Exercise 1.1 5 Questions – Euclid's Division Lemma
Use Euclid's division algorithm to find the HCF of:
(i) 135 and 225
225 > 135. By Euclid's division algorithm:
225 = 135 × 1 + 90
135 = 90 × 1 + 45
90 = 45 × 2 + 0
Remainder = 0. ∴ HCF = 45
(ii) 196 and 38220
38220 = 196 × 195 + 0
Remainder = 0. ∴ HCF = 196
(iii) 867 and 255
867 = 255 × 3 + 102
255 = 102 × 2 + 51
102 = 51 × 2 + 0
∴ HCF = 51
Show that any positive odd integer is of the form 6q + 1, or 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
Possible remainders: r = 0, 1, 2, 3, 4, 5
When r = 0: a = 6q (even) · r = 1: a = 6q + 1 (odd) · r = 2: a = 6q + 2 (even)
r = 3: a = 6q + 3 (odd) · r = 4: a = 6q + 4 (even) · r = 5: a = 6q + 5 (odd)
∴ Any positive odd integer is of the form 6q + 1, 6q + 3, or 6q + 5
An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Maximum columns = HCF(616, 32)
616 = 32 × 19 + 8
32 = 8 × 4 + 0
∴ Maximum columns = 8
Use Euclid's division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Let a be any positive integer. By Euclid's Division Lemma with b = 3:
a = 3q + r, where r = 0, 1, or 2
Case 1 (r=0): a = 3q â‡' a² = 9q² = 3(3q²) = 3m where m = 3q²
Case 2 (r=1): a = 3q+1 â‡' a² = 9q²+6q+1 = 3(3q²+2q)+1 = 3m+1
Case 3 (r=2): a = 3q+2 â‡' a² = 9q²+12q+4 = 3(3q²+4q+1)+1 = 3m+1
∴ Square of any positive integer is of the form 3m or 3m+1
Use Euclid's division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Let a be any positive integer. a = 3q + r, r = 0, 1, 2.
Case 1 (r=0): a = 3q â‡' a³ = 27q³ = 9(3q³) = 9m
Case 2 (r=1): a = 3q+1 â‡' a³ = 27q³+27q²+9q+1 = 9(3q³+3q²+q)+1 = 9m+1
Case 3 (r=2): a = 3q+2 â‡' a³ = 27q³+54q²+36q+8 = 9(3q³+6q²+4q)+8 = 9m+8
∴ Cube of any positive integer is of the form 9m, 9m+1 or 9m+8
Exercise 1.2 7 Questions – Fundamental Theorem of Arithmetic
Express each number as a product of its prime factors:
(i) 140 = 2² × 5 × 7
(ii) 156 = 2² × 3 × 13
(iii) 3825 = 3² × 5² × 17
(iv) 5005 = 5 × 7 × 11 × 13
(v) 7429 = 17 × 19 × 23
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
26 = 2 × 13, 91 = 7 × 13
HCF = 13, LCM = 2 × 7 × 13 = 182
Verify: 13 × 182 = 2366 = 26 × 91 âœ"
(ii) 510 and 92
510 = 2 × 3 × 5 × 17, 92 = 2² × 23
HCF = 2, LCM = 2² × 3 × 5 × 17 × 23 = 23460
Verify: 2 × 23460 = 46920 = 510 × 92 âœ"
(iii) 336 and 54
336 = 2ⴠ× 3 × 7, 54 = 2 × 3³
HCF = 2 × 3 = 6, LCM = 2ⴠ× 3³ × 7 = 3024
Verify: 6 × 3024 = 18144 = 336 × 54 âœ"
Find the LCM and HCF of 17, 23 and 29 by the prime factorisation method.
17, 23, 29 are all prime numbers.
HCF(17, 23, 29) = 1
LCM(17, 23, 29) = 17 × 23 × 29 = 11339
Given that HCF(306, 657) = 9, find LCM(306, 657).
HCF × LCM = Product of numbers
9 × LCM = 306 × 657
LCM = (306 × 657) / 9 = 201042 / 9 = 22338
Check whether 6â¿ can end with the digit 0 for any natural number n.
If a number ends with 0, it must be divisible by 10 = 2 × 5.
6â¿ = (2 × 3)â¿ = 2⿠× 3â¿
6â¿ has prime factors 2 and 3 only. It does NOT have 5 as a prime factor.
∴ 6⿠cannot end with digit 0 for any natural number n.
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
(i) 7×11×13 + 13 = 13(7×11 + 1) = 13(77+1) = 13×78 = 13×2×3×13 = 2×3×13², which has factors other than 1 and itself. ∴ Composite.
(ii) 7×6×5×4×3×2×1 + 5 = 5(7×6×4×3×2×1 + 1) = 5(1008+1) = 5×1009, which has factors 5 and 1009. ∴ 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 same time and go in the same direction. After how many minutes will they meet again at the starting point?
They meet at the starting point after LCM(18, 12) minutes.
18 = 2 × 3², 12 = 2² × 3
LCM = 2² × 3² = 36
∴ They meet after 36 minutes.
Exercise 1.3 3 Questions – Irrational Numbers
Prove that √5 is irrational.
Assume, to the contrary, that √5 is rational.
So √5 = p/q, where p and q are coprime integers, q ≠0.
Squaring: 5 = p²/q² â‡' p² = 5q²
This means 5 divides p². Hence 5 divides p. Let p = 5r.
(5r)² = 5q² â‡' 25r² = 5q² â‡' q² = 5r²
This means 5 divides q², hence 5 divides q.
So 5 divides both p and q, contradicting that p and q are coprime.
∴ √5 is irrational.
Prove that 3 + 2√5 is irrational.
Assume 3 + 2√5 is rational, say r.
Then 2√5 = r âˆ' 3 â‡' √5 = (r âˆ' 3)/2
RHS is rational (rational numbers are closed under subtraction and division by non-zero).
But √5 is irrational (proved in Q1). This is a contradiction.
∴ 3 + 2√5 is irrational.
Prove that the following are irrationals: (i) 1/√2 (ii) 7√5 (iii) 6 + √2
(i) 1/√2
Assume 1/√2 = p/q (rational). Then √2 = q/p, which would be rational. But √2 is irrational. ∴ 1/√2 is irrational.
(ii) 7√5
Assume 7√5 = p/q. Then √5 = p/7q, which would be rational. But √5 is irrational. ∴ 7√5 is irrational.
(iii) 6 + √2
Assume 6+√2 is rational. Then √2 = (rational) âˆ' 6 would be rational. But √2 is irrational. ∴ 6+√2 is irrational.
Exercise 1.4 3 Questions – Decimal Expansions
Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:
A rational number p/q (in simplest form) has terminating decimal if q = 2â¿Â·5áµ only.
(i) 13/3125 = 13/5âµ â†' denominator = 5âµ only â†' Terminating
(ii) 17/8 = 17/2³ â†' Terminating
(iii) 64/455 = 64/(5×7×13) â†' has factor 7, 13 â†' Non-terminating repeating
(iv) 15/1600 = 15/(2â¶Ã—5²) â†' Terminating
(v) 29/343 = 29/7³ â†' Non-terminating repeating
(vi) 23/(2³×5²) â†' Terminating
(vii) 129/(2²×5â·Ã—7âµ) â†' has factor 7 â†' Non-terminating repeating
(viii) 6/15 = 2/5 â†' Terminating
(ix) 35/50 = 7/10 = 7/(2×5) â†' Terminating
(x) 77/210 = 11/30 = 11/(2×3×5) â†' has factor 3 â†' Non-terminating repeating
Write down the decimal expansions of those rational numbers in Question 1 which have terminating decimal expansions.
(i) 13/3125 = 0.00416
(ii) 17/8 = 2.125
(iv) 15/1600 = 0.009375
(vi) 23/200 = 0.115
(viii) 6/15 = 0.4
(ix) 35/50 = 0.7
The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form p/q, what can you say about the prime factors of q?
(i) 43.123456789 (ii) 0.120120012000120000... (iii) 43.123456789 (with bar over 123456789)
(i) Terminating decimal â†' Rational. q = 2â¹Ã—5â¹ (9 decimal places).
(ii) Non-terminating non-repeating â†' Irrational.
(iii) Non-terminating but repeating (bar) â†' Rational. q has prime factors other than 2 and 5.
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