• Rate of exchange: 2 bags of spices = 15 copper ingots.
• Number of sets of 2 bags in 12 bags: 12 / 2 = 6 sets.
• Total copper ingots received: 6 × 15 = 90 copper ingots.

• Common Property: They are all consecutive prime numbers (numbers greater than 1 with no positive divisors other than 1 and themselves). Specifically, they are the prime numbers between 10 and 20.
• Next Three Numbers: The next three prime numbers following 19 are 23, 29, and 31.

No, natural numbers are not closed under subtraction.

• Subtracting a larger natural number from a smaller one or subtracting a number from itself does not yield a natural number:
  • 3 - 5 = -2 (which is a negative integer, not a natural number).
  • 4 - 4 = 0 (zero is a whole number/integer, not a natural number).

• Count on one hand: The four fingers (index, middle, ring, pinky) each have 3 distinct segments/joints. Using the thumb as a pointer gives:
  4 fingers × 3 joints = 12 distinct counts on a single hand.
• Relation to base-12 system: This physiological method is the natural foundation for duodecimal (base-12) systems. It explains the historical division of days and nights into 12 hours each (total 24 hours), the 12 signs of the zodiac, and units like dozens (12) and grosses (144).

• Initial temperature = 4°C
• Temperature drop = 15°C
• Midnight temperature = 4 - 15 = -11°C

• Integer Equation: (-850) + (+1200) + (-450)
• Step 1: (-850) + 1200 = +350
• Step 2: 350 + (-450) = -100
• Final Financial Standing: He has a net debt of ₹100 (standing is -₹100).

• (i) (-12) × 5: Product of a debt and a fortune is a debt ⇒ -60
• (ii) (-8) × (-7): Product of two debts is a fortune ⇒ 56
• (iii) 0 - (-14): Subtracting a debt leaves a positive fortune ⇒ 0 + 14 = 14
• (iv) (-20) ÷ 4: Dividing a debt by a positive quantity gives a debt ⇒ -5

• Suppose you have ₹10 in cash, but you also hold a formal debt (negative value) of ₹5 owed to a storekeeper.
• If someone generously waives or removes that ₹5 debt for you, they are subtracting a debt of 5: -(-5).
• Relieving you of that ₹5 obligation increases your financial net worth by ₹5. Therefore, removing a ₹5 debt is identical to being handed ₹5 in cash:
  10 - (-5) = 10 + 5 = 15.

Two rational numbers a/b and c/d are equal if their cross-products satisfy ad = bc:

• (i) 2/3 and 4/6: 2 × 6 = 12 and 3 × 4 = 12. Since 12 = 12, 2/3 = 4/6.
• (ii) 5/4 and 10/8: 5 × 8 = 40 and 4 × 10 = 40. Since 40 = 40, 5/4 = 10/8.
• (iii) -3/5 and -6/10: (-3) × 10 = -30 and 5 × (-6) = -30. Since -30 = -30, -3/5 = -6/10.
• (iv) 9/3 and 3/1: 9 × 1 = 9 and 3 × 3 = 9. Since 9 = 9, 9/3 = 3.

• (i) 2/5 + 3/10:
  LCM(5, 10) = 10 ⇒ (4 + 3) / 10 = 7/10
• (ii) 7/12 + 5/8:
  LCM(12, 8) = 24 ⇒ (14 + 15) / 24 = 29/24 (or 1 5/24)
• (iii) -4/7 + 3/14:
  LCM(7, 14) = 14 ⇒ (-8 + 3) / 14 = -5/14

• (i) 5/6 - 1/4:
  LCM(6, 4) = 12 ⇒ (10 - 3) / 12 = 7/12
• (ii) 11/8 - 3/4:
  LCM(8, 4) = 8 ⇒ (11 - 6) / 8 = 5/8
• (iii) -7/9 - (-2/3):
  -7/9 + 2/3; LCM(9, 3) = 9 ⇒ (-7 + 6) / 9 = -1/9

• (i) (2/3) × (3/10): (2 × 3) / (3 × 10) = 6/30 = 1/5
• (ii) (7/11) × (5/8): (7 × 5) / (11 × 8) = 35/88
• (iii) (-4/7) × (5/14): (-4 × 5) / (7 × 14) = -20/98 = -10/49

• (i) (2/3) ÷ (3/10): (2/3) × (10/3) = (2 × 10) / (3 × 3) = 20/9
• (ii) (7/11) ÷ (5/8): (7/11) × (8/5) = (7 × 8) / (11 × 5) = 56/55
• (iii) (-4/7) ÷ (5/14): (-4/7) × (14/5) = (-4 × 2) / 5 = -8/5

• Left Hand Side (LHS):
  (1/2 + 3/4) = (2/4 + 3/4) = 5/4
  LHS = (5/4) × (8/3) = 40/12 = 10/3
• Right Hand Side (RHS):
  (1/2 × 8/3) = 8/6 = 4/3
  (3/4 × 8/3) = 24/12 = 2 = 6/3
  RHS = 4/3 + 6/3 = 10/3
• Since LHS = RHS = 10/3, the equality is verified.

Using distributive property: a(b - c) = ab - ac

• = (7/9 × 6/7) - (7/9 × 3/4)
• = 6/9 - 21/36
• = 2/3 - 7/12
• LCM(3, 12) = 12 ⇒ (8 - 7) / 12 = 1/12

1. Apply the distributive law to the left-hand side:
   (5/6)x + (5/6 × 3/5) = (5/6)x + 1/2
2. Simplify the constant term on the left:
   5/6 × 3/5 = 15/30 = 1/2
3. The equation becomes:
   (5/6)x + 1/2 = (5/6)x + 1/2
4. Subtract (5/6)x + 1/2 from both sides:
   0 = 0
5. This is an identity true for all values. Therefore, x can be any rational number (x ∈ Q).

• -5/4 (= -1 1/4): Divide the interval between -2 and -1 into 4 equal segments; mark the point 1 segment to the left of -1 (or 5 units left of 0).
• 2/3: Divide the unit interval between 0 and 1 into 3 equal parts; mark the second tick mark to the right of 0.
• 1 1/2 (= 3/2): Located exactly midway between 1 and 2.

• Convert both fractions to a common denominator:
  -1/2 = -4/8
  1/4 = 2/8
• Pick any three rational numbers whose numerators lie strictly between -4 and 2:
  -1/4 (or -2/8), -1/8, and 0 (or 1/8).

• LCM(4, 12) = 12
• -1/4 = -3/12
• (-3/12) + (5/12) = (-3 + 5) / 12 = 2/12 = 1/6

• Total silk available = 15 3/4 = 63/4 metres
• Silk required per kurta = 2 1/4 = 9/4 metres
• Number of kurtas = (63/4) ÷ (9/4) = (63/4) × (4/9) = 63/9 = 7 kurtas

• Method of Equivalent Fractions: Convert both rational numbers to equivalent fractions with a much larger common denominator, then select any intermediate integer numerator (e.g., between 1/3 = 10/30 and 2/3 = 20/30, choose 11/30, 12/30, etc.).
• Mediant Method: For two positive rational numbers a/b and c/d, the mediant (a + c)/(b + d) always lies strictly between them.
• Decimal Expansion Method: Write both rational numbers as decimals and pick any terminating decimal lying numerically between them.

• 7/20:
  Denominator factorized: 20 = 2² × 5. Since prime factors are only 2 and 5, it is terminating.
  Long division check: 7 ÷ 20 = 0.35
• 4/15:
  Denominator factorized: 15 = 3 × 5. It contains prime factor 3 (other than 2 or 5), so it is repeating.
  Long division check: 4 ÷ 15 = 0.26̅ (0.2666...)
• 13/250:
  Denominator factorized: 250 = 2 × 5³. Since prime factors are only 2 and 5, it is terminating.
  Long division check: 13 ÷ 250 = 0.052

• 1/13: Performing long division gives 1/13 = 0.076923̅ with a 6-digit repeating block: 076923.
• Computing Multiples:
  • 2/13 = 0.153846̅ (repeating block: 153846)
  • 3/13 = 0.230769̅ (cyclic permutation of 076923!)
  • 4/13 = 0.307692̅ (cyclic permutation of 076923!)
  • 9/13 = 0.692307̅ and 10/13 = 0.769230̅ (also cyclic shifts of 076923).
• Observation: The decimal expansion of k/13 splits into two distinct 6-digit cyclic families: one family generated by cyclic shifts of 076923 (for remainders 1, 3, 4, 9, 10, 12) and another generated by cyclic shifts of 153846 (for remainders 2, 5, 6, 7, 8, 11).

• (i) √81: Rational. Since √81 = 9 = 9/1.
• (ii) √12: Irrational. Since √12 = 2√3 and 3 is not a perfect square.
• (iii) 0.33333...: Rational. Non-terminating repeating decimal = 1/3.
• (iv) 0.123451234512345...: Rational. Repeating 5-digit block 12345 = 12345/99999 = 4115/33333.
• (v) 1.01001000100001...: Irrational. The sequence of zeros grows indefinitely, so it is non-terminating and non-repeating.
• (vi) 23.560185612239874790120: Rational. Terminating decimal with 21 decimal places = 23560185612239874790120 / 10²¹ (or 2356018561223987479012 / 10²⁰ in lowest terms).

1. Let x = 0.99999... — (Equation 1)
2. Multiply both sides by 10:
   10x = 9.99999... — (Equation 2)
3. Subtract Equation 1 from Equation 2:
   10x - x = 9.99999... - 0.99999...
   9x = 9
4. Divide both sides by 9:
   x = 9/9 = 1

Thus, 0.9̅ is exactly equal to 1. This demonstrates the non-uniqueness of decimal representations for rational numbers.

Reciprocals 1/n produce full cyclic numbers when n is a full-reptend prime (primes whose decimal periods have maximum possible length n - 1):

• n = 17: 1/17 = 0.0588235294117647̅ (16-digit cyclic number). Multiplying the 16-digit block by 1 through 16 cyclically shifts the same digits.
• n = 19: 1/19 = 0.052631578947368421̅ (18-digit cyclic number).
• n = 23: 1/23 = 0.0434782608695652173913̅ (22-digit cyclic number).
• n = 29: Produces a 28-digit full cyclic repeating block.

• (i) 3/50:
  Dividing 3 by 50 using long division:
  3 ÷ 50 = 0.06 (with remainder 0).
  Since the remainder becomes 0, it is a terminating decimal: 0.06.
• (ii) 2/9:
  Dividing 2 by 9 using long division:
  2 ÷ 9 = 0.2222... where remainder 2 repeats endlessly.
  Since the division loops infinitely, it is a non-terminating and repeating decimal: 0.2̅ (or 0.222...).

Proof by Contradiction:

1. Assume to the contrary that √5 is a rational number.
2. Therefore, it can be expressed in simplest form as √5 = p/q, where p and q are non-zero co-prime integers (sharing no common factor other than 1).
3. Squaring both sides gives: 5 = p² / q² ⇒ p² = 5q².
4. This means p² is divisible by 5, which implies that p must also be divisible by 5 (since 5 is a prime number). Let p = 5k for some integer k.
5. Substitute p = 5k back into the equation:
   (5k)² = 5q² ⇒ 25k² = 5q² ⇒ q² = 5k².
6. This indicates that q² is divisible by 5, which means q must also be divisible by 5.
7. Hence, both p and q have at least 5 as a common factor. This directly contradicts our initial statement that p and q are co-prime.
8. Therefore, our assumption is false, and √5 is an irrational number.

• (i) 0.532:
  Lies strictly between 0 and 1. Locate 0.5 (the midpoint), divide the tenth interval from 0.5 to 0.6 into hundredths to find 0.53, and mark 0.532 just right of 0.53.
• (ii) 1.15̅ (1.1555...):
  In fraction form, 10x = 11.555..., 100x = 115.555... ⇒ 90x = 104 ⇒ x = 104/90 = 52/45 = 1 7/45 ≈ 1.156.
  Lies between 1 and 2, situated slightly past the midpoint of the interval between 1.1 and 1.2 (near 1.16).

To find 6 rational numbers, multiply the numerator and denominator by (6 + 1) = 7:

• 3 = 3 × 7/7 = 21/7
• 4 = 4 × 7/7 = 28/7

Six distinct rational numbers between 21/7 and 28/7 are:

22/7, 23/7, 24/7, 25/7, 26/7, 27/7

Multiply numerator and denominator of both fractions by (5 + 1) = 6:

• 2/5 = (2 × 6) / (5 × 6) = 12/30
• 3/5 = (3 × 6) / (5 × 6) = 18/30

Five distinct rational numbers between 12/30 and 18/30 are:

13/30, 14/30 (or 7/15), 15/30 (or 1/2), 16/30 (or 8/15), 17/30

First find the common denominator of 6 and 5, which is 30:

• 1/6 = 5/30
• 2/5 = 12/30

Since the integer gap between 5 and 12 is large enough (6, 7, 8, 9, 10, 11), five rational numbers are:

6/30 (or 1/5), 7/30, 8/30 (or 4/15), 9/30 (or 3/10), 10/30 (or 1/3)

Step-by-step solution:

1. Combine terms over the common denominator 15:
   (5x + 3x) / 15 = 16/15
2. Simplify the numerator:
   8x / 15 = 16/15
3. Multiply both sides by 15:
   8x = 16
4. Divide by 8:
   x = 16/8 = 2

Solution:

• Given: a + 1/b = 0 ⇒ a = -1/b
• Multiplying both sides by b: ab = -1
• Since -1 is strictly less than 0, the product ab is negative.

• Representation: A number with 4 decimal places can be written as p / 10⁴. If p were divisible by 10, the final 0 would cancel out with 10⁴, leaving fewer than 4 decimal places. Thus, p is not divisible by 10.
• Divisibility by 2⁴ or 5⁴: Yes. In lowest terms, p / (2⁴ × 5⁴) cannot cancel both prime factors 2 and 5 simultaneously because 10 does not divide p. Therefore, at least one of the prime factor powers 2⁴ or 5⁴ must remain fully intact in the denominator.

• Fraction in lowest terms: 18/125 (18 and 125 share no common factors).
• Prime factorization of denominator: 125 = 5³ = 2° × 5³.
• Since the denominator contains only powers of 5, the decimal expansion is terminating.
• The highest power between 2 and 5 is 3 (5³), so it terminates after 3 decimal places (18/125 = 144/1000 = 0.144).

• The denominator is 2³ × 5¹.
• To express the denominator as a power of 10, multiply numerator and denominator by 5²:
  (2³ × 5¹) × 5² = 2³ × 5³ = 10³ = 1000.
• Because the maximum exponent of 2 and 5 is 3, the decimal expansion will terminate after exactly 3 decimal places.

• Equalizing Denominators:
  Choose common denominator m = 24:
  a = 7/12 = 14/24 (k₁ = 14)
  b = 5/6 = 20/24 (k₂ = 20)
  Here, k₂ - k₁ = 20 - 14 = 6 (not strictly > 6). Let's take m = 36:
  a = 21/36 (k₁ = 21)
  b = 30/36 (k₂ = 30)
  Now, k₂ - k₁ = 30 - 21 = 9 > 6.
• Five rational numbers between them:
  22/36, 23/36, 24/36, 25/36, 26/36
• Explanation: Between two integers k₁ and k₂, the number of strictly intermediate integers is (k₂ - k₁ - 1). For this count to be at least n, we must have k₂ - k₁ - 1 ≥ n ⇒ k₂ - k₁ > n (or ≥ n + 1).

Proof:

1. Consider the algebraic identity:
   (x + y + z)² = x² + y² + z² + 2(xy + yz + zx)
2. Substitute the given values x + y + z = 0 and xy + yz + zx = 0:
   0² = x² + y² + z² + 2(0)
   x² + y² + z² = 0
3. Since x, y, z are rational numbers, their squares x², y², z² must all be non-negative real values (≥ 0).
4. The sum of non-negative terms can only equal 0 if each individual term is 0:
   x² = 0 ⇒ x = 0
   y² = 0 ⇒ y = 0
   z² = 0 ⇒ z = 0
5. Thus, x, y, and z must be simultaneously zero.

Let a < b:

1. Add a to both sides of the inequality:
   a + a < a + b ⇒ 2a < a + b ⇒ a < (a + b)/2
2. Add b to both sides of the original inequality:
   a + b < b + b ⇒ a + b < 2b ⇒ (a + b)/2 < b
3. Combining both results gives:
   a < (a + b)/2 < b

Hence, the average (a + b)/2 always lies strictly between a and b.