Chapter 3: The World of Numbers

Solution

Given information:

• 2 bags of spices → 15 copper ingots
• 12 bags of spices → ? copper ingots

Step 1: Find the ratio (rate of exchange)

For every 2 bags, the merchant gets 15 ingots.

$\text{Rate} = \frac{15 \text{ ingots}}{2 \text{ bags}}$

Step 2: Set up the proportion

$\frac{15}{2} = \frac{x}{12}$

Step 3: Solve for x

$x = \frac{15 \times 12}{2} = \frac{180}{2} = 90$

Answer: The merchant will leave with 90 copper ingots.

Solution

Step 1: Analyse the given sequence

The numbers are: 11, 13, 17, 19

Step 2: Identify what they have in common

All four numbers are prime numbers — they are divisible only by 1 and themselves.

• 11 → divisible only by 1 and 11 ✓
• 13 → divisible only by 1 and 13 ✓
• 17 → divisible only by 1 and 17 ✓
• 19 → divisible only by 1 and 19 ✓

Step 3: Find the next three prime numbers after 19

Check numbers after 19:

• 20 = 4 × 5 → Not prime
• 21 = 3 × 7 → Not prime
• 22 = 2 × 11 → Not prime
• 23 → divisible only by 1 and 23 ✓ Prime
• 24, 25, 26, 27, 28 → Not prime
• 29 → divisible only by 1 and 29 ✓ Prime
• 30 → Not prime
• 31 → divisible only by 1 and 31 ✓ Prime

Answer:

• These numbers are all prime numbers.
• The next three numbers in the pattern are: 23, 29, 31.

Solution

Key Definition: A set is closed under an operation if performing that operation on any two members of the set always produces a result that is also in the same set.

Natural Numbers: N = {1, 2, 3, 4, 5, ...}

Claim: Natural Numbers are NOT closed under subtraction.

Step 1: Test with examples where the result is NOT a natural number

Example 1: $5 - 8 = -3$ Here, 5 and 8 are both natural numbers, but −3 is NOT a natural number (it is negative). This breaks closure.

Example 2: $3 - 3 = 0$ Here, 3 and 3 are both natural numbers, but 0 is NOT a natural number (Natural numbers start from 1). This also breaks closure.

Step 2: Conclusion

Since subtracting two natural numbers can result in a negative number or zero — neither of which belongs to the set of natural numbers — we conclude:

Natural Numbers are NOT closed under subtraction.

Note: This is one of the reasons mathematicians extended the number system to include zero and negative integers!

Solution

Step 1: Count the joints on one hand (excluding the thumb)

The thumb is used as the pointer to count. The remaining four fingers are used for counting:

Finger	Number of Joints
Index finger	3
Middle finger	3
Ring finger	3
Little finger	3
Total	12

Step 2: Total count on one hand

$4 \text{ fingers} \times 3 \text{ joints per finger} = 12 \text{ joints}$

So, using one hand, you can count up to 12.

Step 3: Relation to Base-12 (Duodecimal) Counting System

• In a base-12 system, numbers are grouped in twelves (just as our modern system groups in tens).
• Because the human hand naturally allows counting to 12 using finger joints, ancient civilisations found it convenient to use 12 as their base.
• This is why we still see traces of base-12 in everyday life:
  • 12 months in a year
  • 12 hours on a clock face
  • 12 inches in a foot
  • Items sold by the dozen (12)

Answer: You can count 12 numbers on one hand using finger joints, and this natural grouping of 12 is the origin of the ancient base-12 (duodecimal) counting system.

Solution

Given:

• Temperature at noon = 4 °C
• Temperature drop by midnight = 15 °C

Setting up the equation:

A drop in temperature means we subtract:

$\text{Midnight temperature} = 4 - 15$

Calculating:

$4 - 15 = -(15 - 4) = -11$

Answer: The midnight temperature in Ladakh is –11 °C.

Solution

Representing each event as an integer:

• A debt (loan) is a negative amount: –850
• A profit (fortune) is a positive amount: +1200
• A loss is a negative amount: –450

Writing the equation:

$\text{Final Standing} = (-850) + (+1200) + (-450)$

Step-by-step calculation:

• First, add the positive value: $(-850) + 1200 = 350$
• Then, add the loss: $350 + (-450) = 350 - 450 = -100$

Answer: The trader's final financial standing is –₹100, meaning he is in a debt of ₹100.

Solution

Using Brahmagupta's laws for integers:

• Negative × Positive = Negative
• Negative × Negative = Positive
• Subtracting a negative = Adding a positive
• Negative ÷ Positive = Negative

---

(i) (–12) × 5

Rule: Negative × Positive = Negative

$(-12) \times 5 = -(12 \times 5) = -60$

Answer: –60

---

(ii) (–8) × (–7)

Rule: Negative × Negative = Positive

$(-8) \times (-7) = +(8 \times 7) = +56$

Answer: 56

---

(iii) 0 – (–14)

Rule: Subtracting a negative number is the same as adding its positive:

$0 - (-14) = 0 + 14 = 14$

Answer: 14

---

(iv) (–20) ÷ 4

Rule: Negative ÷ Positive = Negative

$(-20) \div 4 = -(20 \div 4) = -5$

Answer: –5

Solution

Real-World Example Using Debt:

Scenario: Suppose you have ₹10 in your pocket. Your friend owes you a debt of ₹5 (i.e., you have a record of –₹5 owed to you by your friend, which counts as a negative liability on your friend's side).

Now, if someone removes (subtracts) that debt of ₹5 from your records — meaning the debt is cancelled and your friend pays you back — you gain that ₹5.

Mathematically:

$10 - (-5) = 10 + 5 = 15$

Why does this work?

• The negative number (–5) represents a debt or a reduction.
• Subtracting a debt means the debt is being taken away, which is the same as gaining money.
• Removing a negative effect results in a positive outcome.

Brahmagupta's Rule:

Brahmagupta stated: "A negative subtracted from a positive gives a positive."

$10 - (-5) = 10 + 5 = \mathbf{15}$

Conclusion: Subtracting a negative number is the same as adding its positive counterpart, because removing a loss is equivalent to making a gain.

To prove: 2/3 = 4/6

Method: Two rational numbers p/q and r/s are equal if p × s = q × r (cross-multiplication).

Step 1: Cross-multiply both sides.

• Left side: 2 × 6 = 12
• Right side: 3 × 4 = 12

Step 2: Since 2 × 6 = 3 × 4 = 12, we conclude:

$\frac{2}{3} = \frac{4}{6}$

Alternatively: Simplify 4/6 by dividing numerator and denominator by 2: $\frac{4}{6} = \frac{4 \div 2}{6 \div 2} = \frac{2}{3}$

Hence proved. ✓

To prove: 5/4 = 10/8

Method: Cross-multiplication.

Step 1: Cross-multiply:

• Left side: 5 × 8 = 40
• Right side: 4 × 10 = 40

Step 2: Since both products are equal: $\frac{5}{4} = \frac{10}{8}$

Alternatively: Simplify 10/8: $\frac{10}{8} = \frac{10 \div 2}{8 \div 2} = \frac{5}{4}$

Hence proved. ✓

To prove: −3/5 = −6/10

Method: Cross-multiplication.

Step 1: Cross-multiply:

• Left side: (−3) × 10 = −30
• Right side: 5 × (−6) = −30

Step 2: Since both products are equal: $\frac{-3}{5} = \frac{-6}{10}$

Alternatively: Simplify −6/10: $\frac{-6}{10} = \frac{-6 \div 2}{10 \div 2} = \frac{-3}{5}$

Hence proved. ✓

To prove: 9/3 = 3

Method: Simplify the fraction.

Step 1: Write 3 as 3/1.

Step 2: Cross-multiply:

• Left side: 9 × 1 = 9
• Right side: 3 × 3 = 9

Step 3: Since both products are equal: $\frac{9}{3} = 3$

Alternatively: Divide numerator by denominator: $\frac{9}{3} = 9 \div 3 = 3$

Hence proved. ✓

Step 1: Find the LCM of denominators 5 and 10. $\text{LCM}(5, 10) = 10$

Step 2: Convert each fraction to have denominator 10. $\frac{2}{5} = \frac{2 \times 2}{5 \times 2} = \frac{4}{10}$ $\frac{3}{10} = \frac{3}{10}$

Step 3: Add the fractions. $\frac{4}{10} + \frac{3}{10} = \frac{4 + 3}{10} = \frac{7}{10}$

$\boxed{\frac{2}{5} + \frac{3}{10} = \frac{7}{10}}$

Step 1: Find the LCM of denominators 12 and 8.

• 12 = 2² × 3
• 8 = 2³ $\text{LCM}(12, 8) = 2^3 \times 3 = 24$

Step 2: Convert each fraction to have denominator 24. $\frac{7}{12} = \frac{7 \times 2}{12 \times 2} = \frac{14}{24}$ $\frac{5}{8} = \frac{5 \times 3}{8 \times 3} = \frac{15}{24}$

Step 3: Add the fractions. $\frac{14}{24} + \frac{15}{24} = \frac{14 + 15}{24} = \frac{29}{24}$

$\boxed{\frac{7}{12} + \frac{5}{8} = \frac{29}{24}}$

Step 1: Find the LCM of denominators 7 and 14. $\text{LCM}(7, 14) = 14$

Step 2: Convert each fraction to have denominator 14. $\frac{-4}{7} = \frac{-4 \times 2}{7 \times 2} = \frac{-8}{14}$ $\frac{3}{14} = \frac{3}{14}$

Step 3: Add the fractions. $\frac{-8}{14} + \frac{3}{14} = \frac{-8 + 3}{14} = \frac{-5}{14}$

$\boxed{\frac{-4}{7} + \frac{3}{14} = \frac{-5}{14}}$

Step 1: Find the LCM of denominators 6 and 4.

• 6 = 2 × 3
• 4 = 2² $\text{LCM}(6, 4) = 2^2 \times 3 = 12$

Step 2: Convert each fraction to have denominator 12. $\frac{5}{6} = \frac{5 \times 2}{6 \times 2} = \frac{10}{12}$ $\frac{1}{4} = \frac{1 \times 3}{4 \times 3} = \frac{3}{12}$

Step 3: Subtract. $\frac{10}{12} - \frac{3}{12} = \frac{10 - 3}{12} = \frac{7}{12}$

$\boxed{\frac{5}{6} - \frac{1}{4} = \frac{7}{12}}$

Step 1: Find the LCM of denominators 8 and 4. $\text{LCM}(8, 4) = 8$

Step 2: Convert each fraction to have denominator 8. $\frac{11}{8} = \frac{11}{8}$ $\frac{3}{4} = \frac{3 \times 2}{4 \times 2} = \frac{6}{8}$

Step 3: Subtract. $\frac{11}{8} - \frac{6}{8} = \frac{11 - 6}{8} = \frac{5}{8}$

$\boxed{\frac{11}{8} - \frac{3}{4} = \frac{5}{8}}$

Step 1: Rewrite the expression by removing double negative. $\frac{-7}{9} - \left(\frac{-2}{3}\right) = \frac{-7}{9} + \frac{2}{3}$

Step 2: Find the LCM of denominators 9 and 3. $\text{LCM}(9, 3) = 9$

Step 3: Convert each fraction to have denominator 9. $\frac{-7}{9} = \frac{-7}{9}$ $\frac{2}{3} = \frac{2 \times 3}{3 \times 3} = \frac{6}{9}$

Step 4: Add. $\frac{-7}{9} + \frac{6}{9} = \frac{-7 + 6}{9} = \frac{-1}{9}$

$\boxed{\frac{-7}{9} - \left(\frac{-2}{3}\right) = \frac{-1}{9}}$

Step 1: Multiply the numerators and denominators. $\frac{2}{3} \times \frac{3}{10} = \frac{2 \times 3}{3 \times 10} = \frac{6}{30}$

Step 2: Simplify by dividing numerator and denominator by their GCD (= 6). $\frac{6}{30} = \frac{6 \div 6}{30 \div 6} = \frac{1}{5}$

$\boxed{\frac{2}{3} \times \frac{3}{10} = \frac{1}{5}}$

Step 1: Multiply the numerators and denominators. $\frac{7}{11} \times \frac{5}{8} = \frac{7 \times 5}{11 \times 8} = \frac{35}{88}$

Step 2: Check if the fraction can be simplified. GCD(35, 88) = 1 (since 35 = 5 × 7 and 88 = 8 × 11, no common factors).

So the fraction is already in its simplest form.

$\boxed{\frac{7}{11} \times \frac{5}{8} = \frac{35}{88}}$

Step 1: Multiply the numerators and denominators. $\frac{-4}{7} \times \frac{5}{14} = \frac{(-4) \times 5}{7 \times 14} = \frac{-20}{98}$

Step 2: Simplify by dividing by GCD(20, 98) = 2. $\frac{-20}{98} = \frac{-20 \div 2}{98 \div 2} = \frac{-10}{49}$

$\boxed{\frac{-4}{7} \times \frac{5}{14} = \frac{-10}{49}}$

Step 1: To divide by a fraction, multiply by its reciprocal. $\frac{2}{3} \div \frac{3}{10} = \frac{2}{3} \times \frac{10}{3}$

Step 2: Multiply numerators and denominators. $= \frac{2 \times 10}{3 \times 3} = \frac{20}{9}$

Step 3: The fraction 20/9 cannot be simplified further (GCD = 1).

$\boxed{\frac{2}{3} \div \frac{3}{10} = \frac{20}{9}}$

Step 1: Multiply by the reciprocal of 5/8. $\frac{7}{11} \div \frac{5}{8} = \frac{7}{11} \times \frac{8}{5}$

Step 2: Multiply numerators and denominators. $= \frac{7 \times 8}{11 \times 5} = \frac{56}{55}$

Step 3: GCD(56, 55) = 1, so the fraction is already in simplest form.

$\boxed{\frac{7}{11} \div \frac{5}{8} = \frac{56}{55}}$

Step 1: Multiply by the reciprocal of 5/14. $\frac{-4}{7} \div \frac{5}{14} = \frac{-4}{7} \times \frac{14}{5}$

Step 2: Multiply numerators and denominators. $= \frac{(-4) \times 14}{7 \times 5} = \frac{-56}{35}$

Step 3: Simplify by dividing by GCD(56, 35) = 7. $\frac{-56}{35} = \frac{-56 \div 7}{35 \div 7} = \frac{-8}{5}$

$\boxed{\frac{-4}{7} \div \frac{5}{14} = \frac{-8}{5}}$

This verifies the Distributive Property of multiplication over addition for rational numbers.

Left Hand Side (LHS):

Step 1: Compute 1/2 + 3/4. $\frac{1}{2} + \frac{3}{4} = \frac{2}{4} + \frac{3}{4} = \frac{5}{4}$

Step 2: Multiply by 8/3. $\frac{5}{4} \times \frac{8}{3} = \frac{5 \times 8}{4 \times 3} = \frac{40}{12} = \frac{10}{3}$

∴ LHS = 10/3

---

Right Hand Side (RHS):

Step 3: Compute 1/2 × 8/3. $\frac{1}{2} \times \frac{8}{3} = \frac{8}{6} = \frac{4}{3}$

Step 4: Compute 3/4 × 8/3. $\frac{3}{4} \times \frac{8}{3} = \frac{24}{12} = 2$

Step 5: Add the results. $\frac{4}{3} + 2 = \frac{4}{3} + \frac{6}{3} = \frac{10}{3}$

∴ RHS = 10/3

---

Since LHS = RHS = 10/3, the statement is proved. ✓

This demonstrates the distributive property: a × (b + c) = a × b + a × c.

Using the Distributive Property: a × (b − c) = a × b − a × c

Here, a = 7/9, b = 6/7, c = 3/4.

Step 1: Distribute 7/9 over the bracket. $\frac{7}{9} \times \left(\frac{6}{7} - \frac{3}{4}\right) = \frac{7}{9} \times \frac{6}{7} - \frac{7}{9} \times \frac{3}{4}$

Step 2: Compute the first term. $\frac{7}{9} \times \frac{6}{7} = \frac{7 \times 6}{9 \times 7} = \frac{42}{63} = \frac{2}{3}$

Step 3: Compute the second term. $\frac{7}{9} \times \frac{3}{4} = \frac{7 \times 3}{9 \times 4} = \frac{21}{36} = \frac{7}{12}$

Step 4: Subtract. $\frac{2}{3} - \frac{7}{12}$

LCM(3, 12) = 12 $= \frac{8}{12} - \frac{7}{12} = \frac{1}{12}$

$\boxed{\frac{7}{9} \left(\frac{6}{7} - \frac{3}{4}\right) = \frac{1}{12}}$

Given equation: $\frac{5}{6}\left(x + \frac{3}{5}\right) = \frac{5}{6} \cdot x + \frac{1}{2}$

Step 1: Expand the left side using the distributive property. $\frac{5}{6} \cdot x + \frac{5}{6} \times \frac{3}{5} = \frac{5}{6} \cdot x + \frac{1}{2}$

Step 2: Compute 5/6 × 3/5. $\frac{5}{6} \times \frac{3}{5} = \frac{15}{30} = \frac{1}{2}$

Step 3: Substitute back. $\frac{5}{6}x + \frac{1}{2} = \frac{5}{6}x + \frac{1}{2}$

Observation: The equation becomes an identity — it holds true for all rational values of x.

This is because the distributive property guarantees: $\frac{5}{6}\left(x + \frac{3}{5}\right) = \frac{5}{6}x + \frac{5}{6} \times \frac{3}{5} = \frac{5}{6}x + \frac{1}{2}$

which matches the RHS for every value of x.

$\boxed{x \text{ can be any rational number.}}$

Representing Rational Numbers on a Number Line

Step 1: Convert all numbers to decimal form for easy placement.

• 2/3 = 0.666... ≈ 0.67 (lies between 0 and 1, closer to 1)
• -5/4 = -1.25 (lies between -2 and -1, closer to -1)
• 1(1/2) = 3/2 = 1.5 (lies between 1 and 2)

Step 2: Draw a number line and mark equal intervals.

Draw a straight line and mark points: ..., -2, -1, 0, 1, 2, ...

Step 3: Locate each number.

• -5/4 = -1.25: Start at -1, move one-quarter unit to the left → mark the point between -1 and -2, one-quarter of the way from -1.
• 2/3 ≈ 0.67: Divide the segment from 0 to 1 into 3 equal parts; mark the second division point from 0.
• 1(1/2) = 1.5: Mark the midpoint between 1 and 2.

Number line representation:

• -5/4 is located 1.25 units to the left of 0
• 2/3 is located approximately 0.67 units to the right of 0
• 3/2 (= 1½) is located 1.5 units to the right of 0

Finding Three Rational Numbers Between -1/2 and 1/4

Method: Convert to equivalent fractions with a common denominator, then find numbers in between.

Step 1: Express both numbers with a common denominator.

$-\frac{1}{2} = -\frac{4}{8} \quad \text{and} \quad \frac{1}{4} = \frac{2}{8}$

Step 2: Identify integers strictly between -4 and 2 (in eighths).

The integers strictly between -4 and 2 are: -3, -2, -1, 0, 1

Step 3: Pick any three values.

$-\frac{3}{8}, \quad -\frac{1}{8}, \quad \frac{1}{8}$

Verification:

$-\frac{1}{2} = -\frac{4}{8} < -\frac{3}{8} < -\frac{1}{8} < \frac{1}{8} < \frac{2}{8} = \frac{1}{4} ✓$

Answer: Three rational numbers strictly between $-\dfrac{1}{2}$ and $\dfrac{1}{4}$ are:

$\boxed{-\frac{3}{8}, \quad -\frac{1}{8}, \quad \frac{1}{8}}$

Most likely intended interpretation: $-\dfrac{1}{4} + \dfrac{5}{12}$

Step 1: Find the LCM of 4 and 12.

$\text{LCM}(4, 12) = 12$

Step 2: Convert -1/4 to an equivalent fraction with denominator 12.

$-\frac{1}{4} = -\frac{3}{12}$

Step 3: Add the fractions.

$-\frac{3}{12} + \frac{5}{12} = \frac{-3 + 5}{12} = \frac{2}{12} = \frac{1}{6}$

Answer: $\boxed{\dfrac{1}{6}}$

Solving the Kurta Problem

Step 1: Convert the mixed numbers to improper fractions.

$15\frac{3}{4} = \frac{15 \times 4 + 3}{4} = \frac{60 + 3}{4} = \frac{63}{4} \text{ metres}$

$2\frac{1}{4} = \frac{2 \times 4 + 1}{4} = \frac{8 + 1}{4} = \frac{9}{4} \text{ metres}$

Step 2: Divide total silk by silk required per kurta.

$\text{Number of kurtas} = \frac{63}{4} \div \frac{9}{4}$

Step 3: Apply the division rule (multiply by the reciprocal).

$= \frac{63}{4} \times \frac{4}{9} = \frac{63 \times 4}{4 \times 9} = \frac{63}{9} = 7$

Answer: The tailor can make exactly $\boxed{7}$ kurtas.

Finding Three Rational Numbers Between 3.1415 and 3.1416

Step 1: Observe the gap between the two numbers.

$3.1416 - 3.1415 = 0.0001$

The two numbers differ in the 4th decimal place. We need to find numbers strictly between them.

Step 2: Express with more decimal places to identify numbers in between.

$3.1415 = 3.14150000...$ $3.1416 = 3.14160000...$

Any number of the form $3.1415x...$ where $x$ is a digit from 1 to 9 will lie strictly between them.

Step 3: Choose three such rational numbers.

$3.14151, \quad 3.14153, \quad 3.14157$

Step 4: Express them as fractions (to confirm they are rational).

$3.14151 = \frac{314151}{100000}, \quad 3.14153 = \frac{314153}{100000}, \quad 3.14157 = \frac{314157}{100000}$

Step 5: Verify the ordering.

$3.1415 < 3.14151 < 3.14153 < 3.14157 < 3.1416 ✓$

Answer: Three rational numbers between 3.1415 and 3.1416 are: $\boxed{3.14151, \quad 3.14153, \quad 3.14157}$

Other Methods to Find a Rational Number Between Two Rational Numbers

Yes! There are several elegant methods to find a rational number between any two rational numbers $a$ and $b$ (where $a < b$).

---

Method 1: The Arithmetic Mean (Average) Method

The simplest and most powerful method: the average of two rational numbers always lies strictly between them.

$\text{Rational number between } a \text{ and } b = \frac{a + b}{2}$

Example: Find a rational number between $\dfrac{1}{3}$ and $\dfrac{1}{2}$.

$\frac{\frac{1}{3} + \frac{1}{2}}{2} = \frac{\frac{2+3}{6}}{2} = \frac{5}{6} \times \frac{1}{2} = \frac{5}{12}$

✓ Indeed, $\dfrac{1}{3} < \dfrac{5}{12} < \dfrac{1}{2}$

Why it works: If $a < b$, then $a < \dfrac{a+b}{2} < b$ always holds.

---

Method 2: The Common Denominator Method

• Convert both fractions to a common denominator.
• Any fraction with the same denominator whose numerator lies strictly between the two numerators is a valid answer.
• If no integer lies between the numerators, multiply both numerator and denominator by 10 (or any integer) to create more room.

Example: Between $\dfrac{1}{4}$ and $\dfrac{1}{3}$:

$\frac{1}{4} = \frac{3}{12}, \quad \frac{1}{3} = \frac{4}{12}$

No integer between 3 and 4, so multiply by 10:

$\frac{30}{120} < \frac{31}{120} < \frac{40}{120}$

✓ So $\dfrac{31}{120}$ lies between $\dfrac{1}{4}$ and $\dfrac{1}{3}$.

---

Method 3: The Decimal Expansion Method

• Write both numbers in decimal form.
• Find a decimal that lies strictly between them.
• Convert back to fraction form $\dfrac{p}{q}$.

Example: Between $0.5$ and $0.6$, choose $0.55 = \dfrac{55}{100} = \dfrac{11}{20}$.

---

Key Insight: Density of Rational Numbers

Between any two rational numbers, there are infinitely many rational numbers. This property is called the density of rational numbers. We can apply the average method repeatedly to generate as many rational numbers as we wish between any two given rationals.

Determining Terminating vs Repeating Decimals

Rule: A rational number p/q (in lowest terms) has a terminating decimal if and only if the denominator q has no prime factors other than 2 and 5.

---

Step 1: Check each denominator

Fraction 7/20:

• Denominator = 20 = 2² × 5
• Prime factors are only 2 and 5
• ✅ Terminating decimal

Fraction 4/15:

• Denominator = 15 = 3 × 5
• Prime factors include 3 (not just 2 and 5)
• 🔁 Repeating decimal

Fraction 13/250:

• Denominator = 250 = 2 × 5³
• Prime factors are only 2 and 5
• ✅ Terminating decimal

---

Step 2: Verify by Long Division

7/20:

7 ÷ 20:
  70 ÷ 20 = 3 remainder 10
  100 ÷ 20 = 5 remainder 0

7/20 = 0.35 ✅ (Terminating)

4/15:

4 ÷ 15:
  40 ÷ 15 = 2 remainder 10
  100 ÷ 15 = 6 remainder 10
  100 ÷ 15 = 6 remainder 10  (repeats)

4/15 = 0.2666... = 0.2̄6̄ 🔁 (Repeating)

13/250:

13 ÷ 250:
  130 ÷ 250 = 0 remainder 130
  1300 ÷ 250 = 5 remainder 50
  500 ÷ 250 = 2 remainder 0

13/250 = 0.052 ✅ (Terminating)

---

Summary Table

Fraction	Denominator Factors	Type	Decimal
7/20	2² × 5	Terminating	0.35
4/15	3 × 5	Repeating	0.2̄6̄
13/250	2 × 5³	Terminating	0.052

Long Division of 1/13 and Cyclic Properties

Step 1: Long Division of 1/13

1 ÷ 13:
  10 ÷ 13 = 0 r 10
  100 ÷ 13 = 7 r 9
  90 ÷ 13 = 6 r 12
  120 ÷ 13 = 9 r 3
  30 ÷ 13 = 2 r 4
  40 ÷ 13 = 3 r 1
  10 ÷ 13 = 0 r 10  ← repeats!

1/13 = 0.076923076923... = 0.̄0̄7̄6̄9̄2̄3̄

The repeating block is 076923 (6 digits long).

---

Step 2: Compute multiples and observe cyclic pattern

Fraction	Decimal Value	Repeating Block
1/13	0.076923...	076923
2/13	0.153846...	153846
3/13	0.230769...	230769
4/13	0.307692...	307692
5/13	0.384615...	384615
6/13	0.461538...	461538
7/13	0.538461...	538461
8/13	0.615384...	615384
9/13	0.692307...	692307
10/13	0.769230...	769230
11/13	0.846153...	846153
12/13	0.923076...	923076

---

Step 3: Observations

• All fractions k/13 (for k = 1 to 12) use the same six digits: 0, 7, 6, 9, 2, 3 arranged cyclically.
• Each fraction starts the repeating block at a different position in the cycle.
• The repeating block length is 6, which equals (13 − 1) = 12 is not the case; rather the block length 6 divides 12.
• 13 is called a full-reptend prime with period 6 (since 10⁶ ≡ 1 mod 13).
• This is a cyclic number property: the repeating block of 1/13 generates all other repeating blocks by cyclic rotation.

Classifying Numbers as Rational or Irrational

---

(i) √81

• 81 = 9² → √81 = 9
• 9 is a whole number, hence a rational number.
• Rational ✅
• Explicit fraction: 9/1 = 9

---

(ii) √12

• 12 = 4 × 3 → √12 = 2√3
• √3 is irrational (3 is not a perfect square)
• 2√3 cannot be expressed as p/q
• Irrational ❌

---

(iii) 0.33333...

• This is a repeating decimal (block: 3)
• Let x = 0.3333...
• 10x = 3.3333...
• 10x − x = 3 → 9x = 3 → x = 3/9 = 1/3
• Rational ✅
• Explicit fraction: 1/3

---

(iv) 0.123451234512345...

• The block 12345 repeats endlessly.
• This is a repeating decimal.
• Let x = 0.̄1̄2̄3̄4̄5̄
• 100000x = 12345.123451234512345...
• 100000x − x = 12345
• 99999x = 12345
• x = 12345/99999
• Simplify: GCD(12345, 99999) = 3 → 4115/33333
• Rational ✅
• Explicit fraction: 4115/33333

---

(v) 1.01001000100001...

• The pattern adds one more zero between each pair of 1s: 1.0101000100001...
• The block of digits is never the same repeating block — it keeps changing (the gap between 1s increases by 1 each time).
• This decimal is non-terminating and non-repeating.
• Irrational ❌

---

(vi) 23.560185612239874790120

• This is a terminating decimal (it ends after finitely many digits).
• Every terminating decimal is rational.
• Rational ✅
• Explicit fraction:
  • 23.560185612239874790120 = 23560185612239874790120 / 10²¹
  • Simplify by dividing by GCD (which is at least 10, since last digit is 0):
  • = 2356018561223987479012 / 10²⁰
  • (Further simplification depends on common factors, but it is a valid rational number.)

---

Summary Table

Number	Type	Reason
√81	Rational (= 9)	Perfect square
√12	Irrational	12 not a perfect square
0.3333...	Rational (= 1/3)	Repeating decimal
0.12345̄	Rational (= 4115/33333)	Repeating block
1.01001000100001...	Irrational	Non-repeating pattern
23.5601...120	Rational	Terminating decimal

Proving that 0.9̄ = 1

Step 1: Set up the equation

Let x = 0.99999... = 0.9̄

---

Step 2: Multiply both sides by 10

10x = 9.99999...

Notice that 9.99999... = 9 + 0.99999... = 9 + x

---

Step 3: Subtract x from both sides

$10x = 9 + x$ $10x - x = 9$ $9x = 9$ $x = 1$

---

Step 4: Conclusion

Since we set x = 0.9̄ and found x = 1:

0.9̄ = 0.99999... = 1 exactly

---

Why this makes sense

• The difference between 1 and 0.9̄ would be 0.000...0001 with infinitely many zeros before the 1.
• But infinitely many zeros means this difference is exactly 0.
• Therefore 0.9̄ and 1 are not approximately equal — they are exactly the same number.
• This illustrates the non-uniqueness of decimal representations: every terminating decimal has an alternative form ending in repeating 9s.
  • Example: 1.000... = 0.999...
  • Example: 2.47000... = 2.46999...

Finding Numbers with Cyclic Reciprocals

Background: What is a cyclic number?

A cyclic number arises when 1/n has a repeating block of length (n − 1). The multiples 1/n, 2/n, ..., (n−1)/n all use the same digits in the repeating block, just starting at different positions (cyclic rotation).

These arise from full-reptend primes — primes p where 10 is a primitive root modulo p.

---

Known full-reptend primes (cyclic reciprocals):

n = 7:

• 1/7 = 0.̄1̄4̄2̄8̄5̄7̄ (block length 6 = 7−1)
• Cyclic number: 142857
• 2/7 = 0.285714..., 3/7 = 0.428571..., etc. — all cyclic rotations ✅

n = 17:

• 1/17 = 0.0588235294117647... (block length 16 = 17−1)
• Cyclic number: 0588235294117647
• All fractions k/17 (k = 1 to 16) are cyclic rotations ✅

n = 19:

• 1/19 = 0.052631578947368421... (block length 18 = 19−1)
• All fractions k/19 are cyclic rotations ✅

n = 23:

• 1/23 has block length 22 = 23−1 → full-reptend ✅

n = 29:

• 1/29 has block length 28 = 29−1 ✅

---

Why these work: Mathematical Reason

For a prime p, the decimal expansion of 1/p has period length equal to the order of 10 modulo p (i.e., the smallest k such that 10^k ≡ 1 mod p).

If this order equals p − 1, then p is a full-reptend prime and 1/p produces a cyclic number.

---

Counter-examples (non-cyclic primes):

n	Block length	n − 1	Cyclic?
7	6	6	✅ Yes
11	2	10	❌ No
13	6	12	❌ No (partial)
17	16	16	✅ Yes
19	18	18	✅ Yes
23	22	22	✅ Yes

---

Conclusion

The numbers n = 7, 17, 19, 23, 29, 47, 59, 61, 97, ... are full-reptend primes whose reciprocals produce cyclic numbers. Whether there are infinitely many such primes is an open question in mathematics!

Solution

(i) $\dfrac{3}{50}$

Perform long division of 3 ÷ 50:

• 3.0000 ÷ 50
• 50 goes into 300 → 6 times (50 × 6 = 300), remainder = 0

$\frac{3}{50} = 0.06$

This is a terminating decimal.

---

(ii) $\dfrac{2}{9}$

Perform long division of 2 ÷ 9:

• 2.000... ÷ 9
• 9 goes into 20 → 2 times (9 × 2 = 18), remainder = 2
• 9 goes into 20 again → 2 times, remainder = 2 (repeats)

$\frac{2}{9} = 0.\overline{2} = 0.2222...$

This is a non-terminating repeating decimal.

Proof that √5 is Irrational

We use proof by contradiction.

Assume that √5 is rational.

Then we can write: $\sqrt{5} = \frac{p}{q}$ where $p$ and $q$ are integers, $q \neq 0$, and $\frac{p}{q}$ is in its lowest form (i.e., $\gcd(p, q) = 1$).

Step 1: Square both sides: $5 = \frac{p^2}{q^2}$ $\Rightarrow p^2 = 5q^2$

Step 2: This means $p^2$ is divisible by 5.

Since 5 is prime, $p$ must also be divisible by 5.

So let $p = 5k$ for some integer $k$.

Step 3: Substitute back: $(5k)^2 = 5q^2$ $25k^2 = 5q^2$ $q^2 = 5k^2$

Step 4: This means $q^2$ is divisible by 5, so $q$ is also divisible by 5.

Step 5: But now both $p$ and $q$ are divisible by 5, which contradicts our assumption that $\gcd(p, q) = 1$.

Conclusion: Our assumption was wrong. Therefore, $\sqrt{5}$ is irrational. $\blacksquare$

Solution

(i) 12.6

$12.6 = \frac{126}{10} = \frac{63}{5}$

---

(ii) 0.0120

$0.0120 = \frac{120}{10000} = \frac{12}{1000} = \frac{3}{250}$

---

(iii) 3.052

$3.052 = \frac{3052}{1000} = \frac{763}{250}$

---

(iv) 1.23̄5̄ (i.e., 1.2353535...)

Let $x = 1.2\overline{35}$

• $10x = 12.\overline{35}$
• $1000x = 1235.\overline{35}$

Subtract: $1000x - 10x = 1235.\overline{35} - 12.\overline{35}$ $990x = 1223$ $x = \frac{1223}{990}$

---

(v) 0.2̄3̄ (i.e., 0.232323...)

Let $x = 0.\overline{23}$

• $100x = 23.\overline{23}$

Subtract: $100x - x = 23$ $99x = 23$ $x = \frac{23}{99}$

---

(vi) 2.0̄5̄ (i.e., 2.0505050... — interpreting as 2.0̄5̄)

Let $x = 2.0\overline{5}$ (i.e., 2.0555...)

• $10x = 20.\overline{5}$
• $100x = 205.\overline{5}$

Subtract: $100x - 10x = 205 - 20 = 185$ $90x = 185$ $x = \frac{185}{90} = \frac{37}{18}$

---

(vii) 2.125

$2.125 = \frac{2125}{1000} = \frac{17}{8}$

---

(viii) 3.125

$3.125 = \frac{3125}{1000} = \frac{25}{8}$

---

(ix) 2.1625

$2.1625 = \frac{21625}{10000} = \frac{865}{400} = \frac{173}{80}$

Solution

(i) Locating 0.532 on the number line

• 0.532 lies between 0 and 1.
• Divide the segment [0, 1] into 10 equal parts; 0.532 lies between 0.5 and 0.6.
• Further divide [0.5, 0.6] into 10 parts; 0.532 lies between 0.53 and 0.54.
• Further divide [0.53, 0.54] into 10 parts; 0.532 is the 2nd mark from 0.53.

Using successive magnification: $0.532 = \frac{532}{1000}$ Mark it as the point that is 532 thousandths of the way from 0 to 1.

---

(ii) Locating 1.1̄5̄ (1.151515...) on the number line

First convert to p/q form:

Let $x = 1.\overline{15}$

$100x = 115.\overline{15}$

$100x - x = 115 - 1 = 114$

$99x = 114 \Rightarrow x = \dfrac{114}{99} = \dfrac{38}{33} \approx 1.1515...$

• This number lies between 1 and 2.
• More precisely, between 1.1 and 1.2.
• Using successive magnification, locate it between 1.15 and 1.16, very close to 1.15.

Mark the point at approximately 1.1515 on the number line between 1 and 2.

Solution

To find 6 rational numbers between 3 and 4, we use the method of writing 3 and 4 with a denominator of at least 8 (n + 2 = 8).

$3 = \frac{24}{8}, \quad 4 = \frac{32}{8}$

The integers between 24 and 32 are: 25, 26, 27, 28, 29, 30, 31.

Picking any 6 of these:

$\frac{25}{8},\ \frac{26}{8},\ \frac{27}{8},\ \frac{28}{8},\ \frac{29}{8},\ \frac{30}{8}$

Simplified: $\frac{25}{8},\ \frac{13}{4},\ \frac{27}{8},\ \frac{7}{2},\ \frac{29}{8},\ \frac{15}{4}$

All six numbers lie strictly between 3 and 4.

Solution

We need 5 rational numbers between $\dfrac{2}{5}$ and $\dfrac{3}{5}$.

Method: Multiply numerator and denominator by 7 (since we need 5 numbers, use a factor ≥ 6).

$\frac{2}{5} = \frac{2 \times 7}{5 \times 7} = \frac{14}{35}$

$\frac{3}{5} = \frac{3 \times 7}{5 \times 7} = \frac{21}{35}$

Integers between 14 and 21 (exclusive): 15, 16, 17, 18, 19, 20

Pick any 5:

$\frac{15}{35},\ \frac{16}{35},\ \frac{17}{35},\ \frac{18}{35},\ \frac{19}{35}$

Simplified: $\frac{3}{7},\ \frac{16}{35},\ \frac{17}{35},\ \frac{18}{35},\ \frac{19}{35}$

All five lie strictly between $\dfrac{2}{5}$ and $\dfrac{3}{5}$.

Solution

Find a common denominator for $\dfrac{1}{6}$ and $\dfrac{2}{5}$.

LCM of 6 and 5 = 30. But we need at least 7 integers between them, so multiply further by 7:

$\frac{1}{6} = \frac{5}{30} = \frac{35}{210}$

$\frac{2}{5} = \frac{12}{60} = \frac{84}{210}$

Integers between 35 and 84 (exclusive): 36, 37, 38, ..., 83

Pick any 5:

$\frac{36}{210},\ \frac{42}{210},\ \frac{49}{210},\ \frac{56}{210},\ \frac{70}{210}$

Simplified: $\frac{6}{35},\ \frac{1}{5},\ \frac{7}{30},\ \frac{4}{15},\ \frac{1}{3}$

Verification: $\dfrac{1}{6} \approx 0.1\overline{6}$ and $\dfrac{2}{5} = 0.4$

All five values lie between 0.167 and 0.4. ✓

Solution

Given: $\frac{x}{3} + \frac{5x}{16} = 15$

Step 1: Find LCM of 3 and 16. $\text{LCM}(3, 16) = 48$

Step 2: Rewrite each fraction with denominator 48: $\frac{16x}{48} + \frac{15x}{48} = 15$

Step 3: Combine: $\frac{31x}{48} = 15$

Step 4: Solve for x: $x = 15 \times \frac{48}{31} = \frac{720}{31}$

$\boxed{x = \frac{720}{31}}$

Verification: $\dfrac{720}{31 \times 3} + \dfrac{5 \times 720}{31 \times 16} = \dfrac{240}{31} + \dfrac{3600}{496} = \dfrac{240}{31} + \dfrac{225}{31} = \dfrac{465}{31} = 15$ ✓

Solution

Given: $a + b^{-1} = 0$, where $a$ and $b$ are non-zero rational numbers.

Step 1: Rewrite the condition. $a + \frac{1}{b} = 0$ $a = -\frac{1}{b}$

Step 2: Multiply both sides by $b$: $ab = -\frac{b}{b} = -1$

Conclusion: $ab = -1$

Since $-1 < 0$, we conclude that $ab$ is negative.

Justification: Since $a = -1/b$, the product $a$ and $b$ have opposite signs (one is positive and the other is negative). The product of two numbers with opposite signs is always negative. In fact, $ab = -1$ exactly.

Solution

Part 1: Writing as p/10⁴

A terminating decimal with last non-zero digit in the 4th decimal place looks like: $0.abcd \quad \text{or} \quad n.abcd$ where $d \neq 0$.

Such a number can be written as: $\frac{\text{integer}}{10^4} = \frac{p}{10^4}$

where $p$ is the integer obtained by removing the decimal point.

Since $d \neq 0$ (the 4th decimal place is non-zero), $p$ is not divisible by 10 (if $p$ were divisible by 10, the last digit would be 0, contradicting that $d \neq 0$).

Part 2: Must the denominator (in lowest form) be divisible by 2⁴ or 5⁴?

Not necessarily both, but let's analyze:

When $\dfrac{p}{10^4}$ is written in lowest form $\dfrac{p'}{q'}$ (where $\gcd(p', q') = 1$):

$\frac{p}{10^4} = \frac{p'}{q'} \Rightarrow q' \text{ divides } 10^4 = 2^4 \times 5^4$

So $q' = 2^a \times 5^b$ where $0 \leq a \leq 4$ and $0 \leq b \leq 4$.

The denominator need not be exactly $2^4$ or $5^4$. For example:

• $\dfrac{1}{10^4} = \dfrac{1}{2^4 \times 5^4}$ → denominator has both $2^4$ and $5^4$
• $\dfrac{5}{10^4} = \dfrac{1}{2^4 \times 5^3}$ → not divisible by $5^4$
• $\dfrac{2}{10^4} = \dfrac{1}{2^3 \times 5^4}$ → not divisible by $2^4$

Conclusion: It is necessary that the denominator (in lowest form) divides $2^4 \times 5^4$, but it is not necessary that it is divisible by $2^4$ or $5^4$ individually.

Solution

A rational number $\dfrac{p}{q}$ (in lowest form) has a terminating decimal if and only if $q$ is of the form $2^m \times 5^n$ for non-negative integers $m$ and $n$.

Step 1: Check if 18/125 is in lowest form. $\gcd(18, 125) = 1 \quad (\text{since } 18 = 2 \times 3^2 \text{ and } 125 = 5^3)$ Yes, it is already in lowest form.

Step 2: Factorize the denominator. $125 = 5^3$ This is of the form $2^0 \times 5^3$, which is a valid form.

Conclusion: The decimal expansion of $\dfrac{18}{125}$ is terminating.

Step 3: Find the number of decimal places.

Multiply numerator and denominator to make denominator $10^3$: $\frac{18}{125} = \frac{18 \times 2^3}{5^3 \times 2^3} = \frac{18 \times 8}{1000} = \frac{144}{1000} = 0.144$

The decimal terminates in 3 decimal places.

Solution

Let the rational number be $\dfrac{p}{q}$ where $q = 2^3 \times 5$ and $\gcd(p, q) = 1$.

Step 1: Since $q = 2^3 \times 5^1$, the denominator is of the form $2^m \times 5^n$ with $m = 3$, $n = 1$.

This guarantees a terminating decimal.

Step 2: To find the number of decimal places, convert to a denominator that is a power of 10.

$\frac{p}{2^3 \times 5} = \frac{p \times 5^2}{2^3 \times 5^3} = \frac{25p}{10^3}$

So the decimal expansion has 3 decimal places.

Explanation: The number of decimal places = $\max(m, n) = \max(3, 1) = 3$, since we need to multiply both numerator and denominator by $5^{3-1} = 5^2$ to make the denominator $10^3$.

Example: If $p = 1$: $\frac{1}{2^3 \times 5} = \frac{1}{40} = \frac{25}{1000} = 0.025$ This has 3 decimal places. ✓

Solution

Step 1: Express $a = \dfrac{7}{12}$ and $b = \dfrac{5}{6}$ with a common denominator.

$b = \frac{5}{6} = \frac{10}{12}$

So: $a = \dfrac{7}{12}$, $b = \dfrac{10}{12}$

Here $k_1 = 7$, $k_2 = 10$, $m = 12$.

But $k_2 - k_1 = 10 - 7 = 3$, which is not greater than 6.

Step 2: Scale to a larger denominator. Let $m = 12 \times 7 = 84$:

$a = \frac{7}{12} = \frac{49}{84}, \quad b = \frac{5}{6} = \frac{70}{84}$

Now $k_1 = 49$, $k_2 = 70$, $k_2 - k_1 = 21 > 6$. ✓

Step 3: Five distinct rational numbers between $\dfrac{49}{84}$ and $\dfrac{70}{84}$ with integer numerators:

$\frac{50}{84},\ \frac{55}{84},\ \frac{60}{84},\ \frac{63}{84},\ \frac{65}{84}$

Simplified: $\frac{25}{42},\ \frac{55}{84},\ \frac{5}{7},\ \frac{3}{4},\ \frac{65}{84}$

Step 4: Why is k₂ – k₁ > n + 1 necessary?

Between $k_1$ and $k_2$ (both exclusive), there are exactly $k_2 - k_1 - 1$ integers.

To find $n$ rational numbers with integer numerators between $\dfrac{k_1}{m}$ and $\dfrac{k_2}{m}$, we need at least $n$ integers strictly between $k_1$ and $k_2$.

This requires: $k_2 - k_1 - 1 \geq n \Rightarrow k_2 - k_1 \geq n + 1$

For a strict inequality (to ensure exactly $n$ can always be found): $k_2 - k_1 > n + 1 \text{ (or at minimum } k_2 - k_1 \geq n + 1)$

If $k_2 - k_1 \leq n$, there are fewer than $n$ integers between $k_1$ and $k_2$, making it impossible to write $n$ rational numbers in this form.

Solution

Given:

• $x + y + z = 0$ ... (1)
• $xy + yz + zx = 0$ ... (2)

Step 1: From equation (1): $x + y + z = 0$

Square both sides: $(x + y + z)^2 = 0$ $x^2 + y^2 + z^2 + 2(xy + yz + zx) = 0$

Step 2: Substitute equation (2): $xy + yz + zx = 0$

$x^2 + y^2 + z^2 + 2(0) = 0$ $x^2 + y^2 + z^2 = 0$

Step 3: Since $x, y, z$ are rational numbers (real numbers), $x^2 \geq 0$, $y^2 \geq 0$, $z^2 \geq 0$.

A sum of non-negative terms equals zero if and only if each term is zero: $x^2 = 0,\quad y^2 = 0,\quad z^2 = 0$

Step 4: Therefore: $x = 0,\quad y = 0,\quad z = 0$

Conclusion: All three rational numbers $x$, $y$, and $z$ must be simultaneously zero. $\blacksquare$

Solution

We need to show that if $a$ and $b$ are rational numbers with $a < b$, then: $a < \frac{a+b}{2} < b$

Step 1: Show $\dfrac{a+b}{2} > a$

$\frac{a+b}{2} - a = \frac{a+b-2a}{2} = \frac{b-a}{2}$

Since $b > a$, we have $b - a > 0$, so: $\frac{b-a}{2} > 0$ $\Rightarrow \frac{a+b}{2} > a \quad \checkmark$

Step 2: Show $\dfrac{a+b}{2} < b$

$b - \frac{a+b}{2} = \frac{2b - a - b}{2} = \frac{b-a}{2}$

Since $b - a > 0$: $\frac{b-a}{2} > 0$ $\Rightarrow \frac{a+b}{2} < b \quad \checkmark$

Conclusion: $a < \frac{a+b}{2} < b$

Therefore, $\dfrac{a+b}{2}$ (the arithmetic mean of $a$ and $b$) always lies between $a$ and $b$. $\blacksquare$

Note: Since $a$ and $b$ are rational, $(a+b)/2$ is also rational (rationals are closed under addition and division by non-zero integers).