The World of Numbers — Detailed Notes
Numbers grew in stages to meet human needs — from counting cattle with pebbles (one-to-one correspondence) and tally marks on the Ishango bone, to the Indian invention of zero and the modern real number line.
3.2–3.3 Zero, integers & Brahmagupta’s rules
The philosophical idea of śhūnya (emptiness) became the mathematical zero in the work of Brahmagupta (628 CE), who defined 0 = a − a and gave: a + 0 = a, a − 0 = a, a × 0 = 0. He also introduced negative numbers (debts, ṛiṇa) opposite to positive numbers (fortunes, dhana), creating the integers ℤ = {…, −2, −1, 0, 1, 2, …}. Sign rules: debt + debt = debt; debt × fortune = debt; debt × debt = fortune (− × − = +).
3.4 Rational numbers
Rational number: any number that can be written as p/q, where p and q are integers and q ≠ 0. This set ℚ includes all natural numbers, whole numbers and integers (e.g. 5 = 5/1). Equivalent fractions (−1/3 = −2/6 = …) represent the same number; we usually keep p, q co-prime.
Rational numbers are closed under +, −, × and ÷ (except ÷ 0); addition and multiplication are commutative and distributive. Absolute value |x| is the distance of x from 0 (always ≥ 0); the distance between a and b is |a − b|. To plot p/q, divide each unit into q equal parts and count p parts from 0.
Density: between any two rational numbers there is always another — e.g. their average (a + b)/2. So there are infinitely many rationals between any two points.
3.5 Irrational numbers
Numbers on the line that cannot be written as p/q are irrational, e.g. √2, √3, π. Proof that √2 is irrational (by contradiction): assume √2 = p/q in lowest terms ⇒ 2q² = p² ⇒ p² even ⇒ p even (p = 2k) ⇒ 2q² = 4k² ⇒ q² = 2k² ⇒ q even. Then p and q are both even — contradicting “lowest terms.” So √2 is irrational. (Hippasus, c. 400 BCE.) Lengths like √n are constructed with ruler and compass (e.g. the square root spiral). π (the circumference÷diameter ratio) is irrational — Mādhava expressed it as an infinite series π = 4(1 − 1/3 + 1/5 − 1/7 + …).
3.6 Real numbers & decimal expansions
The union of all rational and irrational numbers is the continuous set of real numbers ℝ. A decimal expansion tells the type:
| Decimal type | Number | Example |
|---|
| Terminating | Rational | 3/8 = 0.375 |
| Non-terminating, repeating | Rational | 5/11 = 0.͖45 |
| Non-terminating, non-repeating | Irrational | √2 = 1.41421… |
Predicting the type: a rational p/q (lowest terms) terminates exactly when the only prime factors of q are 2 and/or 5 (we can make the denominator a power of 10); otherwise it repeats. The number of decimal places of a terminating decimal = the higher of the powers of 2 and 5 in q.
Converting repeating decimals to p/q (let x = the decimal): • Pure repeating — multiply by 10ⁿ (n = repeating digits) and subtract: 0.͖6 → 9x = 6 → x = 2/3. • General repeating — first shift past the non-repeating digits, then one repeating cycle, and subtract: 0.1͖6 → 90x = 15 → x = 1/6. The block of 1/7 = 0.͖142857 is a famous cyclic number. Also 0.͖9… = 1 exactly.
Number sets (nested)
ℕ (counting) ⊂ ℤ (integers) ⊂ ℚ (rationals) | ℚ + Irrationals = ℝ (real numbers)
(Beyond ℝ lie the imaginary numbers, built on i = √(−1), needed where no real number squares to a negative — a topic for later grades.)
Key Terms
natural numbers, integers, zero (śhūnya), fortunes/debts, rational number, equivalent fractions, closure, absolute value, density, average, irrational number, proof by contradiction, square root spiral, π, real numbers, terminating/repeating decimal, cyclic number.
End-of-Chapter Exercises — Full Solutions
Complete worked solutions to all 16 end-of-chapter problems (starred * = HOTS). In Q3, a bar over digits marks the repeating block.
1Convert by long division to a terminating or repeating decimal: (i) 3/50, (ii) 2/9.
(i) 50 = 2 × 5², only primes 2 and 5 → terminating: 3/50 = 6/100 = 0.06. (ii) 9 = 3² (not 2 or 5) → repeating: 2/9 = 0.2̅ = 0.222…
2Prove that √5 is an irrational number.
Assume, for contradiction, √5 = p/q in lowest terms (p, q co-prime, q ≠ 0). Then 5q² = p², so p² is a multiple of 5 ⇒ p is a multiple of 5; write p = 5k. Then 5q² = 25k² ⇒ q² = 5k², so q² is a multiple of 5 ⇒ q is a multiple of 5. Now both p and q are divisible by 5, contradicting ‘lowest terms.’ Hence √5 cannot be written as p/q — it is irrational.
3Convert to p/q (a bar denotes the repeating block): (i) 12.6̅ (ii) 0.0120 (iii) 3.052 (iv) 1.235 (v) 0.2̅3̅ (vi) 2.0̅5̅ (vii) 2.125 (viii) 3.1̅2̅5̅ (ix) 2.1625
(i) x = 12.66… ⇒ 10x − x = 114 ⇒ x = 38/3. (ii) terminating: 0.0120 = 120/10000 = 3/250. (iii) terminating: 3.052 = 3052/1000 = 763/250. (iv) terminating: 1.235 = 1235/1000 = 247/200. (v) x = 0.2323… ⇒ 100x − x = 23 ⇒ x = 23/99. (vi) x = 2.0505… ⇒ 100x − x = 203 ⇒ x = 203/99. (vii) terminating: 2.125 = 2125/1000 = 17/8. (viii) x = 3.125125… ⇒ 1000x − x = 3122 ⇒ x = 3122/999. (ix) terminating: 2.1625 = 21625/10000 = 173/80.
4Locate on the number line: (i) 0.532, (ii) 1.15.
Use successive magnification. (i) 0.532 lies between 0.5 and 0.6; zoom into [0.53, 0.54] and mark the point 2/10 of the way → 0.532. (ii) 1.15 lies between 1.1 and 1.2; zoom into that interval and mark the point halfway → 1.15. (Each zoom divides the chosen interval into 10 equal parts.)
5Find 6 rational numbers between 3 and 4.
Write 3 = 21/7 and 4 = 28/7. Six rationals between them are 22/7, 23/7, 24/7, 25/7, 26/7, 27/7. (Equivalently 3.1, 3.2, 3.3, 3.4, 3.5, 3.6.)
6Find 5 rational numbers between 2/5 and 3/5.
Make the denominator larger: 2/5 = 12/30 and 3/5 = 18/30. Five rationals between them: 13/30, 14/30 (= 7/15), 15/30 (= 1/2), 16/30 (= 8/15), 17/30.
7Find 5 rational numbers between 1/6 and 2/5.
LCD of 6 and 5 is 30: 1/6 = 5/30 and 2/5 = 12/30. Five rationals between them: 6/30 (= 1/5), 7/30, 8/30 (= 4/15), 9/30 (= 3/10), 10/30 (= 1/3).
8If x/3 + x/5 = 16/15, find the rational number x.
Take LCD 15: (5x + 3x)/15 = 16/15 ⇒ 8x/15 = 16/15 ⇒ 8x = 16 ⇒ x = 2.
9a, b are non-zero rationals with a/b + 1 = 0. Without numbers, is ab positive or negative? Justify.
a/b + 1 = 0 ⇒ a/b = −1 ⇒ a = −b. Then ab = (−b)(b) = −b². Since b ≠ 0, b² > 0, so −b² < 0. Therefore ab is negative.
10A terminating decimal has its last non-zero digit in the 4th decimal place. Show it can be written as p/10⁴ with p not divisible by 10. Must the lowest-form denominator be divisible by 2⁴ or 5⁴?
A decimal that stops at the 4th place equals (that 4-digit string)/10⁴ = p/10⁴; since the last non-zero digit is in the 4th place, p is not divisible by 10. Now 10⁴ = 2⁴ × 5⁴, and in lowest form the denominator is 2^a × 5^b with max(a, b) = 4. It is not necessary that it be divisible by 2⁴ and 5⁴ — e.g. 0.0625 = 1/16 = 1/2⁴ (denominator 2⁴, not divisible by 5⁴). So only the higher power must reach 4.
11Without dividing, is 18/125 terminating or non-terminating? If terminating, how many decimal places?
125 = 5³ (only the prime 5), so 18/125 is terminating. Multiply by 2³: 18/125 = (18×8)/1000 = 144/1000 = 0.144 — so it has 3 decimal places (equal to the higher power, 5³).
12A rational number in lowest form has denominator 2³ × 5. How many decimal places will its expansion have?
The denominator has primes only 2 and 5, so it terminates. The number of decimal places = the higher of the two powers = max(3, 1) = 3 decimal places. (Make the denominator 2³ × 5³ = 10³ by multiplying top and bottom by 5².)
13(*) a = 7/12, b = 5/6. Write both over a common denominator m with k₂ − k₁ > 6, then give 5 rationals between them. Why is k₂ − k₁ > n + 1 needed for n numbers?
Write b = 10/12; the numerators 7 and 10 differ by only 3, too few. Scale up: use m = 120, so a = 70/120 and b = 100/120 (now k₂ − k₁ = 30 > 6). Five rationals between them: 75/120, 80/120, 85/120, 90/120, 95/120 (= 5/8, 2/3, 17/24, 3/4, 19/24). Why: to fit n whole-number numerators strictly between k₁ and k₂ you need at least n gaps, i.e. k₂ − k₁ must exceed n (here, n + 1 to be safe), so choose m large enough.
14(*) Rationals x, y, z satisfy x + y + z = 0 and xy + yz + zx = 0. Show x = y = z = 0.
Use the identity x² + y² + z² = (x + y + z)² − 2(xy + yz + zx) = 0² − 2(0) = 0. A sum of squares of real numbers is zero only if each square is zero. Hence x² = y² = z² = 0, so x = y = z = 0.
15(*) Show that (a + b)/2 lies between the rational numbers a and b.
It is rational (sum and division of rationals). Assume a < b. Then a = (a + a)/2 < (a + b)/2 < (b + b)/2 = b. So (a + b)/2 lies strictly between a and b (and it is rational) — this is why the average always gives a rational number in between.
16Find the hypotenuses of the right triangles in the square root spiral (Fig. 3.14).
Each triangle has one leg of length 1 and the other leg the previous hypotenuse. Starting from a unit square: OP₁ = √(1²+1²) = √2, then √(√2²+1²) = √3, then √4 = 2, then √5, √6, √7, √8, √9 = 3, … In general the n-th hypotenuse has length √(n + 1).