Real Numbers
This chapter covers Euclid's division lemma and algorithm, the Fundamental Theorem of Arithmetic, finding HCF and LCM using prime factorisation, and proving the irrationality of numbers. All exercise-style questions below are solved in full.
For any two positive integers a and b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < b. Here a is the dividend, b is the divisor, q is the quotient, and r is the remainder.
To find HCF(a, b) with a > b: apply the division lemma to a and b to get a = bq + r. If r = 0, b is the HCF. If r ≠ 0, apply the lemma to b and r, and continue until the remainder is 0. The divisor at that last step is HCF(a, b).
Use Euclid's division algorithm to find the HCF of 135 and 225.
225 > 135, so 225 = 135 × 1 + 90. Then 135 = 90 × 1 + 45. Then 90 = 45 × 2 + 0. Since the remainder is now 0, the HCF is 45.
Use Euclid's division algorithm to find the HCF of 196 and 38220.
38220 > 196, so 38220 = 196 × 195 + 0. Since the remainder is 0 on the very first step, the HCF is 196.
Use Euclid's division algorithm to find the HCF of 867 and 255.
867 = 255 × 3 + 102. Then 255 = 102 × 2 + 51. Then 102 = 51 × 2 + 0. The HCF is 51.
Show that any positive odd integer is of the form 6q + 1, 6q + 3, or 6q + 5, where q is some integer.
By Euclid's division lemma, for any positive integer a and b = 6: a = 6q + r where 0 ≤ r < 6, so r can be 0, 1, 2, 3, 4, or 5. So a can be 6q, 6q+1, 6q+2, 6q+3, 6q+4, or 6q+5. Of these, 6q, 6q+2, 6q+4 are even (divisible by 2). So any odd positive integer must be of the form 6q+1, 6q+3, or 6q+5.
An army contingent of 616 members marches behind an army band of 32 members in a parade. The two groups march in the same number of columns. What is the maximum number of columns in which they can march?
The maximum number of columns is HCF(616, 32). 616 = 32 × 19 + 8. Then 32 = 8 × 4 + 0. So HCF = 8. They can march in a maximum of 8 columns.
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. This lets us find HCF and LCM: HCF = product of the smallest powers of common prime factors; LCM = product of the greatest powers of all prime factors involved.
Express 140 as a product of its prime factors.
140 = 2 × 70 = 2 × 2 × 35 = 2 × 2 × 5 × 7. So 140 = 22 × 5 × 7.
Find the LCM and HCF of 26 and 91, and verify that LCM × HCF = product of the two numbers.
26 = 2 × 13. 91 = 7 × 13. HCF = 13 (smallest power of common factor 13). LCM = 2 × 7 × 13 = 182 (product of greatest powers of all primes). Verification: LCM × HCF = 182 × 13 = 2366, and 26 × 91 = 2366. They match.
Find the LCM and HCF of 510 and 92, and verify that LCM × HCF = product of the two numbers.
510 = 2 × 3 × 5 × 17. 92 = 22 × 23. HCF = 2. LCM = 22 × 3 × 5 × 17 × 23 = 23460. Verification: LCM × HCF = 23460 × 2 = 46920, and 510 × 92 = 46920. They match.
Find the LCM and HCF of 336 and 54 by the prime factorisation method.
336 = 24 × 3 × 7. 54 = 2 × 33. HCF = 2 × 3 = 6. LCM = 24 × 33 × 7 = 3024.
Given that HCF(306, 657) = 9, find LCM(306, 657).
Since HCF × LCM = product of the two numbers: LCM = (306 × 657) ÷ 9 = 201042 ÷ 9 = 22338.
Check whether 6n can end with the digit 0 for any natural number n.
If 6n ended in 0, it would be divisible by 10, and hence by 2 and 5. But 6n = (2 × 3)n = 2n × 3n, whose only prime factors are 2 and 3 - by the Fundamental Theorem of Arithmetic this factorisation is unique, so 5 can never appear as a factor. Therefore 6n can never end in 0 for any n.
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × 78. Since it is a product of two factors greater than 1, it is composite. Similarly, 7! + 5 = 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1) = 5 × 1009, a product of two factors greater than 1, so it is composite.
There is a circular path around a sports field. Sonia takes 18 minutes to drive one round, while Ravi takes 12 minutes. 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?
They will meet again at the starting point after LCM(18, 12) minutes. 18 = 2 × 32, 12 = 22 × 3. LCM = 22 × 32 = 36. They will meet after 36 minutes.
HCF(a, b) × LCM(a, b) = a × b holds only for exactly two positive integers - it does not hold directly for three or more numbers.
A number that cannot be written in the form p/q, where p and q are integers and q ≠ 0, is called an irrational number. Examples: √2, √3, √5, π.
Prove that √2 is an irrational number.
Assume, for contradiction, that √2 is rational, so √2 = p/q where p and q are coprime integers (q ≠ 0). Squaring: 2q² = p², so p² is even, meaning p is even. Let p = 2m, so 2q² = 4m², giving q² = 2m², so q is also even. But this contradicts p and q being coprime. Hence √2 is irrational.
Prove that √5 is an irrational number.
Assume √5 = p/q in lowest terms (p, q coprime, q ≠ 0). Squaring: 5q² = p², so 5 divides p², and since 5 is prime, 5 divides p. Let p = 5m, so 5q² = 25m², giving q² = 5m², so 5 divides q² too, meaning 5 divides q. This contradicts p and q being coprime. Hence √5 is irrational.
Prove that 3 + 2√5 is irrational.
Assume 3 + 2√5 is rational, equal to some r. Then 2√5 = r - 3, so √5 = (r-3)/2, which would make √5 rational (since r is rational, (r-3)/2 is rational). This contradicts √5 being irrational. Hence 3 + 2√5 is irrational.
Prove that 1/√2 is irrational.
Assume 1/√2 is rational, equal to p/q (coprime integers, q ≠ 0). Then √2 = q/p, which would make √2 rational. This contradicts the proof that √2 is irrational. Hence 1/√2 is irrational.
The product of a non-zero rational number and an irrational number is:
Correct answer: (B) always irrational.
Every positive integer can be expressed as a product of primes in exactly one way (apart from the order of factors).
True - this is the statement of the Fundamental Theorem of Arithmetic.
For a rational number p/q (in lowest terms) to have a terminating decimal expansion, q must be of the form _______.
2m × 5n, where m and n are non-negative integers.
Without actually performing the long division, state whether 13/3125 will have a terminating or a non-terminating repeating decimal expansion.
3125 = 55, which is of the form 2m × 5n (with m = 0, n = 5). So 13/3125 has a terminating decimal expansion.