Chapter 1 — Real Numbers

Before proving anything new, it helps to see the whole family of numbers laid out in order. The natural numbers N = {1, 2, 3, ...} are the counting numbers. Adding zero gives the whole numbers W = {0, 1, 2, 3, ...}. Adding the negatives gives the integers Z = {..., -3, -2, -1, 0, 1, 2, 3, ...}. Each of these sets is contained in the next one, so N is a subset of W, which is a subset of Z.

A rational number is any number that can be written in the form p/q where p and q are integers and q is not zero. The set of rational numbers is called Q. Every integer is rational because we can write 7 as 7/1. Between any two rational numbers there are infinitely many more rational numbers; for example between 1/2 and 1/3 lies 5/12, and between 1/2 and 5/12 lies 11/24, and so on without end. Yet the rational numbers do not fill the number line. Numbers such as the square root of 2, the square root of 5, pi and the cube root of 3 cannot be written as p/q at all. These are the irrational numbers. In Class 9 you saw that the square root of 2 lies between 1.4 and 1.5, and that its decimal expansion 1.41421356... never ends and never repeats.

The rational and irrational numbers together make up the real numbers R. Every real number corresponds to exactly one point on the number line and every point corresponds to exactly one real number. That is why the number line is also called the real line.

The decimal form of a number tells us which family it belongs to. If the decimal terminates, like 0.375, or repeats in a block, like 0.272727..., the number is rational. If the decimal neither terminates nor repeats, the number is irrational. This chapter will explain the reason behind the first rule and prove the irrationality of several specific numbers, rather than simply asking you to accept these facts.

Two properties of integers will be used again and again. First, division with remainder: dividing 17 by 5 gives quotient 3 and remainder 2, so 17 = 5 × 3 + 2. Second, every integer greater than 1 is either a prime, having exactly two factors, or a composite, having more than two factors. The number 1 is neither prime nor composite. Keeping these definitions sharp makes the proofs that follow easy to read.

A lemma is a small proven statement that is used as a stepping stone to prove larger results. Euclid's division lemma is the mathematical form of what happens when you divide one whole number by another and keep the remainder.

Statement. Given two 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 is the divisor, q is the quotient and r is the remainder.

Two words in the statement deserve attention. Exist means that for every choice of a and b such a pair (q, r) can always be found. Unique means that there is only one such pair once we insist that the remainder r lies between 0 and b − 1. For instance, 23 = 5 × 4 + 3 and also 23 = 5 × 3 + 8, but only the first is allowed, because the remainder must be less than 5.

The remainder can be zero. When r = 0 we have a = bq, which means b divides a exactly. So Euclid's lemma includes the idea of divisibility as a special case.

The lemma is powerful because it lets us classify all integers by their remainder on division by a fixed number. Dividing any positive integer by 2 gives remainder 0 or 1, so every positive integer is of the form 2q or 2q + 1, that is, even or odd. Dividing by 3 gives remainder 0, 1 or 2, so every positive integer is of the form 3q, 3q + 1 or 3q + 2. This idea is used in typical board questions such as: show that the square of any positive integer is of the form 3m or 3m + 1. The proof takes each of the three forms, squares it and rewrites the result. If a = 3q, then a² = 9q² = 3(3q²), which is of the form 3m. If a = 3q + 1, then a² = 9q² + 6q + 1 = 3(3q² + 2q) + 1, of the form 3m + 1. If a = 3q + 2, then a² = 9q² + 12q + 4 = 3(3q² + 4q + 1) + 1, again of the form 3m + 1. Hence no perfect square leaves remainder 2 on division by 3.

Although the lemma is stated for positive integers, it can be extended to all integers with the same condition 0 ≤ r < |b|. In this chapter we stay with positive integers.

An algorithm is a fixed sequence of steps that reaches an answer. Euclid's division algorithm uses the division lemma repeatedly to find the highest common factor (HCF) of two positive integers, that is, the largest positive integer that divides both of them.

The steps. To find HCF(a, b) with a > b:

• Step 1: Apply the division lemma to a and b to get a = bq₁ + r₁.
• Step 2: If r₁ = 0, then b is the HCF. If r₁ ≠ 0, apply the lemma to b and r₁ to get b = r₁q₂ + r₂.
• Step 3: Continue in this way, each time dividing the previous divisor by the previous remainder, until a remainder of zero appears. The divisor at that final stage is the HCF.

Why it works. The key fact is that HCF(a, b) = HCF(b, r), where r is the remainder when a is divided by b. Any number dividing both a and b divides a − bq = r, and any number dividing b and r divides bq + r = a. So the pair (a, b) and the pair (b, r) have exactly the same common factors, and therefore the same highest common factor. Each step replaces the pair with a smaller pair, and since the remainders are decreasing whole numbers, the process must end.

Worked example. Find HCF(455, 42). 455 = 42 × 10 + 35. Then 42 = 35 × 1 + 7. Then 35 = 7 × 5 + 0. The last non-zero remainder is 7, so HCF(455, 42) = 7.

A second example. Find HCF(4052, 12576). 12576 = 4052 × 3 + 420. 4052 = 420 × 9 + 272. 420 = 272 × 1 + 148. 272 = 148 × 1 + 124. 148 = 124 × 1 + 24. 124 = 24 × 5 + 4. 24 = 4 × 6 + 0. So the HCF is 4.

The algorithm has practical uses in word problems. If a hall is 18 m by 12 m and must be paved with the largest possible square tiles, the tile side is HCF(18, 12) = 6 m. If two ropes of lengths 96 m and 120 m are to be cut into pieces of equal maximum length, that length is HCF(96, 120) = 24 m. Questions of this type are common in the two-mark section of the examination, and the marks are given for showing each step of the lemma clearly, not just the final answer.

Every composite number can be broken into factors, and the factors can be broken further until only primes remain. For example 60 = 4 × 15 = 2 × 2 × 3 × 5. If we had started differently, 60 = 6 × 10 = 2 × 3 × 2 × 5, we get the same primes with the same multiplicities, only in a different order. This observation is not an accident; it is a theorem.

Fundamental Theorem of Arithmetic. 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 has two parts. The existence part says a prime factorisation can always be found. The uniqueness part says there is only one such factorisation. Uniqueness is the deeper statement and is what makes the theorem useful. It allows us to talk about the prime factorisation of a number, and to compare two numbers factor by factor.

By convention we write the prime factorisation in ascending order of primes with exponents: 60 = 2² × 3 × 5, 3825 = 3² × 5² × 17, 5005 = 5 × 7 × 11 × 13. To obtain it we use a factor tree or repeated division by the smallest prime.

Using the theorem to reason about divisibility. A frequent examination question asks: can the number 6ⁿ end with the digit 0 for any natural number n? A number ends in 0 only if it is divisible by 10, that is, only if both 2 and 5 appear in its prime factorisation. But 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ, and by the uniqueness of prime factorisation there is no 5 in it. So 6ⁿ can never end in 0. The same reasoning shows that 4ⁿ, 12ⁿ and 9ⁿ never end in zero either.

Another type of question: explain why 7 × 11 × 13 + 13 is composite. Taking 13 as common, the expression is 13 × (77 + 1) = 13 × 78, a product of two numbers greater than 1, hence composite. Similarly 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 × (1008 + 1) = 5 × 1009 is composite.

The theorem also justifies a fact we use silently: if a prime p divides a product ab, then p divides a or p divides b. This property, sometimes called Euclid's lemma on primes, is the tool behind the irrationality proofs later in the chapter. If p divides a² = a × a, then p must divide a.

Once two numbers are written as products of primes, their HCF and their lowest common multiple (LCM) can be read off directly.

Rule for HCF. Take every prime that occurs in both factorisations, with the smallest power in which it occurs, and multiply. Rule for LCM. Take every prime that occurs in either factorisation, with the greatest power in which it occurs, and multiply.

Worked example. Find the HCF and LCM of 96 and 404. 96 = 2⁵ × 3 and 404 = 2² × 101. The only common prime is 2, with smallest power 2², so HCF = 4. For LCM take 2⁵, 3 and 101: LCM = 32 × 3 × 101 = 9696.

The product relation. For any two positive integers a and b, HCF(a, b) × LCM(a, b) = a × b. Check with the example: 4 × 9696 = 38784 and 96 × 404 = 38784. This relation lets us find one quantity when the other three are known. If HCF(306, 657) = 9, then LCM = (306 × 657)/9 = 22338. Note carefully that this relation holds for two numbers only; for three numbers the product of HCF and LCM is not, in general, equal to the product of the numbers.

Three numbers. Find the HCF and LCM of 12, 15 and 21. 12 = 2² × 3, 15 = 3 × 5, 21 = 3 × 7. The only prime common to all three is 3, so HCF = 3. LCM = 2² × 3 × 5 × 7 = 420. Here HCF × LCM = 1260 while 12 × 15 × 21 = 3780, confirming that the two-number relation does not extend.

Word problems. Three bells ring at intervals of 4, 7 and 14 minutes. If they ring together at 6 a.m., when will they next ring together? The answer is after LCM(4, 7, 14) = 28 minutes, at 6:28 a.m. In a circular track, if Sonia takes 18 minutes and Ravi takes 12 minutes for one round and they start together, they meet again at the start after LCM(18, 12) = 36 minutes. When the question asks for the largest quantity that divides given amounts, use HCF; when it asks for the smallest quantity that is a multiple of given amounts, use LCM.

A common error is to include a prime that appears in only one number when finding the HCF, or to add exponents instead of choosing the greater one when finding the LCM. Writing the factorisations one below the other with matching primes aligned prevents both mistakes.

Every rational number p/q has a decimal expansion that either terminates (ends after finitely many digits) or is non-terminating repeating (a block of digits repeats forever). The type of expansion is decided entirely by the prime factors of the denominator, once the fraction is in lowest terms.

Theorem 1. Let x = p/q be a rational number in lowest terms. If the prime factorisation of q is of the form 2ⁿ × 5^m, where n and m are non-negative integers, then x has a terminating decimal expansion. Theorem 2 (converse). If x has a terminating decimal expansion, then x can be written as p/q where q is of the form 2ⁿ × 5^m. Theorem 3. If q is not of the form 2ⁿ × 5^m, then x has a non-terminating repeating expansion.

Why the primes 2 and 5? A terminating decimal like 0.375 is 375/1000, and 1000 = 10³ = 2³ × 5³. Any terminating decimal is a whole number divided by a power of 10, and after cancelling, the denominator can only keep the primes 2 and 5. Conversely, if q = 2ⁿ × 5^m, we can multiply numerator and denominator by enough 2s or 5s to make the denominator a power of 10. For example 7/40 = 7/(2³ × 5) = (7 × 5²)/(2³ × 5³) = 175/1000 = 0.175.

Deciding without dividing. 13/3125: 3125 = 5⁵, so terminating (it is 0.00416). 17/8: 8 = 2³, terminating (2.125). 64/455: 455 = 5 × 7 × 13, so non-terminating repeating. 15/1600: 1600 = 2⁶ × 5², terminating. 29/343: 343 = 7³, non-terminating repeating. 6/15: first reduce to 2/5, then the denominator is 5, so terminating (0.4). Always reduce to lowest terms first; 6/15 would be wrongly judged if we looked at 15 = 3 × 5.

From decimal back to fraction. A terminating decimal is converted by writing it over a power of ten and cancelling: 43.123456789 = 43123456789/10⁹, so the denominator is 2⁹ × 5⁹. A repeating decimal is converted using the shifting method: if x = 0.2333..., then 10x = 2.333... and 100x = 23.333..., so 90x = 21 and x = 7/30. Since 30 = 2 × 3 × 5 contains a 3, the expansion is indeed non-terminating, as expected.

The period of a repeating decimal is the number of digits in the repeating block. 1/7 = 0.142857142857... has period 6. A useful fact for checking answers: the period of 1/q is always less than q.

An irrational number is a real number that cannot be written as p/q with integers p, q and q ≠ 0. Its decimal expansion is non-terminating and non-repeating. The Pythagoreans in ancient Greece discovered such numbers when they realised that the diagonal of a unit square, whose length is the square root of 2, could not be measured by any fraction. This chapter proves that fact rigorously.

The proof uses the method of contradiction: we assume the opposite of what we want to prove, argue logically, and arrive at something impossible. The impossibility shows the assumption was false. It also uses the theorem, which follows from unique factorisation: if a prime p divides a², then p divides a. This is because the primes in a² are exactly the primes in a, each with doubled exponent, so p cannot appear in a² without appearing in a.

Theorem: √2 is irrational. Suppose, on the contrary, that √2 is rational. Then √2 = a/b for some integers a and b with b ≠ 0, and we may assume a and b have no common factor other than 1 (if they had, we would cancel it first). Squaring, 2 = a²/b², so a² = 2b². This means 2 divides a², and since 2 is prime, 2 divides a. Write a = 2c for some integer c. Substituting, (2c)² = 2b², that is, 4c² = 2b², so b² = 2c². Hence 2 divides b² and so 2 divides b. Now both a and b are divisible by 2, which contradicts our choice that they have no common factor. Therefore our assumption was wrong, and √2 is irrational.

The same argument, word for word with 3 in place of 2, proves that √3 is irrational, and in general that √p is irrational for every prime p. Board examinations regularly ask for the proof for √2, √3 or √5, and full marks require every step: the assumption, the coprime condition, the squaring, the use of the prime-divides-square property twice, and the final contradiction.

Care is needed with numbers like √4 or √9. Here the argument breaks down at the first squaring step, because 4 = a²/b² gives a = 2b with no contradiction. Indeed √4 = 2 is rational. The proof only works for numbers that are not perfect squares.

Irrational numbers are not rare curiosities. Between any two real numbers there are infinitely many irrationals, and in a precise sense there are far more irrationals than rationals on the number line. Well-known irrationals include √2, √3, √5, π and e, though the irrationality of π is much harder to prove and is not attempted at this level.

Once we know that √2, √3 and √5 are irrational, we can prove that many other numbers are irrational by using the closure properties of rational numbers. The rational numbers are closed under addition, subtraction, multiplication and division (by non-zero numbers): the result of any such operation on two rationals is again rational. This closure is the tool.

General facts. If r is a non-zero rational number and s is irrational, then r + s, r − s, r × s and r/s are all irrational. Note the condition non-zero for multiplication and division: 0 × √2 = 0 is rational. The sum or product of two irrationals, however, may be rational or irrational: √2 + (−√2) = 0 is rational, √2 × √2 = 2 is rational, but √2 + √3 and √2 × √3 = √6 are irrational.

Proof that 5 − √3 is irrational. Assume, to the contrary, that 5 − √3 is rational. Then 5 − √3 = a/b for coprime integers a, b with b ≠ 0. Rearranging, √3 = 5 − a/b = (5b − a)/b. Since a and b are integers, (5b − a)/b is a rational number. So √3 is rational. But this contradicts the fact that √3 is irrational. Hence 5 − √3 is irrational.

Proof that 3√2 is irrational. Assume 3√2 = a/b with a, b integers, b ≠ 0. Then √2 = a/(3b). Since 3b is a non-zero integer, a/(3b) is rational, so √2 is rational, a contradiction. Hence 3√2 is irrational.

Proof that √2 + √3 is irrational. Assume √2 + √3 = a/b is rational. Then √3 = a/b − √2. Squaring both sides, 3 = a²/b² − 2(a/b)√2 + 2, so 2(a/b)√2 = a²/b² − 1, giving √2 = (a² − b²)/(2ab). The right side is rational, so √2 is rational, a contradiction. This squaring technique is needed whenever two different square roots appear.

The structure of every such proof is the same: (1) assume the number is rational and write it as a/b, (2) isolate the known irrational on one side, (3) observe that the other side is rational because it is built from rationals by the four operations, (4) declare the contradiction. In an examination these proofs carry two or four marks and are graded on the completeness of the four steps. Students often skip step (3), the justification that the expression is rational; write it explicitly.

Simplification of expressions with surds, such as showing that (√2 + 1)(√2 − 1) = 1 or that 1/(√3 − √2) = √3 + √2 after rationalising, also belongs here and gives practice in handling these numbers confidently.

In earlier classes you worked with exponents: 2³ = 8 means 2 multiplied by itself three times is 8. Here 2 is the base, 3 is the exponent and 8 is the result. The laws of exponents that you know are: a^m × aⁿ = a^m+n, a^m ÷ aⁿ = a^m−n, (a^m)ⁿ = a^mn, a⁰ = 1 and a⁻ⁿ = 1/aⁿ. Fractional exponents give roots: a^1/2 = √a and a^1/3 is the cube root of a.

A logarithm answers the reverse question. Instead of asking what 2³ is, it asks: to what power must 2 be raised to give 8? The answer is 3, and we write log₂ 8 = 3, read as the logarithm of 8 to the base 2 is 3.

Definition. For a > 0, a ≠ 1 and x > 0, log_a x = n if and only if aⁿ = x. The two statements aⁿ = x (exponential form) and log_a x = n (logarithmic form) say exactly the same thing. Learning to convert between them instantly is the first skill of this topic.

Why the conditions? The base must be positive and not 1: a base of 1 gives 1ⁿ = 1 for all n, so log₁ x would be meaningless. The number x must be positive because a positive base raised to any real power is always positive; there is no power of 2 that equals −8 or 0. So logarithms of zero and of negative numbers are not defined.

Conversions. 3⁴ = 81 becomes log₃ 81 = 4. 10⁻² = 0.01 becomes log₁₀ 0.01 = −2. 25^1/2 = 5 becomes log₂₅ 5 = 1/2. In the other direction, log₅ 125 = 3 becomes 5³ = 125 and log_x 16 = 2 becomes x² = 16, so x = 4.

Two immediate results. Since a¹ = a, we get log_a a = 1 for every valid base. Since a⁰ = 1, we get log_a 1 = 0 for every valid base. So log₇ 7 = 1 and log₇ 1 = 0.

Finding a logarithm by writing the number as a power of the base. To find log₂ 32, write 32 = 2⁵, so the answer is 5. To find log₃ (1/27), write 1/27 = 3⁻³, so the answer is −3. To find log_√2 8, write 8 = 2³ = (√2)⁶, so the answer is 6. Logarithms to base 10 are called common logarithms and are often written simply as log x; they were invented to turn multiplication into addition and are still used in the pH scale, the decibel scale and the Richter scale.

Because logarithms are exponents, every law of exponents becomes a law of logarithms. The two laws in this section come from the laws a^m × aⁿ = a^m+n and a^m ÷ aⁿ = a^m−n.

Product law. log_a (xy) = log_a x + log_a y, for positive x, y and a valid base a. Proof. Let log_a x = m and log_a y = n. Then x = a^m and y = aⁿ, so xy = a^m × aⁿ = a^m+n. Converting back to logarithmic form, log_a (xy) = m + n = log_a x + log_a y.

Quotient law. log_a (x/y) = log_a x − log_a y. Proof. With the same m and n, x/y = a^m/aⁿ = a^m−n, so log_a (x/y) = m − n.

The examination often asks for these proofs as two-mark questions, so learn the three-line argument: name the logarithms as m and n, convert to exponential form, use the law of exponents, convert back.

What the laws do not say. log (x + y) is not log x + log y, and log x × log y is not log (xy). The product law turns a product inside the logarithm into a sum of logarithms. Confusing these is the most common error in the topic.

Applications. Express log 15 in terms of log 3 and log 5: log 15 = log (3 × 5) = log 3 + log 5. Express log (8/5) as a difference: log 8 − log 5. Write log 2 + log 3 + log 5 as a single logarithm: log (2 × 3 × 5) = log 30. Write log 18 − log 6 as a single logarithm: log (18/6) = log 3.

Evaluating with the laws. Find log₂ 8 + log₂ 4: this is log₂ 32 = 5, or directly 3 + 2 = 5. Find log₁₀ 125 + log₁₀ 8: this is log₁₀ 1000 = 3. Find log₃ 54 − log₃ 2: this is log₃ 27 = 3. Find log₅ 100 − log₅ 4: log₅ 25 = 2.

Extending to three factors. log_a (xyz) = log_a x + log_a y + log_a z, by applying the product law twice. So log 2 + log 5 + log 10 = log 100 = 2 (to base 10). These laws, together with the power law of the next section, are all that is needed to expand or condense any logarithmic expression in this chapter.

The third law of logarithms comes from the exponent law (a^m)ⁿ = a^mn.

Power law. log_a xⁿ = n log_a x, for x > 0 and any real number n. Proof. Let log_a x = m, so x = a^m. Then xⁿ = (a^m)ⁿ = a^mn. Converting to logarithmic form, log_a xⁿ = mn = n log_a x.

Because roots are fractional powers, the law also handles them: log √x = log x^1/2 = (1/2) log x, and log of the cube root of x is (1/3) log x. Because reciprocals are negative powers, log (1/x) = log x⁻¹ = −log x.

Expanding an expression. To expand means to write a single logarithm of a product, quotient or power as a sum, difference or multiple of simpler logarithms. Work from the outside in: first split products and quotients, then bring down powers.

Expand log (x² y³ / z⁴): = log x² + log y³ − log z⁴ = 2 log x + 3 log y − 4 log z.

Expand log √(pq): = (1/2) log (pq) = (1/2)(log p + log q).

Expand log (128 / 625) in terms of log 2 and log 5: 128 = 2⁷ and 625 = 5⁴, so the expression equals 7 log 2 − 4 log 5.

Expand log 343/125 = log 7³ − log 5³ = 3 log 7 − 3 log 5.

Condensing an expression. The reverse process combines several logarithms into one. Write 2 log 3 + 3 log 5 − 5 log 2 as a single logarithm: 2 log 3 = log 9, 3 log 5 = log 125, 5 log 2 = log 32, so the expression is log (9 × 125 / 32) = log (1125/32). Write (1/2) log 25 − log 5 as a single logarithm: (1/2) log 25 = log 5, so the expression is log 5 − log 5 = log 1 = 0.

Evaluating with the power law. Find log₂ 8⁵: = 5 log₂ 8 = 5 × 3 = 15. Find log₅ √125: 125 = 5³, so √125 = 5^3/2, and the answer is 3/2. Find the value of 2 log 5 + log 4 to base 10: log 25 + log 4 = log 100 = 2.

Checking your work. After expanding, substitute simple numbers to see that both sides agree. With x = 10, y = 10, z = 10 in the first example, the left side is log (10² × 10³ / 10⁴) = log 10 = 1, and the right side is 2 + 3 − 4 = 1. This habit catches sign errors, the most common mistake when a quotient is involved.

This section gathers the kinds of logarithm problems the Telangana SSC examination asks, and shows a reliable method for each.

Type 1: Find the value. Write the number as a power of the base. log₂ 32 = log₂ 2⁵ = 5. log₁₀ 0.001 = log₁₀ 10⁻³ = −3. log₁₆ 4: since 16 = 4², 4 = 16^1/2, so the answer is 1/2. log_√2 4: 4 = 2² = (√2)⁴, so the answer is 4. log_a a⁷ = 7 for any valid base. When the base is a fraction, be careful with sign: log_1/2 8: 8 = 2³ = (1/2)⁻³, so the answer is −3.

Type 2: Solve for the unknown. log_x 81 = 4 means x⁴ = 81, so x = 3 (the base must be positive). log₃ x = −2 means x = 3⁻² = 1/9. log₂ (x + 1) = 3 means x + 1 = 8, so x = 7. If log₁₀ x = 2 then x = 100.

Type 3: Express in terms of simpler logs. log 1000 = log 10³ = 3 log 10 = 3. If log 2 = 0.3010, find log 8: log 8 = 3 log 2 = 0.9030. Also log 5 = log (10/2) = 1 − 0.3010 = 0.6990, which is a favourite trick: 5 is 10 divided by 2.

Type 4: Show that. Show that log 2 + log 5 = 1 (base 10): log (2 × 5) = log 10 = 1. Show that log_a b × log_b a = 1 is beyond the SSC laws, but showing that 3 log 2 − log 8 = 0 is standard: 3 log 2 = log 8, so the difference is log 8 − log 8 = 0.

Type 5: Combined evaluation. Evaluate log₂ 8 + log₂ 16 − log₂ 4. Compute each: 3 + 4 − 2 = 5. Or condense: log₂ (8 × 16 / 4) = log₂ 32 = 5.

Connections beyond the chapter. Logarithms are the reason a slide rule works, and they are built into the pH scale (pH = −log₁₀ [H⁺], so a solution with hydrogen ion concentration 10⁻⁷ has pH 7), the decibel scale for sound and the Richter scale for earthquakes, where an increase of 1 unit means ten times the amplitude. In Intermediate mathematics you will meet the natural logarithm with base e and the change-of-base formula, and in physics you will plot log graphs to find power laws. The three laws learnt here are exactly the laws used there.

A checklist. Base positive and not 1; argument positive; product becomes sum, quotient becomes difference, power comes down as a multiplier; and log_a 1 = 0, log_a a = 1. Any answer that gives the logarithm of a negative number or of zero is wrong by definition.

This chapter has three threads: the structure of integers (Euclid's lemma, the Fundamental Theorem of Arithmetic, HCF and LCM), the nature of real numbers (decimal expansions, irrationality proofs) and logarithms (definition and three laws). Here is how each is examined in the Telangana SSC paper and what a complete answer looks like.

One-mark and two-mark questions. These ask for a single application: write 3⁵ = 243 in logarithmic form; find the HCF of 24 and 36 by prime factorisation; state whether 13/3125 has a terminating expansion and why; write log (x²/y) in expanded form; find log₂ 64. Answer in two or three lines with the reason stated. For the decimal question, always show the factorisation of the denominator, because the reason carries the mark.

Four-mark questions. Typical stems: use Euclid's division algorithm to find the HCF of 4052 and 12576 (show every division line); prove that √5 is irrational (all steps of the contradiction, including the coprime assumption); prove that 3 + 2√5 is irrational (isolate √5 and argue the other side is rational); show that the square of any positive integer is of the form 3m or 3m + 1; prove the product law of logarithms; expand or evaluate a longer logarithmic expression.

Reasoning questions. Can 15ⁿ end in zero? Explain why 5 × 7 × 11 + 7 is composite. Is √9 rational? These test whether you can quote the Fundamental Theorem of Arithmetic and apply it in one or two sentences.

Common mistakes to avoid. Writing the remainder greater than the divisor in Euclid's lemma. Adding exponents when finding LCM. Forgetting to reduce a fraction to lowest terms before judging its decimal expansion. Skipping the line that justifies why (a − 3b)/(2b) is rational. Writing log (x + y) = log x + log y. Giving a negative base or negative argument in a logarithm answer.

A revision routine. First, recite the five statements: Euclid's lemma, the Fundamental Theorem, the terminating-decimal theorem, the definition of a logarithm and the three laws. Second, redo one Euclid's algorithm computation, one HCF-LCM check, one irrationality proof and one expansion each day for a week. Third, from the exercises pick the problems you got wrong the first time and repeat only those.

Looking ahead. The idea of unique factorisation returns in the Polynomials chapter, where a polynomial is factorised into linear factors; the idea of a lemma and a contradiction proof appears in Similar Triangles and Tangents to a Circle; and logarithms return in Intermediate mathematics and in every science that deals with quantities spanning many orders of magnitude.