Chapter 1: Real Numbers

Answer:
(i) 140 = 2 × 2 × 5 × 7 = 2² × 5 × 7
(ii) 156 = 2 × 2 × 3 × 13 = 2² × 3 × 13
(iii) 3825 = 3 × 3 × 5 × 5 × 17 = 3² × 5² × 17
(iv) 5005 = 5 × 7 × 11 × 13
(v) 7429 = 17 × 19 × 23

Answer:

(i) 26 and 91
26 = 2 × 13,   91 = 7 × 13
HCF = 13,   LCM = 2 × 7 × 13 = 182
Verification: LCM × HCF = 182 × 13 = 2366;   26 × 91 = 2366 ✓

(ii) 510 and 92
510 = 2 × 3 × 5 × 17,   92 = 2² × 23
HCF = 2,   LCM = 2² × 3 × 5 × 17 × 23 = 23460
Verification: LCM × HCF = 23460 × 2 = 46920;   510 × 92 = 46920 ✓

(iii) 336 and 54
336 = 2⁴ × 3 × 7,   54 = 2 × 3³
HCF = 2 × 3 = 6,   LCM = 2⁴ × 3³ × 7 = 3024
Verification: LCM × HCF = 3024 × 6 = 18144;   336 × 54 = 18144 ✓

Answer:

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

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

(iii) 8, 9, 25
8 = 2³,   9 = 3²,   25 = 5²
HCF = 1,   LCM = 2³ × 3² × 5² = 1800

Answer:
We know that: LCM × HCF = Product of the two numbers
∴ LCM × 9 = 306 × 657
LCM = (306 × 657) / 9
LCM = 306 × 73 = 22338

Answer:
If a number ends with digit 0, it must be divisible by 10, i.e., by 2 and 5.
6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ
The prime factors of 6ⁿ are only 2 and 3. It does not contain 5 as a prime factor.
Therefore, 6ⁿ is not divisible by 5.
Hence, 6ⁿ cannot end with digit 0 for any natural number n.

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

(ii) 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5(7 × 6 × 4 × 3 × 2 × 1 + 1) = 5 × (1008 + 1) = 5 × 1009
It has factors other than 1 and itself, hence it is a composite number.

Answer:
The time after which they meet again at the starting point is the LCM of 18 and 12.
18 = 2 × 3²,   12 = 2² × 3
LCM = 2² × 3² = 36
They will meet again after 36 minutes.

Answer:
Let us assume, to the contrary, that √5 is rational.
Then √5 = a/b, where a and b are co-prime integers (b ≠ 0).
Squaring both sides: 5 = a²/b² ⇒ a² = 5b².
This means 5 divides a², so 5 divides a. Let a = 5c.
Substituting: (5c)² = 5b² ⇒ 25c² = 5b² ⇒ b² = 5c².
Thus, 5 divides b², so 5 divides b.
Therefore, a and b have 5 as a common factor. This contradicts that a and b are co-prime.
Hence, √5 is irrational.

Answer:
Let us assume, to the contrary, that 3 + 2√5 is rational.
Then 3 + 2√5 = a/b, where a and b are integers (b ≠ 0).
2√5 = a/b – 3 = (a – 3b)/b
√5 = (a – 3b)/(2b)
Since a, b are integers, (a – 3b)/(2b) is rational, so √5 is rational.
This contradicts the fact that √5 is irrational.
Therefore, 3 + 2√5 is irrational.

Answer:

(i) 1/√2
Assume 1/√2 is rational. Then 1/√2 = a/b ⇒ √2 = b/a, which is rational. Contradiction.
∴ 1/√2 is irrational.

(ii) 7√5
Assume 7√5 is rational. Then 7√5 = a/b ⇒ √5 = a/(7b), which is rational. Contradiction.
∴ 7√5 is irrational.

(iii) 6 + √2
Assume 6 + √2 is rational. Then 6 + √2 = a/b ⇒ √2 = a/b – 6 = (a – 6b)/b, which is rational. Contradiction.
∴ 6 + √2 is irrational.

Answer:
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.

Answer:
If p is a prime and p divides a², then p divides a, where a is a positive integer.