Chapter 1 — Real Numbers

Numbers grew as human needs grew. The natural numbers N = {1, 2, 3, ...} came first, for counting. Adding zero gave the whole numbers W = {0, 1, 2, 3, ...}. To subtract any number from any other we needed negatives, giving the integers Z = {..., −3, −2, −1, 0, 1, 2, 3, ...}. To divide we needed fractions: a rational number is any number that can be written as p/q where p and q are integers and q ≠ 0, such as 3/4, −7/2, 5 (which is 5/1) and 0.35 (which is 7/20). The set of rationals is called Q. Every integer is rational, but not every rational is an integer.

The Greeks discovered, to their shock, that some lengths cannot be measured by any fraction. The diagonal of a unit square has length √2, and it can be proved (we shall do it in this chapter) that no p/q equals √2. Such numbers are irrational: √2, √3, √5, ∛2, π and e are examples. Their decimal expansions go on for ever without repeating. Rationals and irrationals together make up the real numbers R, and every real number corresponds to exactly one point on the number line, and every point to exactly one real number. That is why the line is called the real number line.

Two facts about decimals separate the families. Every rational number has a decimal expansion that either terminates (1/4 = 0.25) or repeats from some point (1/3 = 0.333..., 1/7 = 0.142857142857...). Every irrational number has a decimal expansion that neither terminates nor repeats. Conversely, a terminating or repeating decimal is always rational; a non-terminating non-repeating decimal is always irrational.

Inside the natural numbers lies the structure this chapter studies. A natural number greater than 1 is prime if its only factors are 1 and itself (2, 3, 5, 7, 11, 13, ...) and composite if it has other factors (4, 6, 8, 9, 10, ...). The number 1 is neither prime nor composite. Two numbers are co-prime if their only common factor is 1, like 8 and 15. The HCF (highest common factor) of two numbers is the largest number dividing both; the LCM (lowest common multiple) is the smallest number both divide. These ideas from earlier classes are the tools we now sharpen with two theorems: Euclid's division lemma and the fundamental theorem of arithmetic.

The examination expects you to place a given number in the right family, to convert between fractions and decimals, and to explain why a number is rational or irrational. Keep the definitions exact: a rational number needs integers p and q with q ≠ 0, and irrational means simply not rational.

When you divide 23 by 5 you get quotient 4 and remainder 3, and you check it as 23 = 5 × 4 + 3. The remainder is always smaller than the divisor: if it were 5 or more you could take one more 5 out. This everyday fact, made precise, is Euclid's division lemma, named after the Greek mathematician Euclid of Alexandria (about 300 BCE), whose book Elements organised the mathematics of his age.

The lemma. Given positive integers a and b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < b. Here a is the dividend, b the divisor, q the quotient and r the remainder. The word lemma means a proved statement used as a step towards other results.

Two things in the statement deserve attention. First, the remainder r may be zero, which happens exactly when b divides a. Second, the integers q and r are unique: for a given a and b there is only one way to write a = bq + r with the remainder in the allowed range. For instance, with a = 23 and b = 5, we could write 23 = 5 × 3 + 8, but that is not the lemma's form because 8 is not less than 5. Only 23 = 5 × 4 + 3 qualifies.

The lemma holds for any positive integers, however large. For a = 1,000,000 and b = 7, division gives q = 142857 and r = 1, so 1,000,000 = 7 × 142857 + 1. When a is smaller than b, the quotient is 0 and the remainder is a itself: 4 = 9 × 0 + 4.

Why call something so obvious a theorem? Because it is the seed from which the rest of the chapter grows. Applied repeatedly it gives an algorithm for the HCF of two numbers that never needs their factors. And it lets us prove facts about all integers at once. Take b = 2: any integer a can be written as 2q + 0 or 2q + 1, which is the precise statement that every integer is even or odd. Take b = 3: every integer is of the form 3q, 3q + 1 or 3q + 2. Take b = 4: every integer is 4q, 4q + 1, 4q + 2 or 4q + 3. These forms are the starting point of many proofs, as the section after the algorithm shows.

In the examination the lemma is usually tested indirectly: through the algorithm for the HCF and through proofs about the forms of integers. But you must be able to state it exactly, including the condition 0 ≤ r < b, and to identify q and r for given numbers.

An algorithm is a step-by-step procedure that is guaranteed to finish and give the right answer. Euclid's algorithm finds the highest common factor of two positive integers by repeated use of the division lemma, without ever factorising them. It rests on one observation: if a = bq + r, then every common factor of a and b is also a factor of r (since r = a − bq), and every common factor of b and r is also a factor of a. So HCF(a, b) = HCF(b, r). The pair (b, r) is smaller than the pair (a, b), so by repeating the step the numbers shrink until the remainder is zero, and the last non-zero remainder is the HCF.

The steps. To find HCF(a, b) with a > b: Step 1, apply the lemma to a and b to get a = bq₁ + r₁. Step 2, if r₁ = 0, then b is the HCF; if not, apply the lemma to b and r₁: b = r₁q₂ + r₂. Step 3, continue until the remainder is zero. The divisor at that stage is the HCF.

Worked example. Find HCF(4052, 12576). Since 12576 > 4052: 12576 = 4052 × 3 + 420. Now divide 4052 by 420: 4052 = 420 × 9 + 272. Divide 420 by 272: 420 = 272 × 1 + 148. Divide 272 by 148: 272 = 148 × 1 + 124. Divide 148 by 124: 148 = 124 × 1 + 24. Divide 124 by 24: 124 = 24 × 5 + 4. Divide 24 by 4: 24 = 4 × 6 + 0. The remainder is now zero and the divisor at this step is 4. So HCF(4052, 12576) = 4. Notice we never needed to know that 4052 = 2² × 1013.

The algorithm has practical uses. A dealer with 420 kg of rice in one bag type and 130 kg in another wants the largest sack that measures both exactly: 420 = 130 × 3 + 30, 130 = 30 × 4 + 10, 30 = 10 × 3 + 0, so the sack is 10 kg. A courtyard 18 m by 12 m to be paved with the largest square tiles: HCF(18, 12) = 6, so 6 m tiles, and (18/6) × (12/6) = 6 tiles. A parade of 616 soldiers and a band of 32 in equal columns: HCF(616, 32) = 8 columns.

The algorithm can be extended to three numbers by finding HCF of two first and then HCF of the result with the third. It also works, with the same steps, for any pair of positive integers however large, which is why computers use it. In the examination write every division as an equation in the form a = bq + r, state clearly which remainder became zero, and box the HCF.

The division lemma gives a method of proving that every integer has a certain property, without checking integers one by one: write the integer in the forms the lemma allows and treat each form separately. This is a standard examination question and the layout matters.

Result 1: every positive even integer is of the form 2q and every positive odd integer is of the form 2q + 1. Let a be any positive integer. By the lemma with b = 2, a = 2q + r with 0 ≤ r < 2, so r = 0 or r = 1. If r = 0, a = 2q, which is even. If r = 1, a = 2q + 1, which is odd. Since these are the only cases, every positive integer is either 2q or 2q + 1, and the even ones are 2q, the odd ones 2q + 1.

Result 2: the square of any positive integer is of the form 3m or 3m + 1. Let a be a positive integer. By the lemma with b = 3, a = 3q, 3q + 1 or 3q + 2. Case 1: a = 3q, so a² = 9q² = 3(3q²) = 3m where m = 3q². Case 2: a = 3q + 1, so a² = 9q² + 6q + 1 = 3(3q² + 2q) + 1 = 3m + 1. Case 3: a = 3q + 2, so a² = 9q² + 12q + 4 = 3(3q² + 4q + 1) + 1 = 3m + 1. In every case a² is 3m or 3m + 1; it is never of the form 3m + 2.

Result 3: the square of any odd positive integer is of the form 8m + 1. An odd integer is 4q + 1 or 4q + 3 (the forms 4q and 4q + 2 are even). If a = 4q + 1, a² = 16q² + 8q + 1 = 8(2q² + q) + 1. If a = 4q + 3, a² = 16q² + 24q + 9 = 8(2q² + 3q + 1) + 1. Both are 8m + 1. Check: 7² = 49 = 8 × 6 + 1.

Result 4: the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8. With a = 3q, a³ = 27q³ = 9(3q³). With a = 3q + 1, a³ = 27q³ + 27q² + 9q + 1 = 9(3q³ + 3q² + q) + 1. With a = 3q + 2, a³ = 27q³ + 54q² + 36q + 8 = 9(3q³ + 6q² + 4q) + 8.

The method is always the same: choose the divisor b that matches the forms in the question, write out all possible remainders, expand the required power in each case using the identities (a + b)² = a² + 2ab + b² and (a + b)³ = a³ + 3a²b + 3ab² + b³, and pull out the common factor to display the form. State at the end that all cases have been covered. A related use is to show that one of any three consecutive integers is divisible by 3: the integers n, n + 1, n + 2 with n = 3q, 3q + 1 or 3q + 2 always include a multiple of 3.

Break 60 into factors: 60 = 6 × 10 = (2 × 3) × (2 × 5) = 2 × 2 × 3 × 5. Or 60 = 4 × 15 = (2 × 2) × (3 × 5). Or 60 = 12 × 5 = 2 × 2 × 3 × 5. However you start, you end with the same primes, 2, 2, 3 and 5, differing only in the order in which you write them. This is no accident. The fundamental theorem of arithmetic states: every composite number can be expressed as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur. The theorem was known in essence to Euclid and was first stated and proved fully by Carl Friedrich Gauss in 1801.

The theorem has two parts. Existence: every composite number can be factored into primes. This is clear, because a composite number has a factor other than 1 and itself; if that factor is composite, factor it again, and so on; the factors get smaller so the process must stop at primes. Uniqueness: there is only one such factorisation. This is the deep part; it is what makes primes the 'atoms' of the integers and allows everything that follows in this chapter.

To find the prime factorisation of a number, divide repeatedly by the smallest prime that goes into it. For 3825: 3825 ÷ 3 = 1275, 1275 ÷ 3 = 425, 425 ÷ 5 = 85, 85 ÷ 5 = 17, and 17 is prime. So 3825 = 3² × 5² × 17. The standard way to write the result is with primes in increasing order and exponents for repeated primes: 32760 = 2³ × 3² × 5 × 7 × 13. A factor tree, splitting a number into two factors at each branch until only primes remain at the tips, gives the same answer.

Uniqueness lets us answer questions that would otherwise be impossible. Can 6ⁿ end with the digit 0 for any natural number n? A number ending in 0 is divisible by 10, so it has 2 and 5 among its prime factors. But 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ, and by uniqueness these are its only primes; 5 never appears. So 6ⁿ never ends in 0. Similarly 4ⁿ = 2²ⁿ has no factor 5 and cannot end in 0, and 12ⁿ cannot end in 5 because it has no prime factor 5 (in fact, it is even).

The theorem also tells us that 7 × 11 × 13 + 13 is composite: it equals 13 × (7 × 11 + 1) = 13 × 78, a product of more than one factor. And 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1) = 5 × 1009 is composite. Look for the common factor and pull it out.

Every result in the rest of this chapter, the HCF-LCM relation, the irrationality of √2, and the test for terminating decimals, is a consequence of uniqueness. Learn the statement word for word.

Once two numbers are written as products of primes, their HCF and LCM can be read off. The HCF is the product of the smallest power of each common prime factor. The LCM is the product of the greatest power of each prime factor that occurs in either number. Think of it this way: the HCF must divide both, so it can contain only what both contain; the LCM must be divisible by both, so it must contain everything either contains.

Worked example. Find HCF and LCM of 96 and 404. Factorise: 96 = 2⁵ × 3 and 404 = 2² × 101. Common primes: only 2, smallest power 2². So HCF = 4. All primes: 2, 3, 101 with greatest powers 2⁵, 3¹, 101¹. So LCM = 32 × 3 × 101 = 9696.

For two numbers there is a beautiful check: HCF(a, b) × LCM(a, b) = a × b. Here 4 × 9696 = 38784 and 96 × 404 = 38784. The relation holds because each prime's smallest power goes into the HCF and its greatest power into the LCM, and smallest × greatest = the product of the two powers in a and b. The relation lets you find one of the four quantities when the other three are known: if HCF(306, 657) = 9 then LCM = 306 × 657 ÷ 9 = 22338.

Three numbers. For 6, 72 and 120: 6 = 2 × 3, 72 = 2³ × 3², 120 = 2³ × 3 × 5. HCF = 2¹ × 3¹ = 6 (smallest power of each prime common to all three). LCM = 2³ × 3² × 5 = 360. Note that HCF × LCM = 2160, which is not equal to 6 × 72 × 120 = 51840; the product relation holds for two numbers only, not three.

The examination often asks for the HCF and LCM of a pair and the verification, or gives one of HCF, LCM and one number and asks for the other number. Write the factorisations in full, mark the common primes, and show the multiplication. When one number is a factor of the other, HCF is the smaller and LCM the larger: HCF(12, 36) = 12, LCM = 36.

A frequent trap: a question saying 'HCF of two numbers is 18 and LCM is 380; find the numbers' has no answer, because the HCF must divide the LCM, and 380 ÷ 18 is not an integer. Always check that HCF divides LCM before working further.

Many practical questions reduce to finding an HCF or an LCM, and the skill is to recognise which. Use the HCF when you need the largest quantity that measures, divides or fits into several given quantities exactly: the largest tile, the longest tape, the greatest number of equal groups, the maximum capacity of a container that fills several tanks exactly. Use the LCM when you need the smallest quantity that several given quantities all divide: the earliest time events coincide again, the least number of steps to cover the same distance, the smallest number divisible by several numbers.

Bells and traffic lights (LCM). Three bells toll at intervals of 9, 12 and 15 minutes. If they toll together at 8:00 a.m., when will they next toll together? Each bell tolls at multiples of its interval, so they coincide at common multiples; the next time is the LCM. 9 = 3², 12 = 2² × 3, 15 = 3 × 5. LCM = 2² × 3² × 5 = 180 minutes = 3 hours. They toll together next at 11:00 a.m.

Circular track (LCM). Sonia takes 18 minutes and Ravi 12 minutes to go round a circular path from the same point in the same direction. They meet again at the start after LCM(18, 12) = 36 minutes, by which time Sonia has done 2 rounds and Ravi 3.

Containers (HCF). Three tanks hold 403 L, 434 L and 465 L. Find the container of maximum capacity that can measure the contents of each an exact number of times. 403 = 13 × 31, 434 = 2 × 7 × 31, 465 = 3 × 5 × 31. HCF = 31 litres.

Steps of equal length (HCF and division). Three persons step off together and their steps measure 40 cm, 42 cm and 45 cm. The minimum distance each should walk so that all can cover it in complete steps is LCM(40, 42, 45): 40 = 2³ × 5, 42 = 2 × 3 × 7, 45 = 3² × 5, LCM = 2³ × 3² × 5 × 7 = 2520 cm = 25.2 m.

Sweets and fruits (HCF). A shopkeeper has 420 sweets and 130 biscuits and wants to make identical packets with none left over. The maximum number of packets is HCF(420, 130) = 10, each packet having 42 sweets and 13 biscuits.

Remainder problems. Find the least number which when divided by 12, 16 and 24 leaves remainder 7 in each case. The number is 7 more than a common multiple, so it is LCM(12, 16, 24) + 7 = 48 + 7 = 55. Find the largest number that divides 245 and 1029 leaving remainder 5 in each case: subtract the remainders, 240 and 1024, and find HCF(240, 1024) = 16.

In each problem write one line saying why HCF or LCM is needed, then factorise, compute, and answer in the units of the question. Examiners award marks for the reasoning as well as the number.

We have said that √2 is irrational, meaning there is no fraction p/q equal to it. But how could anyone be sure? Perhaps some huge fraction, with a million-digit numerator and denominator, equals √2 exactly. The proof that this can never happen is one of the oldest and most beautiful arguments in mathematics, going back to the Pythagoreans, and it depends on the fundamental theorem of arithmetic.

First we need a small result about primes, called a lemma. Theorem: if p is a prime and p divides a², then p divides a, where a is a positive integer. Proof: by the fundamental theorem, write a = p₁p₂...p_n as a product of primes (not necessarily distinct). Then a² = (p₁p₂...p_n)(p₁p₂...p_n) = p₁²p₂²...p_n². By the uniqueness part of the fundamental theorem, the primes of a² are exactly p₁, p₂, ..., p_n, each appearing twice. Since p divides a², p must be one of these primes, say p = p_i. But p_i is a factor of a. Hence p divides a.

The lemma is not true for composite divisors: 4 divides 6² = 36 but 4 does not divide 6. It is the primeness of p that makes it work, through uniqueness of factorisation. Examples of the lemma in action: 3 divides 36 = 6², and indeed 3 divides 6; 5 divides 225 = 15², and 5 divides 15; 7 divides 196 = 14², and 7 divides 14.

We also need the idea of proof by contradiction. To prove a statement is true, we assume it is false and show that this assumption leads to something impossible. Since the assumption produces a contradiction, it cannot be right, so the original statement must be true. In our case we shall assume that √2 is rational, write it as p/q in lowest terms, and show that p and q must both be even, contradicting 'lowest terms'.

One more preparation: any fraction can be reduced to lowest terms by dividing numerator and denominator by their HCF, so if √2 = p/q we may assume p and q are co-prime, having no common factor other than 1. This assumption is the point that the contradiction will attack.

With the lemma and the method in hand, the proofs of the next section are short. What you must remember is the chain: fundamental theorem → 'p prime and p | a² implies p | a' → the assumption √2 = p/q in lowest terms → both p and q divisible by 2 → contradiction. Each link is asked in examinations, and the lemma is sometimes asked on its own with its proof.

Theorem: √2 is irrational. Proof: Suppose, to the contrary, that √2 is rational. Then there exist co-prime positive integers a and b (b ≠ 0) such that √2 = a/b. Squaring both sides, 2 = a²/b², so a² = 2b². Therefore 2 divides a². By the theorem of the previous section (2 is prime), 2 divides a. So we can write a = 2c for some integer c. Substituting, (2c)² = 2b², that is 4c² = 2b², so b² = 2c². Therefore 2 divides b², and again by the theorem, 2 divides b. So both a and b are divisible by 2. But a and b were taken co-prime, with no common factor other than 1. This contradiction shows that our supposition was wrong. Hence √2 is irrational.

Theorem: √3 is irrational. Proof: Suppose √3 = a/b with a, b co-prime positive integers. Then a² = 3b², so 3 divides a², so 3 divides a (3 is prime). Write a = 3c. Then 9c² = 3b², so b² = 3c², so 3 divides b², so 3 divides b. Thus 3 is a common factor of a and b, contradicting co-primeness. Hence √3 is irrational. The proof for √5 is identical with 5 in place of 3, and the same argument works for √p for any prime p.

Once a few square roots are known to be irrational, many more irrationals follow from the fact that the rationals are closed under addition, subtraction, multiplication and division (by a non-zero number): the sum, difference, product or quotient of two rational numbers is rational. So if an expression involving √2 were rational, we could solve for √2 and get a rational, a contradiction.

Show 3 + 2√5 is irrational. Suppose 3 + 2√5 = a/b, rational (a, b integers, b ≠ 0). Then 2√5 = a/b − 3 = (a − 3b)/b, so √5 = (a − 3b)/(2b). The right side is rational since a, b are integers and 2b ≠ 0. So √5 is rational, contradicting the theorem. Hence 3 + 2√5 is irrational.

Show 1/√2 is irrational. Suppose 1/√2 = a/b. Then √2 = b/a, rational (a ≠ 0), contradiction. Show 7√5 is irrational. Suppose 7√5 = a/b; then √5 = a/(7b), rational, contradiction. Show 6 + √2 is irrational. Suppose 6 + √2 = a/b; then √2 = a/b − 6 = (a − 6b)/b, rational, contradiction.

In every such proof the examination expects: the supposition in the form 'let it be rational, equal to a/b'; the algebra isolating the known irrational; the statement that the other side is rational because it is built from integers; the sentence 'this contradicts the fact that √p is irrational'; and the conclusion. Do not skip the co-prime condition in the basic proof for √2 and √3, because the contradiction is exactly that a and b turn out to share a factor.

One caution: the sum or product of two irrationals may be rational. √2 × √2 = 2, and (2 + √3) + (2 − √3) = 4. So the closure argument works only when one of the numbers is rational and the other irrational, or when a specific irrational can be isolated.

Every rational number p/q has a decimal expansion that either stops or repeats. Which of the two it does is decided entirely by the prime factors of the denominator, once the fraction is in lowest terms. This section proves the rule and shows how to apply it without doing the division.

Look at some terminating decimals: 0.375 = 375/1000 = 3/8; 0.104 = 104/1000 = 13/125; 0.0875 = 875/10000 = 7/80; 23.3408 = 233408/10000 = 14588/625. In every case the fraction, in lowest terms, has a denominator that is a product of powers of 2 and 5 only: 8 = 2³, 125 = 5³, 80 = 2⁴ × 5, 625 = 5⁴. The reason is clear: a terminating decimal is a fraction whose denominator is a power of 10 = 2 × 5, and cancelling can only remove 2s and 5s from that denominator.

Theorem 1. Let x be a rational number whose decimal expansion terminates. Then x can be expressed as p/q where p and q are co-prime and the prime factorisation of q is of the form 2ⁿ5ᵐ, where n and m are non-negative integers.

Theorem 2 (the converse). Let x = p/q be a rational number with p, q co-prime and q = 2ⁿ5ᵐ. Then the decimal expansion of x terminates. The reason: multiply numerator and denominator by enough 2s or 5s to make the denominator 10ᵏ for k = the larger of n and m; then the fraction is a whole number over a power of 10, which is a terminating decimal with k decimal places. For example 3/8 = 3/2³ = (3 × 5³)/(2³ × 5³) = 375/1000 = 0.375.

Theorem 3. Let x = p/q be a rational number with p, q co-prime and q not of the form 2ⁿ5ᵐ (that is, q has some prime factor other than 2 and 5). Then the decimal expansion of x is non-terminating repeating (recurring). Example: 1/7 = 0.142857 142857 ... with the block 142857 repeating; 7 is not of the form 2ⁿ5ᵐ. Another: 1/6: 6 = 2 × 3 contains the prime 3, so 1/6 = 0.1666... is non-terminating repeating.

Applying the rule. To decide the nature of 13/3125: 3125 = 5⁵, so terminating. 17/8: 8 = 2³, terminating. 64/455: 455 = 5 × 7 × 13, so non-terminating repeating. 15/1600: 1600 = 2⁶ × 5², terminating. 29/343: 343 = 7³, non-terminating repeating. 23/(2³5²): terminating. 129/(2²5⁷7⁵): the 7 spoils it, non-terminating repeating. 6/15: reduce first to 2/5, then denominator is 5, terminating. Always reduce to lowest terms before looking at the denominator: 6/15 has a 3 in its denominator but it cancels.

To find the number of decimal places of a terminating decimal, take the larger of n and m: 17/8 with 8 = 2³ has 3 decimal places (2.125); 13/3125 with 5⁵ has 5 places (0.00416). Two related questions: a decimal like 43.123456789 is rational with denominator 10⁹ = 2⁹5⁹; the recurring decimal 43.123456789 (bar over 123456789) is rational but its denominator in lowest terms has prime factors other than 2 and 5; and 0.120120012000... with growing blocks of zeros is irrational because it neither terminates nor repeats.

The theorems of the last section say what kind of decimal a fraction gives. This section practises the actual conversions in both directions, which the examination asks for directly.

Fraction to decimal. For a denominator of the form 2ⁿ5ᵐ, either divide, or better, multiply top and bottom to make the denominator a power of ten. 13/3125: 3125 = 5⁵, multiply by 2⁵ = 32: (13 × 32)/(5⁵ × 2⁵) = 416/100000 = 0.00416. 15/1600: 1600 = 2⁶ × 5²; multiply by 5⁴ = 625: 9375/10⁶ = 0.009375. 23/(2³5²) = 23/200; multiply by 5: 115/1000 = 0.115. 6/15 = 2/5 = 4/10 = 0.4. 35/50 = 7/10 = 0.7. For denominators with other primes, long division gives the repeating block: 1/7 → 0.142857 repeating; 2/11 → 0.18 repeating; 5/12 → 0.41666... = 0.416 with bar over the 6.

Terminating decimal to fraction. Write the digits over the matching power of ten and reduce: 0.375 = 375/1000 = 3/8; 2.125 = 2125/1000 = 17/8; 0.00416 = 416/100000 = 13/3125.

Repeating decimal to fraction. Let x be the decimal; multiply by a power of 10 that shifts one full repeating block; subtract to cancel the repeating tail. For x = 0.333...: 10x = 3.333..., so 10x − x = 3, 9x = 3, x = 1/3. For x = 0.181818...: 100x = 18.1818..., 100x − x = 18, 99x = 18, x = 2/11. For x = 0.41666... (the 6 repeats): 10x = 4.1666..., 100x = 41.666..., subtract 100x − 10x = 37.5, so 90x = 37.5, x = 375/900 = 5/12. For x = 2.142857142857...: 10⁶x = 2142857.142857..., subtract x: 999999x = 2142855, x = 2142855/999999 = 15/7 after dividing by 142857. A shortcut for a pure repeating decimal 0.abc(repeating) is abc/999 with as many nines as digits in the block: 0.027027... = 27/999 = 1/37.

Checking the theorem. Having converted 0.41666... to 5/12, note 12 = 2² × 3 contains 3, consistent with non-termination. Having converted 0.00416 to 13/3125 = 13/5⁵, consistent with termination. This cross-check protects against arithmetic slips.

A last point of care: a decimal expansion given as 0.1010010001... is not repeating, because the blocks of zeros grow, so it represents an irrational number and cannot be converted to a fraction at all. Similarly the expansion of π begins 3.14159265... and never repeats; 22/7 = 3.142857 (repeating) and 3.14 are rational approximations, not π itself.

This chapter is examined in every form the Class 10 paper uses: one-mark objective questions, two-mark short answers, four-mark problems and eight-mark proofs. Here is how the ideas combine, with the reasoning that earns full marks.

Objective questions test definitions and quick facts. The HCF of two consecutive integers is 1 (they are co-prime). The LCM of two co-prime numbers is their product. If p and q are primes, HCF(p, q) = 1 and LCM = pq. The HCF of the smallest prime (2) and the smallest composite (4) is 2. The number of decimal places of 17/2³5² is 3. The decimal expansion of 15/1600 terminates after 6 places. 7ⁿ can never end in 0 or 5 because 7ⁿ has no factor 2 or 5. The product of two irrationals may be rational (√2 × √8 = 4). Any number of the form 2q + 1 is odd. HCF of 96 and 404 is 4.

Two-mark questions. Find HCF(52, 117) by Euclid's algorithm: 117 = 52 × 2 + 13, 52 = 13 × 4 + 0, HCF = 13. Express 140 as a product of primes: 2² × 5 × 7. Without division, say whether 77/210 terminates: reduce to 11/30, 30 = 2 × 3 × 5, so it does not terminate. Given HCF(a, b) = 5 and LCM = 100 with a = 20, find b: b = 5 × 100/20 = 25.

Four-mark questions. Show that any positive odd integer is of the form 4q + 1 or 4q + 3 (write a = 4q + r, r = 0, 1, 2, 3, discard the even forms). Find the least number of tiles or the time bells coincide, with reasoning. Find HCF and LCM of 510 and 92 and verify HCF × LCM = product: 510 = 2 × 3 × 5 × 17, 92 = 2² × 23, HCF = 2, LCM = 2² × 3 × 5 × 17 × 23 = 23460, and 2 × 23460 = 46920 = 510 × 92. Prove that 3 + √5 is irrational.

Eight-mark questions. Prove √2 (or √3, √5) is irrational, giving the lemma 'p prime, p | a² implies p | a' as a step; or prove that the square of any positive integer is of the form 3m or 3m + 1, with all three cases. Use the standard layout: statement, supposition or case division, algebra with every line justified, contradiction or conclusion.

Common errors to avoid. Forgetting to reduce a fraction before testing the denominator. Writing a remainder equal to or larger than the divisor in Euclid's lemma. Assuming HCF × LCM = product for three numbers. Omitting 'co-prime' in the irrationality proof. Claiming √2 + √3 = √5. Stopping a factor tree at a composite. Calling 1 a prime. Writing 0.101001000... as rational. In a proof by contradiction, forgetting to state what was contradicted.

The chapter rewards exactness. Learn the three statements, the lemma, the fundamental theorem and the decimal theorems, word for word; practise Euclid's algorithm until each line is automatic; and write proofs as chains of justified steps rather than as answers.