Chapter 3 · Class 9 · Ganita Manjari (NEP 2020)
Long before writing was invented, human beings had to keep track of things they owned. Imagine an ancient herder living on the banks of the now-dried Saraswati river, thousands of years ago. Every morning, as the goats left for pasture, she dropped one small pebble into a clay pot for each goat. In the evening, as the goats returned, she took one pebble out for each goat. If the pot was empty when the last goat came home, all was well. If pebbles remained, some goats had not returned — a clear signal to go searching.
This one-to-one matching between pebbles and goats is the very seed of counting. Over millennia, this idea grew into the system of natural numbers.
Found in the Lebombo Mountains of Africa, this baboon fibula bears 29 distinct notches carved in sequence. It is among the oldest known mathematical artefacts, suggesting early humans were deliberately tallying — perhaps tracking lunar cycles or keeping count of animals.
Discovered near Lake Edward (present-day Democratic Republic of Congo), this carved bone has three columns of grouped notches. Some mathematicians believe the groupings show early multiplication or prime-number awareness, though debate continues. It stands as evidence that numerical thinking is deeply ancient.
In many ancient cultures — including those of the Indian subcontinent and the Middle East — people counted using the phalanges (finger joints) of four fingers, guided by the thumb of the same hand. Each finger has 3 joints, giving 4 × 3 = 12 counts on one hand. This is why we still have 12 in a dozen, 24 hours in a day, 60 minutes in an hour (base 12 × base 5 from the other hand), and 360 degrees in a circle (12 × 30).
The Brahmi numerals of ancient India (inscribed in Ashoka's edicts, ~3rd century BCE) evolved into the Gupta script numerals and eventually into the system described by Brahmagupta and Aryabhata. Arab mathematicians, encountering Indian manuscripts, transmitted this system to Europe — hence the name Hindu-Arabic numerals. The ten symbols 0–9 that you use today are the direct descendants of those ancient Indian glyphs.
ℕ = { 1, 2, 3, 4, 5, … }
The set of natural numbers consists of all positive whole numbers starting from 1. They are the numbers you use when you count objects: one apple, two bananas, three mangoes…
A set is said to be closed under an operation if performing that operation on any two members of the set always gives a result that is also in the set.
For any two natural numbers a and b,
a + b is always a natural number.
Examples: 3 + 5 = 8 ∈ ℕ | 12 + 7 = 19 ∈ ℕ
For any two natural numbers a and b,
a × b is always a natural number.
Examples: 4 × 6 = 24 ∈ ℕ | 3 × 11 = 33 ∈ ℕ
3 − 5 = −2 and −2 ∉ ℕ
One counter-example is enough to show a set is not closed.
This failure of closure is exactly what drives us to invent
integers (Section 3.3).
5 ÷ 2 = 2.5 and 2.5 ∉ ℕ
When we divide two natural numbers, the result is sometimes a
fraction. This motivates rational numbers (Section 3.4).
Enter two positive whole numbers and choose an operation. The tool will tell you the result and whether it belongs to ℕ.
Q1. What is the smallest natural number?
Q2. Is there a largest natural number?
Q3. Is 0 a natural number?
The ancient Indian philosophical tradition had long explored the idea of shunya (शून्य) — the void, nothingness, emptiness. The Sanskrit root śūnya appears in Buddhist and Jain texts long before it entered mathematics. The leap that changed history was treating shunya not just as an absence but as an actual number that could be written, manipulated, and computed with.
This conceptual revolution took place in India, and the circular symbol 0 — representing an empty place in positional notation — is one of India's most profound gifts to world civilisation. The word zero itself traces back through Arabic (sifr, meaning empty) to Sanskrit śūnya.
The Indian mathematician and astronomer Brahmagupta, head of the astronomical observatory at Ujjain, wrote the Brahmasphutasiddhanta in 628–629 CE. In Chapter 18 (Kuttaka), he gave the first explicit arithmetic rules for zero, treating it as a number on equal footing with positive and negative quantities. He called positive numbers dhana (wealth/fortune) and negative numbers rina (debt), with zero being the boundary between them.
| Rule | Example |
|---|---|
| a + 0 = a (additive identity) | 7 + 0 = 7 |
| 0 + a = a | 0 + 13 = 13 |
| a × 0 = 0 (absorption) | 999 × 0 = 0 |
| a − a = 0 (self-difference) | 8 − 8 = 0 |
Formal argument:
Suppose 6 ÷ 0 = k for some number k.
Then, by the definition of division, k × 0 = 6.
But we know k × 0 = 0 for every number k.
So we would need 0 = 6, which is a contradiction.
Therefore, no value of k works, and
6 ÷ 0 is undefined.
𝐖 = { 0, 1, 2, 3, 4, 5, … } = ℕ ∪ {0}
The set of whole numbers is exactly the natural numbers with zero added. Every natural number is a whole number, but 0 is a whole number that is not a natural number.
{ 1, 2, 3, 4, 5, … }
✓ Counting numbers
✗ Does not include 0
Smallest element: 1
{ 0, 1, 2, 3, 4, 5, … }
✓ Includes all natural numbers
✓ Also includes 0
Smallest element: 0
Q. Is 0 ÷ 0 defined? Why or why not?
0 ÷ 0 is indeterminate (not merely undefined).
Using the same reverse-multiplication argument: 0 ÷ 0 = k means k × 0 = 0.
But every number satisfies k × 0 = 0! So any value of k would "work", giving infinitely many answers. Since there is no unique answer, 0 ÷ 0 is called indeterminate.
Comparison: 6 ÷ 0 is undefined (no answer exists), whereas 0 ÷ 0 is indeterminate (too many answers exist). Both are disallowed in arithmetic, but for different reasons.
A number line is a straight line with evenly spaced marks, each labelled with a number. For whole numbers, we start at 0 (the origin) and move to the right as numbers increase.
Notice that 0 is the starting point (origin) of the number line for whole numbers. Moving right means adding; moving left means subtracting. We cannot go left of 0 while staying in 𝐖 — that will change when we introduce integers.
Whole numbers let us count and measure positive quantities, but real life is full of situations where we need to go below zero:
In Shimla in January, the temperature may be −5°C — five degrees below freezing. We cannot describe this with whole numbers alone.
The Dead Sea shore is about −430 m relative to mean sea level. Negative altitude is a real physical measurement.
If you borrow ₹200 more than you have, your balance is −₹200. Banks use negative numbers every day.
To handle these situations, we extend the number line to the left of 0, adding the negative integers: −1, −2, −3, …
Brahmagupta formalised negative numbers in Brahmasphutasiddhanta (628 CE) using a vivid financial metaphor:
This was not merely metaphor — Brahmagupta gave precise computational rules for these quantities that remain valid today. His work predates the widespread use of negative numbers in Europe by nearly a thousand years.
ℤ = { …, −4, −3, −2, −1, 0, 1, 2, 3, 4, … }
The integers include all whole numbers and all their negatives. The number line now extends infinitely in both directions.
How do we multiply integers with different signs? Brahmagupta stated the rules clearly. The table below shows all four cases:
| First Number | Second Number | Product | Example |
|---|---|---|---|
| (+) Positive | (+) Positive | (+) Positive | (+3) × (+4) = +12 |
| (+) Positive | (−) Negative | (−) Negative | (+3) × (−4) = −12 |
| (−) Negative | (+) Positive | (−) Negative | (−3) × (+4) = −12 |
| (−) Negative | (−) Negative | (+) Positive | (−3) × (−4) = +12 |
Q. Why does (−) × (−) = (+)? It seems strange that multiplying two "negatives" gives a positive. Can you explain this using the idea of debt?
The Debt–Removal Analogy:
Let a debt of ₹5 be represented as −5. "Giving someone a debt" means adding a negative — their wealth decreases. But removing a debt means subtracting a negative, which increases their wealth.
Now consider: "Someone's three debts of ₹5 each are cancelled (removed)." Before cancellation they had −₹15 in debt. After cancellation, that debt is gone — they are ₹15 richer. So the act of removing three debts of 5 corresponds to:
More formal argument using patterns:
| (+3) × (−5) = −15 |
| (+2) × (−5) = −10 |
| (+1) × (−5) = −5 |
| (0) × (−5) = 0 |
| (−1) × (−5) = +5 ← pattern continues |
| (−2) × (−5) = +10 |
| (−3) × (−5) = +15 |
Each time the first factor decreases by 1, the product increases by 5 (the pattern of adding 5 continues past zero). Consistency of the pattern demands (−) × (−) = (+).
Click anywhere on the number line below to place a point and read its integer value.
(−3) + (−4) = −7 ∈ ℤ
(−5) + 8 = 3 ∈ ℤ
(ℤ is now closed under subtraction!)
3 − 5 = −2 ∈ ℤ
−7 − (−2) = −5 ∈ ℤ
(−3) × (−4) = 12 ∈ ℤ
(−2) × 7 = −14 ∈ ℤ
7 ÷ 2 = 3.5 ∉ ℤ
This motivates rational numbers ℚ
(Section 3.4)!
Exercise 1. Arrange the following integers on a number line in increasing order: −7, 3, −2, 5, 0
On a number line, integers increase from left to right. A number to the right is greater.
Increasing order: −7 < −2 < 0 < 3 < 5
Exercise 2. Find: (−6) × (−4)
Step 1: Identify the signs. Both factors are negative.
Step 2: Apply the rule. (−) × (−) = (+)
Step 3: Multiply the absolute values. 6 × 4 = 24
Step 4: Attach the sign. (+24)
Exercise 3. Is ℕ ⊂ 𝐖 ⊂ ℤ? Justify your answer.
Yes, this chain of subset relationships is true.
ℕ ⊂ 𝐖: Every natural number {1, 2, 3, …} is also in 𝐖 = {0, 1, 2, 3, …}. The only extra element in 𝐖 is 0.
𝐖 ⊂ ℤ: Every whole number {0, 1, 2, 3, …} is also in ℤ = {…, −2, −1, 0, 1, 2, …}. The extra elements in ℤ are the negative integers.
Therefore, ℕ ⊂ 𝐖 ⊂ ℤ. This is a proper chain: each set is a proper subset of the next (strictly contained, not equal).
The number sets we have studied so far are nested inside each other. Tap or click any coloured region below to highlight it and see example numbers. (This diagram will be expanded in Sections 3.4–3.6 to include ℚ and ℝ.)
← Tap a coloured region in the diagram to explore that number set.
A rational number is any number that can be expressed in the form pq where p, q ∈ ℤ (p and q are integers) and q ≠ 0.
The set of all rational numbers is denoted by ℚ (the symbol comes from the word "quotient").
Multiplying (or dividing) both numerator and denominator by the same non-zero integer gives an equivalent fraction:
pq = p × nq × n for any non-zero integer n
Example:
12 = 24 = 36 = 50100
A rational number pq is in standard form (lowest terms) when:
Example: 68 → GCD(6,8)=2 → divide both by 2 → 34 (standard form)
Worked Example: Find 23 + 14
Step 1: LCM of 3 and 4 is 12.
Step 2: 23 = 812 and 14 = 312
Step 3: 812 + 312 = 1112
Worked Example: Find 34 − 16
Step 1: LCM of 4 and 6 is 12.
Step 2: 34 = 912 and 16 = 212
Step 3: 912 − 212 = 712
Worked Example: Find 25 × 37
Multiply numerators: 2 × 3 = 6
Multiply denominators: 5 × 7 = 35
Result:
635
(already in standard form since GCD(6,35) = 1)
Worked Example: Find 34 ÷ 25
Step 1: Take the reciprocal of the divisor: reciprocal of 25 is 52
Step 2: Multiply: 34 × 52 = 158
For rational numbers a, b, c:
a × (b + c) = (a × b) + (a × c)
Verify: 12 × ( 23 + 14 )
LHS (Left Hand Side):
First find 23 + 14 = 812 + 312 = 1112
Then: 12 × 1112 = 1124
RHS (Right Hand Side):
12 × 23 + 12 × 14 = 26 + 18 = 13 + 18 = 824 + 324 = 1124
✓ LHS = RHS = 1124 Distributive property verified!
Between any two distinct rational numbers, there are infinitely many rational numbers. This is called the Density Property.
Given two rational numbers a and b, the rational number exactly between them is their average:
Repeat on smaller intervals to find more numbers between them.
Example: Between 13 and 12: Average = 13 + 12 2 = 5/62 = 512
Convert both fractions to the same denominator (use a large enough multiple), then pick integers between the two numerators.
Example: Between
12
and
34:
Use denominator 8:
48
and
68
→ pick
58
Step 1: Convert to the same denominator.
12 and 34 with denominator 8: 48 and 68
Only one integer (5) lies between 4 and 6, so we need a larger denominator.
Step 2: Multiply numerator and denominator by 10:
12 = 4080 and 34 = 6080
Step 3: Pick any 5 integers between 40 and 60 (e.g., 41, 43, 50, 55, 59):
4180, 4380, 5080 = 58, 5580 = 1116, 5980
💡 Key insight: We could have chosen any 20 integers between 40 and 60, giving 20 rational numbers. We could scale to denominator 800 and find hundreds more! This is why there are infinitely many rationals between any two rationals.
Enter two fractions and find rational numbers between them:
Step 1: 0.375 has 3 decimal places, so multiply numerator and denominator by 1000:
0.375 = 3751000
Step 2: Find GCD(375, 1000).
375 = 3 × 125 = 3 × 5³ and 1000 = 8 × 125 = 2³ × 5³
GCD = 125
Step 3: Divide both by 125: 375 ÷ 1251000 ÷ 125 = 38
✓ 0.375 = 38
Method: Use common denominator. LCM(4, 3) = 12.
14 = 312 and 13 = 412
No integers between 3 and 4, so scale denominator to 60:
14 = 1560 and 13 = 2060
Three rationals between them: 1660 = 415, 1760, 1860 = 310
Yes. Let ab and cd be any two rational numbers (b, d ≠ 0).
Their product is ab × cd = acbd
Since a, b, c, d ∈ ℤ, their products ac and bd are also integers. Since b ≠ 0 and d ≠ 0, we know bd ≠ 0. Therefore acbd is rational. ✓
LHS: 12 × ( 23 + 14 ) = 12 × ( 812 + 312 ) = 12 × 1112 = 1124
RHS: ( 12 × 23 ) + ( 12 × 14 ) = 13 + 18 = 824 + 324 = 1124
✓ LHS = RHS. Distributive property verified.
Are all integers also rational numbers? Explain.
Yes! Every integer n can be written as n1, which is in the form pq with p = n ∈ ℤ and q = 1 ∈ ℤ, q ≠ 0.
So every integer is automatically a rational number. This means ℤ ⊂ ℚ (the set of integers is a subset of the set of rationals).
For example: −5 = −51, 0 = 01, 100 = 1001
ℕ ⊂ W ⊂ ℤ ⊂ ℚ — Each set contains all the sets inside it.
We have seen that natural numbers, whole numbers, integers, and fractions are all rational. But mathematicians discovered that there exist numbers on the number line that cannot be expressed as a ratio of two integers. These are called Irrational Numbers.
The ancient Greeks were shocked to discover that √2 (the diagonal of a unit square) cannot be written as a fraction. According to legend, the mathematician who revealed this was drowned at sea by fellow Pythagoreans who found it disturbing!
A number is called irrational if it is a real number that cannot be expressed in the form pq where p, q ∈ ℤ and q ≠ 0.
The set of irrational numbers has no standard single-letter symbol, but is sometimes written as ℚ̅ (Q-bar) or ℚʹ or simply described as ℝ \ ℚ.
√n is irrational if and only if n is NOT a perfect square.
A perfect square is a whole number whose square root is also a whole number.
| n | √n | Perfect Square? | √n is… |
|---|---|---|---|
| 1 | 1 | ✓ Yes | Rational (= 1) |
| 2 | 1.4142… | ✗ No | Irrational |
| 3 | 1.7320… | ✗ No | Irrational |
| 4 | 2 | ✓ Yes | Rational (= 2) |
| 5 | 2.2360… | ✗ No | Irrational |
| 9 | 3 | ✓ Yes | Rational (= 3) |
| 10 | 3.1622… | ✗ No | Irrational |
| 16 | 4 | ✓ Yes | Rational (= 4) |
| 25 | 5 | ✓ Yes | Rational (= 5) |
| 49 | 7 | ✓ Yes | Rational (= 7) |
| 100 | 10 | ✓ Yes | Rational (= 10) |
This is one of the most famous and important proofs in all of mathematics. It uses the technique of proof by contradiction (also called reductio ad absurdum). The idea: assume the opposite of what we want to prove, then show that leads to a logical impossibility.
Step 1 — Assume the opposite:
Suppose, for the sake of contradiction, that √2 is rational. Then √2 = pq where p, q ∈ ℤ, q ≠ 0, and pq is in lowest terms (i.e., GCD(p, q) = 1, they are co-prime).
Step 2 — Square both sides:
(√2)² = p²q² ⇒ 2 = p²q² ⇒ p² = 2q²
Step 3 — Conclude p is even:
Since p² = 2q², the number p² is even (it is divisible by 2).
Now, if p were odd, then p² would also be odd (odd × odd = odd). Since p² is even, p must be even.
Step 4 — Write p = 2m:
Since p is even, write p = 2m for some integer m.
Step 5 — Substitute back:
p² = 2q² ⇒ (2m)² = 2q² ⇒ 4m² = 2q² ⇒ q² = 2m²
Step 6 — Conclude q is even:
By the same argument as Step 3: q² = 2m² means q² is even, so q is even.
Step 7 — The Contradiction!
We have shown both p and q are even. This means they share a common factor of 2.
But in Step 1, we assumed p and q are co-prime (GCD = 1). This is a contradiction! ❌
Step 8 — Conclusion:
Our initial assumption that √2 is rational must be false. Therefore, √2 is irrational. ■
Step 1 — Assume the opposite:
Suppose √5 is rational. Then √5 = pq where p, q ∈ ℤ, q ≠ 0, and GCD(p, q) = 1.
Step 2 — Square both sides:
5 = p²q² ⇒ p² = 5q²
Step 3 — Conclude 5 | p:
p² = 5q² means 5 divides p².
Since 5 is prime, if 5 | p² then 5 | p (a prime that divides a square must divide the base).
Step 4 — Write p = 5m:
Since 5 | p, write p = 5m for some integer m.
Step 5 — Substitute back:
(5m)² = 5q² ⇒ 25m² = 5q² ⇒ q² = 5m²
Step 6 — Conclude 5 | q:
q² = 5m² means 5 | q², so 5 | q.
Step 7 — The Contradiction:
Both p and q are divisible by 5, so GCD(p, q) ≥ 5. But we assumed GCD(p, q) = 1. Contradiction! ❌
Step 8 — Conclusion:
Therefore, √5 is irrational. ■
Enter a positive integer to check if it is a perfect square:
1. Is √4 irrational? Why not?
No, √4 is rational. √4 = 2, which is an integer, and every integer is rational (2 = 21).
4 is a perfect square (2² = 4), so its square root is a whole number, hence rational.
2. Is π rational or irrational?
π is irrational. It was proven by Johann Lambert in 1761 that π cannot be expressed as pq for any integers p and q.
π = 3.14159265358979323846… — the decimal never terminates and never repeats.
Important Note: 227 is only an approximation of π. 227 = 3.142857142857… (repeating), whereas π = 3.14159… (non-repeating). They are not equal!
3. Can the sum of two irrational numbers be rational? Give an example.
Yes! Consider √2 and (−√2).
Both √2 and −√2 are irrational numbers.
Their sum: √2 + (−√2) = 0, which is rational (= 01).
Another example: (3 + √2) + (3 − √2) = 6 (rational).
So the sum of two irrationals is not necessarily irrational!
Classify: √7, √16, 0.3333…, √(2/3), π, 22/7
| Number | Classification | Reason |
|---|---|---|
| √7 | Irrational | 7 is not a perfect square |
| √16 | Rational | √16 = 4 (16 is a perfect square) |
| 0.3333… = 0.̅3̅ | Rational | Repeating decimal = 13 |
| √(2/3) | Irrational | √(2/3) = √2/√3; neither simplifies to a rational |
| π | Irrational | Proven non-terminating, non-repeating (Lambert, 1761) |
| 22/7 | Rational | It is p/q with p=22, q=7 ∈ ℤ (it only approximates π) |
Step 1: Assume √3 is rational. Then √3 = pq where p, q ∈ ℤ, q ≠ 0, GCD(p, q) = 1.
Step 2: Square: 3 = p²q² ⇒ p² = 3q²
Step 3: So 3 | p². Since 3 is prime, 3 | p.
Step 4: Write p = 3m for some integer m.
Step 5: (3m)² = 3q² ⇒ 9m² = 3q² ⇒ q² = 3m²
Step 6: So 3 | q² and since 3 is prime, 3 | q.
Step 7: Both p and q divisible by 3 ⇒ GCD(p,q) ≥ 3. Contradicts GCD(p,q) = 1. ❌
Conclusion: √3 is irrational. ■
The number 0.1010010001… is irrational.
Reason: Look at the pattern — the number of 0s between consecutive 1s increases by one each time: 0.101001000100001…
This decimal is non-terminating (it goes on forever) and non-repeating (no fixed block of digits ever repeats periodically, because the gaps between 1s keep growing).
A decimal is rational if and only if it terminates or has a repeating block. Since neither condition holds here, the number is irrational.
Click / tap any coloured region to see examples from that set.
Every real number can be expressed as a decimal. The type of decimal tells us whether the number is rational or irrational.
1/4 = 0.257/8 = 0.8753/5 = 0.61/3 = 0.3 = 0.333…1/7 = 0.1428572/11 = 0.18√2 = 1.41421356…π = 3.14159265…e = 2.71828182…| Fraction | Denominator (lowest form) | Prime factors of denominator | Terminates? | Decimal |
|---|---|---|---|---|
| 3/8 | 8 = 2³ | Only 2 | Yes | 0.375 |
| 7/20 | 20 = 2²×5 | 2 and 5 only | Yes | 0.35 |
| 5/16 | 16 = 2⁴ | Only 2 | Yes | 0.3125 |
| 7/15 | 15 = 3×5 | 3 and 5 (has 3!) | No | 0.46 |
| 1/7 | 7 | 7 (not 2 or 5) | No | 0.142857 |
| 2/11 | 11 | 11 (not 2 or 5) | No | 0.18 |
Any non-terminating recurring decimal can be converted to a fraction. The method uses the idea that multiplying by a power of 10 shifts the decimal so that the recurring parts cancel when we subtract.
Step 1: Let x = 0.333…
Step 2: Multiply both sides by 10 (1 repeating digit):
10x = 3.333…
Step 3: Subtract: 10x − x = 3.333… − 0.333…
9x = 3
Step 4: x = 3/9 = 1/3
Non-repeating digits: 1 (count = 1). Repeating digits: 6 (count = 1).
Step 1: Let x = 0.1666…
Step 2: Shift past non-repeating part (multiply by 101 = 10):
10x = 1.666…
Step 3: Shift past repeating part too (multiply by 101+1 = 100):
100x = 16.666…
Step 4: Subtract: 100x − 10x = 16.666… − 1.666…
90x = 15
Step 5: x = 15/90 = 1/6
Non-repeating part after decimal point: "35" (count = 2). Repeating part: "7" (count = 1).
Step 1: Let x = 2.35777…
Step 2: Multiply by 102 = 100 (shift past 2 non-repeating digits):
100x = 235.777…
Step 3: Multiply by 102+1 = 1000 (shift past repeating part too):
1000x = 2357.777…
Step 4: Subtract: 1000x − 100x = 2357.777… − 235.777…
900x = 2122
Step 5: x = 2122/900 = 1061/450 (after dividing by gcd = 2)
Enter a fraction p/q (e.g. 3/8, 1/7, 5/12) to perform animated long division and classify the decimal.
| Fraction | Decimal | Type | Rational? |
|---|---|---|---|
| 3/8 | 0.375 | Terminating | Yes |
| 1/7 | 0.142857 | Non-term. Recurring (period 6) | Yes |
| 2/11 | 0.18 | Non-term. Recurring (period 2) | Yes |
| 5/16 | 0.3125 | Terminating | Yes |
| 7/12 | 0.583 | Non-term. Recurring (period 1) | Yes |
Any decimal number can be located on the number line by repeatedly zooming in — dividing the interval into 10 equal parts at each step.
Example: Locate 3.765 on the number line. Press Next Step to zoom in.
We can locate irrational numbers on the number line using geometric constructions with a ruler and compass. The key idea is Pythagoras' theorem.
Watch the construction step by step. Use the buttons below.
By repeatedly erecting unit perpendiculars on the previous hypotenuse, we generate a beautiful square root spiral (also called the Wheel of Theodorus).
Press Next √n to add the next triangle. Press Play All to animate all steps.
To locate √x for any positive real number x:
Answer: No, zero is not a natural number.
Reason:
Step 1 — Perform each operation:
Step 2 — Check membership in ℕ:
Since ℕ = {1, 2, 3, …}, all four results (19, 7, 24, 4) are positive integers and therefore belong to ℕ.
Conclusion: All four results are natural numbers.
Definition of Closure: A set S is closed under an operation ⊕ if for every a, b ∈ S, the result a ⊕ b is also in S.
Counterexample:
Choose two natural numbers: a = 3, b = 5. Both 3 ∈ ℕ and 5 ∈ ℕ.
Compute: 3 − 5 = −2
Now, −2 ∉ ℕ (natural numbers contain only positive integers: 1, 2, 3, …).
Conclusion: Since we found two natural numbers whose difference is not a natural number, ℕ is not closed under subtraction. ∴
Context: The Lebombo bone (c. 43,000 years old) is one of the oldest mathematical artifacts, used for tallying with notches.
Given:
Calculation:
Answer: 36 notches would be needed on the bone.
Whole numbers = {0, 1, 2, 3, 4, …}
Whole numbers less than 8 means all w ∈ 𝕎 such that w < 8.
There are 8 whole numbers less than 8 (from 0 to 7 inclusive).
Smallest Whole Number:
Smallest Natural Number:
The predecessor of a number n is n − 1.
For any whole number w, its predecessor would be w − 1.
Consider w = 0: predecessor = 0 − 1 = −1.
But −1 ∉ 𝕎 (whole numbers are non-negative).
For all other whole numbers n ≥ 1, the predecessor n − 1 is also a whole number.
Statement: Every natural number is a whole number.
Truth value: TRUE
Reason: ℕ = {1, 2, 3, …} and 𝕎 = {0, 1, 2, 3, …}. Every element of ℕ is also in 𝕎. So ℕ ⊂ 𝕎.
Converse: Every whole number is a natural number.
Truth value: FALSE
Counterexample: 0 ∈ 𝕎 but 0 ∉ ℕ.
Construction of the number line:
Position of each point:
(a) (−8) + 5:
Adding a positive to a negative: subtract magnitudes, keep sign of larger magnitude.
|−8| = 8, |5| = 5. Since 8 > 5, answer is negative: 8 − 5 = 3 ⇒ −3
(b) (−3) − (−7):
Subtracting a negative = adding its positive: (−3) + 7 = 4
(c) (−4) × (−6):
Rule: negative × negative = positive
4 × 6 = 24, so (−4) × (−6) = 24
(d) (−15) ÷ 3:
Rule: negative ÷ positive = negative
15 ÷ 3 = 5, so (−15) ÷ 3 = −5
Debt Analogy:
Interpretation:
(−3) × (−4) means: "Remove 3 debts of ₹4 each."
When you remove (cancel) 3 debts of ₹4 each, your net worth increases by:
Since this represents a gain (positive), the result is +12.
Verification: (−3) × (−4) = +12 = 12 ✓
Rule: On the number line, numbers increase from left to right. More negative = smaller value.
Step 1 — Identify negatives, zero, positives:
Step 2 — Order negatives (most negative first): −8 < −3 < −1
Step 3 — Order positives: 5 < 7
Answer: No, integers are NOT closed under division.
Counterexample:
Take 7 ∈ ℤ and 2 ∈ ℤ.
But 3.5 ∉ ℤ (it is not an integer).
Conclusion: Since dividing two integers can produce a non-integer result, ℤ is not closed under division.
Method: Multiply step by step, tracking the sign.
Given:
Calculation:
Interpretation: Starting 3 degrees below zero and rising 11 degrees brings us to 8 degrees above zero.
Number line: start at −3, move 11 steps right ⇒ reach +8.
Given:
Step-by-step calculation:
After rising: −200 + 75 = −125 m
After diving: −125 − 40 = −165 m
Answer: The submarine is at a depth of 165 m below sea level.
(a) 0.6
0.6 = 6/10 = 3/5 (dividing numerator and denominator by 2)
(b) −3
Any integer n can be written as n/1. So −3 = −3/1
(c) 2.75
2.75 = 275/100 = 11/4 (dividing by 25)
Verify: 11 ÷ 4 = 2.75 ✓ ⇒ 11/4
(d) 0.333… (0.3̄)
Let x = 0.333…
Then 10x = 3.333…
10x − x = 3.333… − 0.333…
9x = 3
x = 3/9 = 1/3
Method: Convert to equivalent fractions with a common denominator.
We need at least 5 numbers between 1/3 and 1/2.
LCM of 3 and 2 is 6, but we need a larger denominator to fit 5 numbers between them.
Multiply numerator and denominator of each fraction by 10:
1/3 = 10/30 and 1/2 = 15/30
But we need 5 numbers strictly between 10/30 and 15/30:
Available: 11/30, 12/30, 13/30, 14/30 — only 4 numbers.
Use denominator 60: 1/3 = 20/60 and 1/2 = 30/60
Now there are many integers between 20 and 30.
Left-Hand Side (LHS):
First compute 2/3 + 1/4:
LCD of 3 and 4 = 12
2/3 = 8/12 and 1/4 = 3/12
2/3 + 1/4 = 8/12 + 3/12 = 11/12
Now multiply: 1/2 × 11/12 = 11/24
Right-Hand Side (RHS):
1/2 × 2/3 = 2/6 = 1/3
1/2 × 1/4 = 1/8
1/3 + 1/8: LCD = 24
1/3 = 8/24 and 1/8 = 3/24
8/24 + 3/24 = 11/24
Step 1 — Find LCD of 3, 5, 6:
3 = 3, 5 = 5, 6 = 2 × 3
LCM = 2 × 3 × 5 = 30
Step 2 — Convert each fraction:
2/3 = 20/30 (multiply by 10/10)
4/5 = 24/30 (multiply by 6/6)
1/6 = 5/30 (multiply by 5/5)
Step 3 — Add and subtract:
Step 4 — Simplify:
GCD(39, 30) = 3
39/30 = 13/10 or equivalently 1.3
Step 1 — Perform long division of 3 ÷ 13:
3.000000 ÷ 13:
Type: Non-terminating, recurring decimal with period 6 (the block "230769" repeats).
Method: Algebraic approach for repeating decimals
Step 1: Let x = 0.444… = 0.4̄
Step 2: Multiply both sides by 10 (since 1 digit repeats):
10x = 4.444…
Step 3: Subtract original equation from new one:
10x − x = 4.444… − 0.444…
9x = 4
Step 4: Solve for x:
Verification: 4 ÷ 9 = 0.4444… ✓
(a) Every integer is a rational number.
TRUE
Any integer n can be written as n/1, which is of the form p/q with q ≠ 0.
For example: 5 = 5/1, −3 = −3/1, 0 = 0/1. All are rational.
Therefore: ℤ ⊂ ℚ (integers are a subset of rationals).
(b) Every rational number is an integer.
FALSE
Counterexample: 1/2 is rational (p=1, q=2, q≠0) but 1/2 ∉ ℤ.
Other examples: 3/7, −5/4, 0.6 = 3/5 — these are rational but not integers.
Given:
Setup:
(−5/8) × (other) = 15/56
Solve:
other = (15/56) ÷ (−5/8) = (15/56) × (8/−5)
= (15 × 8) / (56 × (−5))
= 120 / (−280)
= −120/280
Simplify: GCD(120, 280) = 40
Verification: (−5/8) × (−3/7) = 15/56 ✓
Method 1: Convert to fractions with denominator 4:
−2 = −8/4 and −1 = −4/4
Numbers between −8/4 and −4/4: −7/4, −6/4, −5/4
Verification on number line:
−2 < −7/4 < −3/2 < −5/4 < −1 ✓
i.e., −2 < −1.75 < −1.5 < −1.25 < −1 ✓
Given: a/b = 2/3, so a = 2k, b = 3k for some non-zero constant k.
Step 1 — Substitute a = 2k, b = 3k:
Numerator: 3a + 4b = 3(2k) + 4(3k) = 6k + 12k = 18k
Denominator: 3a − 4b = 3(2k) − 4(3k) = 6k − 12k = −6k
Step 2 — Compute the ratio:
The k cancels, confirming the result is independent of the value of k.
Recall: A number is rational if it can be expressed as p/q (p, q integers, q ≠ 0). Otherwise it is irrational.
√9: √9 = 3. Since 3 = 3/1 ∈ ℚ, it is RATIONAL.
√7: 7 is not a perfect square. √7 = 2.6457… (non-terminating, non-recurring). IRRATIONAL.
√0.04: √0.04 = √(4/100) = 2/10 = 0.2 = 1/5. RATIONAL.
√18: 18 = 2 × 3². √18 = 3√2. Since √2 is irrational, 3√2 is irrational. IRRATIONAL.
3 + √5: √5 is irrational. Adding 3 (rational) to an irrational gives an irrational number. IRRATIONAL.
π − 3: π is irrational. Subtracting 3 (rational) from π still gives an irrational. IRRATIONAL.
Proof by contradiction:
Step 1 — Assume the contrary: Suppose √3 is rational.
Then √3 = p/q where p, q are integers, q ≠ 0, and GCD(p,q) = 1 (fraction in lowest terms).
Step 2 — Square both sides:
3 = p²/q²
p² = 3q² …(i)
Step 3 — Deduce 3 divides p:
From (i), 3 divides p².
Since 3 is prime, 3 divides p² ⇒ 3 divides p (by Euclid’s lemma).
So p = 3m for some integer m.
Step 4 — Substitute back:
(3m)² = 3q²
9m² = 3q²
q² = 3m² …(ii)
Step 5 — Deduce 3 divides q:
From (ii), 3 divides q² ⇒ 3 divides q.
Step 6 — Contradiction:
We have shown 3 | p AND 3 | q, meaning GCD(p,q) ≥ 3.
But we assumed GCD(p,q) = 1. Contradiction!
Conclusion: Our assumption was wrong. Therefore, √3 is irrational. ∴
Proof by contradiction (same structure as √3):
Step 1: Assume √5 is rational. Then √5 = p/q with GCD(p,q) = 1.
Step 2 — Square: 5 = p²/q² ⇒ p² = 5q² …(i)
Step 3: 5 | p² ⇒ 5 | p (5 is prime, apply Euclid’s lemma).
Write p = 5m.
Step 4: (5m)² = 5q² ⇒ 25m² = 5q² ⇒ q² = 5m² …(ii)
Step 5: 5 | q² ⇒ 5 | q.
Step 6 — Contradiction: 5 | p and 5 | q ⇒ GCD(p,q) ≥ 5, contradicting GCD(p,q) = 1.
Conclusion: √5 is irrational. ∴
Answer: IRRATIONAL
Proof by contradiction:
Assume √2 + √3 = r, where r is rational.
Step 1: Then √3 = r − √2
Step 2 — Square both sides:
3 = r² − 2r√2 + 2
1 = r² − 2r√2
2r√2 = r² − 1
√2 = (r² − 1) / (2r)
Step 3: The right side is rational (r is rational, 2r ≠ 0).
So √2 would be rational — but we know √2 is irrational. Contradiction!
Conclusion: √2 + √3 is irrational.
Computation:
(√2)² = √2 × √2
By definition of square root: √2 × √2 = 2
Since 2 = 2/1 ∈ ℚ, it is rational.
Example:
Take a = √3 and b = −√3
Both √3 and −√3 are irrational (they cannot be expressed as p/q).
Their sum: √3 + (−√3) = 0
Since 0 = 0/1 is rational, we have found two irrationals with a rational sum.
Example:
Take a = √2 and b = √2
Both are irrational.
Product: √2 × √2 = 2
Another example: √3 × √12 = √36 = 6 (rational).
Yet another: √2 × √8 = √16 = 4 (rational).
Analysis of the decimal pattern:
The number is: 0.101001000100001…
Pattern: 1, then 1 zero; then 1, then 2 zeros; then 1, then 3 zeros; then 1, then 4 zeros; …
The number of zeros between consecutive 1s keeps increasing by 1 each time.
Is it terminating? No — it goes on forever.
Is it recurring (periodic)? No — no fixed block repeats. If it were recurring with period k, the pattern of zeros would eventually repeat with fixed length k, but here the zero-gaps grow without bound.
Conclusion: 0.101001000100001… is an irrational number.
Rule: p/q (in lowest terms) terminates iff q = 2m × 5n.
1/11: 11 = 11 (prime, neither 2 nor 5). ⇒ Recurring. 1/11 = 0.090909… = 0.0̄9̄
3/8: 8 = 2³. Only factor is 2. ⇒ Terminating. 3/8 = 0.375
7/25: 25 = 5². Only factor is 5. ⇒ Terminating. 7/25 = 0.28
5/12: 12 = 2² × 3. Has factor 3 (neither 2 nor 5). ⇒ Recurring. 5/12 = 0.41666… = 0.416̄
1/7: 7 = 7 (prime, not 2 or 5). ⇒ Recurring. 1/7 = 0.142857142857… = 0.1̄4̄2̄8̄5̄7̄ (period 6)
Step 1: Let x = 0.666… = 0.6̄
Step 2: Multiply by 10 (1 repeating digit):
10x = 6.666…
Step 3: Subtract:
10x − x = 6.666… − 0.666…
9x = 6
Step 4: x = 6/9 = 2/3 (simplify by dividing by 3)
Verify: 2 ÷ 3 = 0.6666… ✓
Observation: The digit 4 is non-repeating; 7 repeats. So this is of the form 0.47̄
Step 1: Let x = 0.4777…
Step 2: Multiply by 10 to move the non-recurring part past the decimal:
10x = 4.777…
Step 3: Multiply by 100 (one more step to align the recurring part):
100x = 47.777…
Step 4: Subtract the equation in Step 2 from Step 3:
100x − 10x = 47.777… − 4.777…
90x = 43
Verify: 43 ÷ 90 = 0.4777… ✓
Observation: 2.35̄ — "2.3" is non-recurring, "5" recurs.
Step 1: Let x = 2.3555…
Step 2: Multiply by 10:
10x = 23.555…
Step 3: Multiply by 100:
100x = 235.555…
Step 4: Subtract:
100x − 10x = 235.555… − 23.555…
90x = 212
Verify: 106 ÷ 45 = 2.3555… ✓
Rule: A fraction p/q in its lowest terms has a terminating decimal expansion if and only if q can be expressed in the form 2m × 5n where m, n are non-negative integers.
Step 1 — Check if 23/200 is in lowest terms:
GCD(23, 200) = 1 (since 23 is prime and does not divide 200). So it is already in lowest terms.
Step 2 — Factorise the denominator:
200 = 2 × 100 = 2 × 4 × 25 = 8 × 25 = 2³ × 5²
Conclusion: Since 200 = 2³ × 5², the decimal expansion of 23/200 is TERMINATING.
Construction using the Pythagorean theorem:
Stage 1 — Construct √2:
Stage 2 — Construct √3:
Successive Magnification Method:
Step 1 — First zoom (tenths):
4.26 lies between 4 and 5. Divide the segment [4, 5] into 10 equal parts.
Each part = 0.1. The digit in tenths place is 2, so 4.26 lies between 4.2 and 4.3.
Step 2 — Second zoom (hundredths):
Divide [4.2, 4.3] into 10 equal parts. Each part = 0.01.
The digit in hundredths place is 6, so 4.26 lies between 4.26 and 4.27.
More precisely, it is at the 6th mark from 4.2 in this subdivision.
The Square Root Spiral (Spiral of Theodorus):
Start with a right isosceles triangle of legs 1 and 1:
| Triangle | Legs | Hypotenuse |
|---|---|---|
| 1st | 1, 1 | √(1+1) = √2 |
| 2nd | √2, 1 | √(2+1) = √3 |
| 3rd | √3, 1 | √(3+1) = √4 = 2 |
| 4th | √4, 1 | √(4+1) = √5 |
The spiral demonstrates a beautiful geometric construction of successive square roots, each new hypotenuse = √(previous hypotenuse² + 1²) = √(n+1).
(a) Every irrational number is a real number.
TRUE
The set of real numbers ℝ = ℚ ∪ (irrationals). By definition, every irrational number is a real number. Examples: √2, √3, π, e — all are irrational and all are real.
(b) Every point on the number line is the square root of a natural number.
FALSE
The number line represents all real numbers. Consider π ≈ 3.14159… — it is on the number line but π ≠ √n for any natural number n (since π² ≈ 9.87 is not a natural number... well, it is not an integer, so π ≠ √n for n ∈ ℕ).
Also consider 1/2: it’s on the number line but 1/4 ∉ ℕ.
(c) Every real number is rational.
FALSE
ℝ = ℚ ∪ Irrationals. Irrational numbers like √2, π, √3 are real but NOT rational. The irrationals are an uncountably infinite subset of ℝ that are not in ℚ.
Answer: No. The square roots of all positive integers are NOT all irrational.
Examples of rational square roots:
In general, √n is rational if and only if n is a perfect square.
Proof by contradiction:
Step 1 — Assumption: Suppose √7 is rational.
Then √7 = p/q where p, q ∈ ℤ, q ≠ 0, and GCD(p, q) = 1.
Step 2 — Square both sides:
7 = p² / q²
p² = 7q² …(i)
Step 3 — Show 7 | p:
From (i), 7 divides p². Since 7 is prime, by Euclid’s lemma, 7 | p.
So p = 7m for some integer m.
Step 4 — Substitute p = 7m into (i):
(7m)² = 7q²
49m² = 7q²
q² = 7m² …(ii)
Step 5 — Show 7 | q:
From (ii), 7 | q² ⇒ 7 | q.
Step 6 — Contradiction:
We have 7 | p and 7 | q, so GCD(p, q) ≥ 7 ≠ 1. This contradicts GCD(p, q) = 1.
Conclusion: √7 is irrational. ∴
Rule: First reduce to lowest terms, then check if denominator = 2m × 5n.
23/8: GCD(23,8)=1 (lowest terms). 8 = 2³. ⇒ Terminating. 23/8 = 2.875
15/1600: GCD(15,1600). 15=3×5, 1600=2&sup6;×5². GCD=5. 15/1600 = 3/320. 320=2&sup6;×5. ⇒ Terminating. 3/320 = 0.009375
29/343: 343=7³. GCD(29,343)=1. 7 is neither 2 nor 5. ⇒ Non-terminating recurring.
3/15: GCD(3,15)=3. 3/15 = 1/5. 5=5¹. ⇒ Terminating. 1/5 = 0.2
77/210: GCD(77,210). 77=7×11. 210=2×3×5×7. GCD=7. 77/210 = 11/30. 30=2×3×5. Has factor 3. ⇒ Non-terminating recurring.
Non-terminating non-recurring decimals represent irrational numbers.
Three examples:
Approximate values: √2 ≈ 1.414 and √5 ≈ 2.236
We need three irrationals strictly between 1.414… and 2.236…
Method: Look for √n where 2 < n < 5, and n is not a perfect square.
Observation: "58" is the repeating block (2 digits).
Step 1: Let x = 0.585858… = 0.5̄8̄
Step 2: Multiply by 100 (since 2 digits repeat):
100x = 58.5858…
Step 3: Subtract the original:
100x − x = 58.5858… − 0.5858…
99x = 58
Verify: 58 ÷ 99 = 0.585858… ✓
Check if reducible: GCD(58, 99). 58 = 2×29. 99 = 9×11 = 3²×11. No common factors. So 58/99 is already in lowest terms.
Construction method:
To represent √x on the number line, use the following geometric construction:
√9.3 ≈ 3.05, so P lies just past 3 on the number line.
Proof by contradiction:
Step 1: Assume 3 + 2√5 is rational.
Then 3 + 2√5 = r for some rational number r.
Step 2 — Isolate √5:
2√5 = r − 3
√5 = (r − 3) / 2
Step 3 — Analyze the right side:
r is rational (assumed), and 3 is rational, so r − 3 is rational.
2 is rational and non-zero, so (r − 3)/2 is rational.
Therefore √5 is rational.
Step 4 — Contradiction:
But we proved (in Exercise 3.4, Q3) that √5 is irrational. Contradiction!
Conclusion: Our assumption was wrong. Therefore, 3 + 2√5 is irrational. ∴
Part (a): (√5 + √3)²
Use the identity: (a + b)² = a² + 2ab + b²
Here a = √5, b = √3:
= (√5)² + 2(√5)(√3) + (√3)²
= 5 + 2√15 + 3
Part (b): (√5 − √3)(√5 + √3)
Use the identity: (a − b)(a + b) = a² − b²
Here a = √5, b = √3:
= (√5)² − (√3)²
= 5 − 3
❌ Wrong: “0 is a natural number”
✅ Correct: 0 is a WHOLE number but NOT a natural number.
Natural numbers start at 1. Zero was added separately to form whole numbers. In India the NCERT convention: ℕ = {1,2,3,…}, W = {0,1,2,3,…}.
❌ Wrong: “22/7 = π”
✅ Correct: 22/7 ≈ π. 22/7 is rational; π is irrational.
22/7 is only an approximation accurate to 2 decimal places. π = 3.14159265… and cannot be written as any fraction.
❌ Wrong: “√4 is irrational”
✅ Correct: √4 = 2, which is a rational (in fact natural) number.
4 is a perfect square. √n is irrational only when n is not a perfect square.
❌ Wrong: “All decimals are irrational”
✅ Correct: Terminating and recurring decimals are rational.
E.g., 0.5 = 1/2 and 0.3&overline;3 = 1/3 are both rational. Only non-terminating, non-repeating decimals are irrational.
❌ Wrong: “a ÷ 0 = 0”
✅ Correct: Division by zero is UNDEFINED — not zero, not infinity.
There is no number that, multiplied by 0, gives a non-zero result. 0 ÷ 0 is also undefined (indeterminate form).
❌ Wrong: “Negative × Negative = Negative”
✅ Correct: (−) × (−) = Positive (+).
Sign rules: (+)(×)(+) = (+), (+)(×)(−) = (−), (−)(×)(+) = (−), (−)(×)(−) = (+). Two negatives always produce a positive product.
❌ Wrong: “√2 + √3 = √5”
✅ Correct: √2 + √3 ≠ √5. Square roots do not add like ordinary numbers!
(√2 + √3)2 = 5 + 2√6 ≠ 5. The correct value: √2 ≈ 1.414, √3 ≈ 1.732, sum ≈ 3.146, whereas √5 ≈ 2.236.
❌ Wrong: “0.999… < 1”
✅ Correct: 0.999… = 1 exactly — they are the same number.
Let x = 0.999…. Then 10x = 9.999…. Subtract: 9x = 9 ⇒ x = 1. Also: 1/3 = 0.333… ⇒ 3 × (1/3) = 0.999… = 1.
❌ Wrong: “Between 1/3 and 1/2 there are no rational numbers”
✅ Correct: There are INFINITELY many rational numbers between any two distinct rationals.
The mean 5/12 lies between them. Between 1/3 and 5/12 lies 3/8. This process never ends — rationals are dense.
❌ Wrong: “In the proof that √2 is irrational, assume p/q is any fraction”
✅ Correct: You MUST assume p and q are co-prime (in lowest terms, gcd = 1).
The contradiction arises because the algebra forces both p and q to be even, violating the co-prime assumption. Without this assumption there is no contradiction.
| Set | + | − | × | ÷ |
|---|---|---|---|---|
| ℕ | ✓ | ✗ | ✓ | ✗ |
| W | ✓ | ✗ | ✓ | ✗ |
| ℤ | ✓ | ✓ | ✓ | ✗ |
| ℚ | ✓ | ✓ | ✓ | ✓* |
* except division by 0, which is always undefined.
Same sign rules apply for division (÷).
Termination test for p/q (lowest terms):
Examples:
| Number | Approximate Value |
|---|---|
| √2 | 1.41421356… |
| √3 | 1.73205080… |
| √5 | 2.23606797… |
| √7 | 2.64575131… |
| √10 | 3.16227766… |
| π | 3.14159265… |
| e | 2.71828182… |