2Find 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
4Given that HCF(306, 657) = 9, find LCM(306, 657).
Answer & Reference:
Using the relationship HCF(a,b) × LCM(a,b) = a × b (Page 4):
LCM(306, 657) = (306 × 657) / HCF(306, 657) = (306 × 657) / 9
= 34 × 657 = 22338
5Check whether 6n can end with the digit 0 for any natural number n.
Answer & Reference:
If 6n ends with digit 0, it must be divisible by 5 (Page 4, Example 1).
But 6n = (2 × 3)n = 2n × 3n
The prime factorization of 6n contains only primes 2 and 3.
By Fundamental Theorem of Arithmetic, there is no prime factor 5 in 6n.
Therefore, 6n cannot end with digit 0 for any natural number n.
7There 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?
Answer & Reference:
They will meet again at the starting point after LCM of 18 and 12 minutes.
18 = 2 × 3²
12 = 2² × 3
LCM = 2² × 3² = 36 minutes
Therefore, they will meet again at the starting point after 36 minutes.
8Prove that √5 is irrational.
Answer & Reference:
We use proof by contradiction (Page 6-7, Theorem 1.3):
Assume √5 is rational. Then √5 = a/b where a, b are coprime integers (b ≠ 0).
Squaring both sides: 5 = a²/b² ⇒ a² = 5b²
This means 5 divides a². By Theorem 1.2 (Page 6), 5 divides a.
So, a = 5c for some integer c.
Substituting: (5c)² = 5b² ⇒ 25c² = 5b² ⇒ b² = 5c²
This means 5 divides b², so 5 divides b (by Theorem 1.2).
But this contradicts that a and b are coprime (both divisible by 5).
Hence, our assumption is wrong. Therefore, √5 is irrational.
9Prove that 3 + 2√5 is irrational.
Answer & Reference:
We use proof by contradiction (Page 8, Example 6):
Assume 3 + 2√5 is rational. Then 3 + 2√5 = a/b where a, b are coprime integers (b ≠ 0).
Rearranging: 2√5 = a/b - 3 ⇒ √5 = (a - 3b)/(2b)
Since a, b are integers, (a - 3b)/(2b) is rational.
This implies √5 is rational, which contradicts the fact that √5 is irrational (Proved in Q8).
Hence, our assumption is wrong. Therefore, 3 + 2√5 is irrational.
10Prove that the following are irrationals: (i) 1/√2 (ii) 7√5 (iii) 6 + √2
Answer & Reference: (i) 1/√2
Assume 1/√2 is rational. Then 1/√2 = a/b where a, b are coprime integers (b ≠ 0).
Then √2 = b/a, which would mean √2 is rational.
But √2 is irrational (Page 6-7, Theorem 1.3). Contradiction.
Hence, 1/√2 is irrational.
(ii) 7√5
Assume 7√5 is rational. Then 7√5 = a/b where a, b are coprime integers (b ≠ 0).
Then √5 = a/(7b), which would mean √5 is rational.
But √5 is irrational (Proved in Q8). Contradiction.
Hence, 7√5 is irrational.
(iii) 6 + √2
Assume 6 + √2 is rational. Then 6 + √2 = a/b where a, b are coprime integers (b ≠ 0).
Then √2 = a/b - 6 = (a - 6b)/b, which would mean √2 is rational.
But √2 is irrational (Page 6-7, Theorem 1.3). Contradiction.
Hence, 6 + √2 is irrational.
SKey Concepts Summary
Reference: Page 9, Section 1.4 Summary
Concept
Description
Fundamental Theorem of Arithmetic
Every composite number can be expressed as a product of primes, and this factorisation is unique, apart from the order of prime factors.
Theorem 1.2
If p is a prime and p divides a², then p divides a (where a is a positive integer).
Irrational Numbers
Numbers that cannot be expressed as p/q where p and q are integers and q ≠ 0. Examples: √2, √3, √5.
HCF and LCM Relationship
For any two positive integers a and b: HCF(a, b) × LCM(a, b) = a × b