The World of Numbers

1. The Dawn of Mathematics: The Human Need to Count

Long before cities, laws, or star maps existed, humans had one basic need: keeping count. Mathematics did not start in classrooms with blackboards. It started in dirt, on tree bark, and on bones.

Table of Contents

Imagine living thousands of years ago near the Saraswati river. You have cattle. Every morning they go out to graze, and every evening they return. How do you make sure none are missing?

Without number words or written symbols, early humans used one-to-one correspondence. For every cow that left, the herder placed one pebble in a clay pot. In the evening, for every cow that returned, one pebble was removed. If the pot was empty, the herd was safe. If pebbles remained, cows were missing. This simple matching of objects was the birth of Natural Numbers (N = {1, 2, 3, 4, ...}).

A History Written in Bone

The earliest physical proof of humans recording numbers comes from Africa.

Imp: The Lebombo Bone, found between South Africa and Swaziland, is about 35,000 years old. It has 29 carefully carved notches, probably used as a lunar calendar or menstrual tracker.

Even more amazing is the Ishango Bone, found near the Nile River in Congo, dating to about 20,000 BCE. It has three columns of notches. One column groups notches into 11, 13, 17, and 19 — the prime numbers between 10 and 20. Another column shows doubling. This proves that the idea of "number" is tens of thousands of years old.

Fig 3.1: Prime number tally groupings found on the Ishango bone

The Indian Context: Trade and Astronomy

As civilisations grew, so did the need for larger numbers. In ancient Indian cities like Lothal and Harappa, standard weights and measures were needed for trade. Indian philosophers in Vedic times gave names to powers of 10 up to 10¹² (called parardha). In the 4th century BCE, Buddha described names up to 10⁵³ (called tallakshana). This set the stage for the decimal place-value system used worldwide today.

Exercise Set 3.1 — Solutions

1. A merchant in Lothal exchanges bags of spices for copper ingots. He receives 15 ingots for every 2 bags. If he brings 12 bags, how many ingots does he get?

For 2 bags → 15 ingots
For 1 bag → 15/2 = 7.5 ingots
For 12 bags → 7.5 × 12 = 90 ingots.

2. The Ishango bone column shows: 11, 13, 17, 19. What do these have in common? List the next three numbers.

These are all prime numbers (numbers with exactly two factors: 1 and themselves). The next three primes are 23, 29, and 31.

3. Natural numbers are closed under addition. Are they closed under subtraction? Give examples.

No. Closure means the result stays in the same set.
5 − 3 = 2 (natural number ✓)
But 3 − 5 = −2 (not a natural number ✗)
So natural numbers are not closed under subtraction.

4. Each finger has 3 joints, and the thumb counts them. How many can you count on one hand? How does this relate to base-12?

On one hand: 4 fingers × 3 joints = 12 counts.
Using the other hand to track complete sets of 12 gives 12 × 12 = 144.
This is why ancient cultures used base-12 (dozen) and base-60 systems — they came from finger-joint counting!

2. The Revolution of Shunya: When Nothing Became Something

For thousands of years, the number line started at 1. If you had five apples and gave all five away, you had no number for "nothing." Civilisations like the Babylonians used placeholders, but they did not treat "nothing" as a real number.

It was Brahmagupta (628 CE) who transformed the void into a number. This leap was inspired by Indian philosophy.

From Philosophy to Mathematics

In the Upanishads and Buddhist texts, Shunyata (emptiness) described the goal of meditation — emptying the mind of all thoughts. Indian thinkers revered this state of "nothingness." This concept moved into mathematics through Aryabhata and finally Brahmagupta.

Imp: The philosophical idea of emptiness became the mathematical number zero.

The Bakhshali Manuscript and Brahmagupta's Rules

The Bakhshali Manuscript (early centuries CE) used a bold dot (bindu) for zero. But a symbol is just a mark until it has rules. Brahmagupta, in his book Brahmasphutasiddhanta (628 CE), defined zero as a − a = 0 and gave these laws:

a + 0 = a (Adding zero changes nothing)
a − 0 = a (Subtracting zero changes nothing)
a × 0 = 0 (Any number times zero is zero)

3. Debts and Fortunes: Negative Numbers

Brahmagupta asked: If 5 − 5 = 0, what is 3 − 5? To answer this, he used real-life ideas from commerce:

Fortunes (Dhana) = Positive numbers (wealth, assets)
Debts (Rina) = Negative numbers (loans, losses)

Moving left of zero on the number line gave us negative numbers. Together with positive numbers and zero, they form the Integers (Z).

Fig 3.2: The integer number line showing debts, zero, and fortunes

Brahmagupta's Rules for Integers

A fortune plus a fortune is a fortune: 5 + 4 = 9
A debt plus a debt is a debt: (−5) + (−4) = −9
A fortune minus zero is a fortune: 7 − 0 = 7
A debt times a fortune is a debt: (−3) × 4 = −12
A debt times a debt is a fortune: (−3) × (−4) = 12

Imp: Negative × Negative = Positive. Think of it as "removing a debt." If someone takes away 4 debts of ₹3 each, you are ₹12 richer!

Exercise Set 3.2 — Solutions

1. Temperature in Ladakh is 4°C at noon. By midnight it drops by 15°C. What is the midnight temperature?

4 − 15 = −11°C.

2. A trader takes a loan of ₹850, makes a profit of ₹1,200, then incurs a loss of ₹450. Write as an equation and find the final standing.

Equation: −850 + 1200 − 450
= 350 − 450 = −100
The trader has a debt of ₹100.

3. Calculate using Brahmagupta's laws:
(i) (−12) × 5 (ii) (−8) × (−7) (iii) 0 − (−14) (iv) (−20) ÷ 4

(i) −60 (debt × fortune = debt)
(ii) 56 (debt × debt = fortune)
(iii) 14 (subtracting a negative = adding a positive)
(iv) −5

4. Explain, using a real-world debt example, why 10 − (−5) = 15.

If you owe ₹10 and someone forgives (removes) a debt of ₹5, your new position is: you now owe only ₹5, which is ₹15 better than before. So 10 − (−5) = 10 + 5 = 15.

4. Filling the Spaces: Fractions and Rational Numbers

If a farmer divides a field among three children, how much does each get? Numbers that represent parts of a whole are called fractions.

Every positive fraction has a negative twin: −3/4 for 3/4, −19/7 for 19/7. When we combine all integers and all fractions (positive and negative), we get the set of Rational Numbers (Q, from "quotient").

A rational number = p/q where p and q are integers, and q ≠ 0

Imp: We need q ≠ 0 because division by zero is undefined. You cannot share something among zero people!

Important Observations

All integers are rational numbers (write them as p/1).
Rational numbers do not have a unique form. For example: −1/3 = −2/6 = −3/9 = −10/30.
We usually write them in lowest terms (co-prime numerator and denominator).

Rules for Rational Numbers

Equality: a/b = c/d if and only if ad = bc
Addition/Subtraction: a/b ± c/b = (a ± c)/b (same denominator first)
Multiplication: (a/b) × (c/d) = (ac)/(bd)
Division: (a/b) ÷ (c/d) = (a/b) × (d/c) = (ad)/(bc), where c ≠ 0

Addition and multiplication are commutative and follow the distributive law: p(q + r) = pq + pr.

Rational numbers are closed under addition, subtraction, and multiplication. They are also closed under division, provided we do not divide by zero.

Exercise Set 3.3 — Solutions

1. Prove these rational numbers are equal:
(i) 2/3 and 4/6 (ii) 5/4 and 10/8 (iii) −3/5 and −6/10 (iv) 9/3 and 3

(i) 2 × 6 = 12 and 3 × 4 = 12. Since ad = bc, they are equal.
(ii) 5 × 8 = 40 and 4 × 10 = 40. Equal.
(iii) (−3) × 10 = −30 and 5 × (−6) = −30. Equal.
(iv) 9/3 = 3/1. 9 × 1 = 9 and 3 × 3 = 9. Equal.

2. Find the sum:
(i) 2/5 + 3/10 (ii) 7/12 + 5/8 (iii) −4/7 + 3/14

(i) 4/10 + 3/10 = 7/10
(ii) 14/24 + 15/24 = 29/24
(iii) −8/14 + 3/14 = −5/14

3. Find the difference:
(i) 5/6 − 1/4 (ii) 11/8 − 3/4 (iii) −7/9 − (−2/3)

(i) 10/12 − 3/12 = 7/12
(ii) 11/8 − 6/8 = 5/8
(iii) −7/9 + 6/9 = −1/9

4. Find the product:
(i) 2/3 × 3/10 (ii) 7/11 × 5/8 (iii) −4/7 × 5/14

(i) 6/30 = 1/5
(ii) 35/88
(iii) −20/98 = −10/49

5. Find the quotient:
(i) 2/3 ÷ 3/10 (ii) 7/11 ÷ 5/8 (iii) −4/7 ÷ 5/14

(i) 2/3 × 10/3 = 20/9
(ii) 7/11 × 8/5 = 56/55
(iii) −4/7 × 14/5 = −56/35 = −8/5

6. Show that: (1/2 + 3/4) × 8/3 = (1/2 × 8/3) + (3/4 × 8/3)

LHS: (2/4 + 3/4) × 8/3 = 5/4 × 8/3 = 40/12 = 10/3
RHS: 8/6 + 24/12 = 4/3 + 2 = 4/3 + 6/3 = 10/3
LHS = RHS. Verified.

7. Simplify using distributive property: 7/9 × (6/7 − 3/4)

= (7/9 × 6/7) − (7/9 × 3/4)
= 42/63 − 21/36
= 2/3 − 7/12
= 8/12 − 7/12 = 1/12

8. Find x such that: 5/6(x + 3/5) = 5/6x + 1/2

5/6x + 5/6 × 3/5 = 5/6x + 1/2
5/6x + 15/30 = 5/6x + 1/2
5/6x + 1/2 = 5/6x + 1/2
This is true for all rational numbers x. The equation is an identity.

5. Rational Numbers on the Number Line

To place integers on a number line, we mark 0 as the origin. Moving right gives positive numbers; moving left gives negative numbers.

Rational numbers sit between integers. To place p/q, divide the unit interval into q equal parts, then count p parts from 0 (right for positive, left for negative).

Fig 3.4: Rational numbers between integers on the number line

Absolute Value

The absolute value of a number, written |x|, is its distance from 0 on the number line. It is always non-negative.

|5/3| = 5/3, |−5/3| = 5/3, |0| = 0

Distance between two rational numbers a and b = |a − b|.

The Density of Rational Numbers

Rational numbers are dense. Between any two rational numbers, you can always find another one by taking their average: (a + b)/2.

Fig 3.9: 5/4 lies between 1 and 3/2

Exercise Set 3.4 — Solutions

1. Represent 3/3, −5/4, and 1½ on a single number line.

3/3 = 1 (at the point 1)
−5/4 = −1.25 (one quarter left of −1)
1½ = 1.5 (midway between 1 and 2)

2. Find three distinct rational numbers between −1/2 and 1/4.

Average of −1/2 and 1/4 = (−2/4 + 1/4)/2 = −1/8
Between −1/2 and −1/8: average = −5/16
Between −1/8 and 1/4: average = 1/16
So three numbers are: −5/16, −1/8, 1/16 (answers may vary).

3. Simplify: −(−1/4) + (5/12)

= 1/4 + 5/12 = 3/12 + 5/12 = 8/12 = 2/3

4. A tailor has 15¾ metres of silk. One kurta needs 2¼ metres. How many kurtas can he make?

15¾ = 63/4 metres
2¼ = 9/4 metres
Number of kurtas = (63/4) ÷ (9/4) = 63/4 × 4/9 = 7 kurtas

5. Find three rational numbers between 3.1415 and 3.1416.

3.1415 = 31415/10000, 3.1416 = 31416/10000
Multiply by 10: 314150/100000 and 314160/100000
Three numbers: 314152/100000, 314155/100000, 314158/100000
Or in decimal: 3.14152, 3.14155, 3.14158

6. Can you think of other ways to find a rational number between any two rational numbers?

Yes. Besides the average, you can:
1. Convert both to decimals and pick any decimal in between.
2. Find a common denominator and pick a numerator between the two.
3. Use weighted averages: (a + 2b)/3, (2a + b)/3, etc.

6. Irrational Numbers

For centuries, mathematicians believed every length could be written as a fraction. But when Baudhayana built fire altars (around 800 BCE), he found lengths that refused to be fractions.

Consider a square with side = 1 unit. By the Baudhayana–Pythagoras Theorem, the diagonal d satisfies:

1² + 1² = d² → d² = 2 → d = √2

Fig 3.10: The diagonal of a unit square is √2

Imp: √2 cannot be written as p/q. Numbers that cannot be expressed as fractions are called Irrational Numbers.

Proof that √2 is Irrational (Proof by Contradiction)

We assume the opposite of what we want to prove, then show this leads to a logical disaster.

Step 1 (Assume): Suppose √2 = p/q in lowest terms (p and q share no common factors).

Step 2: Square both sides: 2 = p²/q²

Step 3: Multiply by q²: 2q² = p²

Step 4: So p² is even, which means p is even. Let p = 2k.

Step 5: Substitute: 2q² = (2k)² = 4k² → q² = 2k²

Step 6: So q² is even, which means q is even.

Step 7 (Contradiction): Both p and q are even, so they share factor 2. But we said p/q was in lowest terms!

Conclusion: Our assumption was wrong. √2 is irrational.

Fig: Flow of the proof by contradiction for √2

Constructing Irrational Lengths

We can mark √2 on the number line using a compass:

Mark OA = 1 unit on the number line.
Draw a perpendicular at A, mark AB = 1 unit.
Join O to B. By Pythagoras, OB = √2.
With compass at O and radius OB, draw an arc cutting the number line at P. Then OP = √2.

Fig 3.11: Constructing √2 on the number line

The Story of Pi (π)

π is the ratio of a circle's circumference to its diameter. Aryabhata (499 CE) gave π ≈ 3.1416, but said it was only an approximation. In 1761, Lambert proved π is irrational — no fraction can ever give its exact value.

In the 14th century, Madhava of Sangamagrama discovered an infinite series for π:

π = 4 × (1 − 1/3 + 1/5 − 1/7 + 1/9 − ...)

Adding more and more terms gets us closer and closer to π, but we never reach it exactly with a finite number of terms.

Fig 3.12: Rational and irrational numbers together on the number line

7. Real Numbers: Terminating and Repeating Decimals

When we unite rational and irrational numbers, we get Real Numbers (R) — the complete, unbroken number line.

Rational Decimals

When you divide numerator by denominator, exactly one of two things happens:

Terminating: The division ends with remainder 0.
Example: 3/8 = 0.375
Repeating: The digits loop forever.
Example: 5/11 = 0.454545... = 0.4̄5̄

Imp: A rational number p/q (in lowest terms) has a terminating decimal if and only if the prime factors of q are only 2, 5, or both.

Converting Decimals to p/q Form

Case 1: Terminating

0.35 = 35/100 = 7/20

Case 2: Pure repeating (repeats immediately after decimal)

Let x = 0.6̄. Multiply by 10: 10x = 6.6̄. Subtract: 9x = 6, so x = 2/3

Case 3: General repeating (some digits, then repeating block)

Let x = 0.16̄. Multiply by 10: 10x = 1.6̄. Multiply by 10 again: 100x = 16.6̄. Subtract: 90x = 15, so x = 15/90 = 1/6

Decimal Type	Steps to Follow
Pure repeating	Let x = decimal. Multiply by 10ⁿ where n = repeating digits. Subtract original, solve for x.
General repeating	Multiply by 10ᵐ (m = non-repeating digits), then by 10ⁿ (n = repeating digits). Subtract and solve.

Cyclic Numbers

The decimal for 1/7 = 0.142857142857... The repeating block 142857 is a cyclic number:

142857 × 1 = 142857
142857 × 2 = 285714
142857 × 3 = 428571
142857 × 4 = 571428
142857 × 5 = 714285
142857 × 6 = 857142

The same digits simply rotate! This is the beautiful hidden structure inside rational numbers.

Irrational Decimals

Irrational numbers have decimals that never end and never repeat:

√2 = 1.414213562... π = 3.141592653...

Exercise Set 3.5 — Solutions

1. Without long division, determine which have terminating decimals: 7/20, 4/15, 13/250. Then verify by division.

7/20: 20 = 2² × 5. Only 2s and 5s → Terminating. 7/20 = 0.35
4/15: 15 = 3 × 5. Has factor 3 → Non-terminating repeating. 4/15 = 0.2666...
13/250: 250 = 2 × 5³. Only 2s and 5s → Terminating. 13/250 = 0.052

2. Perform long division for 1/13. Identify the repeating block. Check 2/13, 3/13, 4/13. What do you notice?

1/13 = 0.076923076923... Repeating block: 076923
2/13 = 0.153846153846... Block: 153846
3/13 = 0.230769230769... Block: 230769
Notice: Each block is a cyclic rotation of the digits! The same six digits appear in a rotating pattern.

3. Classify as rational or irrational:
(i) √81 (ii) √12 (iii) 0.33333... (iv) 0.1234512345... (v) 1.010010001... (vi) 23.560185612239...

(i) √81 = 9 = Rational (9/1)
(ii) √12 = 2√3 = Irrational
(iii) 0.333... = 1/3 = Rational
(iv) Repeating block "12345" = Rational (12345/99999)
(v) Pattern grows (1, 01, 001, 0001...) but not repeating = Irrational
(vi) Given as non-repeating, non-terminating = Irrational

4. Show that 0.999... = 1 using algebra.

Let x = 0.999...
Multiply by 10: 10x = 9.999...
Subtract: 10x − x = 9.999... − 0.999...
9x = 9
x = 1
Therefore, 0.999... = 1 exactly.

5. Find more numbers n whose reciprocals 1/n produce cyclic repeating blocks.

1/7 → cyclic (142857)
1/13 → cyclic (076923)
1/17 → cyclic (0588235294117647)
These occur when n is a full reptend prime — a prime where the repeating block has length n−1.

8. The Complete Family of Numbers

Our number system grew over thousands of years:

Natural Numbers (N): {1, 2, 3, ...} — for counting
Integers (Z): {..., −2, −1, 0, 1, 2, ...} — adds zero and negatives
Rational Numbers (Q): p/q where q ≠ 0 — adds fractions
Irrational Numbers (I): √2, π, etc. — fills the gaps
Real Numbers (R): Q ∪ I — the complete continuous line

Fig 3.13: The family of Real Numbers

Imaginary Numbers: A Glimpse Beyond

What is the square root of −1? No real number multiplied by itself gives a negative answer. Mathematicians invented i (imaginary unit) where i² = −1. While they sound like fiction, imaginary numbers power modern electrical engineering, quantum mechanics, and your mobile phone!

9. End-of-Chapter Exercises — Solutions

1. Convert by long division:
(i) 3/50 (ii) 2/3

(i) 3/50 = 0.06 (terminating)
(ii) 2/3 = 0.666... = 0.6̄ (non-terminating repeating)

2. Prove that √5 is irrational.

Proof by contradiction:
Assume √5 = p/q in lowest terms.
Then 5 = p²/q² → 5q² = p².
So p² is divisible by 5, meaning p is divisible by 5. Let p = 5k.
Then 5q² = 25k² → q² = 5k².
So q² is divisible by 5, meaning q is divisible by 5.
Both p and q share factor 5 — contradiction!
Therefore, √5 is irrational.

3. Convert to p/q form:
(i) 12.6 (ii) 0.0120 (iii) 3.052 (iv) 1.235 (v) 0.2̄3̄ (vi) 2.0̄5̄ (vii) 2.125 (viii) 3.125 (ix) 2.162̄5̄

(i) 126/10 = 63/5
(ii) 120/10000 = 3/250
(iii) 3052/1000 = 763/250
(iv) Let x = 1.235̄. 1000x = 1235.5̄, 100x = 123.5̄. 900x = 1112. x = 1112/900 = 278/225
(v) Let x = 0.2323... 100x = 23.23... 99x = 23. x = 23/99
(vi) Let x = 2.0505... 100x = 205.05... 99x = 203. x = 203/99
(vii) 2125/1000 = 17/8
(viii) 3125/1000 = 25/8
(ix) Let x = 2.162525... 100x = 216.2525..., 10000x = 21625.25... 9900x = 21409. x = 21409/9900

4. Locate on the number line:
(i) 0.532 (ii) 1.15

(i) 0.532 lies between 0 and 1, closer to 0.5. Divide the 0–1 interval into 1000 parts and mark the 532nd part.
(ii) 1.15 lies between 1 and 2, at 15/100 of the way from 1 to 2.

5. Find 6 rational numbers between 3 and 4.

Write 3 = 30/10 and 4 = 40/10.
Six numbers: 31/10, 32/10, 33/10, 34/10, 35/10, 36/10
Or: 3.1, 3.2, 3.3, 3.4, 3.5, 3.6

6. Find 5 rational numbers between 2/5 and 3/5.

Convert: 2/5 = 8/20, 3/5 = 12/20.
Five numbers: 9/20, 10/20, 11/20, 21/40, 23/40
Or: 0.45, 0.475, 0.5, 0.525, 0.55

7. Find 5 rational numbers between 1 and 2/5.

1 = 5/5, 2/5 = 0.4. Wait — 2/5 < 1. So we need numbers between 0.4 and 1.
Convert: 2/5 = 4/10, 1 = 10/10.
Five numbers: 5/10, 6/10, 7/10, 8/10, 9/10 = 1/2, 3/5, 7/10, 4/5, 9/10

8. If x + 3/5 = 16/15, find x.

x = 16/15 − 3/5 = 16/15 − 9/15 = 7/15

9. Let a and b be non-zero rationals with a + b = 0. Is ab positive or negative?

If a + b = 0, then b = −a.
So ab = a × (−a) = −a².
Since a ≠ 0, a² is always positive. Therefore ab is negative.

10. A rational number has a terminating decimal whose last non-zero digit is in the 4th decimal place. Show it can be written as p/10⁴ where p is not divisible by 10.

If the last non-zero digit is in the 4th decimal place, the decimal has form 0.abcd where d ≠ 0.
This equals abcd/10000 = p/10⁴ where p = abcd.
Since d ≠ 0, p is not divisible by 10.
The denominator in lowest terms need not be divisible by 2⁴ or 5⁴. Example: 0.1250 = 1250/10000 = 1/8, where denominator is 8 = 2³, not 2⁴ or 5⁴.

11. Without division, determine if 18/125 has a terminating decimal. If yes, how many places?

125 = 5³. The denominator has only factor 5.
To make it a power of 10, multiply by 2³: 18/125 = (18 × 8)/(125 × 8) = 144/1000 = 0.144.
Terminating, 3 decimal places.

12. A rational in lowest form has denominator 2³ × 5. How many decimal places?

Denominator = 8 × 5 = 40. To make power of 10, multiply by 5²: 40 × 25 = 1000 = 10³.
So the decimal will have 3 decimal places (the maximum of the powers of 2 and 5, which is 3).

13. Let a = 7/12 and b = 5/6. Express both as k₁/m and k₂/m where k₂ − k₁ > 6. Write exactly 5 distinct rationals between a and b with integer numerator.

a = 7/12 = 14/24, b = 5/6 = 20/24. Here m = 24, k₁ = 14, k₂ = 20, and 20 − 14 = 6. We need k₂ − k₁ > 6.
Use m = 36: a = 21/36, b = 30/36. k₂ − k₁ = 9 > 6. ✓
Five rationals with integer numerators: 22/36, 23/36, 24/36, 25/36, 26/36
= 11/18, 23/36, 2/3, 25/36, 13/18

Why k₂ − k₁ > n + 1 is needed: To find n numbers between them, we need at least n+1 gaps between k₁ and k₂.

14. If x + y + z = 0 and xy + yz + zx = 0, show that x = y = z = 0.

We know (x + y + z)² = x² + y² + z² + 2(xy + yz + zx).
Substitute: 0² = x² + y² + z² + 2(0)
So x² + y² + z² = 0.
Since squares are always ≥ 0, the only way their sum is 0 is if x = y = z = 0.

15. Show that (a + b)/2 lies between a and b.

Assume a < b. Then a + a < a + b < b + b → 2a < a + b < 2b.
Dividing by 2: a < (a + b)/2 < b.
So the average always lies strictly between the two numbers.

16. Find the hypotenuses of all right triangles in the square root spiral (Fig 3.14).

The square root spiral starts with a right triangle with both legs = 1.
Hypotenuse 1: √(1² + 1²) = √2
Hypotenuse 2: √((√2)² + 1²) = √(2 + 1) = √3
Hypotenuse 3: √((√3)² + 1²) = √(3 + 1) = √4 = 2
Hypotenuse 4: √(4 + 1) = √5
And so on... The nth hypotenuse = √(n + 1).