RS Aggarwal Class 10 Mathematics Solutions Chapter 1 Real Numbers

Question 1. State Euclid's division algorithm.
Answer: Euclid's division algorithm tells us that when you divide any two positive whole numbers a and b, you can always find two other whole numbers q and r such that a equals b times q plus r, with the condition that r is greater than or equal to zero and less than or equal to b.

Exam Tip: Remember the formula a = bq + r where 0 ≤ r ≤ b. This is the foundation for finding HCF using repeated division.

Question 2. Find the dividend when divisor is 61, quotient is 27, and remainder is 32.
Answer: Using the relationship Dividend = Divisor × Quotient + Remainder:
Dividend = 61 × 27 + 32
= 1647 + 32
= 1679
Therefore, the required number is 1679.

Exam Tip: Always check your answer by verifying that the remainder is smaller than the divisor and that all values are correctly substituted into the formula.

Question 3. If a dividend is 1365, quotient is 31, and remainder is 32, find the divisor.
Answer: Using the relationship Dividend = Divisor × Quotient + Remainder:
1365 = x × 31 + 32

\( \implies \) 1365 - 32 = 31x

\( \implies \) 1333 = 31x

\( \implies \) x = \( \frac{1333}{31} \) = 43
Therefore, 1365 should be divided by 43 to get 31 as the quotient and 32 as the remainder.

Exam Tip: Rearrange the division formula to isolate the unknown. Check your work by confirming that remainder is less than divisor.

Question 4. Use Euclid's algorithm to find the HCF of the following pairs.
(i) 2520 and 405
(ii) 1188 and 504
(iii) 1575 and 960
Answer:
(i) Applying Euclid's algorithm by dividing 2520 by 405:
Quotient = 6, Remainder = 90

\( \implies \) 2520 = 405 × 6 + 90
Next, dividing 405 by 90:
Quotient = 4, Remainder = 45

\( \implies \) 405 = 90 × 4 + 45
Next, dividing 90 by 45:

\( \implies \) 90 = 45 × 2 + 0
Therefore, the HCF of 2520 and 405 is 45.

(ii) Applying Euclid's algorithm by dividing 1188 by 504:
Quotient = 2, Remainder = 180

\( \implies \) 1188 = 504 × 2 + 180
Next, dividing 504 by 180:
Quotient = 2, Remainder = 144

\( \implies \) 504 = 180 × 2 + 144
Next, dividing 180 by 144:
Quotient = 1, Remainder = 36

\( \implies \) 180 = 144 × 1 + 36
Next, dividing 144 by 36:

\( \implies \) 144 = 36 × 4 + 0
Therefore, the HCF of 1188 and 504 is 36.

(iii) Applying Euclid's algorithm by dividing 1575 by 960:
Quotient = 1, Remainder = 615

\( \implies \) 1575 = 960 × 1 + 615
Next, dividing 960 by 615:
Quotient = 1, Remainder = 345

\( \implies \) 960 = 615 × 1 + 345
Next, dividing 615 by 345:
Quotient = 1, Remainder = 270

\( \implies \) 615 = 345 × 1 + 270
Next, dividing 345 by 270:
Quotient = 1, Remainder = 75

\( \implies \) 345 = 270 × 1 + 75
Next, dividing 270 by 75:
Quotient = 3, Remainder = 45

\( \implies \) 270 = 75 × 3 + 45
Next, dividing 75 by 45:
Quotient = 1, Remainder = 30

\( \implies \) 75 = 45 × 1 + 30
Next, dividing 45 by 30:
Quotient = 1, Remainder = 15

\( \implies \) 45 = 30 × 1 + 15
Next, dividing 30 by 15:
Quotient = 2, Remainder = 0

\( \implies \) 30 = 15 × 2 + 0
Therefore, the HCF of 1575 and 960 is 15.

Exam Tip: Continue dividing until the remainder is zero. The last non-zero remainder is always your HCF. Write each step clearly showing the division statement.

Question 5. Prove that every positive integer is either even or odd.
Answer: Assume that a smallest positive integer exists that is neither odd nor even, and call it n. Since n is the smallest such integer, then n - 1 must be either odd or even.

Case 1: If n - 1 is even, then n - 1 = 2k for some value of k. This means n = 2k + 1, which shows n is odd. This contradicts our assumption.

Case 2: If n - 1 is odd, then n - 1 = 2k + 1 for some value of k. This means n = 2k + 2 = 2(k + 1), which shows n is even. This contradicts our assumption.

In both cases, we reach a contradiction. Therefore, every positive integer must be either even or odd.

Exam Tip: Use proof by contradiction - assume the opposite of what you want to prove, then show this leads to a logical contradiction, proving the original statement.

Question 6. Show that any positive odd integer can be expressed in one of the forms 6m + 1, 6m + 3, or 6m + 5.
Answer: Let n be any positive odd integer. When you divide n by 6, using Euclid's division lemma:
n = 6m + r, where 0 ≤ r < 6

Since r is an integer and must satisfy 0 ≤ r < 6, the possible values are r = 0, 1, 2, 3, 4, 5.

This gives us: n = 6m or n = 6m + 1 or n = 6m + 2 or n = 6m + 3 or n = 6m + 4 or n = 6m + 5

However, n ≠ 6m, n ≠ 6m + 2, or n ≠ 6m + 4, because 6m, 6m + 2, and 6m + 4 are all multiples of 2, making them even. Since n is an odd integer, these cases are ruled out.

Therefore, n = 6m + 1 or n = 6m + 3 or n = 6m + 5

Any positive odd integer takes one of these three forms, where m is some non-negative integer.

Exam Tip: After applying Euclid's division lemma, eliminate cases that contradict the given condition (odd in this case). This systematic elimination proves only the required forms are possible.

Question 7. Show that any positive odd integer can be expressed in one of the forms 4m + 1 or 4m + 3.
Answer: Let n be any positive odd integer. When you divide n by 4, using Euclid's division lemma:
n = 4m + r, where 0 ≤ r < 4

Since r is an integer and must satisfy 0 ≤ r < 4, the possible values are r = 0, 1, 2, 3.

This gives us: n = 4m or n = 4m + 1 or n = 4m + 2 or n = 4m + 3

However, n ≠ 4m or n ≠ 4m + 2, because 4m and 4m + 2 are both multiples of 2, making them even. Since n is an odd integer, these cases are ruled out.

Therefore, n = 4m + 1 or n = 4m + 3

Any positive odd integer takes one of these two forms, where m is some non-negative integer.

Exam Tip: Apply Euclid's division lemma with the appropriate divisor, then eliminate cases incompatible with the given property (odd number).

Question 1. Find the HCF and LCM of the following pairs using prime factorization.
(i) 36 and 84
(ii) 23 and 31
(iii) 96 and 404
(iv) 144 and 198
(v) 396 and 1080
(vi) 1152 and 1664
Answer:
(i) Prime factorization:
36 = 2² × 3
84 = 2² × 3 × 7
HCF = product of lowest power of each common prime factor = 2² × 3 = 12
LCM = product of highest power of each prime factor = 2² × 3² × 7 = 252

(ii) Prime factorization:
23 = 23
31 = 31
HCF = product of lowest power of each common prime factor = 1
LCM = product of highest power of each prime factor = 23 × 31 = 713

(iii) Prime factorization:
96 = 2⁵ × 3
404 = 2² × 101
HCF = product of lowest power of each common prime factor = 2² = 4
LCM = product of highest power of each prime factor = 2⁵ × 3 × 101 = 9696

(iv) Prime factorization:
144 = 2⁴ × 3²
198 = 2 × 3² × 11
HCF = product of lowest power of each common prime factor = 2 × 3² = 18
LCM = product of highest power of each prime factor = 2⁴ × 3² × 11 = 1584

(v) Prime factorization:
396 = 2² × 3² × 11
1080 = 2³ × 3³ × 5
HCF = product of lowest power of each common prime factor = 2² × 3² = 36
LCM = product of highest power of each prime factor = 2³ × 3³ × 5 × 11 = 11880

(vi) Prime factorization:
1152 = 2⁷ × 3²
1664 = 2⁷ × 13
HCF = product of lowest power of each common prime factor = 2⁷ = 128
LCM = product of highest power of each prime factor = 2⁷ × 3² × 13 = 14976

Exam Tip: For HCF, take the smallest power of common factors only. For LCM, take the largest power of all factors that appear. Always express results in exponential form first, then multiply to find the final answer.

Question 2. Find the HCF and LCM of the following sets of three numbers using prime factorization.
(i) 8, 9, and 25
(ii) 12, 15, and 21
(iii) 17, 23, and 29
(iv) 24, 36, and 40
(v) 30, 72, and 432
(vi) 21, 28, 36, and 45
Answer:
(i) Prime factorization:
8 = 2³
9 = 3²
25 = 5²
HCF = product of lowest power of each common prime factor = 1 (no common factors)
LCM = product of highest power of each prime factor = 2³ × 3² × 5² = 1800

(ii) Prime factorization:
12 = 2² × 3
15 = 3 × 5
21 = 3 × 7
HCF = product of lowest power of each common prime factor = 3
LCM = product of highest power of each prime factor = 2² × 3 × 5 × 7 = 420

(iii) Prime factorization:
17 = 17
23 = 23
29 = 29
HCF = product of lowest power of each common prime factor = 1 (no common factors)
LCM = product of highest power of each prime factor = 17 × 23 × 29 = 11339

(iv) Prime factorization:
24 = 2³ × 3
36 = 2² × 3²
40 = 2³ × 5
HCF = product of lowest power of each common prime factor = 2² = 4
LCM = product of highest power of each prime factor = 2³ × 3² × 5 = 360

(v) Prime factorization:
30 = 2 × 3 × 5
72 = 2³ × 3²
432 = 2⁴ × 3³
HCF = product of lowest power of each common prime factor = 2 × 3 = 6
LCM = product of highest power of each prime factor = 2⁴ × 3³ × 5 = 2160

(vi) Prime factorization:
21 = 3 × 7
28 = 2² × 7
36 = 2² × 3²
45 = 3² × 5
HCF = product of lowest power of each common prime factor = 1 (no common factor in all four)
LCM = product of highest power of each prime factor = 2² × 3² × 5 × 7 = 1260

Exam Tip: When finding HCF of multiple numbers, identify prime factors that appear in ALL numbers. For LCM, include every prime factor that appears in ANY number, using its highest power.

Question 3. If one number is 161, HCF is 23, and LCM is 1449, find the other number.
Answer: Let the two numbers be a and b, where a = 161.
Given: HCF = 23 and LCM = 1449

Using the fundamental relationship: a × b = HCF × LCM

\( \implies \) 161 × b = 23 × 1449

\( \implies \) b = \( \frac{23 \times 1449}{161} \) = \( \frac{33327}{161} \) = 207

Therefore, the other number is 207.

Exam Tip: Always use the relationship a × b = HCF × LCM when two quantities and any two of the four values are known. Verify by checking that HCF divides both numbers and both numbers divide the LCM.

Question 4. Two numbers have HCF 145 and LCM 2175. If one number is 725, find the other.
Answer: Using the relationship: product of two numbers = HCF × LCM

725 × x = 145 × 2175

\( \implies \) x = \( \frac{145 \times 2175}{725} \) = 435

Therefore, the other number is 435.

Exam Tip: Simplify your fractions by cancelling common factors before multiplying to get cleaner calculations.

Question 5. Two numbers have HCF 18 and product 12960. Find their LCM.
Answer: Using the relationship: product of two numbers = HCF × LCM

12960 = 18 × LCM

\( \implies \) LCM = \( \frac{12960}{18} \) = 720

Therefore, their LCM is 720.

Exam Tip: Rearrange the fundamental formula to isolate the unknown quantity. Always check that HCF divides the LCM evenly.

Question 6. Is it possible to have two numbers whose HCF is 18 and LCM is 760? Justify your answer.
Answer: No, it is not possible to have two numbers whose HCF is 18 and LCM is 760. The reason is that the HCF must always be a factor of the LCM. In this case, 18 does not divide 760 evenly, so this combination is impossible.

Exam Tip: Remember the key principle: HCF must always divide the LCM without remainder. Check this condition before accepting any given pair of HCF and LCM values.

Question 7. Reduce each fraction to its simplest form using prime factorization.
(i) \( \frac{69}{92} \)
(ii) \( \frac{473}{645} \)
(iii) \( \frac{1095}{1168} \)
(iv) \( \frac{368}{496} \)
Answer:
(i) Prime factorization:
69 = 3 × 23
92 = 2² × 23
\( \frac{69}{92} \) = \( \frac{3 \times 23}{2^2 \times 23} \) = \( \frac{3}{4} \)

(ii) Prime factorization:
473 = 11 × 43
645 = 3 × 5 × 43
\( \frac{473}{645} \) = \( \frac{11 \times 43}{3 \times 5 \times 43} \) = \( \frac{11}{15} \)

(iii) Prime factorization:
1095 = 3 × 5 × 73
1168 = 2⁴ × 73
\( \frac{1095}{1168} \) = \( \frac{3 \times 5 \times 73}{2^4 \times 73} \) = \( \frac{15}{16} \)

(iv) Prime factorization:
368 = 2⁴ × 23
496 = 2⁴ × 31
\( \frac{368}{496} \) = \( \frac{2^4 \times 23}{2^4 \times 31} \) = \( \frac{23}{31} \)

Exam Tip: Find prime factors of both numerator and denominator, then cancel all common factors. Always verify your reduced fraction has no common factors left.

Question 8. Find the largest number which divides 438 and 606, leaving remainder 6 in each case.
Answer: The largest number that divides both 438 and 606, leaving remainder 6, is actually the largest number that divides (438 - 6) = 432 and (606 - 6) = 600, leaving remainder 0. Therefore, we need to find the HCF of 432 and 600.

Prime factorization:
432 = 2⁴ × 3³
600 = 2³ × 3 × 5²

HCF = product of lowest power of each common prime factor = 2³ × 3 = 24

Thus, the largest number which divides 438 and 606, leaving remainder 6, is 24.

Exam Tip: When remainder is constant for all numbers, subtract it from each number first, then find HCF. This converts the problem to finding a common divisor without remainder.

Question 9. Find the largest number that divides 320 leaving remainder 5 and also divides 457 leaving remainder 7.
Answer: The required number divides (320 - 5) = 315 and (457 - 7) = 450 with remainder 0.

Required number = HCF(315, 450)

Applying Euclid's algorithm:
450 = 315 × 1 + 135
315 = 135 × 2 + 45
135 = 45 × 3 + 0

Therefore, the HCF of 315 and 450 is 45.
Hence, the required number is 45.

Exam Tip: Subtract the given remainders from their respective numbers to convert into a straightforward HCF problem. Apply Euclid's algorithm step by step.

Question 10. Find the least number which when divided by 35, 56, and 91 leaves the same remainder 7.
Answer: The least number that can be divided by 35, 56, and 91 is the LCM of these three numbers.

Prime factorization:
35 = 5 × 7
56 = 2³ × 7
91 = 7 × 13

LCM = product of highest power of each prime factor = 2³ × 5 × 7 × 13 = 3640

The least number which when divided by 35, 56, and 91 leaves the same remainder 7 is 3640 + 7 = 3647.

Thus, the required number is 3647.

Exam Tip: When a constant remainder applies to all divisors, find LCM first, then add that remainder to get your final answer.

Question 11. Find the smallest number which when divided by 28 leaves remainder 8 and when divided by 32 leaves remainder 12.
Answer: Let the required number be x. Using Euclid's lemma:
x = 28p + 8 and x = 32q + 12, where p and q are quotients

\( \implies \) 28p + 8 = 32q + 12

\( \implies \) 28p = 32q + 4

\( \implies \) 7p = 8q + 1 ... (1)

Here p = 8n - 1 and q = 7n - 1 satisfy equation (1), where n is a natural number

Substituting n = 1:
p = 8 - 1 = 7 and q = 7 - 1 = 6

Thus, x = 28p + 8
= 28 × 7 + 8
= 196 + 8
= 204

Hence, the smallest number which when divided by 28 and 32 leaves remainders 8 and 12 is 204.

Exam Tip: When remainders differ, use the method of solving simultaneous congruences. First set up equations using Euclid's formula, simplify to get a single linear equation, then find its solution.

Question 12. Find the smallest number which when increased by 17 is exactly divisible by both 468 and 520.
Answer: The smallest number which when increased by 17 becomes divisible by both 468 and 520 can be found by subtracting 17 from the LCM of 468 and 520.

Prime factorization:
468 = 2² × 3² × 13
520 = 2³ × 5 × 13

LCM = product of highest power of each prime factor = 2³ × 3² × 5 × 13 = 4680

The required number is 4680 - 17 = 4663.

Hence, the smallest number which when increased by 17 is exactly divisible by both 468 and 520 is 4663.

Exam Tip: When a condition states "increased/decreased by a constant," find LCM then adjust by that constant in the opposite direction.

Question 13. Find the greatest number of four digits which is exactly divisible by 15, 24, and 36.
Answer: Prime factorization:
15 = 3 × 5
24 = 2³ × 3
36 = 2² × 3²

LCM = product of highest power of each prime factor = 2³ × 3² × 5 = 360

The greatest four digit number is 9999. When 9999 is divided by 360, the remainder is 279.

Thus, 9999 - 279 = 9720 is exactly divisible by 360.

Hence, the greatest number of four digits which is exactly divisible by 15, 24 and 36 is 9720.

Exam Tip: To find the greatest number of n digits divisible by given numbers, find LCM, divide the largest n-digit number by it, subtract the remainder from that number.

Question 14. Three groups have 60, 84, and 108 participants respectively. Rooms must be arranged so each room has the same number of participants and no group is split. What is the minimum number of rooms needed?
Answer: Minimum number of rooms required = \( \frac{\text{Total number of participants}}{\text{HCF}(60, 84, 108)} \)

Prime factorization:
60 = 2² × 3 × 5
84 = 2² × 3 × 7
108 = 2² × 3³

HCF = product of lowest power of each common prime factor = 2² × 3 = 12

Total participants = 60 + 84 + 108 = 252

Therefore, minimum number of rooms required = \( \frac{252}{12} \) = 21

Thus, the minimum number of rooms required is 21.

Exam Tip: In problems involving grouping without splitting, always find HCF of the group sizes. This represents the maximum size for each room.

Question 15. There are 336 English books, 240 mathematics books, and 96 science books. Stack them so each stack contains the same number of books and each stack has books of only one type. How many books are in each stack? How many stacks are there in total?
Answer: Total number of English books = 336
Total number of mathematics books = 240
Total number of science books = 96

Number of books in each stack = HCF(336, 240, 96)

Prime factorization:
336 = 2⁴ × 3 × 7
240 = 2⁴ × 3 × 5
96 = 2⁵ × 3

HCF = Product of lowest power of each common prime factor = 2⁴ × 3 = 48

Hence, stacks have 48 books each.

Number of stacks = \( \frac{336}{48} \) + \( \frac{240}{48} \) + \( \frac{96}{48} \) = (7 + 5 + 2) = 14

Exam Tip: Find HCF to determine stack size, then divide each quantity by this to find the number of stacks. Always verify that HCF divides all quantities evenly.

Question 16. Three logs of timber measure 42m, 49m, and 63m. They must be cut into planks of equal length with no leftover. What is the greatest possible length of each plank?
Answer: The lengths of the timber pieces are 42m, 49m, and 63m. To divide them into equal length planks:

Greatest possible length of each plank = HCF(42, 49, 63)

Prime factorization:
42 = 2 × 3 × 7
49 = 7 × 7
63 = 3² × 7

HCF = Product of lowest power of each common prime factor = 7

Hence, the greatest possible length of each plank is 7m.

Exam Tip: When items must be cut into equal pieces with no waste, find HCF of all measurements. This ensures maximum plank length while using all material.

Question 17. Three lengths measure 7m (700cm), 3m 85cm (385cm), and 12m 95cm (1295cm). Find the greatest possible length of rope that can measure all three lengths exactly (converting to cm first).
Answer: Converting to cm first:
7m = 700cm
3m 85cm = 385cm
12m 95cm = 1295cm

Required length = HCF(700, 385, 1295)

Prime factorization:
700 = 2² × 5² × 7
385 = 5 × 7 × 11
1295 = 5 × 7 × 37

HCF = 5 × 7 = 35

Hence, the greatest possible length is 35cm.

Exam Tip: Always convert all measurements to the same unit before finding HCF. This ensures accuracy and consistency.

Question 18. There are 1001 pens and 910 pencils to be distributed equally among the maximum number of students. Each student gets the same number of pens and pencils. How many students can receive them?
Answer: Total number of pens = 1001
Total number of pencils = 910

Maximum number of students who get the same number of pens and pencils = HCF(1001, 910)

Prime factorization:
1001 = 11 × 91
910 = 10 × 91

HCF = 91

Hence, 91 students receive the same number of pens and pencils.

Exam Tip: To find the maximum number of equal groups with no remainder, always find HCF of the quantities. This maximizes group size.

Question 19. A rectangular ceiling measures 15m 17m (1517cm) by 9m 2m (902cm). Square tiles are used to cover it exactly. Find the side length of each tile and the total number of tiles needed.
Answer: Given:
Length of ceiling = 15m 17m = 1517cm (∵ 1m = 100cm)
Breadth of ceiling = 9m 2m = 902cm

Side of each square tile = HCF(1517, 902)

Prime factorization:
1517 = 37 × 41
902 = 2² × 41

HCF = product of lowest power of each common prime factor = 41

Required number of tiles = \( \frac{\text{Area of ceiling}}{\text{Area of one tile}} \) = \( \frac{1517 \times 902}{41 \times 41} \) = 37 × 22 = 814

Exam Tip: When square tiles cover a rectangle exactly, find HCF of length and width for tile size. Then divide total area by tile area to get number of tiles.

Question 20. Three measuring rods have lengths 64cm, 80cm, and 96cm. What is the greatest length of cloth that can be measured an exact number of times using all three rods?
Answer: Length of the three measuring rods are 64cm, 80cm and 96cm, respectively.

Length of cloth that can be measured an exact number of times = LCM(64, 80, 96)

Prime factorization:
64 = 2⁶
80 = 2⁴ × 5
96 = 2⁵ × 3

LCM = product of highest power of each prime factor = 2⁶ × 3 × 5 = 960cm = 9.6m

Hence, the required length of cloth is 9.6m.

Exam Tip: When measuring cloth using multiple measuring rods, find LCM of their lengths. This gives the smallest cloth length measurable by all rods without remainder.

Question 21. Two devices beep at intervals of 60 and 62 seconds respectively. Both beep together at 10:00 a.m. At what time will they beep together again?
Answer: Beep duration of first device = 60 seconds
Beep duration of second device = 62 seconds

Interval of beeping together = LCM(60, 62)

Prime factorization:
60 = 2² × 3 × 5
62 = 2 × 31

LCM = 2² × 3 × 5 × 31 = 1860 seconds = 1860/60 = 31 minutes

Hence, they will beep together again at 10:31 a.m.

Exam Tip: For periodic events, find LCM of the time intervals to determine when they coincide again. Convert to appropriate units (minutes, hours).

Question 22. Six bells toll at intervals of 2, 4, 6, 8, 10, and 12 minutes respectively. After how much time will they toll together again?
Answer: Six bells toll at intervals of 2, 4, 6, 8, 10, and 12 minutes respectively.

Prime factorization:
2 = 2
4 = 2²
6 = 2 × 3
8 = 2³
10 = 2 × 5
12 = 2² × 3

LCM(2, 4, 6, 8, 10, 12) = 2³ × 3 × 5 = 120 minutes

Hence, after every 120 minutes (i.e. 2 hours), they will toll together.

Exam Tip: For multiple periodic events, find LCM of all intervals. This represents the time when all events coincide. Convert final answer to the most appropriate time unit.

Question 1. Write the repeating decimal for each of the following and use a bar to show the repetend.
(i) \( \frac { 23 }{ 2^3 \times 5^2 } \)
(ii) \( \frac { 24 }{ 125 } \)
(iii) \( \frac { 171 }{ 800 } \)
(iv) \( \frac { 15 }{ 1600 } \)
(v) \( \frac { 17 }{ 320 } \)
(vi) \( \frac { 19 }{ 3125 } \)
Answer:
(i) \( \frac { 23 }{ 2^3 \times 5^2 } = \frac { 23 \times 5 }{ 2^3 \times 5^3 } = \frac { 115 }{ 1000 } = 0.115 \)
Either 2 or 5 is not a factor of 23, so it remains in simplest form. Moreover, it takes the form \( 2^m \times 5^n \). Thus, the given rational number is terminating.
(ii) \( \frac { 24 }{ 125 } = \frac { 24 }{ 5^3 } = \frac { 24 \times 2^3 }{ 5^3 \times 2^3 } = \frac { 192 }{ 1000 } = 0.192 \)
5 is not a factor of 24, so it stays in simplest form. Moreover, it takes the form \( 2^m \times 5^n \). Thus, the given rational number is terminating.
(iii) \( \frac { 171 }{ 800 } = \frac { 171 }{ 2^5 \times 5^2 } = \frac { 171 \times 5^3 }{ 2^5 \times 5^5 } = \frac { 21375 }{ 100000 } = 0.21375 \)
Either 2 or 5 is not a factor of 171, so it remains in simplest form. Moreover, it takes the form \( 2^m \times 5^n \). Thus, the given rational number is terminating.
(iv) \( \frac { 15 }{ 1600 } = \frac { 15 }{ 2^6 \times 5^2 } = \frac { 15 \times 5^4 }{ 2^6 \times 5^6 } = \frac { 9375 }{ 1000000 } = 0.009375 \)
Either 2 or 5 is not a factor of 15, so it stays in simplest form. Moreover, it takes the form \( 2^m \times 5^n \). Thus, the given rational number is terminating.
(v) \( \frac { 17 }{ 320 } = \frac { 17 }{ 2^6 \times 5 } = \frac { 17 \times 5^5 }{ 2^6 \times 5^6 } = \frac { 53125 }{ 1000000 } = 0.053125 \)
Either 2 or 5 is not a factor of 17, so it remains in simplest form. Moreover, it takes the form \( 2^m \times 5^n \). Thus, the given rational number is terminating.
(vi) \( \frac { 19 }{ 3125 } = \frac { 19 }{ 5^5 } = \frac { 19 \times 2^5 }{ 5^5 \times 2^5 } = \frac { 608 }{ 100000 } = 0.00608 \)
Either 2 or 5 is not a factor of 19, so it remains in simplest form. Moreover, it takes the form \( 2^m \times 5^n \). Thus, the given rational number is terminating.

Exam Tip: For terminating decimals, check that the denominator (after simplification) has only factors of 2 and 5 in any power. Always verify the decimal expansion is correct by multiplication.

Question 2. Identify whether the following rational numbers have terminating or non-terminating repeating decimal expansions.
(i) \( \frac { 11 }{ 2^3 \times 3 } \)
(ii) \( \frac { 73 }{ 2^3 \times 3^3 \times 5 } \)
(iii) \( \frac { 129 }{ 2^2 \times 5^7 \times 7^5 } \)
(iv) \( \frac { 9 }{ 35 } \)
(v) \( \frac { 77 }{ 210 } \)
(vi) \( \frac { 32 }{ 147 } \)
(vii) \( \frac { 29 }{ 343 } \)
(viii) \( \frac { 64 }{ 455 } \)
Answer:
(i) \( \frac { 11 }{ 2^3 \times 3 } \)
Either 2 or 3 is not a factor of 11, so it is in its simplest form. Moreover, \( (2^3 \times 3) \neq (2^m \times 5^n) \)
Thus, the given rational number is non-terminating repeating decimal.
(ii) \( \frac { 73 }{ 2^3 \times 3^3 \times 5 } \)
2, 3 or 5 is not a factor of 73, so it remains in simplest form. Moreover, \( (2^2 \times 3^3 \times 5) \neq (2^m \times 5^n) \)
Thus, the given rational number is non-terminating repeating decimal.
(iii) \( \frac { 129 }{ 2^2 \times 5^7 \times 7^5 } \)
2, 5 or 7 is not a factor of 129, so it is in its simplest form. Moreover, \( (2^2 \times 5^7 \times 7^5) \neq (2^m \times 5^n) \)
Thus, the given rational number is non-terminating repeating decimal.
(iv) \( \frac { 9 }{ 35 } = \frac { 9 }{ 5 \times 7 } \)
Either 5 or 7 is not a factor of 9, so it is in its simplest form. Moreover, \( (5 \times 7) \neq (2^m \times 5^n) \)
Thus, the given rational number is non-terminating repeating decimal.
(v) \( \frac { 77 }{ 210 } = \frac { 77 \div 7 }{ 210 \div 7 } = \frac { 11 }{ 30 } = \frac { 11 }{ 2 \times 3 \times 5 } \)
2, 3 or 5 is not a factor of 11, so \( \frac { 11 }{ 30 } \) remains in simplest form. Moreover, \( (2 \times 3 \times 5) \neq (2^m \times 5^n) \)
Thus, the given rational number is non-terminating repeating decimal.
(vi) \( \frac { 32 }{ 147 } = \frac { 32 }{ 3 \times 7^2 } \)
Either 3 or 7 is not a factor of 32, so it is in its simplest form. Moreover, \( (3 \times 7^2) \neq (2^m \times 5^n) \)
Thus, the given rational number is non-terminating repeating decimal.
(vii) \( \frac { 29 }{ 343 } = \frac { 29 }{ 7^3 } \)
7 is not a factor of 29, so it is in its simplest form. Moreover, \( 7^3 \neq (2^m \times 5^n) \)
Thus, the given rational number is non-terminating repeating decimal.
(viii) \( \frac { 64 }{ 455 } = \frac { 64 }{ 5 \times 7 \times 13 } \)
5, 7 or 13 is not a factor of 64, so it remains in simplest form. Moreover, \( (5 \times 7 \times 13) \neq (2^m \times 5^n) \)
Thus, the given rational number is non-terminating repeating decimal.

Exam Tip: A rational number in lowest terms has a terminating decimal only if its denominator contains no prime factors other than 2 and 5. If any other prime appears, the decimal is non-terminating and repeating.

Question 3. Convert the following repeating decimals into the form p/q.
(i) \( 0.\overline{8} \)
(ii) \( 2.\overline{4} \)
(iii) \( 0.\overline{24} \)
(iv) \( 0.\overline{12} \)
(v) \( 2.2\overline{4} \)
(vi) \( 0.3\overline{65} \)
Answer:
(i) Let \( x = 0.\overline{8} \)
\( \therefore x = 0.888 \) \(...(1)\)
\( 10x = 8.888 \) \(...(2)\)
Subtracting equation (1) from (2), we get
\( 9x = 8 \Rightarrow x = \frac{8}{9} \)
\( \therefore 0.\overline{8} = \frac{8}{9} \)
(ii) Let \( x = 2.\overline{4} \)
\( \therefore x = 2.444 \) \(...(1)\)
\( 10x = 24.444 \) \(...(2)\)
Subtracting equation (1) from (2), we get
\( 9x = 22 \Rightarrow x = \frac{22}{9} \)
\( \therefore 2.\overline{4} = \frac{22}{9} \)
(iii) Let \( x = 0.\overline{24} \)
\( \therefore x = 0.2424 \) \(...(1)\)
\( 100x = 24.2424 \) \(...(2)\)
Subtracting equation (1) from (2), we get
\( 99x = 24 \Rightarrow x = \frac{8}{33} \)
\( \therefore 0.\overline{24} = \frac{8}{33} \)
(iv) Let \( x = 0.\overline{12} \)
\( \therefore x = 0.1212 \) \(...(1)\)
\( 100x = 12.1212 \) \(...(2)\)
Subtracting equation (1) from (2), we get
\( 99x = 12 \Rightarrow x = \frac{4}{33} \)
\( \therefore 0.\overline{12} = \frac{4}{33} \)
(v) Let \( x = 2.2\overline{4} \)
\( \therefore x = 2.2444 \) \(...(1)\)
\( 10x = 22.444 \) \(...(2)\)
\( 100x = 224.444 \) \(...(3)\)
Subtracting equation (2) from (3), we get
\( 90x = 202 \Rightarrow x = \frac{202}{90} = \frac{101}{45} \)
\( \therefore 2.2\overline{4} = \frac{101}{45} \)
(vi) Let \( x = 0.3\overline{65} \)
\( \therefore x = 0.3656565 \) \(...(1)\)
\( 10x = 3.656565 \) \(...(2)\)
\( 1000x = 365.656565 \) \(...(3)\)
Subtracting equation (2) from (3), we get
\( 990x = 362 \Rightarrow x = \frac{362}{990} = \frac{181}{495} \)
\( \therefore 0.3\overline{65} = \frac{181}{495} \)

Exam Tip: The number of 9's in the denominator equals the number of repeating digits. For mixed repeating decimals, subtract the equation where the non-repeating part ends from the equation where one full cycle of repetition is complete.

Question 1. Define rational numbers and irrational numbers with examples.
Answer: Rational numbers are quantities that can be written in the form \( \frac{p}{q} \) where p and q are integers and \( q \neq 0 \). An example is \( \frac{2}{3} \). Irrational numbers are quantities which, when written as a decimal, appear as non-terminating and non-repeating decimals. An example is \( \sqrt{2} \).
In simple words: A rational number is any number you can write as a fraction with two whole numbers. An irrational number cannot be written this way - it is a decimal that never ends and never repeats the same digits.

Exam Tip: Real numbers are composed of both rational and irrational numbers. Remember that terminating decimals and repeating decimals are always rational.

Question 2. Classify the following as rational or irrational numbers.
(i) \( \frac{22}{7} \)
(ii) 3.1416
(iii) \( \pi \)
(iv) \( 3.\overline{142857} \)
(v) 5.636363...
(vi) 2.040040004...
(vii) 1.535335333...
(viii) 3.121221222...
(ix) \( \sqrt{21} = \sqrt{3} \times \sqrt{7} \)
(x) \( \sqrt[3]{3} \)
Answer:
(i) \( \frac{22}{7} \) is a rational number because it is of the form \( \frac{p}{q} \) where \( q \neq 0 \).
(ii) 3.1416 is a rational number because it is a terminating decimal.
(iii) \( \pi \) is an irrational number because it is a non-repeating and non-terminating decimal.
(iv) \( 3.\overline{142857} \) is a rational number because it is a repeating decimal.
(v) 5.636363... is a rational number because it is a non-terminating and repeating decimal.
(vi) 2.040040004... is an irrational number because it is a non-terminating and non-repeating decimal.
(vii) 1.535335333... is an irrational number because it is a non-terminating and non-repeating decimal.
(viii) 3.121221222... is an irrational number because it is a non-terminating and non-repeating decimal.
(ix) \( \sqrt{21} = \sqrt{3} \times \sqrt{7} \) is an irrational number because \( \sqrt{3} \) and \( \sqrt{7} \) are irrational and prime numbers.
(x) \( \sqrt[3]{3} \) is an irrational number because 3 is a prime number. So, \( \sqrt[3]{3} \) is an irrational number.

Exam Tip: Repeating decimals (with a bar notation) are always rational, while non-repeating non-terminating decimals are always irrational. Check carefully whether the decimal pattern repeats.

Question 3. Prove that the following numbers are irrational using the method of contradiction.
(i) \( \sqrt{6} \)
(ii) \( 2 - \sqrt{3} \)
(iii) \( 3 + \sqrt{2} \)
(iv) \( 2 + \sqrt{5} \)
(v) \( 5 + 3\sqrt{2} \)
(vi) \( 3\sqrt{7} \)
(vii) \( \frac{3}{\sqrt{5}} \)
(viii) \( 2 - 3\sqrt{5} \)
(ix) \( \sqrt{3} + \sqrt{5} \)
Answer:
(i) Assume \( \sqrt{6} = \sqrt{2} \times \sqrt{3} \) is rational. Then both \( \sqrt{2} \) and \( \sqrt{3} \) are rational. This contradicts the fact that \( \sqrt{2} \) and \( \sqrt{3} \) are irrational. The contradiction arises from assuming \( \sqrt{6} \) is rational. Hence, \( \sqrt{6} \) is irrational.
(ii) Assume \( 2 - \sqrt{3} \) is rational. Then both 2 and \( 2 - \sqrt{3} \) are rational. Since the difference of two rationals is rational, \( (2 - 2 + \sqrt{3}) = \sqrt{3} \) must be rational. This contradicts that \( \sqrt{3} \) is irrational. The contradiction arises from assuming \( 2 - \sqrt{3} \) is rational. Hence, \( 2 - \sqrt{3} \) is irrational.
(iii) Assume \( 3 + \sqrt{2} \) is rational. Then both 3 and \( 3 + \sqrt{2} \) are rational. Since the difference of two rationals is rational, \( 3 + \sqrt{2} - 3 = \sqrt{2} \) must be rational. This contradicts that \( \sqrt{2} \) is irrational. The contradiction arises from assuming \( 3 + \sqrt{2} \) is rational. Hence, \( 3 + \sqrt{2} \) is irrational.
(iv) Assume \( 2 + \sqrt{5} \) is rational. Then both \( 2 + \sqrt{5} \) and 2 are rational. Since the difference of two rationals is rational, \( (2 + \sqrt{5}) - 2 = \sqrt{5} \) must be rational. This contradicts that \( \sqrt{5} \) is irrational. The contradiction arises from assuming \( 2 + \sqrt{5} \) is rational. Hence, \( 2 + \sqrt{5} \) is irrational.
(v) Assume \( 5 + 3\sqrt{2} \) is rational. Then both 5 and \( 5 + 3\sqrt{2} \) are rational. Since the difference of two rationals is rational, \( (5 + 3\sqrt{2} - 5) = 3\sqrt{2} \) must be rational. Since the product of two rationals is rational, \( \frac{1}{3} \times 3\sqrt{2} = \sqrt{2} \) must be rational. This contradicts that \( \sqrt{2} \) is irrational. The contradiction arises from assuming \( 5 + 3\sqrt{2} \) is rational. Hence, \( 5 + 3\sqrt{2} \) is irrational.
(vi) Assume \( 3\sqrt{7} \) is rational. Since the product of two rationals is rational, \( \frac{1}{3} \times 3\sqrt{7} = \sqrt{7} \) must be rational. This contradicts that \( \sqrt{7} \) is irrational. The contradiction arises from assuming \( 3\sqrt{7} \) is rational. Hence, \( 3\sqrt{7} \) is irrational.
(vii) Assume \( \frac{3}{\sqrt{5}} \) is rational. Since the product of two rationals is rational, \( \frac{1}{3} \times \frac{3}{\sqrt{5}} = \frac{1}{\sqrt{5}} \) must be rational. This contradicts that \( \frac{1}{\sqrt{5}} \) is irrational. Rationalizing, \( \frac{1 \times \sqrt{5}}{\sqrt{5} \times \sqrt{5}} = \frac{1}{5}\sqrt{5} \). If \( \frac{1}{\sqrt{5}} \) is irrational, then \( \frac{1}{5}\sqrt{5} \) is rational. Since the product of two rationals is rational, \( 5 \times \frac{1}{5}\sqrt{5} = \sqrt{5} \) must be rational. This contradicts that \( \sqrt{5} \) is irrational. The contradiction arises from assuming \( \frac{3}{\sqrt{5}} \) is rational. Hence, \( \frac{3}{\sqrt{5}} \) is irrational.
(viii) Assume \( 2 - 3\sqrt{5} \) is rational. Then both 2 and \( 2 - 3\sqrt{5} \) are rational. Since the difference of two rationals is rational, \( 2 - (2 - 3\sqrt{5}) = 3\sqrt{5} \) must be rational. Since the product of two rationals is rational, \( \frac{1}{3} \times 3\sqrt{5} = \sqrt{5} \) must be rational. This contradicts that \( \sqrt{5} \) is irrational. The contradiction arises from assuming \( 2 - 3\sqrt{5} \) is rational. Hence, \( 2 - 3\sqrt{5} \) is irrational.
(ix) Assume \( \sqrt{3} + \sqrt{5} \) is rational, say equal to \( a \). Then \( \sqrt{3} = a - \sqrt{5} \). Squaring both sides, \( 3 = a^2 + 5 - 2\sqrt{5}a \), which gives \( \sqrt{5} = \frac{a^2 + 2}{2a} \). The right side is rational while the left side is irrational. This is a contradiction. Hence, \( \sqrt{3} + \sqrt{5} \) is irrational.

Exam Tip: In proof by contradiction, clearly state your assumption, follow the logical steps, and show where the contradiction appears. Always conclude that the original statement must be true.

Question 4. Prove that 1/√3 is irrational.
Answer: Assume \( \frac{1}{\sqrt{3}} \) is rational. Then \( \frac{1}{\sqrt{3}} = \frac{a}{b} \), where a and b are positive integers with no common factor other than 1. This gives \( \sqrt{3} = \frac{b}{a} \). Since a and b are non-zero integers, \( \frac{b}{a} \) is rational. Thus, this shows \( \sqrt{3} \) is rational. This contradicts the fact that \( \sqrt{3} \) is irrational. The contradiction arises from assuming \( \sqrt{3} \) is rational. So, if \( \frac{1}{\sqrt{3}} \) is irrational, then \( \frac{1}{5}\sqrt{5} \) is rational. Since the product of two rationals is rational, \( 5 \times \frac{1}{5}\sqrt{5} = \sqrt{5} \) must be rational. This contradicts that \( \sqrt{5} \) is irrational. The contradiction arises from assuming \( \frac{3}{\sqrt{5}} \) is rational. Hence, \( \frac{3}{\sqrt{5}} \) is irrational.
In simple words: If 1/√3 were rational (a fraction), then √3 would also have to be rational. But we know √3 is irrational, so 1/√3 cannot be rational either.

Exam Tip: Use the property that if a rational number times an irrational number equals a rational number, you reach a contradiction. This helps prove irrationality elegantly.

Question 5. Prove that the sum of two irrational numbers can be rational. Give examples.
Answer:
(i) Let the two irrationals be \( (4 - \sqrt{5}) \) and \( (4 + \sqrt{5}) \). Then \( (4 - \sqrt{5}) + (4 + \sqrt{5}) = 8 \), which is a rational number.
(ii) Let the two irrationals be \( 2\sqrt{3} \) and \( 3\sqrt{3} \). Then \( 2\sqrt{3} \times 3\sqrt{3} = 18 \), which is a rational number.

Exam Tip: Remember that while sums and products of two irrationals can be rational, this is not always true. Choose examples carefully - conjugate forms work well for sums.

Question 6. State whether the following statements are true or false. If false, give a counter-example.
(i) The sum of two rationals is always rational
(ii) The product of two rationals is always rational
(iii) The sum of two irrationals is an irrational
(iv) The product of two irrationals is an irrational
(v) The sum of a rational and an irrational is irrational
(vi) The product of a rational and an irrational is irrational
Answer:
(i) The sum of two rationals is always rational - True
(ii) The product of two rationals is always rational - True
(iii) The sum of two irrationals is an irrational - False. Counter-example: \( 2 + \sqrt{3} \) and \( 2 - \sqrt{3} \) are two irrational numbers. But their sum equals 4, which is a rational number.
(iv) The product of two irrationals is an irrational - False. Counter-example: \( 2\sqrt{3} \) and \( 4\sqrt{3} \) are two irrational numbers. But their product is 24, which is a rational number.
(v) The sum of a rational and an irrational is irrational - True
(vi) The product of a rational and an irrational is irrational - True

Exam Tip: A key insight is that operations with irrational numbers can sometimes yield rational results. Always test with specific counter-examples when unsure about general statements.

Question 7. Prove that (2√3 - 1) is an irrational number.
Answer: Assume \( x = 2\sqrt{3} - 1 \) is a rational number. Then:
\( x = 2\sqrt{3} - 1 \)
\( \Rightarrow x^2 = (2\sqrt{3} - 1)^2 \)
\( \Rightarrow x^2 = (2\sqrt{3})^2 + (1)^2 - 2(2\sqrt{3})(1) \)
\( \Rightarrow x^2 = 12 + 1 - 4\sqrt{3} \)
\( \Rightarrow x^2 - 13 = -4\sqrt{3} \)
\( \Rightarrow \frac{13 - x^2}{4} = \sqrt{3} \)
Since x is a rational number, \( x^2 \) is also rational. This means \( 13 - x^2 \) is rational, so \( \frac{13 - x^2}{4} \) is rational. But this implies \( \sqrt{3} \) is rational, which contradicts the fact that \( \sqrt{3} \) is irrational. Hence, \( (2\sqrt{3} - 1) \) is an irrational number.
In simple words: If you square the expression and rearrange, you can isolate √3. But a rational number cannot equal an irrational number, so the original assumption must be wrong.

Exam Tip: When proving expressions of the form \( a + b\sqrt{c} \) are irrational, squaring both sides and isolating the radical is a powerful technique.

Question 8. Prove that (4 - 5√2) is an irrational number.
Answer: Assume \( x = 4 - 5\sqrt{2} \) is a rational number. Then:
\( x = 4 - 5\sqrt{2} \)
\( \Rightarrow x^2 = (4 - 5\sqrt{2})^2 \)
\( \Rightarrow x^2 = 4^2 + (5\sqrt{2})^2 - 2(4)(5\sqrt{2}) \)
\( \Rightarrow x^2 = 16 + 50 - 40\sqrt{2} \)
\( \Rightarrow x^2 - 66 = -40\sqrt{2} \)
\( \Rightarrow \frac{66 - x^2}{40} = \sqrt{2} \)
Since x is a rational number, \( x^2 \) is also rational. This means \( 66 - x^2 \) is rational, so \( \frac{66 - x^2}{40} \) is rational. But this implies \( \sqrt{2} \) is rational, which contradicts the fact that \( \sqrt{2} \) is irrational. Hence, \( (4 - 5\sqrt{2}) \) is an irrational number.

Exam Tip: The method of assuming the contrary and deriving a contradiction is consistent across all such proofs. Systematically square, rearrange, and isolate the radical.

Question 9. Prove that (5 - 2√3) is an irrational number.
Answer: Assume \( x = 5 - 2\sqrt{3} \) is a rational number. Then:
\( x = 5 - 2\sqrt{3} \)
\( \Rightarrow x^2 = (5 - 2\sqrt{3})^2 \)
\( \Rightarrow x^2 = 5^2 + (2\sqrt{3})^2 - 2(5)(2\sqrt{3}) \)
\( \Rightarrow x^2 = 25 + 12 - 20\sqrt{3} \)
\( \Rightarrow x^2 - 37 = -20\sqrt{3} \)
\( \Rightarrow \frac{37 - x^2}{20} = \sqrt{3} \)
Since x is a rational number, \( x^2 \) is also rational. This means \( 37 - x^2 \) is rational, so \( \frac{37 - x^2}{20} \) is rational. But this implies \( \sqrt{3} \) is rational, which contradicts the fact that \( \sqrt{3} \) is irrational. Hence, \( (5 - 2\sqrt{3}) \) is an irrational number.

Exam Tip: Pay careful attention to arithmetic when expanding the square. Small errors in coefficients can lead to wrong conclusions, so verify all steps.

Question 10. Prove that 5√2 is irrational.
Answer: Let \( 5\sqrt{2} \) be a rational number. Then \( 5\sqrt{2} = \frac{p}{q} \), where p and q are integers and HCF(p, q) = 1.
\( \Rightarrow 5\sqrt{2}q = p \)
\( \Rightarrow (5\sqrt{2}q)^2 = p^2 \)
\( \Rightarrow 2(25q^2) = p^2 \)
\( \Rightarrow p^2 \) is divisible by 2
\( \Rightarrow p \) is divisible by 2
Let \( p = 2m \), where m is an integer. Then:
\( 5\sqrt{2}q = 2m \)
\( \Rightarrow (5\sqrt{2}q)^2 = (2m)^2 \)
\( \Rightarrow 2(25q^2) = 4m^2 \)
\( \Rightarrow 25q^2 = 2m^2 \)
\( \Rightarrow q^2 \) is divisible by 2
\( \Rightarrow q \) is divisible by 2
From both results, 2 is a common factor of both p and q, which contradicts HCF(p, q) = 1. Hence, \( 5\sqrt{2} \) is irrational.

Exam Tip: The Euclidean algorithm approach shows that if a prime divides the square, it must divide the number. Use this property to establish common factors and derive contradictions.

Question 11. Prove that 2/√7 is irrational.
Answer: \( \frac{2}{\sqrt{7}} = \frac{2}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{2\sqrt{7}}{7} \)
Let \( \frac{2}{7}\sqrt{7} \) be a rational number. Then \( \frac{2}{7}\sqrt{7} = \frac{p}{q} \), where p and q are integers and HCF(p, q) = 1.
\( \Rightarrow 2\sqrt{7}q = 7p \)
\( \Rightarrow (2\sqrt{7}q)^2 = (7p)^2 \)
\( \Rightarrow 7(4q^2) = 49p^2 \)
\( \Rightarrow 4q^2 = 7p^2 \)
\( \Rightarrow q^2 \) is divisible by 7
\( \Rightarrow q \) is divisible by 7
Let \( q = 7m \), where m is an integer. Then:
\( 2\sqrt{7}q = 7p \)
\( \Rightarrow [2\sqrt{7}(7m)]^2 = (7p)^2 \)
\( \Rightarrow 343(4m^2) = 49p^2 \)
\( \Rightarrow 7(4m^2) = p^2 \)
\( \Rightarrow p^2 \) is divisible by 7
\( \Rightarrow p \) is divisible by 7
From both results, 7 is a common factor of both p and q, which contradicts HCF(p, q) = 1. Hence, \( \frac{2}{\sqrt{7}} \) is irrational.

Exam Tip: Always rationalize the denominator first before applying the standard proof by contradiction technique with the HCF approach.

Question 1. State Euclid's division lemma.
Answer: Euclid's division lemma states that for any two positive integers a and b, there exist unique whole numbers q and r, such that \( a = b \times q + r \) where \( 0 \leq r < b \).
In simple words: When you divide one number by another, you get a quotient and a remainder. The remainder is always smaller than the divisor.

Exam Tip: This lemma forms the foundation for the Euclidean algorithm used to find HCF. Always ensure that the remainder satisfies the given constraint.

Question 2. State the fundamental theorem of arithmetic.
Answer: The fundamental theorem of arithmetic states that every integer greater than 1 either is prime itself or is the product of prime numbers, and this product is unique (up to the order of factors).
In simple words: Every whole number bigger than 1 can be broken down into prime factors in exactly one way (ignoring the order you write them).

Exam Tip: This theorem guarantees the uniqueness of prime factorization, which is crucial for solving problems involving HCF, LCM, and divisibility.

Question 3. Find the prime factorization of 360.
Answer: Prime factorization of 360 is \( 360 = 2^3 \times 3^2 \times 5 \)

Exam Tip: Organize prime factors in ascending order and use exponent notation to write the factorization concisely. Verify by multiplying back.

Question 4. If the prime factorizations of two numbers are a = a and b = b, find their HCF.
Answer: When the prime factorizations show that a and b share no common prime factors (or have no common factors), the HCF is found as the product of the smallest power of each common factor. In this case, HCF equals 1. Thus, HCF(a, b) = 1.

Exam Tip: Two numbers with HCF equal to 1 are called coprime or relatively prime. Always check all prime factors systematically.

Question 5. If the prime factorizations of two numbers are a = a and b = b, find their LCM.
Answer: The LCM is found as the product of the greatest power of each prime factor involved in the factorizations. Thus, LCM equals \( a \times b \). Therefore, LCM(a, b) = ab.

Exam Tip: Use the formula HCF × LCM = Product of the two numbers to verify your answer.

Question 6. The HCF of two numbers is 25 and their product is 1050. Find their LCM.
Answer: HCF of two numbers = 25
Product of two numbers = 1050
Let their LCM be x.
Using the formula: Product of two numbers = HCF × LCM
We conclude that:
\( 1050 = 25 \times x \)
\( x = \frac{1050}{25} = 42 \)
Hence, their LCM is 42.

Exam Tip: Remember this fundamental relationship: HCF(a, b) × LCM(a, b) = a × b. It provides a quick way to find missing values.

Question 7. What is a composite number?
Answer: A composite number is a positive integer which is not prime (i.e., which has factors other than 1 and itself).
In simple words: A composite number is a number that can be divided evenly by numbers other than just 1 and itself. For example, 6 can be divided by 1, 2, 3, and 6.

Exam Tip: Every integer greater than 1 is either prime or composite. The number 1 is neither prime nor composite.

Question 8. What are relatively prime numbers?
Answer: If two numbers are relatively prime then their greatest common factor will be 1. Thus, HCF(a, b) = 1.
In simple words: Two numbers are relatively prime if they share no common factors except 1. They don't have to be prime themselves, just coprime to each other.

Exam Tip: Examples of relatively prime pairs: (5, 7), (8, 15), (9, 10). Their HCF is always 1.

Question 9. State the condition for a rational number to have a terminating decimal expansion.
Answer: Let x be a rational number whose decimal expansion terminates. Then, we can express x in the form \( \frac{a}{b} \), where a and b are coprime, and the prime factorization of b is of the form \( (2^m \times 5^n) \), where m and n are non-negative integers.
In simple words: A fraction in lowest terms gives a terminating decimal only if its denominator has only 2 and 5 as prime factors (in any powers).

Exam Tip: If the denominator has any prime factor other than 2 or 5, the decimal will be non-terminating and repeating.

Question 10. Simplify: \( \frac { 2\sqrt{45} + 3\sqrt{20} }{ 2\sqrt{5} } \)
Answer: \( \frac { 2\sqrt{45} + 3\sqrt{20} }{ 2\sqrt{5} } = \frac { 2\sqrt{3 \times 3 \times 5} + 3\sqrt{2 \times 2 \times 5} }{ 2\sqrt{5} } \)
\( = \frac { 2 \times 3\sqrt{5} + 3 \times 2\sqrt{5} }{ 2\sqrt{5} } \)
\( = \frac { 6\sqrt{5} + 6\sqrt{5} }{ 2\sqrt{5} } \)
\( = \frac { 12\sqrt{5} }{ 2\sqrt{5} } \)
\( = 6 \)
Thus, the simplified form is 6.

Exam Tip: Always simplify radicals by factoring perfect squares first. Like radicals can then be combined easily.

Question 11. Find the decimal expansion of \( \frac{73}{2^4 \times 5^3} \)
Answer: Decimal expansion:
\( \frac{73}{2^4 \times 5^3} = \frac{73 \times 5^2}{2^4 \times 5^4} = \frac{365}{(2 \times 5)^4} = \frac{365}{(10)^4} = \frac{365}{10000} = 0.0365 \)
Thus, the decimal expansion is 0.0365.

Exam Tip: To create powers of 10 in the denominator, multiply numerator and denominator by appropriate powers of 2 or 5 until both exponents match.

Question 12. Show that there is no value of n for which (2^n × 5^n) ends in 5.
Answer: We can write: \( (2^n \times 5^n) = (2 \times 5)^n = 10^n \)
For any value of n, we get 0 in the end (i.e., the number ends in 0, not 5). Thus, there is no value of n for which \( (2^n \times 5^n) \) ends in 5.

Exam Tip: Any number of the form \( 10^n \) will always end in 0, regardless of n. This is because \( 10^n = 1\underbrace{00...0}_{n \text{ zeros}} \)

Question 13. Is it possible to have two numbers whose HCF is 25 and LCM is 520?
Answer: No, it is not possible to have two numbers whose HCF is 25 and LCM is 520. Since HCF must be a factor of LCM, but 25 is not a factor of 520 (as \( 520 \div 25 = 20.8 \), which is not an integer).

Exam Tip: A key property: HCF(a, b) is always a divisor of LCM(a, b). If this relationship fails, the pair of values is impossible.

Question 14. Give an example of two irrational numbers whose sum is rational.
Answer: Let the two irrationals be \( 4 - \sqrt{5} \) and \( 4 + \sqrt{5} \). Then:
\( (4 - \sqrt{5}) + (4 + \sqrt{5}) = 8 \)
Thus, the sum (i.e., 8) is a rational number.

Exam Tip: Conjugate pairs of the form \( (a - \sqrt{b}) \) and \( (a + \sqrt{b}) \) always add to give the rational number 2a.

Question 15. Give an example of two irrational numbers whose product is rational.
Answer: Let the two irrationals be \( 4\sqrt{5} \) and \( 3\sqrt{5} \). Then:
\( (4\sqrt{5}) \times (3\sqrt{5}) = 60 \)
Thus, the product (i.e., 60) is a rational number.

Exam Tip: The product of numbers of the form \( a\sqrt{c} \) and \( b\sqrt{c} \) equals \( ab \times c \), which is always rational.

Question 16. If two numbers are relatively prime, find their LCM in terms of a and b.
Answer: If two numbers are relatively prime, then their greatest common factor is 1. Thus, HCF(a, b) = 1. Using the formula: Product of two numbers = HCF × LCM, we conclude that:
\( a \times b = 1 \times \text{LCM} \)
\( \therefore \text{LCM} = ab \)
Thus, LCM(a, b) is ab.

Exam Tip: For coprime numbers, the LCM is simply their product. This makes calculations much faster.

Question 17. If the LCM of two numbers is 1200, can their HCF be 500?
Answer: If the LCM of two numbers is 1200 then, it is not possible to have their HCF equal to 500. Since HCF must be a factor of LCM, but 500 is not a factor of 1200 (as \( 1200 \div 500 = 2.4 \), which is not an integer).

Exam Tip: Always verify that HCF divides LCM evenly. If not, the pair of values cannot exist.

Question 18. Express 0.̄4̄ as a fraction.
Answer: Let \( x = 0.\overline{4} \)
\( x = 0.4 \) \(...(1)\)
Multiplying both sides by 10, we get
\( 10x = 4.4 \) \(...(2)\)
Subtracting (1) from (2), we get
\( 10x - x = 4.4 - 0.4 \)
\( 9x = 4 \Rightarrow x = \frac{4}{9} \)
\( \therefore 0.\overline{4} = \frac{4}{9} \)

Exam Tip: For a single repeating digit, multiply by 10. For two repeating digits, multiply by 100. Always subtract to eliminate the repeating part.

Question 19. Express 0.4̄ as a rational number in simplest form.
Answer: Set x to equal 0.4̄ (where the bar indicates the digit 4 repeats). So x = 0.4444…. When we multiply both sides by 10, we get 10x = 4.4444…. Now subtract the first equation from this new equation: 10x - x = 4.4444… - 0.4444…, which gives us 9x = 4. Therefore, x = 4/9. The simplest form of 0.4̄ as a rational number is 4/9.

Exam Tip: When converting repeating decimals, always multiply by an appropriate power of 10 based on the period of repetition - for single repeating digits, multiply by 10.

Question 20. Express 0.23̄ as a rational number in simplest form.
Answer: Let x equal 0.23̄ (where both digits repeat). From the first equation, x = 0.2323…. Multiplying both sides by 100 gives us 100x = 23.2323…. Subtracting equation (1) from equation (2): 100x - x = 23.2323… - 0.2323…, which simplifies to 99x = 23. Dividing both sides by 99 gives us x = 23/99. The simplest form of 0.23̄ as a rational number is 23/99.

Exam Tip: For decimals where two digits repeat, multiply by 100; for three digits, multiply by 1000. The multiplier's power of 10 must match the repetition period.

Question 21. Identify whether 0.15015001500015… is rational or irrational.
Answer: Irrational numbers are expressed as non-terminating and non-repeating decimals. The decimal 0.15015001500015… does not repeat in a fixed pattern, nor does it come to an end. This makes 0.15015001500015… an irrational number.

Exam Tip: Look carefully at the decimal expansion - if you cannot identify a repeating block of digits that continues forever, the number is likely irrational.

Question 22. Prove that √2/3 is irrational.
Answer: Assume √2/3 is a rational number. Then √2/3 = p/q, where p and q are integers with HCF(p,q) = 1. From this, √2q = 3p. Squaring both sides yields 2q² = 9p², which means p² is divisible by 2, so p is divisible by 2. Let p = 2m where m is an integer. Substituting: √2q = 3(2m), so (√2q)² = [3(2m)]², giving us 2q² = 4(9m²) = 36m². This simplifies to q² = 2(9m²), meaning q² is divisible by 2, hence q is divisible by 2. Since both p and q are divisible by 2, they share a common factor of 2, which contradicts our assumption that HCF(p,q) = 1. Our assumption was wrong. Therefore, √2/3 is irrational.

Exam Tip: In proof by contradiction for irrational numbers, always establish the coprimality condition (HCF = 1) at the start, then show how divisibility creates a contradiction.

Question 23. Find a rational number between √3 and 2.
Answer: Since √3 ≈ 1.732…, we can select 1.8 as the required rational number between √3 and 2. The number 1.8 is rational because it can be written as the fraction 18/10, and it clearly lies between 1.732… and 2. Thus, the required rational number is 1.8.

Exam Tip: When finding a rational number between two values, first approximate the irrational number, then choose any terminating or repeating decimal that falls within the range.

Question 24. Classify 3.1̄4̄1̄6̄ as rational or irrational.
Answer: Since 3.1416̄ is a non-terminating and repeating decimal (the block 1416 repeats), it belongs to the category of rational numbers.

Exam Tip: Any non-terminating but repeating decimal can always be expressed as a fraction, making it rational. The presence of a repeating pattern is the key identifier.

Exercise MCQ

Question 1. Determine the decimal nature of the fractions 1351/(5⁴ × 2), 2017/(5³ × 2), 3219/1800, and 1723/625.
Answer: For 1351/(5⁴ × 2): The numerator 1351 shares no common factors with 5 or 2. The fraction is already in simplest form and can be expressed as (2^m × 5^n) for integers m = 1, n = 4. So it is a terminating decimal. Multiplying numerator and denominator by 2³: (1351 × 2³)/(5⁴ × 2⁴) = 10808/10000 = 1.0808.

For 2017/(5³ × 2): The numerator 2017 shares no common factors with 5 or 2. In simplest form, it can be written as (2^m × 5^n) with m = 1, n = 3. So it is a terminating decimal. Multiplying by 2²: (2017 × 2²)/(5³ × 2³) = 8068/1000 = 8.068.

For 3219/1800 = 3219/(2³ × 5² × 3²): Since 2, 3, and 5 are not factors of 3219, the fraction is in simplest form. As (2³ × 5² × 3²) ≠ (2^m × 5^n), it is not a terminating decimal but rather a repeating decimal: 3219/1800 = 1.78833333….

For 1723/625 = 1723/5⁴: Since 5 is not a factor of 1723, the fraction is in simplest form. Since it is not of the form (2^m × 5^n), it is not a terminating decimal.

Exam Tip: A fraction in simplest form yields a terminating decimal if and only if the denominator has only 2 and 5 as prime factors.

Question 2. If a = (2² × 3³ × 5⁴) and b = (2³ × 3² × 5), find the HCF(a, b).
Answer: (b) 180
The HCF equals the product of the smallest powers of each common prime factor. The common prime factors are 2, 3, and 5. Taking the smallest power of each: 2² × 3² × 5 = 4 × 9 × 5 = 180.

Exam Tip: For HCF, always select the lowest exponent of each prime factor that appears in all the numbers.

Question 3. Find the HCF of (2³ × 3² × 5), (2² × 3³ × 5²), and (2⁴ × 3 × 5³ × 7).
Answer: (c) 60
The HCF is calculated by taking the product of the smallest power of each common prime factor across all three numbers. The primes common to all three are 2, 3, and 5. Their smallest powers are 2², 3¹, and 5¹. Therefore, HCF = 2² × 3 × 5 = 4 × 3 × 5 = 60.

Exam Tip: When finding HCF of three or more numbers, ensure the prime factor is present in every single number before including it in the HCF.

Question 4. Calculate the LCM of (2³ × 3 × 5) and (2⁴ × 5 × 7).
Answer: (c) 1680
The LCM is the product of the greatest powers of all prime factors that appear in either number. From the two expressions, the primes involved are 2, 3, 5, and 7. Their greatest powers are 2⁴, 3¹, 5¹, and 7¹. Therefore, LCM = 2⁴ × 3 × 5 × 7 = 16 × 3 × 5 × 7 = 1680.

Exam Tip: For LCM, take the highest power of each prime factor that appears in any of the numbers being compared.

Question 5. Two numbers have HCF = 27 and LCM = 162. If one number is 54, find the other number.
Answer: (d) 81
Using the fundamental relationship: Product of two numbers = HCF × LCM. Let the unknown number be y. Then 54 × y = 27 × 162. So 54y = 4374, which gives y = 4374/54 = 81.

Exam Tip: Remember that for any two numbers, x × y = HCF(x, y) × LCM(x, y). This relationship helps find an unknown number when others are given.

Question 6. If the product of two numbers is 1600 and their HCF is 5, what is their LCM?
Answer: (c) 320
Using the relationship HCF × LCM = Product of the two numbers, we have 5 × LCM = 1600. Solving for LCM: LCM = 1600/5 = 320.

Exam Tip: This fundamental formula can rearranged to solve for any unknown when the other three quantities are provided.

Question 7. Find the largest number that divides both 1152 and 1664 exactly.
Answer: (c) 128
The largest number that divides both numbers is their HCF. Finding prime factorizations: 1152 = 2⁷ × 3² and 1664 = 2⁷ × 13. The HCF is 2⁷ = 128.

Exam Tip: The largest number that divides two or more numbers is precisely their highest common factor, which you can find using prime factorization.

Question 8. What is the largest number that divides 70 - 5 = 65 and 125 - 8 = 117?
Answer: (a) 13
The required number divides both 65 and 117. Finding factorizations: 65 = 13 × 5 and 117 = 13 × 3². The HCF is 13.

Exam Tip: When a problem mentions a number that leaves specific remainders, find the differences and then calculate the HCF of those differences.

Question 9. What number divides both 245 - 5 = 240 and 1029 - 5 = 1024?
Answer: (b) 16
The required number divides both 240 and 1024. Prime factorizations: 240 = 2⁴ × 3 × 5 and 1024 = 2¹⁰. The HCF equals 2⁴ = 16.

Exam Tip: When numbers share a common remainder, subtract that remainder from each to get the numbers whose HCF you need to find.

Question 10. Express 1095/1168 in simplest form.
Answer: (d) 15/16
To simplify, we find the HCF of 1095 and 1168. Using division: 1095 ÷ 73 = 15 and 1168 ÷ 73 = 16. Since HCF(1095, 1168) = 73, the simplest form is 15/16.

Exam Tip: Always verify that the numerator and denominator of your simplified fraction share no common factors (other than 1).

Question 11. State Euclid's Division Lemma.
Answer: Euclid's Division Lemma asserts that for any two positive integers a and b, unique integers q and r exist such that a = bq + r, where the remainder r must satisfy the condition 0 ≤ r < b.

Exam Tip: This lemma is the foundation for the Euclidean algorithm used to find HCF and is essential for divisibility arguments.

Question 12. A number when divided by 143 leaves remainder 31. What remainder does this number leave when divided by 13?
Answer: (d) 5
From the given information, the number can be written as 143x + 31 for some quotient x. We can rewrite this as 143x + 31 = 13(11x) + (13 × 2) + 5 = 13(11x + 2) + 5. Therefore, when this number is divided by 13, the remainder is 5.

Exam Tip: When dealing with remainders, express the number in the form that relates to the new divisor, then identify the final remainder.

Question 13. Which of the following is irrational: 3.141141114…, π, 2.35̄, or 1.732?
Answer: (d) 3.141141114…
The decimal 3.141141114… is irrational because it is non-repeating and non-terminating. While the digits seem somewhat patterned, the increasing number of 1s prevents any true repetition of a fixed block.

Exam Tip: Be careful with decimals that appear patterned - true irrationals never settle into a repeating block, no matter how regular they initially seem.

Question 14. Identify the nature of π.
Answer: (c) π is an irrational number
π is irrational because its decimal expansion is non-repeating and non-terminating. It cannot be expressed as a ratio of two integers.

Exam Tip: Notable irrational numbers like π, e, √2, and √3 appear frequently in exams - memorize their classification.

Question 15. Classify 2.35̄ as rational or irrational.
Answer: (b) 2.35̄ is a rational number
The decimal 2.35̄ is rational because it is a non-terminating repeating decimal. The digits 35 repeat indefinitely, which means it can be expressed as a fraction.

Exam Tip: Any repeating decimal, no matter how long the repetition cycle, is always rational.

Question 16. Is the decimal 0.232332333… rational or irrational?
Answer: (c) an irrational number
The decimal 0.232332333… is irrational because it is non-terminating and non-repeating. Each block has an increasing number of 3s, preventing any true repetition.

Exam Tip: Decimals with increasing patterns (even if organized) do not settle into a repeating cycle, making them irrational.

Question 17. What type of number is 1.2348̄?
Answer: (b) 1.2348̄ is a rational number
The decimal 1.2348̄ is rational because it is a repeating decimal. The block 2348 repeats indefinitely, allowing it to be expressed as a fraction p/q.

Exam Tip: Look for the repeating bar - its presence guarantees rationality.

Question 18. Which of these fractions terminates as a decimal: 124/165, 131/30, 2027/625, or 1625/462?
Answer: (c) 2027/625
For 124/165 = 124/(5 × 33): Since 5 and 33 are not factors of 124, it is in simplest form and does not equal (2^m × 5^n), so it does not terminate.

For 131/30 = 131/(5 × 6): Since 5 and 6 are not factors of 131, it is in simplest form and does not equal (2^m × 5^n), so it does not terminate.

For 2027/625 = 2027/5⁴: This is of the form (2^m × 5^n) where m = 0, n = 4. Multiplying by 2⁴: (2027 × 2⁴)/(5⁴ × 2⁴) = 32432/10000 = 3.2432. This terminates after four decimal places.

For 1625/462 = 1625/(2 × 7 × 33): Since 2, 7, and 33 are not factors of 1625, it is in simplest form and does not equal (2^m × 5^n), so it does not terminate.

Exam Tip: A simplified fraction yields a terminating decimal if and only if its denominator has exactly 2 and/or 5 as its only prime factors.

Question 19. The decimal expansion of 37/(2⁵ × 5) terminates after how many decimal places?
Answer: (b) two decimal places
Rewriting the denominator: 37/(2⁵ × 5) = (37 × 5)/(2² × 5²) = 185/100 = 1.85. The expansion terminates after exactly two decimal places.

Exam Tip: For a fraction of the form p/(2^m × 5^n), the number of decimal places equals max(m, n).

Question 20. After how many decimal places does 14753/1250 terminate?
Answer: (d) four decimal places
Rewriting: 14753/1250 = 14753/(5⁴ × 2) = (14753 × 2³)/(5⁴ × 2⁴) = 118024/10000 = 11.8024. The expansion terminates after exactly four decimal places.

Exam Tip: Count the number of decimal places by finding the highest power of 2 or 5 in the simplified denominator.

Question 21. Is 1.732 rational or irrational?
Answer: (b) rational
The number 1.732 is a terminating decimal, which makes it rational. It can be expressed as the fraction 1732/1000.

Exam Tip: All terminating decimals are rational because they can always be written as fractions with powers of 10 in the denominator.

Question 22. If a = 5 and b = 3, what is the least prime factor of a + b?
Answer: (a) 2
Since 5 + 3 = 8, the sum is even. The least (and smallest) prime factor of 8 is 2. When the sum of two odd numbers is even, 2 must be a factor.

Exam Tip: The sum of two odd numbers is always even, so 2 is always a factor unless the sum happens to be a prime greater than 2 itself.

Question 23. Prove that √2 is irrational.
Answer: Suppose √2 is a rational number. Then √2 = p/q, where p and q are integers with HCF(p, q) = 1. From this equation, √2q = p. Squaring both sides: 2q² = p². This means p² is divisible by 2, so p must be divisible by 2. Let p = 2m for some integer m. Substituting back: 2q² = 4m², which simplifies to q² = 2m². This shows q² is divisible by 2, so q is divisible by 2. Since both p and q are divisible by 2, they have 2 as a common factor, contradicting the condition HCF(p, q) = 1. Our assumption was false. Therefore, √2 is irrational.

Exam Tip: Structure your proof of irrationality clearly: assume the opposite, derive a contradiction about the HCF, then conclude the original claim.

Question 24. Is 1/√2 rational or irrational?
Answer: (c) an irrational number
Since √2 is irrational, the reciprocal 1/√2 is also irrational. The quotient of a rational (1) and an irrational (√2) yields an irrational result.

Exam Tip: If an operation involves an irrational number and does not cancel it out completely, the result typically remains irrational.

Question 25. Is 2 + √2 rational or irrational?
Answer: (c) an irrational number
The sum of 2 and √2 is irrational. If 2 + √2 were rational, then subtracting the rational number 2 would yield √2 as the difference of two rationals, making √2 rational. But √2 is irrational, creating a contradiction. Therefore, 2 + √2 must be irrational.

Exam Tip: Adding a rational to an irrational always gives an irrational, since the irrational "survives" the operation.

Question 26. Find the least number divisible by all integers from 1 to 10.
Answer: (c) 2520
The least number divisible by all numbers from 1 to 10 is their LCM. Computing: LCM(1 to 10) = 2³ × 3² × 5 × 7 = 8 × 9 × 5 × 7 = 2520. This is the smallest positive integer that all these numbers divide into evenly.

Exam Tip: For consecutive integers starting from 1, find their LCM by taking the highest power of each prime up to the largest number in the range.

Exercise - Formative Assessment

Question 1. Determine whether 71/150 is a terminating or non-terminating decimal.
Answer: (b) a non-terminating, repeating decimal
Express the denominator in prime factors: 71/150 = 71/(2 × 3 × 5²). Since 2, 3, and 5 are not factors of 71, the fraction is in simplest form. The denominator (2 × 3 × 5²) does not match the form (2^m × 5^n) because of the factor 3. Therefore, 71/150 = 0.473̄ is a non-terminating repeating decimal.

Exam Tip: If the denominator (after simplification) has any prime factor other than 2 or 5, the decimal will be non-terminating.

Question 2. Is 19/80 a terminating or non-terminating decimal?
Answer: (b) 19/80
Express as: 19/80 = 19/(2⁴ × 5). Since 2 and 5 are not factors of 19, it is in simplest form. The denominator equals (2^m × 5^n) where m = 4, n = 1. Therefore, 19/80 is a terminating decimal.

Exam Tip: Once you confirm the fraction is in simplest form, check if the denominator is a product of powers of only 2 and 5.

Question 3. When n is divided by 9, the remainder is 7. What is the remainder when 3n - 1 is divided by 9?
Answer: (b) 2
From the given condition, n = 9q + 7 for some quotient q. Then 3n = 3(9q + 7) = 27q + 21. So 3n - 1 = 27q + 20 = 9(3q) + 9(2) + 2 = 9(3q + 2) + 2. When 3n - 1 is divided by 9, the remainder is 2.

Exam Tip: Use the division lemma systematically: express the original divisibility condition, then manipulate algebraically to match the new divisor.

Question 4. Express 1.42̄ as a fraction.
Answer: (a) 1.42̄
Let x = 1.42̄ = 1.4242…. Multiplying by 100: 100x = 142.4242…. Subtracting the original: 100x - x = 141, so 99x = 141. Simplifying: x = 141/99 = 47/33.

Exam Tip: When converting repeating decimals to fractions, multiply by an appropriate power of 10 based on the period, then subtract to eliminate the repeating part.

Question 5. Explain why 4ⁿ can never end with the digit 0.
Answer: If 4ⁿ ended with 0, then 5 would be a factor of 4ⁿ. However, the only prime factor of 4ⁿ = (2²)ⁿ = 2²ⁿ is 2. By the Fundamental Theorem of Arithmetic, every number has a unique prime factorization. Since 5 cannot appear in the prime factorization of 4ⁿ, the number 4ⁿ can never be divisible by 10 and thus can never end in 0.

Exam Tip: Numbers ending in 0 must be divisible by 10 = 2 × 5, requiring both 2 and 5 as prime factors.

Question 6. Two numbers have HCF = 27 and LCM = 162. If one is 81, find the other.
Answer: Using the product formula: x × y = HCF × LCM. So 81 × y = 27 × 162 = 4374. Therefore, y = 4374/81 = 54. The other number is 54.

Exam Tip: This relationship is one of the most useful in number theory - practice applying it in different scenarios.

Question 7. Is 17/30 a terminating decimal?
Answer: Express as: 17/30 = 17/(2 × 3 × 5). Since 2, 3, and 5 are not factors of 17, it is in simplest form. The denominator contains the factor 3, so it is not of the form (2^m × 5^n). Therefore, 17/30 is a non-terminating decimal.

Exam Tip: The presence of any prime factor other than 2 or 5 in the simplified denominator guarantees a non-terminating decimal.

Question 8. Reduce 148/185 to simplest form.
Answer: Find HCF(148, 185) = 37. Then 148/185 = (148 ÷ 37)/(185 ÷ 37) = 4/5. The simplest form is 4/5.

Exam Tip: Always verify your answer by checking that the numerator and denominator of the simplified form share no common factors.

Question 9. Classify each number as rational or irrational: (a) √2, (b) ∛6, (c) 3.142857, (d) 2.3̄, (e) π, (f) 22/7, (g) 0.232332333…, (h) 5.2741̄
Answer: (a) √2 is irrational (since p is prime, √p is always irrational).
(b) ∛6 = ∛2 × ∛3 is irrational (products and quotients of rationals with irrationals remain irrational).
(c) 3.142857 is rational (it is a terminating decimal).
(d) 2.3̄ is rational (it is a non-terminating repeating decimal).
(e) π is irrational (it is a non-repeating, non-terminating decimal).
(f) 22/7 is rational (it is expressed in the form p/q where q ≠ 0).
(g) 0.232332333… is irrational (it is non-terminating and non-repeating, with increasing numbers of 3s).
(h) 5.2741̄ is rational (it is non-terminating but repeating).

Exam Tip: Memorize the classification of common constants and rules: any terminating or repeating decimal is rational; any non-repeating, non-terminating decimal is irrational.

Question 10. Prove that 2 + √3 is irrational.
Answer: Assume 2 + √3 is rational. Then both (2 + √3) and the rational number 2 are rational. Since the difference of two rationals is rational, (2 + √3) - 2 = √3 would be rational. But √3 is irrational, which is a contradiction. Therefore, our assumption is false, and 2 + √3 must be irrational.

Exam Tip: For sums and differences involving irrational numbers, use indirect reasoning: assume it's rational, show the contradiction, and conclude it's irrational.

Question 11. Find HCF and LCM of 12, 15, 18, and 27.
Answer: Prime factorizations:
12 = 2² × 3
15 = 3 × 5
18 = 2 × 3²
27 = 3³

HCF = Product of smallest powers of common prime factors = 3

LCM = Product of greatest powers of all prime factors = 2² × 3³ × 5 = 4 × 27 × 5 = 540

Exam Tip: For multiple numbers, always write out all prime factorizations first, then systematically identify the smallest and greatest powers of each prime.

Question 12. If (2 + √2) and (2 - √2) are two irrational numbers, what is their sum?
Answer: Sum = (2 + √2) + (2 - √2) = 2 + √2 + 2 - √2 = 4. The sum equals 4, which is a rational number. This demonstrates that the sum of two irrational numbers can be rational.

Exam Tip: Remember that operations on irrational numbers can yield either rational or irrational results - always work through the algebra carefully.

Question 13. Find the prime factorization of 4620.
Answer: 4620 = 2 × 2310 = 2 × 2 × 1155 = 2² × 3 × 385 = 2² × 3 × 5 × 77 = 2² × 3 × 5 × 7 × 11

Exam Tip: Always divide by the smallest prime first and continue systematically - this reduces the chance of missing a factor.

Question 14. Find the HCF of 1008 and 1080.
Answer: Prime factorizations:
1008 = 2⁴ × 3² × 7
1080 = 2³ × 3³ × 5

HCF = Product of smallest powers of each common prime factor = 2³ × 3² = 8 × 9 = 72

Exam Tip: Compare the exponents carefully - use only the common primes and their smallest powers.

Question 15. Find HCF and LCM of the fractions 8/9, 10/18, and 16/27.
Answer: HCF of fractions = (HCF of numerators) / (LCM of denominators)

Prime factorizations of numerators: 8 = 2³, 10 = 2 × 5, 16 = 2⁴
HCF of numerators = 2

Prime factorizations of denominators: 9 = 3², 18 = 2 × 3², 27 = 3³
LCM of denominators = 2 × 3³ = 2 × 27 = 54

HCF of fractions = 2/54 = 1/27

LCM of fractions = (LCM of numerators) / (HCF of denominators)

LCM of numerators = 2⁴ × 5 = 80
HCF of denominators = 3² = 9

LCM of fractions = 80/9

Exam Tip: For fractions, remember: HCF uses smallest powers of numerators but largest powers of denominators (and vice versa for LCM).

Question 16. Find the largest number that leaves remainder 6 when dividing 546 and remainder 8 when dividing 764.
Answer: The required number divides (546 - 6) = 540 and (764 - 8) = 756. Find HCF(540, 756):

Prime factorizations:
540 = 2² × 3³ × 5
756 = 2² × 3³ × 7

HCF = 2² × 3² = 4 × 9 = 36

Wait, let me recalculate: 540 = 2² × 3² × 3 × 5, 756 = 2² × 3³ × 7. Actually: 540 = 2² × 135 = 2² × 27 × 5 = 2² × 3³ × 5; 756 = 4 × 189 = 4 × 27 × 7 = 2² × 3³ × 7. So HCF = 2² × 3² = 4 × 9 = 36. But this doesn't match the expected answer. Let me verify: 540/36 = 15 ✓, 756/36 = 21 ✓. So HCF = 36. However, checking 2² × 3³: that would be 4 × 27 = 108. Then 540/108 = 5 ✓, 756/108 = 7 ✓. So HCF = 108. The largest number is 108.

Exam Tip: When a number leaves specific remainders, subtract those remainders first to find what both numbers divide into, then compute the HCF.

Question 17. Prove that √3 is irrational.
Answer: Assume √3 is rational and can be written as a/b in simplest form, where a and b are integers with HCF(a, b) = 1 and b ≠ 0. Then √3 = a/b, which gives us 3 = a²/b² after squaring both sides. This means 3b² = a². Since 3 divides 3b², it must divide a², and therefore 3 divides a (since 3 is prime). Let a = 3c for some integer c. Substituting: 3b² = 9c², which simplifies to b² = 3c². This shows 3 divides b² and hence 3 divides b. Now both a and b are divisible by 3, contradicting HCF(a, b) = 1. This contradiction arose from assuming √3 is rational. Therefore, √3 is irrational.

Exam Tip: Use the prime divisibility property: if a prime p divides p², then p divides the original number - this is key to proofs of irrationality.

Question 18. Prove that every positive odd integer is of the form (4q + 1) or (4q + 3) for some integer q.
Answer: Let a represent the given positive odd integer. When we divide a by 4, we get a quotient q and a remainder r. By Euclid's division algorithm,

\( a = 4q + r \) where \( 0 \leq r < 4 \)

\( \implies a = 4q + r \) where \( r = 0, 1, 2, 3 \)

\( \implies a = 4q \) or \( a = 4q + 1 \) or \( a = 4q + 2 \) or \( a = 4q + 3 \)

Now, \( 4q \) and \( 4q + 2 = 2(2q + 1) \) are both even numbers. Since a is odd, it cannot take either of these forms. Therefore, when a is odd, it must be of the form \( (4q + 1) \) or \( (4q + 3) \) for some integer q.
In simple words: When you divide any odd number by 4, the remainder is always either 1 or 3 - never 0 or 2. So every odd number looks like 4q + 1 or 4q + 3.

Exam Tip: Always verify by checking a few odd numbers (5, 7, 9, 11) and confirm they fit one of these two forms - examiners value explicit verification before the general proof.

Question 19. Show that one and only one out of n, (n + 2) and (n + 4) is divisible by 3.
Answer: Let q be the quotient and r be the remainder. When we apply Euclid's division algorithm by dividing n by 3,

\( n = 3q + r \) where \( 0 \leq r < 3 \)

\( \implies n = 3q + r \) where \( r = 0, 1 \) or \( 2 \)

\( \implies n = 3q \) or \( n = (3q + 1) \) or \( n = (3q + 2) \)

Case 1: If \( n = 3q \), then n is divisible by 3.

Case 2: If \( n = (3q + 1) \), then \( (n + 2) = 3q + 3 = 3(q + 1) \), which is clearly divisible by 3. In this case, (n + 2) is divisible by 3.

Case 3: If \( n = (3q + 2) \), then \( (n + 4) = 3q + 6 = 3(q + 2) \), which is clearly divisible by 3. In this case, (n + 4) is divisible by 3.

Hence, exactly one and only one out of n, (n + 2) and (n + 4) is divisible by 3.
In simple words: Any whole number has a remainder of 0, 1, or 2 when divided by 3. Depending on which remainder it has, either the number itself, or the number plus 2, or the number plus 4 will be divisible by 3 - but never more than one.

Exam Tip: Always test with concrete examples like n = 3, 4, 5 to verify all three cases work - this demonstrates full understanding of the proof.

Question 20. Prove that (4 + 3√2) is irrational.
Answer: Suppose (4 + 3√2) is a rational number. Then both (4 + 3√2) and 4 must be rational. This gives us

\( (4 + 3\sqrt{2}) - 4 = 3\sqrt{2} \) is rational [since the difference of two rational numbers is rational]

\( \implies 3\sqrt{2} \) is rational

\( \implies \frac{1}{3}(3\sqrt{2}) = \sqrt{2} \) is rational [since the product of two rational numbers is rational]

\( \implies \sqrt{2} \) is rational

But this contradicts the established fact that √2 is irrational (since 2 is prime, √2 must be irrational). Our assumption was wrong. Therefore, (4 + 3√2) must be irrational.
In simple words: If (4 + 3√2) were rational, then by subtracting 4 we'd get 3√2 is rational. But then √2 would also have to be rational. This is a contradiction, since we know √2 is irrational. So (4 + 3√2) cannot be rational.

Exam Tip: Always state "assume the opposite" at the start of a proof by contradiction, and clearly mark the contradiction at the end - examiners look for this logical structure.