Comprehensive Class 9 Maths Chapter 3 notes covering rational, irrational numbers, number lines, and the real number system.
Long before humanity built cities or wrote equations, early humans needed to track their livestock and passage of time. They solved this using One-to-One Correspondence: matching physical objects (like pebbles in a clay pot along the Saraswati river) to animals in a herd.
Matching one pebble for every returning cow created the fundamental set of Natural Numbers: \( \mathbb{N} = \{1, 2, 3, 4, \dots\} \).
Discovered in the Lebombo Mountains (South Africa / Swaziland). Features 29 deliberate notches, believed to be the earliest lunar phase counter or time tracking calendar.
Found near the headwaters of the Nile (DR Congo). Features columns grouping prime numbers (11, 13, 17, 19) and multiplication by 2 (doubling patterns).
In Lothal and Harappa (Indus Valley), standardized cubic weights and measures enabled international trade. In Vedic literature, Indian scholars named immense powers of 10:
Scenario: A merchant in Lothal exchanges 2 bags of spices for 15 copper ingots. How many copper ingots will he receive for 12 bags of spices?
Solution:
Rate of exchange = \( \frac{15 \text{ ingots}}{2 \text{ bags}} \).
For 12 bags: \( \text{Ingots} = 12 \times \frac{15}{2} = 6 \times 15 = \mathbf{90 \text{ copper ingots}} \).
While Babylonians and Mayans used blank spaces or placeholder symbols to indicate empty columns, they did not treat "nothing" as an operational number. In India, philosophical traditions transformed emptiness into mathematics.
The Bakhśhālī Manuscript (early centuries CE) shows the earliest physical transition from blank space to a symbol: a solid dot (bindu) representing zero.
In his Brāhmasphuṭasiddhānta, Brahmagupta explicitly defined zero as \( a - a = 0 \) and established the arithmetic laws:
Brahmagupta expanded the number line to the left of zero by defining two financial realities:
Positive numbers representing assets, wealth, or moves to the right of zero.
Negative numbers representing debts, losses, or moves to the left of zero.
The set of Integers is denoted by \( \mathbb{Z} \) (from German Zahlen, meaning numbers):
Problem: A trader takes a loan (debt) of ₹850. The next day he makes a profit (fortune) of ₹1200. The following week he incurs a loss of ₹450. Calculate his final financial standing.
Solution:
Represent as signed integers: Debt = \( -850 \), Profit = \( +1200 \), Loss = \( -450 \).
\[ \text{Net Standing} = (-850) + (+1200) + (-450) = 1200 - 1300 = \mathbf{-100 \text{ rupees}} \]
His final standing is a debt of ₹100.
A Rational Number is any number that can be expressed in the form \( \frac{p}{q} \), where \( p \) and \( q \) are integers and \( q \neq 0 \). The set is denoted by \( \mathbb{Q} \) (for Quotient).
Division by zero is undefined in mathematics. Splitting a quantity into zero parts has no mathematical meaning!
Rational numbers are infinitely dense. Between any two rational numbers \( a \) and \( b \), there is ALWAYS another rational number given by their average:
Problem: Find 3 distinct rational numbers lying strictly between \( 1 \) and \( \frac{3}{2} \).
Solution:
1. First average between 1 and \( \frac{3}{2} \):
\[ m_1 = \frac{1 + \frac{3}{2}}{2} = \frac{\frac{5}{2}}{2} = \mathbf{\frac{5}{4}} \]
2. Second average between 1 and \( \frac{5}{4} \):
\[ m_2 = \frac{1 + \frac{5}{4}}{2} = \frac{\frac{9}{4}}{2} = \mathbf{\frac{9}{8}} \]
3. Third average between \( \frac{5}{4} \) and \( \frac{3}{2} \):
\[ m_3 = \frac{\frac{5}{4} + \frac{6}{4}}{2} = \frac{\frac{11}{4}}{2} = \mathbf{\frac{11}{8}} \]
Three rational numbers are \( \frac{9}{8}, \frac{5}{4}, \frac{11}{8} \).
Around 800 BCE, Baudhāyana (in his Śulba Sūtras) encountered lengths that could not be expressed as fractions: the diagonal of a unit square \( d^2 = 1^2 + 1^2 = 2 \implies d = \sqrt{2} \). Numbers that cannot be written as \( \frac{p}{q} \) are Irrational Numbers.
Greek mathematician Hippasus (c. 400 BCE) proved that \( \sqrt{2} \) is irrational using Proof by Contradiction:
Step 1 (Assumption): Assume \( \sqrt{2} \) is rational. Then \( \sqrt{2} = \frac{p}{q} \), where \( p, q \in \mathbb{Z}, q \neq 0 \), and \( \gcd(p, q) = 1 \) (co-prime).
Step 2 (Square both sides): \( 2 = \frac{p^2}{q^2} \implies p^2 = 2q^2 \).
Step 3 (Deduction for p): Since \( p^2 \) is 2 times an integer, \( p^2 \) is even, which implies \( p \) is even. Let \( p = 2k \).
Step 4 (Substitute p): \( (2k)^2 = 2q^2 \implies 4k^2 = 2q^2 \implies q^2 = 2k^2 \).
Step 5 (Deduction for q): Since \( q^2 \) is 2 times an integer, \( q^2 \) is even, which implies \( q \) is even.
Step 6 (The Contradiction!): Both \( p \) and \( q \) are even, meaning they share a common factor of 2. But Step 1 stated \( \gcd(p, q) = 1 \)!
Conclusion: Our initial assumption is FALSE. Therefore, \( \sqrt{2} \) is Irrational.
Āryabhaṭa (499 CE) gave the famous approximation \( \pi \approx \frac{3927}{1250} = 3.1416 \), noting it was an āsanna (approximation). In the 14th century, Mādhava of Sangamagrama (Kerala School of Mathematics) discovered that an exact representation of an irrational number requires an Infinite Series:
Uniting dense Rational Numbers (\( \mathbb{Q} \)) with Irrational Numbers (\( \mathbb{I} \)) forms the unbroken, continuous line of Real Numbers (\( \mathbb{R} \)).
| Number Type | Decimal Behavior | Example |
|---|---|---|
| Rational (\( \mathbb{Q} \)) | Terminating OR Repeating Block | \( \frac{3}{8} = 0.375 \) or \( \frac{1}{7} = 0.\overline{142857} \) |
| Irrational (\( \mathbb{I} \)) | Non-Terminating AND Non-Repeating | \( \sqrt{2} = 1.41421356\dots \) or \( \pi = 3.14159265\dots \) |
A rational fraction \( \frac{p}{q} \) (in lowest terms) has a terminating decimal if and only if the prime factorization of denominator \( q \) contains only 2s and/or 5s (\( q = 2^m \times 5^n \)).
The repeating block 142857 is a mathematical gem! When multiplied by digits 1 through 6, the digits simply rotate in a cyclic circle:
(A) Convert \( 0.\overline{6} \) to \( p/q \):
Let \( x = 0.6666\dots \)
Multiply by 10: \( 10x = 6.6666\dots \)
Subtract: \( 10x - x = 6 \implies 9x = 6 \implies x = \mathbf{\frac{2}{3}} \).
(B) Convert \( 0.1\overline{6} \) to \( p/q \):
Let \( x = 0.1666\dots \)
Shift 1 non-repeating digit: \( 10x = 1.6666\dots \)
Shift repeating cycle: \( 100x = 16.6666\dots \)
Subtract: \( 100x - 10x = 16.6 - 1.6 \implies 90x = 15 \implies x = \mathbf{\frac{1}{6}} \).
(C) Proof that \( 0.99999\dots = 1 \):
Let \( x = 0.9999\dots \)
\( 10x = 9.9999\dots \)
\( 10x - x = 9.9999\dots - 0.9999\dots \implies 9x = 9 \implies \mathbf{x = 1} \)!