Chapter 3 Important Questions: The World of Numbers

(B) √12. √12 = 2√3, which cannot be expressed as p/q where p and q are integers and q ≠ 0. √9 = 3 (rational), 0.333... = 1/3 (rational), and 22/7 is already a fraction (rational).

35. According to Brahmagupta's rule, 'the product of two debts is a fortune', meaning a negative number multiplied by a negative number gives a positive number. Therefore, (–7) × (–5) = +35.

(C) 13/250. The denominator 250 = 2 × 5³. Since the prime factors of the denominator are only 2 and 5, the decimal expansion will terminate. The other denominators (15 = 3×5, 12 = 2²×3, 7) contain prime factors other than 2 and 5.

The sum of a rational number and an irrational number is always irrational. Example: 3 + √2 = 3 + 1.4142... = 4.4142..., which is non-terminating and non-repeating, hence irrational.

(i) Let x = 0.666... → 10x = 6.666... → 10x – x = 6 → 9x = 6 → x = 6/9 = 2/3.

(ii) Let x = 0.454545... → 100x = 45.454545... → 100x – x = 45 → 99x = 45 → x = 45/99 = 5/11.

(iii) Let x = 0.1666... → 10x = 1.666... → 100x = 16.666... → 100x – 10x = 15 → 90x = 15 → x = 15/90 = 1/6.

Starting temperature = –4°C

Rise by 19°C: –4 + 19 = 15°C

Drop by 23°C: 15 – 23 = –8°C

Equation: (–4) + 19 + (–23) = –8°C

The final temperature is –8°C. This illustrates Brahmagupta's rules for integer arithmetic — combining fortunes (dhana) and debts (ṛiṇa).

Method: To find a rational number between a and b, compute their average = (a + b)/2.

Step 1: Average of 3/5 and 4/5 = (3/5 + 4/5)/2 = (7/5)/2 = 7/10.

Step 2: Average of 3/5 and 7/10 = (6/10 + 7/10)/2 = (13/10)/2 = 13/20.

Step 3: Average of 7/10 and 4/5 = (7/10 + 8/10)/2 = (15/10)/2 = 15/20 = 3/4.

Three rational numbers between 3/5 and 4/5 are: 13/20, 7/10, and 3/4.

General Method: The average (a+b)/2 of any two rational numbers a and b is always a rational number lying strictly between them. This process can be repeated infinitely, demonstrating the density of rational numbers.

Total silk = 15¾ = 63/4 metres.

Silk required per kurta = 2¼ = 9/4 metres.

Number of kurtas = (63/4) ÷ (9/4) = (63/4) × (4/9) = 63/9 = 7.

The tailor can make exactly 7 complete kurtas.

Silk used = 7 × 9/4 = 63/4 = 15¾ metres.

Leftover silk = 63/4 – 63/4 = 0 metres.

Therefore, the tailor can make 7 kurtas with no silk remaining.

(i) √81 = 9 = 9/1. It can be expressed as p/q (p=9, q=1). Hence RATIONAL.

(ii) √5: Using proof by contradiction (similar to √2), if √5 = p/q in lowest terms, then 5q² = p², so 5 divides p. Let p = 5k, then q² = 5k², so 5 divides q — contradiction. Hence IRRATIONAL.

(iii) 1.01001000100001...: The pattern shows increasing zeros between 1s (1, 0, 1, 00, 1, 000, 1...). This is non-terminating and non-repeating (no fixed repeating block). Hence IRRATIONAL.

(iv) 2.3575757...: The digits '57' repeat after the non-repeating digit '3'. It is a non-terminating but repeating decimal. Hence RATIONAL. (It equals 2334/990 = 1167/495 simplified.)

(i) The numbers 11, 13, 17, and 19 are all prime numbers between 10 and 20. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. This grouping on the Ishango bone (c. 20,000 BCE) suggests early humans recognised the concept of prime numbers.

(ii) The next three prime numbers after 19 are: 23, 29, and 31.

(Check: 20=4×5, 21=3×7, 22=2×11 — not prime; 23 is prime; 24,25,26,27,28 — not prime; 29 is prime; 30 — not prime; 31 is prime.)

(iii) Natural numbers are NOT closed under subtraction.

Closure means: for any two natural numbers a and b, a – b must also be a natural number.

Counter-examples:

• 3 – 5 = –2, and –2 is NOT a natural number.

• 7 – 7 = 0, and 0 is NOT a natural number (natural numbers start from 1).

Therefore, natural numbers are not closed under subtraction.

(i) (–12) × 5: A debt multiplied by a fortune is a debt. (–12) × 5 = –60.

(ii) (–8) × (–7): The product of two debts is a fortune. (–8) × (–7) = +56.

(iii) 0 – (–14): Subtracting a negative number is the same as adding a positive number. 0 – (–14) = 0 + 14 = 14.

(iv) (–20) ÷ 4: A debt divided by a fortune remains a debt. (–20) ÷ 4 = –5.

Real-world explanation for (iii): Suppose someone owes you ₹14 (a debt of –14 to them means +14 to you). If that debt is cancelled or removed (subtracted), you gain ₹14. So removing a debt of ₹14 is the same as gaining ₹14: 0 – (–14) = +14.

Proof by Contradiction:

Step 1 – Assumption: Assume √2 is rational. Then it can be written as √2 = p/q, where p and q are integers, q ≠ 0, and p/q is in its lowest terms (i.e., p and q are co-prime, sharing no common factor other than 1).

Step 2 – Square both sides:

2 = p²/q²

Step 3 – Rearrange:

2q² = p²

Step 4 – Deduction about p: Since p² = 2q², p² is even (it equals 2 times an integer). If the square of an integer is even, the integer itself must be even. Therefore, p is even. Let p = 2k for some integer k.

Step 5 – Substitute p = 2k:

2q² = (2k)² = 4k²

Step 6 – Simplify:

q² = 2k²

Step 7 – Deduction about q: Since q² = 2k², q² is even. Therefore, q must also be even.

Step 8 – The Contradiction: We have shown that both p and q are even, meaning they share a common factor of 2. But this contradicts our assumption in Step 1 that p and q are co-prime (share no common factors other than 1).

Conclusion: Since our assumption leads to a logical contradiction, and each algebraic step was valid, the assumption itself must be false. Therefore, √2 cannot be expressed as a rational number. Hence, √2 is irrational. ∎

(i) Converting 2.3575757...:

Let x = 2.3575757...

'35' is non-repeating (2 digits), '57' repeats (2 digits).

Multiply by 10² = 100: 100x = 235.7575...

Multiply by 10² = 100 again: 10000x = 23575.7575...

Subtract: 10000x – 100x = 23575.7575... – 235.7575...

9900x = 23340

x = 23340/9900 = 2334/990 = 1167/495

Verification: 1167 ÷ 495 = 2.35757... ✓

(ii) Converting 0.9999...:

Let x = 0.9999...

Multiply by 10: 10x = 9.9999...

Subtract: 10x – x = 9.9999... – 0.9999...

9x = 9

x = 9/9 = 1

Therefore 0.9̄ = 1 exactly.

Explanation: This remarkable result shows that decimal representations of numbers are not always unique. Every terminating decimal has an equivalent form ending in repeating 9s. For example, 1.000... = 0.999..., 2.47000... = 2.46999.... This is not a paradox but a property of our number system — two different decimal expressions can represent the exact same rational number. The density of real numbers ensures there is no 'gap' between 0.999... and 1.

Construction of √2 on the number line:

Step 1: Draw a number line and mark the origin O. Mark point A at 1 unit to the right of O (OA = 1 unit).

Step 2: At point A, draw a perpendicular to the number line. On this perpendicular, mark point B such that AB = 1 unit.

Step 3: Join O to B. By the Baudhāyana–Pythagoras Theorem:

OB² = OA² + AB² = 1² + 1² = 2

Therefore OB = √2 units.

Step 4: With O as centre and OB as radius, draw an arc that intersects the number line at point P (to the right of O). Then OP = √2 units. P represents √2 on the number line.

Diagram: O----A (1 unit), perpendicular AB (1 unit) at A, hypotenuse OB = √2, arc from B to P on number line.

Construction of √3 on the number line:

Step 1: Using the point P located above where OP = √2, draw a perpendicular at P.

Step 2: Mark point Q on this perpendicular such that PQ = 1 unit.

Step 3: Join O to Q. By Pythagoras:

OQ² = OP² + PQ² = (√2)² + 1² = 2 + 1 = 3

Therefore OQ = √3 units.

Step 4: With O as centre and OQ as radius, draw an arc intersecting the number line at R. Then OR = √3 units. R represents √3.

General Method: To construct √n, first construct √(n–1) as a hypotenuse, then erect a unit perpendicular at its tip, and the new hypotenuse gives √n. This forms the square root spiral. This method can locate any √n for any positive integer n.

(i) A decimal terminates when the denominator of the fraction (in lowest terms) can be expressed as a power of 10. Since 10 = 2 × 5, any denominator of the form 2^m × 5^n can be converted to 10^k by multiplying numerator and denominator by the appropriate power of 2 or 5. This gives a fraction with denominator as a power of 10, which is a terminating decimal. If q has any prime factor other than 2 or 5 (like 3, 7, 11...), the denominator can never become a power of 10, so the division never terminates.

Examples:

• 3/8: 8 = 2³. Multiply by 5³: 3×125/(8×125) = 375/1000 = 0.375 ✓ (terminates)

• 5/11: 11 is prime (not 2 or 5). Division never terminates → 0.454545... = 0.4̄5̄

(ii) 7/20: 20 = 2² × 5. Prime factors are only 2 and 5. → TERMINATING.

Conversion: 7/20 = 7/(2²×5) = (7×5)/(2²×5²) = 35/100 = 0.35

4/15: 15 = 3 × 5. Prime factor 3 is present (not just 2 and 5). → NON-TERMINATING REPEATING.

(4/15 = 0.2666... = 0.2̄6̄)

(iii) Mādhava's infinite series π = 4(1 – 1/3 + 1/5 – 1/7 + ...) reveals a fundamental truth about irrational numbers: they cannot be expressed as any single fraction (finite ratio), but can be approached as closely as desired through an infinite process. Since π is irrational, its decimal expansion is non-terminating and non-repeating — it goes on forever without any repeating block. No finite sum of terms gives the exact value of π; only the infinite sum does. This illustrates why irrational numbers require infinite, non-repeating decimals for their exact representation, and why Āryabhaṭa (499 CE) correctly called his value 3927/1250 an 'asanna' (approximation), anticipating that π could not be captured by any single fraction.