Learning Objectives
After studying this chapter you should be able to:
- explain how counting began with one-to-one correspondence and natural numbers;
- describe the historical journey of numbers from the Lebombo and Ishango bones to the Indian numeral system;
- state how the idea of śhūnyatā (emptiness) grew into the mathematical number zero;
- state and apply Brahmagupta’s rules for zero and for adding and multiplying integers;
- define rational numbers as \(\dfrac{p}{q}\) with integers \(p\), \(q\) and \(q \neq 0\);
- represent rational numbers on the number line and use the absolute value \(|x|\) and distance \(|a - b|\);
- use the density property \((\frac{a+b}{2})\) to find rational numbers between any two rational numbers;
- explain why \(\sqrt{2}\) is irrational using proof by contradiction;
- construct the length \(\sqrt{2}\) and other roots on the number line using a compass;
- distinguish terminating, repeating and non-terminating non-repeating decimals;
- convert terminating and repeating decimals into the form \(\dfrac{p}{q}\);
- recognise cyclic numbers such as the block \(142857\) of \(\frac{1}{7}\);
- explain why \(0.\overline{9} = 1\); and
- describe real numbers as the union of rational and irrational numbers.
1. The Dawn of Mathematics: The Human Need to Count
Long before buildings, laws or the study of stars, humans had a basic practical need: to keep count. Mathematics began not in a classroom but in the dirt, on tree bark and on bones.
Imagine a herder living thousands of years ago near the Saraswati river with a herd of cattle. Every morning the cattle leave to graze; every evening they return. How does the herder make sure no calf has wandered off? Without number words or written symbols, early humans used one-to-one correspondence: for every cow that left, one pebble was placed in a clay pot; for every cow that returned, one pebble was removed. An empty pot at day’s end meant the herd was safe; leftover pebbles meant some cattle were missing.
Matching one object to another was the birth of the Natural Numbers, written \(\mathbb{N} = \{1, 2, 3, 4, \dots\}\). They are the counting numbers.
1.1 A History Written in Bone
The decimal place-value system we use today was perfected in the Indian subcontinent. Yet the earliest physical evidence of humans recording natural numbers comes from Africa — as tally marks carved into bone.
- Lebombo Bone: discovered in the Lebombo Mountains between South Africa and Swaziland, it is about 35,000 years old and carries 29 deliberately carved, uniformly sized notches. It was probably used as a lunar phase counter or a menstrual calendar — early humans were tracking time using natural numbers.
- Ishango Bone: found near the headwaters of the Nile River in the Democratic Republic of Congo, it dates to around 20,000 BCE. One of its three columns of notches groups tallies into 11, 13, 17, 19 — exactly the prime numbers between 10 and 20. Another column appears to show doubling (multiplying by 2). This tells us the abstract idea of a ‘number’ is tens of thousands of years old.
| Column on the bone | Grouping of notches | Mathematical idea |
|---|---|---|
| One column | 11, 13, 17, 19 | The prime numbers between 10 and 20 |
| Another column | Tallies grouped in pairs | The concept of multiplication by 2 (doubling) |
1.2 The Indian Context: Trade and Astronomy
As civilisations advanced, trade demanded larger numbers. In the ancient urban centres of the Indus Valley Civilisation — such as Lothal and Harappa — standardised weights and measures were essential. A merchant exchanging terracotta pottery, lapis lazuli or cotton with Mesopotamia needed a dependable system of accounting.
During Vedic times, Indian thinkers were fascinated by large numbers:
- The Vedas gave names for all powers of 10 up to \(10^{12}\), which was called parārdha.
- In the Lalitavistara (4th century BCE), the Buddha is described naming numbers up to \(10^{53}\), called tallakṣhaṇa.
- The Ṛigveda explicitly expressed quantities using powers of 10.
This way of thinking — quantities built from powers of 10 and place value — was developed into the Indian numeral system we still use worldwide today. It also prepared the ground for perhaps the most important mathematical invention in human history: the concept of zero.
Why is expressing quantities in powers of 10 so powerful? Think of writing \(10^{12}\) instead of 1,000,000,000,000.
2. The Revolution of Śhūnya: When Nothing Became Something
For millennia the number line started at 1. If you had five apples and gave all five away, you had no number for your state — only a void. The Babylonians and Mayans used placeholders (symbols for an empty column in a number), but they did not treat ‘nothing’ as a number with which one could add, subtract or multiply.
It was in the work of Brahmagupta that the void was formally transformed into a number — a leap that truly transformed mathematics. This was inspired by Indian philosophical traditions.
2.1 From Philosophy to Mathematics: The Concept of Śhūnyatā
In the Upanishads and the vast Buddhist literature beginning well before the 7th century BCE, śhūnyatā (emptiness or nothingness) was a profound state that was the goal of yoga and meditation. The word śhūnya means zero, and śhūnyatā (zeroness) described the state of emptying the mind of all vṛttis (mental fluctuations) to reach perfect stillness. Patanjali, in the Yoga Sutras (around the 3rd century BCE), describes how śhūnyatā can lead to control over the mind, body and senses.
Because Indian thinkers revered this state of emptiness, they were able to welcome ‘nothingness’ as a concept. Zeroness travelled into architecture, literature and linguistics, and finally into mathematics in the works of Āryabhaṭa and then Brahmagupta. The philosophical concept of emptiness crystallised into the mathematical zero.
2.2 The Bakhśhālī Manuscript and Brahmagupta’s Rules
In the Hindu number system we use today, the change from a blank space to a symbol can be seen in the Bakhśhālī Manuscript (dated to the early centuries CE), which uses a bold dot (bindu) to represent zero.
A symbol, however, is only a mark until it has rules. Brahmagupta, in his work the Brāhmasphuṭasiddhānta (628 CE), defined zero explicitly as the result of subtracting a number from itself, \(a - a = 0\), and then laid down the fundamental laws of arithmetic with śhūnya:
- Zero added to a number leaves it unchanged: \(a + 0 = a\).
- Zero subtracted from a number leaves it unchanged: \(a - 0 = a\).
- Any number multiplied by zero gives zero: \(a \times 0 = 0\).
There is no rule for dividing by zero. \(\dfrac{a}{0}\) is undefined: there is no number which, when multiplied by 0, gives \(a\). So whenever you write a fraction, the denominator must not be 0.
3. Integers: Expanding the Horizon
Brahmagupta did not stop at zero. He asked: if \(5 - 5 = 0\), what about \(3 - 5\)? To answer, he grounded mathematics in reality — commerce and life. He recognised two states:
- Fortunes (Dhana): positive numbers, representing wealth or assets;
- Debts (Ṛiṇa): negative numbers, representing debts.
By moving to the left of zero on the number line, Brahmagupta formally introduced negative numbers to the world. Positive natural numbers, their negatives and zero together make the set of Integers, written \(\mathbb{Z}\) (from the German word Zahlen, meaning numbers).
Figure 1: The integers. Debt (ṛiṇa) lies to the left of zero and fortune (dhana) to the right; zero itself is śhūnya.
3.1 The Arithmetic of Integers
Brahmagupta gave explicit rules for adding and multiplying integers, which we still use exactly as he wrote them over 1,300 years ago:
- A fortune plus a fortune is a fortune: \(5 + 4 = 9\).
- A debt plus a debt is a debt: \((-5) + (-4) = -9\). If you owe ₹5 and borrow ₹4 more, you owe ₹9.
- A fortune minus zero is a fortune, and a debt minus zero is a debt: \(7 - 0 = 7\), and \((-6) - 0 = -6\).
- The product of a debt and a fortune is a debt: \((-3) \times 4 = -12\). If you take on 4 debts of ₹3, your total debt is ₹12.
- The product of two debts is a fortune: \((-3) \times (-4) = 12\).
| Sign of first number | Sign of second number | Sign of product | Example |
|---|---|---|---|
| positive | positive | positive | \((3) \times (4) = 12\) |
| positive | negative | negative | \((3) \times (-4) = -12\) |
| negative | positive | negative | \((-3) \times (4) = -12\) |
| negative | negative | positive | \((-3) \times (-4) = 12\) |
Think in terms of action and debt. If a negative number represents a debt, then multiplying by a negative number represents the removal of that debt. If someone takes away (−) four of your debts that are each worth ₹3 (that is, \(-3\)), you are effectively ₹12 richer. Therefore \((-3) \times (-4) = +12\).
Real-life example: the temperature in the high-altitude desert of Ladakh is recorded as \(4^{\circ}\)C at noon. If it drops by \(15^{\circ}\)C by midnight, the midnight temperature is
\[4 - 15 = -11^{\circ}\text{C}\]
The negative sign means the temperature is 11 degrees below zero.
4. Filling the Spaces: Fractions and Rational Numbers
As society grew more complex, measuring became as important as counting. If a farmer divides a wheat field among three children, how much does each get? If a recipe calls for half a cup of ghee, how do we represent that?
Numbers that represent parts of a whole are called fractions. Just as every natural number has an additive inverse (\(3\) has \(-3\), \(19\) has \(-19\)), every positive fraction has an additive inverse: \(-\dfrac{3}{4}\) for \(\dfrac{3}{4}\), \(-\dfrac{19}{7}\) for \(\dfrac{19}{7}\). These are negative fractions. In a negative fraction the minus sign may be placed with the numerator or the denominator:
\[-\frac{1}{5} = \frac{-1}{5} = \frac{1}{-5}\]
When we combine all integers and all fractions (positive and negative), we get the set of Rational Numbers, denoted \(\mathbb{Q}\) (for quotient).
A rational number is any number that can be expressed in the form
\[\dfrac{p}{q}\]
where \(p\) and \(q\) are integers and \(\boxed{q \neq 0}\).
Because dividing by zero is undefined. A fraction \(\dfrac{a}{0}\) would ask “what number multiplied by 0 gives \(a\)?” — and no such number exists. So we simply forbid \(q = 0\).
Important observations
- Every rational number can be written as \(\dfrac{p}{q}\) with integers \(p\), \(q\) and \(q \neq 0\). For example, \(5 = \dfrac{5}{1}\) and \(-10 = \dfrac{-10}{1}\). So rational numbers include the natural numbers, whole numbers and integers.
- Rational numbers do not have a unique \(\dfrac{p}{q}\) representation: \(-\dfrac{1}{3} = -\dfrac{2}{6} = -\dfrac{3}{9} = -\dfrac{10}{30} = -\dfrac{2026}{6078}\). These are equivalent rational numbers (equivalent fractions). This lets us cancel common factors: \(\dfrac{12}{30} = \dfrac{2}{5}\) (divide numerator and denominator by 6).
- By convention, when we write a rational number \(\dfrac{p}{q}\) we take \(q \neq 0\) and \(p\), \(q\) with no common factor other than 1 (that is, co-prime). Among the infinitely many fractions equivalent to \(\dfrac{1}{2}\) (\(\dfrac{1}{2}, \dfrac{2}{4}, \dfrac{3}{6}, \dfrac{6}{12}, \dots\)), we choose \(\dfrac{1}{2}\) to represent them all.
4.1 The Laws of Rational Numbers
As you remember from Grades 6 and 7, Brahmagupta gave rules for the arithmetic of fractions that apply to both positive and negative fractions — the rational numbers:
- Equality: \(\dfrac{a}{b}\) and \(\dfrac{c}{d}\) are equal if and only if \(ad = bc\).
- Addition and subtraction: express the numbers with the same denominator \(b\), then \(\dfrac{a}{b} + \dfrac{c}{b} = \dfrac{a + c}{b}\) and \(\dfrac{a}{b} - \dfrac{c}{b} = \dfrac{a - c}{b}\).
- Multiplication: \(\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{ac}{bd}\), provided \(b \neq 0\) and \(d \neq 0\).
- Division: \(\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \times \dfrac{d}{c} = \dfrac{ad}{bc}\), provided \(b \neq 0\), \(d \neq 0\) and \(c \neq 0\).
- Commutativity: \(\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{c}{d} + \dfrac{a}{b}\) and \(\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{c}{d} \times \dfrac{a}{b}\).
- Distributivity: for rational numbers \(p\), \(q\), \(r\), \(p(q + r) = pq + pr\).
- Closure: rational numbers are closed under addition, subtraction and multiplication. They are also closed under division, provided we do not divide by zero.
4.2 Representation of Rational Numbers on the Number Line
We already know how to place integers on a number line. We first choose a point and mark it as 0, called the origin. One unit to the right is 1, two units is 2, and so on; one unit to the left is \(-1\), and so on. Each integer lies at an equal distance from the next. Rational numbers may lie between integers: \(\dfrac{1}{2}\) sits exactly halfway between 0 and 1, and \(-\dfrac{3}{4}\) lies between \(-1\) and 0.
To represent \(\dfrac{p}{q}\) (\(q \neq 0\)):
For example, to mark \(\dfrac{3}{4}\): divide the interval between 0 and 1 into four equal parts and move three parts to the right of 0. For \(\dfrac{9}{4} = 2\dfrac{1}{4}\): it lies between 2 and 3, so divide the interval between 2 and 3 into four equal parts and move one part to the right of 2.
Figure 2: \(\frac{1}{2}\) and \(\frac{3}{4}\) lie inside the interval 0 to 1, while \(\frac{9}{4}\) lies between 2 and 3.
Absolute value and distance
- The absolute value of a rational number \(x\), written \(|x|\), is its distance from 0 on the number line. So \(|x| \geq 0\) always. Example: \(\left|\dfrac{5}{3}\right| = \dfrac{5}{3}\), \(\left|-\dfrac{5}{3}\right| = \dfrac{5}{3}\), and \(|0| = 0\).
- The distance between two rational numbers \(a\) and \(b\) on the number line is \(\boxed{|a - b|}\). Example: the distance between \(-4\) and \(3\) is \(|3 - (-4)| = |7| = 7\).
Figure 3: The distance between \(a = -4\) and \(b = 3\) is \(|a - b| = |3 - (-4)| = 7\) units.
4.3 The Density of Rational Numbers
One of the most magical properties of rational numbers is that they are dense. Between the integers 1 and 2 there is \(\dfrac{3}{2}\); between 1 and \(\dfrac{3}{2}\) there is \(\dfrac{5}{4}\). No matter how close two rational numbers are, you can always find another one strictly between them by taking their average:
\[\frac{1 + \frac{3}{2}}{2} = \frac{\frac{5}{2}}{2} = \frac{5}{4}\]
For any two rational numbers \(a\) and \(b\) with \(a < b\), their average \(\dfrac{a + b}{2}\) is a rational number that lies strictly between \(a\) and \(b\). Repeating this process gives infinitely many rational numbers between any two points.
Why is \(\dfrac{a + b}{2}\) always between \(a\) and \(b\)? Because half of their sum sits exactly halfway along the segment joining them. Since \(a\) and \(b\) are rational, their sum and half of it are rational too — so the density argument can be repeated forever.
Figure 4: \(\frac{5}{4}\) is the average of 1 and \(\frac{3}{2}\), so it lies between them.
5. Irrational Numbers
For centuries, mathematicians believed every measurable length could be written as a ratio of two integers. When Baudhāyana composed his Śhulbasūtra (a manual for constructing geometric fire altars) around 800 BCE, he came across lengths that defied fractions. The ancient Greeks met the same crisis a few centuries later.
Consider a square whose side is exactly 1 unit. By the Baudhāyana–Pythagoras Theorem, if \(d\) is the diagonal, then
\[1^2 + 1^2 = d^2 \quad\Rightarrow\quad d^2 = 2\]
So the diagonal has length \(\sqrt{2}\). Can \(\sqrt{2}\) be written as a rational \(\dfrac{p}{q}\)? As you learned in Grade 8, the answer is no.
Numbers on the number line that cannot be expressed as a ratio of integers are called Irrational Numbers. Examples: \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{5}\), \(\pi\).
5.1 The Proof of the Irrationality of \(\sqrt{2}\)
The first proof of the irrationality of \(\sqrt{2}\) is credited to Hippasus, a member of the Pythagorean school (c. 400 BCE). He used a brilliant technique called proof by contradiction: assume the opposite of what you want to prove, and show that this assumption leads to a logical disaster.
Because every logical step was flawless, our initial assumption must have been wrong. Therefore \(\sqrt{2}\) cannot be written as a fraction — it is irrational.
Try to prove that \(\sqrt{3}\) is irrational using the same pattern. The same approach works for \(\sqrt{5}\), \(\sqrt{7}\) and \(\sqrt{10}\). Think: which fact about even numbers did the proof rely on?
5.2 Construction of the Length \(\sqrt{2}\)
We can construct a line segment of length \(\sqrt{2}\) and mark its position on the number line with a ruler and a compass:
Figure 5: \(OA = AB = 1\) unit, so \(OB = \sqrt{2}\) units. The arc from \(B\) down to the number line locates \(P\), the point \(\sqrt{2}\).
From \(P\) (=\(\sqrt{2}\)) draw a perpendicular of length 1 unit; the distance from \(O\) to that top point is \(\sqrt{3}\). Continuing, you can construct \(\sqrt{4}, \sqrt{5}, \sqrt{6}, \dots\) — any length of the form \(\sqrt{n}\) where \(n\) is a positive integer. This staircase of right triangles is called the square root spiral.
5.3 The Story of \(\pi\) and Mādhava’s Infinite Series
Another famous irrational number is \(\pi\), the ratio of a circle’s circumference to its diameter. For centuries mathematicians looked for better and better fractional approximations:
- Āryabhaṭa (499 CE) gave the highly accurate approximation \(\dfrac{3927}{1250} = 3.1416\), but called it asanna (an approximation), indicating that an exact fraction could probably never be found.
- π is irrational, as formally proven by Johann Lambert in 1761. No single fraction (or finitely many fractions) can ever give \(\pi\) exactly.
- An exact formula for \(\pi\) was unlocked in the 14th century by Mādhava of Sangamagrama, who founded the Kerala School of Mathematics. He realised that an irrational number cannot be expressed by one fraction — it needs an infinite sum: \(\pi = 4\left(1 - \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{7} + \cdots\right)\).
Adding infinitely many terms means finding the value we approach more and more closely as we add more and more terms, starting from the first. You will study infinite series in higher grades.
The number line is therefore filled not only with rational numbers but also with irrational numbers.
6. Real Numbers: Decimals and Cyclic Patterns
When we unite the dense web of rational numbers with the unfillable gaps of irrational numbers, we obtain the unbroken, continuous line of the Real Numbers, written \(\mathbb{R}\).
\(\mathbb{R} =\) (all rational numbers) \(\cup\) (all irrational numbers). Every real number has a point on the number line, and every point on the number line is a real number.
The easiest way to tell a rational number from an irrational number is to look at its decimal expansion.
6.1 Rational Decimals: Terminating and Repeating
If you divide the numerator of a rational number by its denominator, exactly one of two things happens:
- It terminates: the division eventually leaves a remainder of 0 and the decimal stops. Example: \(\dfrac{3}{8} = 0.375\).
- It repeats: the division never leaves a remainder of 0, but the digits begin to loop infinitely. Example: \(\dfrac{5}{11} = 0.454545\ldots = 0.\overline{45}\).
Try \(\dfrac{1}{7}\). In each step of long division, the possible remainders are only \(1, 2, 3, 4, 5, 6\). Because there are only finitely many possible remainders, a remainder must eventually repeat. Once a remainder repeats, the whole division process loops — producing a repeating decimal!
Predicting the type of decimal expansion
We can predict whether the decimal of \(\dfrac{p}{q}\) terminates without doing the long division. Analyse the prime factors of the denominator (after writing the fraction in lowest terms):
The decimal expansion of \(\dfrac{p}{q}\) (in lowest terms, \(q \neq 0\)) is terminating precisely when the prime factors of \(q\) are only 2, only 5, or both 2 and 5. If \(q\) has any other prime factor, the decimal repeats.
Why? Because with only 2s and 5s we can multiply numerator and denominator by a suitable number to make the denominator a power of 10. Example: \(\dfrac{3}{20}\). Since \(20 = 2^2 \times 5\), choose the factor 5 to make the denominator 100:
\[\frac{3}{20} = \frac{3 \times 5}{20 \times 5} = \frac{15}{100} = 0.15\]
Converting decimals into \(\dfrac{p}{q}\)
Case 1 — Terminating decimals. Write the digits over the appropriate power of 10 and simplify. Example: \(0.35 = \dfrac{35}{100} = \dfrac{7}{20}\).
Case 2 — Pure repeating decimals. The repeating block begins immediately after the decimal point. Let \(x\) equal the decimal, multiply by \(10^n\) where \(n\) is the number of repeating digits, subtract, and solve.
Worked: \(0.\overline{6} = ?\)
Let \(x = 0.\overline{6}\). One digit repeats, so multiply by \(10^1 = 10\): \(10x = 6.\overline{6}\). Subtract: \(10x - x = 6.\overline{6} - 0.\overline{6} = 6\). So \(9x = 6\) and \(x = \dfrac{6}{9} = \dfrac{2}{3}\).
Worked: \(0.\overline{45} = ?\)
Let \(x = 0.\overline{45}\). Two digits repeat, so multiply by \(10^2 = 100\): \(100x = 45.\overline{45}\). Subtract: \(99x = 45\), so \(x = \dfrac{45}{99} = \dfrac{5}{11}\).
Case 3 — General repeating decimals. There are some non-repeating digits just after the decimal point, followed by a repeating block. First shift the non-repeating digits (multiply by \(10^m\)), then shift a full repeating cycle (multiply by \(10^n\)), subtract, and solve.
Worked: \(0.1\overline{6} = ?\)
Let \(x = 0.1\overline{6}\). One digit (\(1\)) is non-repeating and one digit (\(6\)) repeats. Multiply by \(10^1 = 10\): \(10x = 1.\overline{6}\). Multiply by 10 again: \(100x = 16.\overline{6}\). Subtract: \(100x - 10x = 16.\overline{6} - 1.\overline{6}\), so \(90x = 15\) and \(x = \dfrac{15}{90} = \dfrac{1}{6}\).
Worked: \(2.35\overline{7} = ?\)
Here ‘35’ is non-repeating (2 digits) and ‘7’ repeats (1 digit). Let \(x = 2.35\overline{7}\). Shift the non-repeating part: \(100x = 235.\overline{7}\). Shift the full cycle: \(1000x = 2357.\overline{7}\). Subtract: \(1000x - 100x = 2357.\overline{7} - 235.\overline{7}\), so \(900x = 2122\) and \(x = \dfrac{2122}{900} = \dfrac{1061}{450}\).
Worked: \(2.4\overline{37} = ?\)
Here the non-repeating digit is ‘4’ (1 digit) and the repeating block is ‘37’ (2 digits). Let \(x = 2.4\overline{37}\). Shift the non-repeating part: \(10x = 24.\overline{37}\). Shift the full cycle: \(1000x = 2437.\overline{37}\). Subtract: \(990x = 2413\), so \(x = \dfrac{2413}{990}\). Check by division: this equals \(2.4373737\ldots\).
| Decimal type | Steps to follow |
|---|---|
| Terminating | Write the digits over a power of 10 and simplify (e.g. \(0.35 = \frac{35}{100} = \frac{7}{20}\)). |
| Pure repeating | Let \(x\) = the decimal. Multiply by \(10^n\) where \(n\) = number of repeating digits. Subtract the original equation and solve for \(x\). |
| General repeating | Let \(x\) = the decimal. Multiply by \(10^m\) (non-repeating digits), then by \(10^n\) (repeating digits). Subtract the two shifted equations and solve for \(x\). |
6.2 The Magic of Cyclic Numbers
When we calculate \(\dfrac{1}{7}\) we get \(0.142857142857\ldots = 0.\overline{142857}\). The repeating block \(142857\) is a cyclic number. Watch what happens when we multiply it by 1 through 6:
| Product | Result |
|---|---|
| \(142857 \times 1\) | 142857 |
| \(142857 \times 2\) | 285714 |
| \(142857 \times 3\) | 428571 |
| \(142857 \times 4\) | 571428 |
| \(142857 \times 5\) | 714285 |
| \(142857 \times 6\) | 857142 |
The same six digits appear each time — they simply shift around in a cyclic circle. This beautiful internal structure is a hallmark of the rational number \(\dfrac{1}{7}\).
6.3 Irrational Decimals: Chaos and Infinity
Irrational numbers have decimal expansions that never end and never repeat — there is no cyclic block and no repeating pattern:
- \(\sqrt{2} = 1.41421356237\ldots\)
- \(\pi = 3.14159265358\ldots\)
| Kind of number | Decimal expansion | Example |
|---|---|---|
| Rational | terminating, or non-terminating and repeating | \(\frac{3}{8} = 0.375\), \(\frac{5}{11} = 0.\overline{45}\) |
| Irrational | non-terminating and non-repeating | \(\sqrt{2} = 1.41421356237\ldots\), \(\pi = 3.14159\ldots\) |
6.4 The Surprising Case: \(0.\overline{9} = 1\)
Just as \(1 = \dfrac{10}{10} = \dfrac{100}{100}\), rational numbers can have two decimal forms. Any terminating decimal has an alternative ending in repeating 9s:
\[1.000\ldots = 0.\overline{9}, \qquad 2.47000\ldots = 2.46999\ldots\]
Worked: prove \(0.\overline{9} = 1\)
Let \(x = 0.\overline{9}\). One digit repeats, so multiply by 10: \(10x = 9.\overline{9}\). Subtract: \(10x - x = 9.\overline{9} - 0.\overline{9} = 9\). Hence \(9x = 9\), so \(x = 1\). Therefore \(0.\overline{9}\) is exactly equal to 1 — not “a little less than 1”.
6.5 Conclusion: The Never-Ending Journey
Our world of numbers has grown in stages over thousands of years, each stage meeting a need of humanity:
- Natural numbers (\(\mathbb{N}\)): the counting numbers \(\{1, 2, 3, \dots\}\), contained in the integers.
- Integers (\(\mathbb{Z}\)): \(\{\dots, -2, -1, 0, 1, 2, \dots\}\), contained in the rational numbers.
- Rational numbers (\(\mathbb{Q}\)): all fractions \(\dfrac{p}{q}\) with integers \(p, q\), \(q \neq 0\); those with terminating or repeating decimals, like \(0.135\) and \(0.\overline{142857}\).
- Irrational numbers: numbers that cannot be written as fractions (e.g. \(\sqrt{2}\), \(\pi\), \(\sqrt{10}\)) and whose decimals never terminate or repeat.
- Real numbers (\(\mathbb{R}\)): rational numbers together with irrational numbers make the entire real number line — every length, every temperature, every physical measurement has a home on it.
Is the journey over? Consider this puzzle: what is \(\sqrt{-1}\)? Since \(1 \times 1 = 1\) and \((-1) \times (-1) = 1\), no real number multiplies by itself to give a negative number. So \(\sqrt{-1}\) cannot lie on the real number line. Mathematicians stepped completely off the line and invented a new kind of number denoted \(i\), standing for Imaginary Numbers. They are essential for electrical engineering, quantum mechanics and the technology in your mobile phone — but that is a journey for another year. For now, master the real numbers.
7. Key Results and Formulas
- Rational number: \(\dfrac{p}{q}\) where \(p, q \in \mathbb{Z}\) and \(q \neq 0\). \(\mathbb{Q}\) includes \(\mathbb{N}\), whole numbers and \(\mathbb{Z}\).
- Equality: \(\dfrac{a}{b} = \dfrac{c}{d} \iff ad = bc\).
- Addition and subtraction: \(\dfrac{a}{b} \pm \dfrac{c}{b} = \dfrac{a \pm c}{b}\).
- Multiplication: \(\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{ac}{bd}\) (\(b, d \neq 0\)).
- Division: \(\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \times \dfrac{d}{c} = \dfrac{ad}{bc}\) (\(b, d, c \neq 0\)).
- Commutativity & distributivity: \(p + q = q + p\), \(p \times q = q \times p\), \(p(q + r) = pq + pr\).
- Absolute value: \(|x|\) is the distance of \(x\) from 0; always \(|x| \geq 0\).
- Distance: distance between \(a\) and \(b\) on the number line is \(|a - b|\).
- Density: \(\dfrac{a + b}{2}\) is a rational number strictly between \(a\) and \(b\); there are infinitely many rationals between any two rationals.
- Terminating decimals: \(\dfrac{p}{q}\) (lowest terms) terminates iff the prime factors of \(q\) are only 2, only 5, or both.
- Repeating decimals: every repeating decimal equals a rational number; it converts to \(\dfrac{p}{q}\) by shifting the decimal.
- Irrational: \(\sqrt{2}, \sqrt{3}, \sqrt{5}, \pi, \ldots\) cannot be written as \(\dfrac{p}{q}\); their decimals never terminate or repeat.
- Special result: \(0.\overline{9} = 1\).
- Real numbers: \(\mathbb{R} = \mathbb{Q} \cup\) irrational numbers — the complete number line.
8. Solved Examples
Each example is worked out in four parts: Question, Concept, Solution (step by step) and Final answer.
A. Counting, history and number systems
Example 1
Question: A merchant in the port city of Lothal exchanges bags of spices for copper ingots. He receives 15 ingots for every 2 bags of spices. If he brings 12 bags to the market, how many copper ingots will he leave with?
Concept: ratios — set up the comparison 15 ingots per 2 bags and scale it.
Solution: In 12 bags there are \(12 \div 2 = 6\) groups of 2 bags. Each group gives 15 ingots, so he gets \(6 \times 15 = 90\) ingots.
Final answer: 90 copper ingots.
Example 2
Question: Look at the numbers on one column of the Ishango bone: 11, 13, 17, 19. What do they have in common? List the next three numbers that fit the pattern.
Concept: identifying primes between 10 and 20.
Solution: Each is a prime number (divisible only by 1 and itself), and together they are exactly the prime numbers between 10 and 20. Continuing after 19, the next three primes are 23, 29, 31.
Final answer: primes between 10 and 20; next three are 23, 29, 31.
Example 3
Question: Natural numbers are closed under addition (the sum of any two natural numbers is a natural number). Are they closed under subtraction? Justify with examples.
Concept: closure of a set under an operation requires the answer to stay in the set.
Solution: No. For example, \(3 - 5 = -2\), and \(-2\) is not a natural number; likewise \(1 - 2 = -1\). So subtracting can leave the set of natural numbers.
Final answer: not closed under subtraction, since e.g. \(3 - 5 = -2 \notin \mathbb{N}\).
Example 4
Question: Ancient Indians counted with the joints of their fingers, using the thumb to count them. Each finger has 3 joints. How many can you count on one hand, and how does this relate to the ancient base-12 counting systems?
Concept: combinatorial counting with the four fingers and the thumb as pointer.
Solution: On one hand there are 4 fingers, each with 3 joints, counted with the thumb: \(4 \times 3 = 12\) counts. Twelve is exactly the base of the ancient base-12 (duodecimal) counting systems, which is why 12-based groupings appear in so many old units (dozen, hour, months in a year).
Final answer: 12 counts on one hand — a natural link to base-12 numbers.
B. Zero and the arithmetic of integers
Example 5
Question: Using Brahmagupta’s rules, compute: (a) \(8 - 8\) (b) \(8 + 0\) (c) \(8 - 0\) (d) \(8 \times 0\). Also state what is not defined.
Concept: the rules for zero: \(a - a = 0\), \(a + 0 = a\), \(a - 0 = a\), \(a \times 0 = 0\).
Solution: (a) \(8 - 8 = 0\) (a number minus itself is zero). (b) \(8 + 0 = 8\). (c) \(8 - 0 = 8\). (d) \(8 \times 0 = 0\). Division by zero (\(\dfrac{8}{0}\)) is undefined.
Final answer: 0, 8, 8, 0; \(\dfrac{8}{0}\) is undefined.
Example 6
Question: The temperature in the high-altitude desert of Ladakh is recorded as \(4^{\circ}\)C at noon. By midnight it drops by \(15^{\circ}\)C. What is the midnight temperature?
Concept: a drop is a subtraction; going below zero gives a negative temperature.
Solution: \(4 - 15 = -11\). Starting from 4 and counting down 15 units lands 11 units below zero.
Final answer: \(-11^{\circ}\)C.
Example 7
Question: A spice trader takes a loan (debt) of ₹850. The next day he makes a profit (fortune) of ₹1,200. The following week he incurs a loss of ₹450. Write this sequence as an integer equation and find his final financial standing.
Concept: debts are negative and fortunes positive; add them all.
Solution: Loan \(= -850\), profit \(= +1200\), loss \(= -450\). Total: \(-850 + 1200 - 450\). First \(-850 + 1200 = 350\); then \(350 - 450 = -100\).
Final answer: \(-100\), i.e. he is still ₹100 in debt.
Example 8
Question: Using Brahmagupta’s laws, compute: (i) \((-12) \times 5\) (ii) \((-8) \times (-7)\) (iii) \(0 - (-14)\) (iv) \((-20) \div 4\).
Concept: sign rules for products; subtracting a negative adds; dividing by a positive keeps the sign of the dividend.
Solution: (i) debt × fortune = debt: \((-12) \times 5 = -60\). (ii) debt × debt = fortune: \((-8) \times (-7) = 56\). (iii) \(0 - (-14) = 0 + 14 = 14\). (iv) \((-20) \div 4 = -5\).
Final answer: (i) \(-60\) (ii) \(56\) (iii) \(14\) (iv) \(-5\).
Example 9
Question: Using a real-world example of debt, explain why subtracting a negative number is the same as adding a positive number, e.g. \(10 - (-5) = 15\).
Concept: subtracting a debt removes it, making you richer.
Solution: You have ₹10 in cash. Someone cancels (removes) your debt of ₹5. You have not lost anything — you are effectively ₹5 richer. So \(10 - (-5) = 10 + 5 = 15\).
Final answer: subtracting a debt of 5 adds 5; \(10 - (-5) = 15\).
C. Rational numbers: equality and operations
Example 10
Question: Prove that \(\dfrac{2}{3}\) and \(\dfrac{4}{6}\) are equal.
Concept: \(\dfrac{a}{b} = \dfrac{c}{d}\) iff \(ad = bc\).
Solution: Cross-multiply: \(2 \times 6 = 12\) and \(3 \times 4 = 12\). Since the cross products are equal, the fractions are equal.
Final answer: \(\dfrac{2}{3} = \dfrac{4}{6}\) because \(2 \times 6 = 3 \times 4 = 12\).
Example 11
Question: Find the sum: \(\dfrac{2}{5} + \dfrac{3}{10}\).
Concept: express both fractions with the same denominator, then add numerators.
Solution: LCM of 5 and 10 is 10. \(\dfrac{2}{5} = \dfrac{4}{10}\). So \(\dfrac{4}{10} + \dfrac{3}{10} = \dfrac{7}{10}\).
Final answer: \(\dfrac{7}{10}\).
Example 12
Question: Find the difference: \(\dfrac{5}{6} - \dfrac{1}{4}\).
Concept: same denominator, then subtract numerators.
Solution: LCM of 6 and 4 is 12. \(\dfrac{5}{6} = \dfrac{10}{12}\), \(\dfrac{1}{4} = \dfrac{3}{12}\). So \(\dfrac{10}{12} - \dfrac{3}{12} = \dfrac{7}{12}\).
Final answer: \(\dfrac{7}{12}\).
Example 13
Question: Find the product: \(\dfrac{2}{3} \times \dfrac{3}{10}\).
Concept: multiply numerators and denominators, then simplify.
Solution: \(\dfrac{2 \times 3}{3 \times 10} = \dfrac{6}{30} = \dfrac{1}{5}\).
Final answer: \(\dfrac{1}{5}\).
Example 14
Question: Find the quotient: \(\dfrac{2}{3} \div \dfrac{3}{10}\).
Concept: multiply by the reciprocal of the second fraction.
Solution: \(\dfrac{2}{3} \div \dfrac{3}{10} = \dfrac{2}{3} \times \dfrac{10}{3} = \dfrac{20}{9}\).
Final answer: \(\dfrac{20}{9}\).
Example 15
Question: Show that \(\left(\dfrac{1}{2} + \dfrac{3}{4}\right) \times \dfrac{8}{3} = \dfrac{1}{2} \times \dfrac{8}{3} + \dfrac{3}{4} \times \dfrac{8}{3}\).
Concept: the distributive law \(p(q + r) = pq + pr\).
Solution: LHS: \(\dfrac{1}{2} + \dfrac{3}{4} = \dfrac{5}{4}\), so \(\dfrac{5}{4} \times \dfrac{8}{3} = \dfrac{40}{12} = \dfrac{10}{3}\). RHS: \(\dfrac{1}{2} \times \dfrac{8}{3} = \dfrac{8}{6} = \dfrac{4}{3}\), and \(\dfrac{3}{4} \times \dfrac{8}{3} = \dfrac{24}{12} = 2\); sum \(= \dfrac{4}{3} + 2 = \dfrac{10}{3}\). LHS = RHS.
Final answer: both sides equal \(\dfrac{10}{3}\); distributivity verified.
Example 16
Question: Simplify \(\dfrac{7}{9} \times \left(\dfrac{6}{7} - \dfrac{3}{4}\right)\) using the distributive property.
Concept: distribute the multiplier over the difference.
Solution: \(= \dfrac{7}{9} \times \dfrac{6}{7} - \dfrac{7}{9} \times \dfrac{3}{4} = \dfrac{6}{9} - \dfrac{21}{36} = \dfrac{2}{3} - \dfrac{7}{12} = \dfrac{8}{12} - \dfrac{7}{12} = \dfrac{1}{12}\).
Final answer: \(\dfrac{1}{12}\).
D. Number line, absolute value and density
Example 17
Question: Represent \(\dfrac{3}{4}\) and \(\dfrac{9}{4}\) on a number line.
Concept: divide unit intervals into 4 equal parts.
Solution: For \(\dfrac{3}{4}\), divide the interval [0, 1] into four equal parts and move 3 parts right of 0. For \(\dfrac{9}{4} = 2\dfrac{1}{4}\), divide the interval [2, 3] into four equal parts and move 1 part right of 2. (See Figure 2 above.)
Final answer: \(\dfrac{3}{4}\) inside [0, 1]; \(\dfrac{9}{4}\) inside [2, 3].
Example 18
Question: Give \(\left|\dfrac{5}{3}\right|\), \(\left|-\dfrac{5}{3}\right|\) and \(|0|\), and find the distance between \(-4\) and \(3\).
Concept: \(|x|\) is the distance from 0; distance between \(a\) and \(b\) is \(|a - b|\).
Solution: \(\left|\dfrac{5}{3}\right| = \dfrac{5}{3}\), \(\left|-\dfrac{5}{3}\right| = \dfrac{5}{3}\), \(|0| = 0\). Distance: \(|3 - (-4)| = |3 + 4| = 7\).
Final answer: \(\dfrac{5}{3}, \dfrac{5}{3}, 0\); distance \(= 7\) units.
Example 19
Question: Find three distinct rational numbers strictly between \(-\dfrac{1}{2}\) and \(\dfrac{1}{4}\).
Concept: density — use a common denominator or averages.
Solution: With denominator 8, \(-\dfrac{1}{2} = -\dfrac{4}{8}\) and \(\dfrac{1}{4} = \dfrac{2}{8}\). Between them lie \(-\dfrac{3}{8}\), \(0 = \dfrac{0}{8}\), and \(\dfrac{1}{8}\). (Many answers are possible.)
Final answer: e.g. \(-\dfrac{3}{8}\), 0 and \(\dfrac{1}{8}\).
Example 20
Question: Using the average method, find a rational number between 1 and \(\dfrac{3}{2}\).
Concept: \(\dfrac{a + b}{2}\) always lies between \(a\) and \(b\).
Solution: \(\dfrac{1 + \frac{3}{2}}{2} = \dfrac{\frac{5}{2}}{2} = \dfrac{5}{4}\). Since \(1 < \dfrac{5}{4} < \dfrac{3}{2}\), it works. Repeating with 1 and \(\dfrac{5}{4}\) would give \(\dfrac{9}{8}\), and so on forever.
Final answer: \(\dfrac{5}{4}\) (and infinitely many more).
Example 21
Question: Find three rational numbers between 3.1415 and 3.1416.
Concept: use digits between the two values (density of rationals).
Solution: 3.14151, 3.14152 and 3.14155 all lie strictly between 3.1415 and 3.1416. Each is rational: e.g. \(3.14151 = \dfrac{314151}{100000}\).
Final answer: e.g. 3.14151, 3.14152, 3.14155.
Example 22
Question: Find six rational numbers between 3 and 4.
Concept: pick terminating decimals (rationals) inside the interval.
Solution: 3.1, 3.2, 3.3, 3.5, 3.7, 3.9 — each can be written as a fraction with a power of 10 in the denominator (e.g. \(3.1 = \dfrac{31}{10}\)), so each is rational and lies strictly between 3 and 4.
Final answer: e.g. 3.1, 3.2, 3.3, 3.5, 3.7, 3.9.
E. Irrational numbers and classification
Example 23
Question: Classify as rational or irrational, and give the fraction where the number is rational: (i) \(\sqrt{81}\) (ii) \(\sqrt{12}\) (iii) \(0.33333\ldots\) (iv) \(0.123451234512345\ldots\) (v) \(1.01001000100001\ldots\) (vi) \(23.560185612239874790120\).
Concept: simplify roots; check whether the decimal terminates or repeats.
Solution:
- \(\sqrt{81} = 9\), which is rational, \(= \dfrac{9}{1}\).
- \(\sqrt{12}\) cannot be simplified to an integer, so it is irrational.
- \(0.33333\ldots = 0.\overline{3}\) repeats, so it is rational \(= \dfrac{1}{3}\).
- \(0.123451234512345\ldots = 0.\overline{12345}\) repeats, so it is rational \(= \dfrac{12345}{99999} = \dfrac{4115}{33333}\).
- \(1.01001000100001\ldots\) never ends and never repeats a block, so it is irrational.
- \(23.560185612239874790120\) terminates, so it is rational \(= \dfrac{23560185612239874790120}{10^{21}}\).
Final answer: (i), (iii), (iv), (vi) rational; (ii), (v) irrational.
F. Decimals: predicting and converting
Example 24
Question: Without performing long division, decide which of \(\dfrac{7}{20}\), \(\dfrac{4}{15}\), \(\dfrac{13}{250}\) have terminating decimals. Then write each as a decimal to confirm.
Concept: prime factors of the denominator decide termination.
Solution: \(20 = 2^2 \times 5\) → terminating; \(15 = 3 \times 5\) → repeating; \(250 = 2 \times 5^3\) → terminating. Confirm: \(\dfrac{7}{20} = 0.35\), \(\dfrac{4}{15} = 0.2666\ldots = 0.2\overline{6}\), \(\dfrac{13}{250} = 0.052\).
Final answer: \(\dfrac{7}{20}\) and \(\dfrac{13}{250}\) terminate; \(\dfrac{4}{15}\) repeats.
Example 25
Question: Convert \(0.35\), \(0.\overline{6}\) and \(0.\overline{45}\) into the form \(\dfrac{p}{q}\).
Concept: terminating → divide by power of 10; pure repeating → multiply, subtract, solve.
Solution: \(0.35 = \dfrac{35}{100} = \dfrac{7}{20}\). For \(0.\overline{6}\): \(10x = 6.\overline{6}\), subtract \(x\): \(9x = 6\), \(x = \dfrac{6}{9} = \dfrac{2}{3}\). For \(0.\overline{45}\): \(100x = 45.\overline{45}\), subtract: \(99x = 45\), \(x = \dfrac{45}{99} = \dfrac{5}{11}\).
Final answer: \(\dfrac{7}{20}\), \(\dfrac{2}{3}\), \(\dfrac{5}{11}\).
Example 26
Question: Convert the general repeating decimal \(0.1\overline{6}\) into \(\dfrac{p}{q}\).
Concept: shift the non-repeating part first, then the full cycle.
Solution: Let \(x = 0.1\overline{6}\). \(10x = 1.\overline{6}\) and \(100x = 16.\overline{6}\). Subtract: \(90x = 15\), so \(x = \dfrac{15}{90} = \dfrac{1}{6}\).
Final answer: \(\dfrac{1}{6}\).
Example 27
Question: Convert \(2.35\overline{7}\) into \(\dfrac{p}{q}\).
Concept: ‘35’ non-repeating (2 digits), ‘7’ repeating (1 digit).
Solution: Let \(x = 2.35\overline{7}\). \(100x = 235.\overline{7}\); \(1000x = 2357.\overline{7}\). Subtract: \(900x = 2122\), so \(x = \dfrac{2122}{900} = \dfrac{1061}{450}\).
Final answer: \(\dfrac{1061}{450}\).
G. Cyclic numbers and 0.9̅ = 1
Example 28
Question: \(\dfrac{1}{7} = 0.\overline{142857}\). Write the repeating block for \(142857 \times 4\) and \(142857 \times 6\).
Concept: cyclic number — multiplication rotates the digits.
Solution: \(142857 \times 4 = 571428\) and \(142857 \times 6 = 857142\). Each product is a rotation of the block 142857.
Final answer: 571428 and 857142.
Example 29
Question: The number \(0.\overline{9}\) looks “a little less than 1”. Use algebra to show it is exactly equal to 1.
Concept: let \(x = 0.\overline{9}\), multiply by 10, subtract.
Solution: Let \(x = 0.\overline{9}\). Then \(10x = 9.\overline{9}\). Subtract the first equation: \(10x - x = 9.\overline{9} - 0.\overline{9} = 9\), so \(9x = 9\) and \(x = 1\).
Final answer: \(0.\overline{9} = 1\).
Example 30
Question: A rational number has a terminating decimal whose last non-zero digit is in the 4th decimal place. Show that it can be written in the form \(\dfrac{p}{10^4}\) where \(p\) is an integer not divisible by 10.
Concept: a terminating decimal with its last non-zero digit at the 4th place has exactly the form of tenths, hundredths, thousandths and ten-thousandths.
Solution: Such a number is a sum of at most four decimal terms: \(p_1 \times 10^{-1} + p_2 \times 10^{-2} + p_3 \times 10^{-3} + p_4 \times 10^{-4}\), which is \(\dfrac{p}{10^4}\) for some integer \(p\) whose last digit is non-zero (so \(p\) is not divisible by 10). Example: \(0.1357 = \dfrac{1357}{10^4}\).
Final answer: any such number equals \(\dfrac{p}{10^4}\) with 10 \(\nmid p\).
9. Applications in Real Life
Application 1 — Bank passbook (money)
Question: A passbook starts with ₹2,500. Deposits of ₹300 and ₹1,500 are made, then withdrawals of ₹800 and ₹450. What is the final balance?
Concept: deposits are positive fortunes, withdrawals are negative.
Solution: \(2500 + 300 + 1500 - 800 - 450 = 3050\).
Final answer: ₹3,050.
Application 2 — Recipe measurement (fractions)
Question: A halwa recipe needs \(1\frac{1}{2}\) cups of sugar for one batch. How much sugar is needed for 3 batches?
Concept: repeated addition or multiplication of a fraction.
Solution: \(3 \times 1\frac{1}{2} = 3 \times \dfrac{3}{2} = \dfrac{9}{2} = 4\frac{1}{2}\) cups.
Final answer: \(4\frac{1}{2}\) cups.
Application 3 — Stations on a straight track (distance)
Question: Two stations are marked at 2.4 km and 7.9 km from the start of a straight rail line. How far apart are they?
Concept: distance on the number line is \(|a - b|\).
Solution: \(|7.9 - 2.4| = 5.5\) km.
Final answer: 5.5 km.
Application 4 — Jeweller’s bangles (fraction division)
Question: A jeweller has \(10\frac{1}{2}\) g of gold wire; each bangle needs \(1\frac{3}{4}\) g. How many bangles can be made?
Concept: divide by the per-bangle requirement.
Solution: \(\dfrac{21}{2} \div \dfrac{7}{4} = \dfrac{21}{2} \times \dfrac{4}{7} = \dfrac{84}{14} = 6\) bangles.
Final answer: 6 bangles.
Application 5 — Weather (temperature integers)
Question: The morning temperature is \(6^{\circ}\)C and the night temperature drops to \(-3^{\circ}\)C. How much has the temperature fallen?
Concept: the fall is the difference \(6 - (-3)\).
Solution: \(6 - (-3) = 6 + 3 = 9^{\circ}\)C.
Final answer: 9 degrees of fall.
Application 6 — Market sale (trade and decimals)
Question: Rice costs ₹40 per kg. A customer buys 2.5 kg for ₹100 and pays with a ₹500 note. What change is due?
Concept: decimal multiplication and integer subtraction.
Solution: Cost \(= 2.5 \times 40 = ₹100\). Change \(= 500 - 100 = ₹400\).
Final answer: ₹400.
Application 7 — Parking tower levels (integers)
Question: A parking building has a basement at level \(-2\) and a rooftop restaurant at level \(12\). How many storeys apart are they?
Concept: vertical distance is \(|12 - (-2)|\).
Solution: \(|12 - (-2)| = 14\) storeys.
Final answer: 14 storeys.
Application 8 — Cutting rope (terminating decimals)
Question: A 4.5 m rope is cut into pieces of 0.25 m each. How many pieces are obtained?
Concept: dividing terminating decimals.
Solution: \(4.5 \div 0.25 = 18\) pieces (exact, since 0.25 is a terminating decimal).
Final answer: 18 pieces.
10. Competency-Based Questions
Competency 1
Question: A friend says every number between 0.1 and 0.2 is rational. Do you agree? Justify with a counterexample if not.
Concept: rational ↔ terminating or repeating decimal.
Solution: Disagree. A number such as \(0.12112111211112\ldots\) never terminates and never repeats a block, so it is irrational — even though it lies between 0.1 and 0.2. So not every number in that interval is rational.
Final answer: false; e.g. \(0.12112111211112\ldots\) is irrational.
Competency 2
Question: Justify why \(|a - b| = |b - a|\) for any rationals \(a, b\), and use it to find the distance between \(-\dfrac{7}{8}\) and \(\dfrac{7}{8}\).
Concept: absolute value removes the sign of a difference.
Solution: \(|a - b|\) and \(|b - a|\) are both the non-negative size of the same difference \((a - b) = -(b - a)\). Distance: \(\left|\dfrac{7}{8} - \left(-\dfrac{7}{8}\right)\right| = \left|\dfrac{14}{8}\right| = \dfrac{7}{4}\).
Final answer: distance \(= \dfrac{7}{4}\) units.
Competency 3
Question: Without dividing, decide whether \(\dfrac{1}{20}, \dfrac{3}{250}\) and \(\dfrac{1}{6}\) terminate, and generalise your rule.
Concept: termination depends only on the prime factors of the denominator.
Solution: \(20 = 2^2 \times 5\) → terminates (\(0.05\)); \(250 = 2 \times 5^3\) → terminates (\(0.012\)); \(6 = 2 \times 3\) contains 3 → repeats (\(0.1666\ldots\)). General rule: in lowest terms, the decimal of \(\dfrac{p}{q}\) terminates exactly when the prime factors of \(q\) are only 2 and/or 5.
Final answer: \(\dfrac{1}{20}\) and \(\dfrac{3}{250}\) terminate; \(\dfrac{1}{6}\) repeats.
Competency 4
Question: Using the average method twice, generate two more rational numbers between 3.1415 and 3.1416.
Concept: \(\dfrac{a + b}{2}\) lies between \(a\) and \(b\) and is rational.
Solution: Average of 3.1415 and 3.1416 is 3.14155. Average of 3.1415 and 3.14155 is 3.141525. Both are rational (terminating decimals) and lie strictly between the two given values.
Final answer: e.g. 3.14155 and 3.141525.
Competency 5
Question: Explain, using the idea of debt, why the product of two negative integers is positive.
Concept: a negative multiplier means “remove that many debts”.
Solution: If a negative number is a debt, multiplying by a negative means removing debts. Removing 3 debts of ₹4 each makes you ₹12 richer: \((-3) \times (-4) = +12\).
Final answer: removing debts adds wealth, so a negative × negative is positive.
Competency 6
Question: Why must \(q \neq 0\) in the definition of a rational number \(\dfrac{p}{q}\)?
Concept: division by zero is undefined.
Solution: \(\dfrac{p}{0}\) would be a number \(n\) such that \(n \times 0 = p\). If \(p \neq 0\), no such \(n\) exists; if \(p = 0\), every \(n\) works, which is also meaningless. So division by zero is simply undefined and excluded by \(q \neq 0\).
Final answer: because division by zero is not defined.
Competency 7
Question: Using the idea of remainders, explain why \(\dfrac{1}{7}\) must have a repeating decimal.
Concept: only finitely many remainders are possible in division by 7.
Solution: When dividing 1 by 7, the possible remainders are only 1, 2, 3, 4, 5, 6. After at most 6 steps a remainder must occur a second time, and then the whole pattern repeats forever. So the decimal is \(0.\overline{142857}\).
Final answer: finite possible remainders force repetition.
11. Case-Based Questions
Case 1 — The grain cooperative ledger
A village cooperative records daily changes in a grain stock: Day 1: receive \(+\,45\) bags; Day 2: sell \(-\,30\) bags; Day 3: receive \(+\,20\) bags; Day 4: sell \(-\,35\) bags; Day 5: receive \(+\,25\) bags. The opening stock was 50 bags.
Q1. What is the net change over the five days?
Solution: \(45 - 30 + 20 - 35 + 25 = 45 + 20 + 25 - 30 - 35 = 90 - 65 = 25\) bags.
Q2. What is the closing stock?
Solution: opening 50 + net change 25 = 75 bags.
Q3. After which day is the stock first below the opening stock?
Solution: after Day 2 the stock is \(50 + 45 - 30 = 65\) (still above); after Day 4 it is \(50 + 45 - 30 + 20 - 35 = 50\) (equal); so never below. Days 1–5 stay at or above the opening stock. (If Day 4 had been a sale of 40 instead, the stock would drop to 45 — the ledger “goes below”.)
Final answer: net +25 bags; closing stock 75 bags; never below opening stock.
Case 2 — Mountain base-camp weather
A base camp at high altitude records the temperature at noon as \(7^{\circ}\)C. By midnight it has fallen by \(12^{\circ}\)C, and the next noon it reads \(3^{\circ}\)C.
Q1. Find the midnight temperature.
Solution: \(7 - 12 = -5^{\circ}\)C.
Q2. What was the total fall from the first noon to the second noon?
Solution: \(7 - 3 = 4^{\circ}\)C.
Q3. By how much did the temperature rise from midnight to the second noon?
Solution: \(3 - (-5) = 8^{\circ}\)C.
Final answer: \(-5^{\circ}\)C; fall of \(4^{\circ}\)C; rise of \(8^{\circ}\)C.
Case 3 — The tailor’s silk
A tailor has \(15\frac{3}{4}\) metres of fine silk. Making one kurta requires \(2\frac{1}{4}\) metres of silk.
Q1. Exactly how many kurtas can he make?
Solution: \(\dfrac{63}{4} \div \dfrac{9}{4} = \dfrac{63}{4} \times \dfrac{4}{9} = \dfrac{63}{9} = 7\) kurtas.
Q2. How much silk does he use in total?
Solution: \(7 \times \dfrac{9}{4} = \dfrac{63}{4} = 15\frac{3}{4}\) m — exactly all the silk.
Q3. A customer orders one more kurta. Is there enough silk left?
Solution: Silk left \(= 15\frac{3}{4} - 15\frac{3}{4} = 0\) m, which is less than \(2\frac{1}{4}\) m. Not enough.
Final answer: 7 kurtas; all \(15\frac{3}{4}\) m used; no silk remains for an extra kurta.
Case 4 — Metro stations on a straight line
Three metro stations lie on one straight line marked from the terminal: A at \(2\) km, B at \(7\) km and C at \(-4\) km (one station lies before the terminal on the opposite side).
Q1. Find the distance between A and B.
Solution: \(|7 - 2| = 5\) km.
Q2. Find the distance between B and C.
Solution: \(|7 - (-4)| = 11\) km.
Q3. A food court is placed exactly midway between A and B. Where is it?
Solution: halfway distance \((2 + 7)/2 = \dfrac{9}{2} = 4.5\) km — a rational number, as expected from the density of rationals.
Final answer: 5 km; 11 km; at 4.5 km.
Case 5 — The magical 142857 wristband
It is known that \(\dfrac{1}{7} = 0.\overline{142857}\). A magic wristband displays the six digits 1, 4, 2, 8, 5, 7 in a circle.
Q1. What is \(142857 \times 5\)?
Solution: \(714285\).
Q2. What do you observe about the digits of all multiples from \(1 \times 142857\) to \(6 \times 142857\)?
Solution: every multiple is just the block 142857 rotated — the same six digits in a cyclic order.
Q3. Compute \(142857 \times 7\). What special number do you get, and why is it connected to \(\dfrac{1}{7}\)?
Solution: \(142857 \times 7 = 999999\). Because \(1 = 7 \times \dfrac{1}{7} = 7 \times \dfrac{142857}{999999}\), we must have \(7 \times 142857 = 999999 = 10^6 - 1\). Letting the six digits wrap around one full circle brings the block back to the all-9s number.
Final answer: 714285; rotations of the same block; \(142857 \times 7 = 999999\).
Case 6 — Marks on a classroom board (bonus)
A teacher marks the board number line from \(-6\) to \(6\). A student must plot the numbers \(-\dfrac{5}{2}\), \(\dfrac{1}{3}\) and \(\sqrt{2}\).
Q1. Where does \(-\dfrac{5}{2}\) lie?
Solution: \(-\dfrac{5}{2} = -2\frac{1}{2}\), between \(-3\) and \(-2\).
Q2. Where does \(\dfrac{1}{3}\) lie?
Solution: between 0 and 1, at one-third of the interval [0, 1].
Q3. Why can \(\sqrt{2}\) be placed on the line but not written as a fraction?
Solution: \(\sqrt{2} \approx 1.414\) lies between 1 and 2 and has a point on the number line (constructed with the compass method), but it cannot be written as \(\dfrac{p}{q}\) because it is irrational.
Final answer: \(-\frac{5}{2}\) between \(-3\) and \(-2\); \(\frac{1}{3}\) inside [0, 1]; \(\sqrt{2}\) lies on the line at about 1.414 yet is irrational.
12. Common Mistakes to Avoid
Wrong idea: “Zero is not really a number, it only fills columns.”
Correct idea: Zero is a fully-fledged number with arithmetic rules (\(a + 0 = a\), \(a - 0 = a\), \(a \times 0 = 0\)), formalised by Brahmagupta.
Wrong idea: “A rational number is a fraction with a smaller numerator.”
Correct idea: \(5 = \dfrac{5}{1}\), \(\dfrac{9}{4}\) and \(\dfrac{20}{9}\) are all rational. Any number of the form \(\dfrac{p}{q}\), \(q \neq 0\), is rational.
Wrong idea: “\(\dfrac{5}{0} = 0\)” or “\(\dfrac{0}{0} = 1\).”
Correct idea: Division by zero is undefined. Always state \(q \neq 0\) in a rational number \(\dfrac{p}{q}\).
Wrong idea: \(-3 - (-8) = -11\).
Correct idea: subtracting a negative adds it: \(-3 - (-8) = -3 + 8 = 5\).
Wrong idea: the distance from \(-4\) to \(3\) is \(-7\).
Correct idea: distance \(= |3 - (-4)| = 7\). Use the absolute value, so the answer is never negative.
Wrong idea: “0.333… never ends, so it is irrational.”
Correct idea: \(0.333\ldots = 0.\overline{3} = \dfrac{1}{3}\) repeats, so it is rational. Only decimals that never end and never repeat are irrational.
Wrong idea: “Every square root is a fraction in disguise.”
Correct idea: \(\sqrt{81} = 9\) is rational, but \(\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{12}\) are irrational numbers that cannot be written as \(\dfrac{p}{q}\).
Wrong idea: \(0.\overline{9}\) is just below 1.
Correct idea: the algebra \(10x - x = 9\) shows \(x = 1\); \(0.\overline{9}\) is exactly 1.
Wrong idea: a negative × a negative is negative.
Correct idea: \((-3) \times (-4) = 12\). Same signs give a positive product; different signs give a negative product.
13. Exam Focus
- Always state the definition of a rational number with the condition \(q \neq 0\) — that half-mark is easy to lose.
- When asked to prove \(\sqrt{2}\) is irrational, present the six steps in order and end with the contradiction (p and q share factor 2).
- When classifying decimals, look for repetition of a block: \(0.1234512345\ldots\) repeats the block 12345, so it is rational even though it never terminates.
- For conversions of repeating decimals, show each step: let \(x = \dots\), multiply by \(10^n\), subtract, solve. Never jump straight to the fraction.
- Use absolute value for distances: \(|a - b|\), never \(a - b\) alone.
- For “find a rational between”, the average \(\dfrac{a + b}{2}\) is the quickest and safest route — and it always works.
- Remember the terminating-decimal rule: prime factors of the denominator (in lowest terms) must be only 2 and/or 5.
- Read signs carefully in integer arithmetic — one sign slip loses the entire product or sum.
14. Quick Revision
Natural numbers
\(\mathbb{N} = \{1, 2, 3, \dots\}\): came from counting, one-to-one correspondence, tally marks.
Zero (śhūnya)
\(a - a = 0\), \(a + 0 = a\), \(a - 0 = a\), \(a \times 0 = 0\). No division by zero.
Integers
\(\mathbb{Z}\) (from Zahlen): \(\{\dots, -2, -1, 0, 1, 2, \dots\}\); fortune (dhana) and debt (ṛiṇa).
Signs
Same signs → positive product; different signs → negative product. Subtracting a negative adds.
Rational number
\(\dfrac{p}{q}\), \(p, q \in \mathbb{Z}\), \(q \neq 0\). Equal iff \(ad = bc\).
Absolute value
\(|x|\) is the distance from 0; \(|x| \geq 0\). Distance between \(a\) and \(b\) is \(|a - b|\).
Density
\(\dfrac{a + b}{2}\) is rational and lies between \(a\) and \(b\); infinitely many rationals exist between any two rationals.
Irrational
\(\sqrt{2}, \sqrt{3}, \sqrt{5}, \pi\) cannot be written as \(\dfrac{p}{q}\). Proof by contradiction for \(\sqrt{2}\) (Hippasus).
Terminating rule
\(\dfrac{p}{q}\) in lowest terms terminates iff the prime factors of \(q\) are only 2 and/or 5.
Cyclic 142857
\(\dfrac{1}{7} = 0.\overline{142857}\); multiplying the block by 1–6 rotates its digits.
0.9̅ = 1
\(10x - x = 9\) gives \(x = 1\). Terminating decimals have an alternative with repeating 9s.
Real numbers
\(\mathbb{R} = \mathbb{Q} \;\cup\;\) irrationals — the complete, uninterrupted number line.
From counting pebbles and carved bones, humans built the natural numbers; India gave the world zero and negative numbers; fractions and irrationals meet in the real numbers \(\mathbb{R}\), where \(0.\overline{9} = 1\) and \(\sqrt{2}\) has its own point on the line.
15. Self-Practice Questions
Attempt these yourself before checking Section 16. They are practice questions based on the concepts of this chapter.
S1.
Write the four smallest natural numbers. Is 0 a natural number? Explain briefly.
S2.
Illustrate one-to-one correspondence with an example from daily life not used in these notes.
S3.
How old is the Lebombo bone, and what do its 29 notches probably record?
S4.
Where was the Ishango bone found, about how old is it, and which primes does one of its columns display?
S5.
A Lothal merchant receives 15 ingots per 2 bags of spices. How many ingots would he receive for 10 bags?
S6.
In Indian tradition, what is the name for \(10^{12}\) and for \(10^{53}\), and in which text does the larger number appear?
S7.
Using Brahmagupta’s rules, compute \(9 - 9\), \(9 + 0\), \(9 - 0\), \(9 \times 0\).
S8.
Why is \(\dfrac{5}{0}\) not a rational number?
S9.
Compute: \((-7) + (-9)\); \(7 - (-3)\); \((-12) \div (-4)\); \(8 \times (-6)\); \((-15) \times 0\); \(0 - (-18)\).
S10.
The temperature at Leh is \(6^{\circ}\)C at noon and falls by \(14^{\circ}\)C by midnight. Find the midnight temperature.
S11.
A shopkeeper earns ₹600 in the morning, spends ₹350 by afternoon and earns ₹200 in the evening. Find his final amount for the day.
S12.
Find the sum: \(\dfrac{11}{12} + \dfrac{5}{18}\).
S13.
Find the difference: \(\dfrac{11}{8} - \dfrac{3}{4}\).
S14.
Find the product: \(\dfrac{7}{11} \times \dfrac{5}{8}\).
S15.
Find the quotient: \(\left(-\dfrac{4}{7}\right) \div \dfrac{5}{14}\).
S16.
Represent \(\dfrac{8}{5}\) on a number line.
S17.
Represent \(-\dfrac{7}{4}\) on a number line.
S18.
Find \(\left|-\dfrac{7}{8}\right|\), \(\left|\dfrac{7}{8}\right|\) and \(|0|\), and then the distance between \(-\dfrac{7}{8}\) and \(\dfrac{7}{8}\).
S19.
Using the average method, find a rational number between \(\dfrac{1}{3}\) and \(\dfrac{1}{2}\), and then another between your answer and \(\dfrac{1}{2}\).
S20.
Find five distinct rational numbers between \(\dfrac{2}{5}\) and \(\dfrac{3}{5}\).
S21.
Find five distinct rational numbers between \(\dfrac{1}{6}\) and \(\dfrac{2}{5}\).
S22.
Classify as rational or irrational: \(\sqrt{25}\); \(\sqrt{10}\); \(\sqrt{\dfrac{4}{9}}\); \(-\sqrt{36}\).
S23.
Without performing long division, predict terminating or repeating: \(\dfrac{9}{25}\), \(\dfrac{11}{12}\), \(\dfrac{17}{40}\), \(\dfrac{5}{7}\).
S24.
Convert to \(\dfrac{p}{q}\): \(0.75\), \(0.125\), \(0.02\).
S25.
Convert the pure repeating decimals \(0.\overline{7}\) and \(0.\overline{12}\) to \(\dfrac{p}{q}\).
S26.
Convert the general repeating decimal \(0.1\overline{6}\) to \(\dfrac{p}{q}\).
S27.
Evaluate \(142857 \times 4\) and \(142857 \times 5\), and write the decimal form of \(\dfrac{1}{7}\).
S28.
Give two irrational numbers between 2 and 3, other than \(\sqrt{2}\) or \(\pi\).
S29.
Prove that \(0.\overline{9} = 1\) using \(x = 0.\overline{9}\).
S30.
Find three rational numbers between \(\dfrac{1}{4}\) and \(\dfrac{1}{2}\).
S31.
Using the Baudhāyana–Pythagoras Theorem, find the diagonal of a square of side 1, and say how the length \(\sqrt{2}\) can be marked on a number line.
S32.
Show that \(\dfrac{a + b}{2}\) lies strictly between \(a\) and \(b\) when \(a = \dfrac{12}{7}\) and \(b = \dfrac{13}{8}\), and verify with decimals.
S33.
Reduce \(\dfrac{12345}{99999}\) to lowest terms (it equals the repeating decimal \(0.\overline{12345}\)).
S34.
A number has decimal expansion \(23.560185612239874790120\). Is it rational? If so, write it in the form \(\dfrac{p}{q}\).
16. Answers and Hints for Self-Practice
| Question | Answer / hint |
|---|---|
| S1 | \(\{1, 2, 3, 4\}\); 0 is not a natural number (it is a whole number and an integer). |
| S2 | e.g. matching students to chairs: if every student sits and no chair is empty, the counts match. |
| S3 | About 35,000 years; a lunar phase counter or menstrual calendar. |
| S4 | Near the headwaters of the Nile, Democratic Republic of Congo; about 20,000 BCE; the primes 11, 13, 17, 19. |
| S5 | \(10 \div 2 = 5\) groups; \(5 \times 15 = 75\) ingots. |
| S6 | \(10^{12}\) = parārdha; \(10^{53}\) = tallakṣhaṇa; the Lalitavistara. |
| S7 | 0, 9, 9, 0 (a − a = 0; a + 0 = a; a − 0 = a; a × 0 = 0). |
| S8 | No number multiplied by 0 gives 5, so \(\dfrac{5}{0}\) is undefined; rational numbers require \(q \neq 0\). |
| S9 | \(-16\); \(10\); \(3\); \(-48\); \(0\); \(18\). |
| S10 | \(6 - 14 = -8^{\circ}\)C. |
| S11 | \(600 - 350 + 200 = ₹450\). |
| S12 | \(\dfrac{43}{36}\) (LCM 36). |
| S13 | \(\dfrac{5}{8}\) (\(\dfrac{11}{8} - \dfrac{6}{8}\)). |
| S14 | \(\dfrac{35}{88}\). |
| S15 | \(-\dfrac{4}{7} \times \dfrac{14}{5} = -\dfrac{8}{5}\). |
| S16 | \(\dfrac{8}{5} = 1\dfrac{3}{5}\): divide [1, 2] into 5 parts, move 3 parts right of 1. |
| S17 | \(-\dfrac{7}{4} = -1\dfrac{3}{4}\): divide [−2, −1] into 4 parts, move 3 parts left of −1. |
| S18 | All equal \(\dfrac{7}{8}\); distance \(= \left|\dfrac{7}{8} - \left(-\dfrac{7}{8}\right)\right| = \dfrac{7}{4}\). |
| S19 | First average \(\dfrac{5}{12}\); second average \(\dfrac{11}{24}\). |
| S20 | e.g. \(\dfrac{1}{2}, \dfrac{9}{20}, \dfrac{11}{20}, \dfrac{19}{40}, \dfrac{21}{40}\) (any five strictly between 0.4 and 0.6). |
| S21 | e.g. \(\dfrac{1}{5}, \dfrac{1}{4}, \dfrac{3}{10}, \dfrac{1}{3}, \dfrac{7}{20}\) (each between \(\approx 0.1667\) and 0.4). |
| S22 | \(\sqrt{25} = 5\) rational; \(\sqrt{10}\) irrational; \(\sqrt{\tfrac{4}{9}} = \tfrac{2}{3}\) rational; \(-\sqrt{36} = -6\) rational. |
| S23 | \(\dfrac{9}{25}\) terminates (25 = 5²); \(\dfrac{11}{12}\) repeats (12 has factor 3); \(\dfrac{17}{40}\) terminates (40 = 2³ × 5); \(\dfrac{5}{7}\) repeats. |
| S24 | \(\dfrac{3}{4}; \dfrac{1}{8}; \dfrac{1}{50}\). |
| S25 | \(\dfrac{7}{9}; \dfrac{12}{99} = \dfrac{4}{33}\). |
| S26 | \(\dfrac{1}{6}\) (as in Example 26). |
| S27 | \(142857 \times 4 = 571428\); \(142857 \times 5 = 714285\); \(\dfrac{1}{7} = 0.\overline{142857}\). |
| S28 | e.g. \(\sqrt{5}\) and \(\sqrt{6}\) (both irrational, both between 2 and 3). |
| S29 | \(10x = 9.\overline{9}\), \(9x = 9\), \(x = 1\). |
| S30 | e.g. \(\dfrac{3}{8}, \dfrac{5}{16}, \dfrac{7}{16}\). |
| S31 | \(d^2 = 1^2 + 1^2 = 2\), so \(d = \sqrt{2}\); construct with perpendicular 1 unit at point 1 and an arc of radius \(\sqrt{2}\) (Figure 5). |
| S32 | Average \(= \dfrac{187}{112} \approx 1.67\), between \(1.625 = \dfrac{13}{8}\) and \(1.714\ldots = \dfrac{12}{7}\). |
| S33 | \(\dfrac{4115}{33333}\) (divide numerator and denominator by 3). |
| S34 | Yes — terminating, so rational: \(\dfrac{23560185612239874790120}{10^{21}}\). |
17. Chapter Checklist
- I can explain how counting began with one-to-one correspondence.
- I can describe the Lebombo bone and the Ishango bone and their significance.
- I can explain the Indian contribution to large numbers and the place-value system.
- I understand how the concept of emptiness (śhūnyatā) became the number zero.
- I can state Brahmagupta’s rules for zero and apply them.
- I know why division by zero is undefined.
- I can add, subtract and multiply integers using the sign rules.
- I can define a rational number with the condition \(q \neq 0\).
- I can check equality of rational numbers using \(ad = bc\).
- I can perform the four operations on rational numbers.
- I can represent \(\dfrac{p}{q}\) on the number line.
- I can compute \(|x|\) and the distance \(|a - b|\).
- I can use the density property to find rationals between two rationals.
- I can explain why \(\sqrt{2}\) is irrational using proof by contradiction.
- I can construct \(\sqrt{2}\) on the number line.
- I know the story of \(\pi\) and Mādhava’s infinite series.
- I can predict terminating versus repeating decimals from the denominator.
- I can convert terminating and repeating decimals into \(\dfrac{p}{q}\).
- I can recognise cyclic numbers like 142857.
- I can prove that \(0.\overline{9} = 1\).
- I understand real numbers as the union of rational and irrational numbers.