
= 2 × 2 × 3 × 13
= 2² × 3 × 13

(iii) 3825
3825 = 3 × 1275
= 3 × 3 × 425
= 3 × 3 × 5 × 85
= 3 × 3 × 5 × 5 × 17
= 3² × 5² × 17

(iv) 5005
5005 = 5 × 1001
= 5 × 7 × 143
= 5 × 7 × 11 × 13
= 5 × 7 × 11 × 13

(v) 7429
7429 = 17 × 437
= 17 × 19 × 23
= 17 × 19 × 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

Solution:

(i) 26 and 91
26 = 2 × 13
91 = 7 × 13
HCF = 13
LCM = 2 × 7 × 13 = 182
Verification:
Product of numbers = 26 × 91 = 2366
HCF × LCM = 13 × 182 = 2366
Hence Verified.

(ii) 510 and 92
510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23 = 2² × 23
HCF = 2
LCM = 2² × 3 × 5 × 17 × 23 = 23460
Verification:
Product = 510 × 92 = 46920
HCF × LCM = 2 × 23460 = 46920
Hence Verified.

(iii) 336 and 54
336 = 2⁴ × 3 × 7
54 = 2 × 3³
HCF = 2 × 3 = 6
LCM = 2⁴ × 3³ × 7 = 3024
Verification:
Product = 336 × 54 = 18144
HCF × LCM = 6 × 3024 = 18144
Hence Verified.

Q3. Find the LCM and HCF of the following integers by applying the prime factorization method. (i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25

Solution:

(i) 12, 15 and 21
12 = 2² × 3
15 = 3 × 5
21 = 3 × 7
HCF = 3
LCM = 2² × 3 × 5 × 7 = 420

(ii) 17, 23 and 29
All are prime numbers.
HCF = 1
LCM = 17 × 23 × 29 = 11339

(iii) 8, 9 and 25
8 = 2³
9 = 3²
25 = 5²
No common factors other than 1.
HCF = 1
LCM = 2³ × 3² × 5² = 8 × 9 × 25 = 1800

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

Solution:

We know that LCM × HCF = Product of two numbers
LCM × 9 = 306 × 657
LCM = (306 × 657) / 9
LCM = 34 × 657
LCM = 22338

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

Solution:

If a number ends with digit 0, it must be divisible by 10, meaning it must have 2 and 5 as prime factors.
Prime factorization of 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ.
It contains the prime factor 2 but not 5.
By the Fundamental Theorem of Arithmetic, this factorization is unique.
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:

Case 1: 7 × 11 × 13 + 13
= 13 (7 × 11 + 1)
= 13 (77 + 1)
= 13 × 78
Since it has factors other than 1 and itself (13 and 78), it is a composite number.

Case 2: 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 5 (7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 (1008 + 1)
= 5 × 1009
Since it has factors other than 1 and itself (5 and 1009), it 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?

Solution:

They will meet again at the LCM of their times.
LCM of 18 and 12:
18 = 2 × 3²
12 = 2² × 3
LCM = 2² × 3² = 4 × 9 = 36
Therefore, they will meet again after 36 minutes.

Exercise 1.2: Irrational Numbers

Q1. Prove that √5 is irrational.

Solution:

Let us assume, to the contrary, that √5 is rational.
Then we can find coprime integers a and b (b ≠ 0) such that √5 = a/b.
So, √5 b = a.
Squaring both sides:
5b² = a² ... (i)
Therefore, 5 divides a². By Fundamental Theorem, 5 divides a.
Let a = 5c for some integer c.
Substituting a = 5c in (i):
5b² = (5c)²
5b² = 25c²
b² = 5c²
Therefore, 5 divides b², which means 5 divides b.
Thus, a and b have at least 5 as a common factor.
This contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that √5 is rational.
So, we conclude that √5 is irrational.

Q2. Prove that 3 + 2√5 is irrational.

Solution:

Let us assume that 3 + 2√5 is rational.
So we can find coprime integers a and b (b ≠ 0) such that:
3 + 2√5 = a/b
2√5 = a/b - 3
2√5 = (a - 3b)/b
√5 = (a - 3b) / 2b
Since a and b are integers, (a - 3b)/2b is rational.
So, √5 is rational.
But this contradicts the fact that √5 is irrational.
Therefore, our assumption is false.
Hence, 3 + 2√5 is irrational.

Q3. Prove that the following are irrationals: (i) 1/√2 (ii) 7√5 (iii) 6 + √2

Solution:

(i) 1/√2
Assume 1/√2 is rational, say a/b.
b/a = √2.
Since a,b are integers, b/a is rational, so √2 is rational, which is a contradiction.
Thus, 1/√2 is irrational.

(ii) 7√5
Assume 7√5 is rational, say a/b.
√5 = a/7b.
Since a/7b is rational, √5 is rational, which is a contradiction.
Thus, 7√5 is irrational.

(iii) 6 + √2
Assume 6 + √2 is rational, say a/b.
√2 = a/b - 6 = (a-6b)/b.
Since RHS is rational, √2 is rational, which is a contradiction.
Thus, 6 + √2 is irrational.

Real Numbers - RD Sharma Important Questions

Q1. Find the HCF of 96 and 404 by prime factorization satisfying HCF × LCM = Product of numbers.

96 = 2⁵ × 3
404 = 2² × 101
HCF = 2² = 4
LCM = (96 × 404) / 4 = 96 × 101 = 9696.

Q2. Prove that if x and y are odd positive integers, then x² + y² is even but not divisible by 4.

Let x = 2m + 1 and y = 2n + 1.
x² + y² = (2m+1)² + (2n+1)²
= 4m² + 4m + 1 + 4n² + 4n + 1
= 4(m² + m + n² + n) + 2
= 4k + 2.
Clearly, it is even (divisible by 2) but leaves a remainder 2 when divided by 4, so not divisible by 4.

Q3. Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.

By Euclid's division lemma, a = bq + r.
Let b = 4, then r = 0, 1, 2, 3.
a can be 4q, 4q+1, 4q+2, 4q+3.
Since a is odd, it cannot be 4q or 4q+2 (as they are divisible by 2).
Therefore, any odd integer is of the form 4q + 1 or 4q + 3.

Q4. Find the largest number which divides 70 and 125, leaving remainders 5 and 8, respectively.

Numbers are 70 - 5 = 65 and 125 - 8 = 117.
65 = 5 × 13
117 = 3² × 13
HCF = 13.
The required largest number is 13.

Q5. If HCF(a, b) = 12 and a × b = 1800, then find LCM(a, b).

LCM = Product / HCF = 1800 / 12 = 150.

Q6. Can the number 4ⁿ, n being a natural number, end with the digit 0?

4ⁿ = (2²)ⁿ = 2²ⁿ.
The only prime factor is 2. To end with 0, it needs 2 and 5.
Since 5 is missing, it cannot end with 0.

Q7. Prove that √3 is irrational.

(Similar to √5 proof)
Assume √3 = a/b.
3b² = a² ⇒ 3 divides a ⇒ a = 3c.
3b² = 9c² ⇒ b² = 3c² ⇒ 3 divides b.
Common factor 3 makes a,b not coprime. Contradiction.

Q8. Prove that 2√3 - 1 is an irrational number.

Let 2√3 - 1 = r (rational).
2√3 = r + 1
√3 = (r + 1)/2
RHS is rational, LHS is irrational. Contradiction.

Q9. Two tankers contain 850 litres and 680 litres of petrol respectively. Find the maximum capacity of a container which can measure the petrol of either tanker in exact number of times.

We need HCF of 850 and 680.
850 = 2 × 5² × 17
680 = 2³ × 5 × 17
HCF = 2 × 5 × 17 = 170.
Max capacity = 170 litres.

Q10. Three bells toll at intervals of 12, 15 and 18 minutes. If they start tolling together, after what time will they next toll together?

We need LCM of 12, 15, 18.
12 = 4 × 3, 15 = 3 × 5, 18 = 2 × 9.
LCM = 180 minutes = 3 hours.

Q11. Explain why 3 × 5 × 7 + 7 is a composite number.

= 7(3 × 5 + 1) = 7(16). Factors are 1, 7, 16, etc. Composite.

Q12. Show that n² - 1 is divisible by 8, if n is an odd positive integer.

Let n = 4q + 1 or 4q + 3.
If n = 4q+1: n²-1 = (4q+1)²-1 = 16q²+8q = 8(2q²+q). Divisible by 8.
If n = 4q+3: n²-1 = (4q+3)²-1 = 16q²+24q+9-1 = 16q²+24q+8 = 8(2q²+3q+1). Divisible by 8.

Q13. Prove that √p is irrational for any prime p.

Standard proof: p|a² ⇒ p|a. Lead to contradiction.

Q14. Find HCF of 378, 180 and 420.

378 = 2 × 3³ × 7
180 = 2² × 3² × 5
420 = 2² × 3 × 5 × 7
HCF = 2 × 3 = 6.

Q15. The traffic lights at three different road crossings change after every 48, 72 and 108 seconds respectively. If they change simultaneously at 7 a.m., at what time will they change simultaneously again?

LCM(48, 72, 108) = 432 seconds.
432 sec = 7 min 12 sec.
Time = 7:07:12 a.m.

Q16. Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468.

LCM(520, 468) - 17.
LCM = 4680.
Ans = 4680 - 17 = 4663.

Q17. Prove that square of any positive integer is of the form 3m or 3m + 1.

Let a = 3q, 3q+1, 3q+2.
(3q)² = 9q² = 3(3q²) = 3m.
(3q+1)² = 3(3q²+2q)+1 = 3m+1.
(3q+2)² = 9q²+12q+4 = 3(3q²+4q+1)+1 = 3m+1.

Q18. If HCF of 144 and 180 is expressed in the form 13m - 3, find m.

HCF(144, 180) = 36.
13m - 3 = 36 ⇒ 13m = 39 ⇒ m = 3.

Q19. Find the greatest number of 6 digits exactly divisible by 24, 15 and 36.

LCM(24, 15, 36) = 360.
Greatest 6 digit no = 999999.
999999 / 360 leaves remainder 279.
Ans = 999999 - 279 = 999720.

Q20. Prove that one of any three consecutive positive integers must be divisible by 3.

n, n+1, n+2.
If n=3k, n is divisible.
If n=3k+1, n+2=3k+3 is divisible.
If n=3k+2, n+1=3k+3 is divisible.

Real Numbers - Formulas & PYQs

Key Formulas & Concepts

1. Fundamental Theorem of Arithmetic

Every composite number can be expressed (factorized) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.

Composite Number = Product of Primes

2. Relation between HCF and LCM

For any two positive integers a and b:

HCF(a, b) × LCM(a, b) = a × b

Note: This applies only to two numbers, not three.

3. Irrational Numbers

A number is called irrational if it cannot be written in the form p/q, where p and q are integers and q ≠ 0.

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

Examples: √2, √3, √5, π, 0.1010010001...

Previous Year Questions (CBSE/JKBOSE)

Q1. Find the ratio of LCM and HCF of the least composite and the least prime numbers. (CBSE 2023)

Least composite number = 4.
Least prime number = 2.
LCM(4, 2) = 4.
HCF(4, 2) = 2.
Ratio = LCM : HCF = 4 : 2 = 2 : 1.

Q2. Prove that √2 is an irrational number. (CBSE 2019, 2023 / JKBOSE Trend)

(Standard proof as covered in Notes). Assumed p/q form, squared, found common factor 2, contradiction.

Q3. If HCF(336, 54) = 6, find LCM(336, 54). (CBSE 2019)

LCM = (336 × 54) / 6
= 336 × 9 = 3024.

Q4. Explain why 13233343563715 is a composite number. (Board Model)

The number ends with 5, so it is divisible by 5.
Since it has a factor 5 other than 1 and itself, it is composite.

Q5. The LCM of two numbers is 182 and their HCF is 13. If one of the numbers is 26, find the other. (JKBOSE)

Other number = (LCM × HCF) / Given number
= (182 × 13) / 26
= 182 / 2 = 91.

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