Real Numbers
1. What This Chapter Covers
The chapter opens with a line from Leopold Kronecker: "God made the integers. All else is the work of man."
Then it opens with a puzzle, which is a better introduction than any definition.
In a garden, a swarm of bees settles in equal numbers on the same flowers. On two flowers, one bee is left out. On three flowers, two bees are left out. On four flowers, three are left out. On five flowers, none is left out. If there are at most fifty bees, how many bees are in the swarm?
Let the number of bees be x. Working backwards, x < 50. Each condition becomes an equation:
- five equal groups, none left: x = 5a + 0
- four equal groups, three left: x = 4b + 3
- three equal groups, two left: x = 3c + 2
- two equal groups, one left: x = 2d + 1
Look for multiples of 5 first. Since the number leaves remainder 1 on division by 2, it must be an odd multiple: 5, 15, 25, 35, 45. Checking the remaining two conditions on those five numbers leaves 35 as the only possibility.
Verifying: 35 = 2 × 17 + 1, 35 = 3 × 11 + 2, 35 = 4 × 8 + 3, and 35 = 5 × 7 + 0.
The textbook's own comment is the point of the whole section: "In the process of writing above equations, we have used division algorithm unknowingly." Every one of those four lines is the same statement with different numbers.
The chapter is allotted 15 periods in June and runs from textbook page 1 to page 27.
2. The Division Algorithm
Generalise the four bee equations. For each pair of positive integers a and b — the dividend and the divisor — we can find whole numbers q and r, the quotient and the remainder, satisfying a relation.
Theorem-1.1 (Division Algorithm). Given positive integers a and b, there exist a unique pair of whole numbers q and r satisfying a = bq + r, 0 ≤ r < b.
Two things in that statement do the work, and both are easy to read past.
The remainder is allowed to be zero but never reaches the divisor. That is what 0 ≤ r < b says. If the remainder ever equalled the divisor you could fit one more group in, so the quotient was wrong.
The pair is unique. For a given a and b there is exactly one q and one r, not several.
One more group of 4 will not fit inside a leftover of 3 — that is the whole content of the condition 0 ≤ r < b.
The textbook notes that this result was first recorded in Book VII of Euclid's Elements, and that Euclid's algorithm is built on it. It also warns, in a remark, that the two names get mixed up: "Euclid's algorithm and division algorithm are so closely interlinked that people often call former as the division algorithm also." The book itself calls it Euclid's Division Lemma inside the algorithm's steps.
A second remark: although the theorem is stated for positive integers only, it can be extended to all integers a and b with b ≠ 0 — the book says so and then declines to discuss it.
3. Euclid's Algorithm for the HCF
Recall that the HCF of two positive integers a and b is the greatest positive integer d that divides both.
The book finds the HCF of 60 and 100 by an activity with paper before it finds it by arithmetic, and the activity is the better memory.
Take two paper strips of equal width, of lengths 60 cm and 100 cm. Find the greatest length of strip that measures both completely.
Measure the 100 cm strip with the 60 cm one; cut off the left-over 40 cm. Measure the 60 cm strip with that 40 cm piece; cut off the left-over 20 cm. Measure the 40 cm with the 20 cm — nothing is left over. So 20 cm is the longest strip that measures both without leaving any part.
The paper strips and the three equations are the same procedure; the cut-off piece is the remainder.
Written out, the algorithm for two positive integers c and d with c > d is:
- Step 1. Apply Euclid's Division Lemma to c and d, giving the unique whole numbers q and r with c = dq + r, 0 ≤ r < d.
- Step 2. If r = 0, then d is the HCF of c and d. If r ≠ 0, apply the lemma to d and r.
- Step 3. Continue until the remainder is zero. The divisor at that stage is the required HCF.
The reason it works is one identity, and the book states it plainly: HCF (c, d) = HCF (d, r). Every step replaces a pair by a smaller pair with the same HCF, and the remainders strictly decrease, so the process has to stop.
The book adds that Euclid's algorithm is useful for very large numbers, and that it was one of the earliest examples of an algorithm a computer was programmed to carry out.
A Think and discuss box asks whether you can find the HCF of 1.2 and 0.12 by Euclid's algorithm, and to justify the answer. The theorem was stated for integers, which is the hinge of that question.
4. What the Division Algorithm Proves
The division algorithm is not only a way to divide; it is a way to split all integers into cases and then argue about each case. Fix the divisor, list the possible remainders, and you have covered every integer.
Example 1. Show that every positive even integer is of the form 2q, and every positive odd integer of the form 2q + 1.
Take b = 2. Then a = 2q + r with r = 0 or r = 1, because 0 ≤ r < 2. So a = 2q or 2q + 1. The first is even. Since a positive integer is either even or odd, every positive odd integer is 2q + 1.
Example 2. Show that every positive odd integer is of the form 4q + 1 or 4q + 3.
Take b = 4. The possible remainders are 0, 1, 2 and 3, so a is 4q, 4q + 1, 4q + 2 or 4q + 3. But a is odd, and 4q = 2(2q) and 4q + 2 = 2(2q + 1) are both even. Only 4q + 1 and 4q + 3 survive.
Exercise 1.1 runs the same move with larger divisors: any positive odd integer is of the form 6q + 1, 6q + 3 or 6q + 5; the square of any positive integer is of the form 3p or 3p + 1; the cube is of the form 9m, 9m + 1 or 9m + 8.
The last question — that exactly one of n, n + 2, n + 4 is divisible by 3 — is the same idea with the cases taken modulo 3.
The book's printed answers for the three Euclid HCF computations in Exercise 1.1 are 90, 196 and 127.
5. The Fundamental Theorem of Arithmetic
Take any collection of primes — say 2, 3, 7, 11 and 23 — and multiply some or all of them, repeating as often as you like. You get infinitely many composite numbers: 2 × 3 × 11 = 66, 7 × 11 × 23 = 1771, 2³ × 3 × 7³ = 8232, 2² × 3 × 7 × 11 × 23 = 21252.
Now reverse the question. Given a composite number, can it always be written as a product of primes?
Theorem-1.2 (Fundamental Theorem of Arithmetic). Every composite number can be expressed (factorised) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.
Uniqueness is the part that carries weight later. Factorising 210 as 2 × 3 × 5 × 7 and as 3 × 5 × 7 × 2 is the same factorisation written in a different order; there is no second, genuinely different way. Written in general, a composite x factorises as x = p₁p₂p₃ … pₙ with the primes in ascending order, and collecting equal primes gives powers: 27300 = 2² × 3 × 5² × 7 × 13.
Example 3 is the first thing the theorem buys you. Can 4ⁿ end in zero for any natural number n? To end in zero a number must be divisible by 2 and by 5, so 5 must appear in its prime factorisation. But 4ⁿ = 2²ⁿ, so 2 is the only prime there. By uniqueness no 5 can appear. No such n exists. The same argument in Exercise 1.2 settles 6ⁿ, and the Try this box settles 3ⁿ × 4ᵐ.
Example 4 recalls HCF and LCM by prime factorisation, with 12 = 2² × 3¹ and 18 = 2¹ × 3²:
| Quantity | Value | Rule |
|---|---|---|
| HCF (12, 18) | 2¹ × 3¹ = 6 | product of the smallest power of each common prime factor |
| LCM (12, 18) | 2² × 3² = 36 | product of the greatest power of each prime factor |
From the example, HCF (12, 18) × LCM [12, 18] = 6 × 36 = 216 = 12 × 18. In general, for any two positive integers, HCF (a, b) × LCM [a, b] = a × b — which is how you get the LCM cheaply once you have run Euclid's algorithm for the HCF.
That identity is stated for two integers. Exercise 1.2 asks for the LCM and HCF of three numbers such as 12, 15 and 21, and the product rule does not extend to three.
Exercise 1.2 also asks why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite. Both have a common factor staring out of them once you take it outside the bracket.
6. When Does a Decimal Stop?
In class IX you learned that every rational number is either a terminating decimal or a non-terminating repeating one. This section says which rational numbers fall on which side, and the Fundamental Theorem of Arithmetic is what decides it.
Start from terminating decimals and put them over powers of ten:
| Decimal | Over a power of 10 | In lowest terms |
|---|---|---|
| 0.375 | 375 / 10³ | (3 × 5³) / (2³ × 5³) = 3 / 2³ = 3/8 |
| 1.04 | 104 / 10² | (2³ × 13) / (2² × 5²) = 26 / 5² = 26/25 |
| 0.0875 | 875 / 10⁴ | (5³ × 7) / (2⁴ × 5⁴) = 7 / (2⁴ × 5) = 7/80 |
| 12.5 | 125 / 10¹ | 5³ / (2 × 5) = 25/2 |
The pattern is in the denominators. Once the fraction is in lowest terms, the denominator has only powers of 2, or of 5, or both — because 2 and 5 are the only prime factors of any power of 10.
Theorem-1.3. Let x be a rational number whose decimal form terminates. Then x can be expressed as p/q, where p and q are coprime and the prime factorisation of q is of the form 2ⁿ5ᵐ, with n, m non-negative integers.
The converse is true too, and the book proves it by running the four examples backwards — multiplying top and bottom by whatever makes the denominator a power of 10. For instance 7/80 = 7/(2⁴ × 5) = (7 × 5³)/(2⁴ × 5⁴) = 875/10⁴ = 0.0875.
Theorem-1.4. Let x = p/q be a rational number such that the prime factorisation of q is of the form 2ⁿ5ᵐ. Then x has a decimal expansion which terminates.
And the other side of the fence. One-seventh, worked out by long division, is 0.1428571428571…, and the block 142857 repeats forever. 7 cannot be written as 2ⁿ5ᵐ.
Theorem-1.5. Let x = p/q be a rational number such that the prime factorisation of q is not of the form 2ⁿ5ᵐ. Then x has a decimal expansion which is non-terminating repeating (recurring).
Reducing to lowest terms before you look at the denominator is the step that is easiest to skip and easiest to get wrong.
Together the three theorems say: the decimal form of every rational number is either terminating or non-terminating repeating, and which one it is depends on nothing but the primes in the reduced denominator.
Exercise 1.3 works both directions. Its printed answers include 229/400 = 0.5725 terminating, 2/11 = 0.18 recurring, 23/(2³ × 5²) = 0.115, and 7218/(3² × 5²) = 32.08 — that last one terminating only because the 3² cancels against the numerator.
7. Irrational Numbers
A real number is called irrational — the book writes the set as Q' — if it cannot be written as p/q where p and q are integers and q ≠ 0. Familiar examples: √2, √3, √15, π, and 0.10110111011110… .
The rationals and the irrationals do not overlap; between them they account for every real number.
Class IX asserted that these numbers are irrational. This chapter proves it, and the proof needs one lemma.
Theorem-1.6. Let p be a prime number. If p divides a², where a is a positive integer, then p divides a.
The proof is the Fundamental Theorem of Arithmetic doing its work. Write a = p₁p₂ … pₙ as a product of primes, not necessarily distinct. Then a² = p₁²p₂² … pₙ². If p divides a², then by uniqueness of factorisation p must be one of p₁, p₂, …, pₙ. But those are exactly the primes making up a, so p divides a.
Example 7: √2 is irrational. The method is proof by contradiction.
Assume √2 is rational. Then there are integers r and s (s ≠ 0) with √2 = r/s. Divide both by their HCF to get √2 = a/b with a and b coprime. So b√2 = a.
Every step is forced; the only thing that can be wrong is the assumption at the top, so that is what is false.
Squaring and rearranging gives 2b² = a², so 2 divides a². By Theorem 1.6 with p = 2, 2 divides a. Write a = 2c. Then 2b² = 4c², so b² = 2c². So 2 divides b², and again by Theorem 1.6, 2 divides b.
Now 2 is a common factor of a and b — but they were chosen coprime. The contradiction came from the assumption, so √2 is irrational.
The book generalises: √d is irrational whenever d is a positive integer that is not the square of another integer, so √6, √8, √15 and √24 are all irrational.
Two facts from class IX are then proved in particular cases:
- the sum or difference of a rational and an irrational is irrational (Example 8: 5 − √3);
- the product or quotient of a non-zero rational and an irrational is irrational (Example 9: 3√2).
Example 10 proves √2 + √3 irrational by squaring: from √2 = a/b − √3 it follows that √3 = (a² + b²)/2ab, which would make √3 rational.
Two cautions close the section, and both are places students over-generalise:
- The sum of two irrationals need not be irrational. Take a = √2 and b = −√2; both are irrational but a + b = 0.
- The product of two irrationals need not be irrational. Take a = √2 and b = 3√2; then ab = 6.
8. Exponents, and the Logarithm as the Missing Exponent
The power aⁿ is the product of n factors each equal to a. When 81 is written as 3⁴, the 4 is the exponent or index and 3 is the base. The laws, for real a, b not zero and integers m, n:
| Law | Statement |
|---|---|
| Product | aᵐ · aⁿ = aᵐ⁺ⁿ |
| Quotient | aᵐ / aⁿ = aᵐ⁻ⁿ |
| Power of a product | (ab)ᵐ = aᵐ · bᵐ |
| Power of a quotient | (a/b)ᵐ = aᵐ/bᵐ |
| Power of a power | (aᵐ)ⁿ = aᵐⁿ |
| Zero exponent | a⁰ = 1 |
| Negative exponent | a⁻ᵐ = 1/aᵐ |
Now the question that forces something new. Solving 2ˣ = 4 is easy, because 4 = 2². Solving 3ʸ = 81 is easy, because 81 = 3⁴. But 2ˣ = 5? Five is not a power of 2 that anyone can write down.
Since 2¹ = 2, 2² = 4 and 2³ = 8, the answer lies between 2 and 3. To pin it down, draw the graph of y = 2ˣ.
| x | −3 | −2 | −1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|---|
| y = 2ˣ | 1/8 | 1/4 | 1/2 | 1 | 2 | 4 | 8 |
Locate 5 on the Y-axis at P, go across to the curve at Q, then straight down to R; the length OR is the exponent you were looking for.
The value of x at R is called the logarithm of 5 to the base 2, written log₂5. In general:
For positive real numbers a and N with a ≠ 1, we define logₐ N = x if and only if aˣ = N.
Exponential form and logarithmic form are two ways of writing the same fact, read in opposite directions:
| x | −2 | −1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|
| y = 2ˣ | 1/4 | 1/2 | 1 | 2 | 4 | 8 |
Reading that table right to left gives x = log₂y: log₂(1/4) = −2, log₂(1/2) = −1, log₂2 = 1, log₂4 = 2, log₂8 = 3.
Logarithms to base 10 are called common logarithms, and the base is usually dropped: log₁₀25 is written log 25.
The book's own Think and discuss box asks whether log₂0 exists — worth pausing on, since 2ˣ is never zero for any x, which is exactly what the graph shows. It also asks you to check the claim that logₓ16 = 2 gives x = ±4. Since a logarithm's base has to be positive, only one of those two survives.
Every positive real number has a unique logarithm, because a horizontal line cuts the graph of y = 2ˣ at exactly one point.
9. The Laws of Logarithms
Each law of logarithms is a law of exponents in disguise, and each proof is the same three lines: name the logarithms, convert to exponential form, use the exponent law, convert back.
Product Rule. For positive real a, x, y with a ≠ 1: logₐ(xy) = logₐ x + logₐ y.
Proof. Let logₐ x = m and logₐ y = n, so aᵐ = x and aⁿ = y. Then xy = aᵐaⁿ = aᵐ⁺ⁿ, and converting back, logₐ(xy) = m + n = logₐ x + logₐ y.
Quotient Rule. logₐ(x/y) = logₐ x − logₐ y.
Proof. With the same m and n, x/y = aᵐ/aⁿ = aᵐ⁻ⁿ, so logₐ(x/y) = m − n.
Power Rule. logₐ(xⁿ) = n logₐ x.
Proof. Let aᵐ = x, so m = logₐ x. Then xⁿ = aᵐⁿ, so logₐ(xⁿ) = mn = n logₐ x.
The power rule is what makes logarithms useful for equations where the unknown is an exponent. Take 2ˣ = 3⁵. Logs to base 2 on both sides give x log₂2 = 5 log₂3, and since logₐ a = 1, x = 5 log₂3.
Two worked examples show the rules running in both directions.
Example 11 expands: log(343/125) = log 343 − log 125 = log 7³ − log 5³ = 3 log 7 − 3 log 5 = 3(log 7 − log 5).
Example 12 contracts: 2 log 3 + 3 log 5 − 5 log 2 = log 9 + log 125 − log 32 = log 1125 − log 32 = log (1125/32).
Example 13 solves 3ˣ = 5ˣ⁻². Taking logs, x log 3 = (x − 2) log 5. Collecting the x terms, x(log 5 − log 3) = 2 log 5, so x = 2 log₁₀5 / (log₁₀5 − log₁₀3).
Example 14 finds x from 2 log 5 + ½ log 9 − log 3 = log x. The left side is log 25 + log √9 − log 3 = log 25 + log 3 − log 3 = log 25, so x = 25.
Exercise 1.5 ends with two questions that are about the definition rather than the rules: whether log 2 is rational or irrational, and whether log 100 is. The second has an exact answer and the first does not, which is the whole point of asking them together.
The chapter's closing note in the Optional Exercise — find the number of digits in 4²⁰¹³ given log₁₀2 = 0.3010 — is the one classical use of logarithms the chapter keeps: they turn the size of a number into something you can add up.
10. Summary
Everything in this chapter grows out of two theorems about whole numbers.
The division algorithm says that for positive integers a and b there is a unique pair of whole numbers q, r with a = bq + r and 0 ≤ r < b. Iterating it is Euclid's algorithm, which finds the HCF because HCF (c, d) = HCF (d, r) at every step; the divisor when the remainder first hits zero is the answer.
The Fundamental Theorem of Arithmetic says every composite number factorises into primes in exactly one way, order aside. That uniqueness is not decoration — it is what proves 4ⁿ can never end in zero, what makes HCF and LCM computable from prime powers, and what proves Theorem 1.6, that a prime dividing a² must divide a.
Theorem 1.6 is then the engine of the irrationality proofs: assume √2 = a/b in lowest terms, derive that 2 divides both a and b, and the assumption collapses.
The same factorisation idea decides decimal expansions. Reduce p/q to lowest terms and look at q. If q = 2ⁿ5ᵐ the decimal terminates; if any other prime appears, it recurs. Nothing else about the fraction matters.
The last two sections change subject: 2ˣ = 5 has no answer among the exponents you can write down, so the logarithm is defined to be that missing exponent, logₐN = x exactly when aˣ = N. Its three laws — product to sum, quotient to difference, power to multiple — are the three exponent laws read backwards.
