Solution — Step by step

1. Assumption: Assume √2 is rational. So we can write √2 = p/q, where p and q are integers, q ≠ 0, and p/q is in its simplest form (i.e., p and q are co-prime, sharing no common factors).
2. Square both sides: 2 = p²/q²
3. Rearrange: p² = 2q² — this means p² is even.
4. Deduction for p: If p² is even, then p must be even (if an odd number were squared, the result would be odd). So let p = 2k for some integer k.
5. Substitute p = 2k into Step 3: (2k)² = 2q² → 4k² = 2q² → q² = 2k²
6. Deduction for q: q² is even, so q is also even.
7. Contradiction: Both p and q are even — they share the common factor 2. But this contradicts our assumption that p/q was in simplest form (co-prime).
8. Conclusion: Our assumption was wrong.

Solution — Step by step

1. Assumption: Assume √5 is rational. Write √5 = p/q in lowest terms (p, q integers, q ≠ 0, gcd(p,q) = 1).
2. Square both sides: 5 = p²/q², so p² = 5q².
3. 5 divides p²: Since p² = 5q², we see that 5 divides p². Since 5 is prime, 5 must also divide p. Write p = 5m.
4. Substitute: (5m)² = 5q² → 25m² = 5q² → q² = 5m².
5. 5 divides q²: So 5 divides q as well.
6. Contradiction: Both p and q are divisible by 5 — they share common factor 5. This contradicts the assumption that p/q is in lowest terms.
7. Conclusion: The assumption is false.

Solution — Step by step

Part (i): Convert 0.1̄6̄ (i.e., 0.16666…)

1. Let x = 0.16666… — here ‘1’ is non-repeating (1 digit) and ‘6’ repeats (1 digit).
2. Multiply by 10 (to move non-repeating part): 10x = 1.6666…
3. Multiply by another 10 (to move repeating cycle): 100x = 16.6666…
4. Subtract: 100x – 10x = 16.666… – 1.666… → 90x = 15
5. Solve: x = 15/90 = 1/6

Part (ii): Convert 2.35̄7̄ (i.e., 2.35757…)

1. Let x = 2.35757… — ’35’ is non-repeating (2 digits), ’57’ repeats (2 digits).
2. Multiply by 100 (2 non-repeating digits): 100x = 235.757…
3. Multiply by 100 again (2 repeating digits): 10000x = 23575.757…
4. Subtract: 10000x – 100x = 23575.757… – 235.757… → 9900x = 23340
5. Solve: x = 23340/9900 = 2334/990 = 1167/495 = 389/165

Solution — Step by step

1. Draw the number line with origin O at 0. Mark point A at 1 unit to the right of O (so OA = 1 unit).
2. Erect a perpendicular at point A, standing straight up from the number line.
3. Mark point B on the perpendicular such that AB = 1 unit. Now we have a right triangle OAB with OA = AB = 1 unit.
4. Find OB: By the Baudhāyana–Pythagoras Theorem: OB² = OA² + AB² = 1² + 1² = 2, so OB = √2.
5. Open the compass to the length OB. With O as centre, draw an arc that cuts the number line at point P.
6. Since OP = OB = √2, the point P represents √2 on the number line.

Solution — Full explanation

1. Philosophical origin: In the Upanishads and Buddhist literature (before 7th century BCE), Śhūnyatā (from the Sanskrit word śhūnya, meaning zero) described the state of emptying the mind of all fluctuations (vṛttis) to achieve perfect stillness. Patanjali, in the Yoga Sutras (~3rd century BCE), also described this state.
2. Cultural acceptance: Because Indian thinkers deeply valued emptiness as a profound philosophical concept, they had the mental framework to accept ‘nothingness’ as a legitimate mathematical entity — unlike other civilisations that only used zero as a placeholder (e.g., Babylonians, Mayans).
3. Path to mathematics: The concept moved from philosophy → architecture → linguistics → finally into mathematics through the works of Āryabhaṭa and then Brahmagupta.
4. Physical transition: The Bakhśhālī Manuscript (early centuries CE) used a bold dot (bindu) as a symbol for zero — the first physical zero symbol.
5. Brahmagupta’s formal definition (628 CE): In his Brāhmasphuṭasiddhānta, he defined zero as the result of subtracting a number from itself: a – a = 0. He then established full arithmetic rules for zero.

Solution — Step by step

1. Terminating decimal rule: The decimal expansion of p/q (in lowest terms) terminates if and only if the prime factorisation of q contains only powers of 2 and/or 5, i.e., q = 2^a × 5^b for non-negative integers a, b.
2. Check 18/125: gcd(18, 125) = 1 (co-prime). Prime factors of denominator: 125 = 5³. Only powers of 5 — satisfies the rule.
3. Conclusion: The decimal expansion of 18/125 terminates.
4. Number of decimal places: Since denominator = 5³, multiply by 2³ to get denominator 10³ = 1000:
5. 18/125 = (18 × 8)/(125 × 8) = 144/1000 = 0.144

Solution — Left-hand side first, then right-hand side

Left-Hand Side (LHS):

1. Add the fractions inside: 1/2 + 3/4 = 2/4 + 3/4 = 5/4
2. Multiply: (5/4) × (8/3) = 40/12 = 10/3

Right-Hand Side (RHS):

1. First product: 1/2 × 8/3 = 8/6 = 4/3
2. Second product: 3/4 × 8/3 = 24/12 = 2
3. Add: 4/3 + 2 = 4/3 + 6/3 = 10/3

Solution — Step by step

1. Convert mixed numbers to improper fractions: 15¾ = 63/4 metres, and 2¼ = 9/4 metres per kurta.
2. Number of kurtas = Total cloth ÷ cloth per kurta = (63/4) ÷ (9/4)
3. Division rule: (63/4) × (4/9) = 252/36 = 7
4. So the tailor can make exactly 7 complete kurtas.
5. Cloth used = 7 × 9/4 = 63/4 = 15¾ metres.
6. Cloth remaining = 63/4 – 63/4 = 0 metres.

Solution — Full explanation

1. Natural Numbers ℕ = {1, 2, 3, …}: Born from the need to count animals, crops, and trade goods. Evidenced by tally marks on the Ishango bone (≈20,000 BCE). Served for counting but could not express subtraction results like 3 – 5.
2. Zero (Śhūnya): The philosophical concept of emptiness (Śhūnyatā) in Indian thought was formalised by Brahmagupta (628 CE) as the number 0, defined as a – a = 0. This solved the problem of representing “nothing.”
3. Integers ℤ = {…, –2, –1, 0, 1, 2, …}: Brahmagupta’s insight — subtracting a larger number from a smaller one requires negative numbers. These were grounded in real life as “debts” (Ṛiṇa) vs. “fortunes” (Dhana). Solved problems like 3 – 5 = –2.
4. Rational Numbers ℚ: Complex trade (dividing fields, measuring ingredients) needed fractions. All numbers p/q (q ≠ 0) including positive and negative fractions. Brahmagupta gave the formal arithmetic rules.
5. Irrational Numbers: Baudhāyana (~800 BCE) encountered lengths like the diagonal of a unit square (= √2) that could not be expressed as fractions. Later, Mādhava discovered that π also has no exact fractional form. These “gap-fillers” cannot be written as p/q.
6. Real Numbers ℝ: The union of rational and irrational numbers fills the entire number line continuously. Every physical measurement has a corresponding real number.

Solution — Step by step

Part 1: Six rational numbers between 3 and 4

1. Write 3 and 4 with denominator 7: 3 = 21/7 and 4 = 28/7.
2. The integers 22, 23, 24, 25, 26, 27 lie between 21 and 28.
3. So six rational numbers are: 22/7, 23/7, 24/7, 25/7, 26/7, 27/7.

Part 2: Five rational numbers between 2/5 and 3/5

1. Make denominators equal and large enough: multiply by 6/6 → 2/5 = 12/30 and 3/5 = 18/30.
2. Integers between 12 and 18 (exclusive): 13, 14, 15, 16, 17.
3. Five rational numbers: 13/30, 14/30, 15/30, 16/30, 17/30 which simplify to 13/30, 7/15, 1/2, 8/15, 17/30.