Chapter 2 — Real Numbers Optional

The most common optional problem asks you to show that the square of any positive integer can only take certain forms. The method is always the same: use Euclid's division lemma to list the possible forms of the integer itself, square each form, and rearrange the result.

Problem 1. Show that the square of any positive integer is of the form 3m or 3m + 1. Let a be any positive integer. Dividing by 3, a = 3q + r with r = 0, 1 or 2. Case r = 0: a² = 9q² = 3(3q²) = 3m with m = 3q². Case r = 1: a² = 9q² + 6q + 1 = 3(3q² + 2q) + 1 = 3m + 1. Case r = 2: a² = 9q² + 12q + 4 = 9q² + 12q + 3 + 1 = 3(3q² + 4q + 1) + 1 = 3m + 1. In every case the square is 3m or 3m + 1, never 3m + 2.

Problem 2. Show that the square of any positive integer is of the form 4q or 4q + 1. Write a = 2k or a = 2k + 1. If a = 2k, then a² = 4k², of the form 4q. If a = 2k + 1, then a² = 4k² + 4k + 1 = 4(k² + k) + 1, of the form 4q + 1. So no perfect square leaves remainder 2 or 3 on division by 4. This is a quick way to see that numbers like 1234567 (which ends in ...67, remainder 3 on dividing by 4) cannot be perfect squares.

Problem 3. Show that the square of any positive integer is of the form 5q, 5q + 1 or 5q + 4. Write a = 5k + r, r = 0, 1, 2, 3, 4. Then a² = 25k² + 10kr + r² = 5(5k² + 2kr) + r². The remainder is decided by r²: r² = 0, 1, 4, 9, 16, which leave remainders 0, 1, 4, 4, 1 on division by 5. So the square is of the form 5q, 5q + 1 or 5q + 4, and never 5q + 2 or 5q + 3. This is why a perfect square can end only in 0, 1, 4, 5, 6 or 9.

Problem 4. Show that the square of an odd positive integer is of the form 8q + 1. An odd integer is 4k + 1 or 4k + 3. (4k + 1)² = 16k² + 8k + 1 = 8(2k² + k) + 1. (4k + 3)² = 16k² + 24k + 9 = 8(2k² + 3k + 1) + 1. Both are of the form 8q + 1.

The general trick. Writing a = nk + r and expanding, the square is n(nk² + 2kr) + r², so only r² matters. Compute r² for each remainder r, reduce it modulo n, and the list of possible forms appears. This one observation solves every problem of this family and is worth remembering as a pattern rather than as separate results.

The cube problems use exactly the same structure as the square problems but need more careful expansion. The identity (x + y)³ = x³ + 3x²y + 3xy² + y³ is used in every case.

Problem 1. Show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8. Let a = 3q + r with r = 0, 1, 2. Case r = 0: a³ = 27q³ = 9(3q³) = 9m. Case r = 1: a³ = (3q + 1)³ = 27q³ + 27q² + 9q + 1 = 9(3q³ + 3q² + q) + 1 = 9m + 1. Case r = 2: a³ = (3q + 2)³ = 27q³ + 54q² + 36q + 8 = 9(3q³ + 6q² + 4q) + 8 = 9m + 8. Hence every cube is of the form 9m, 9m + 1 or 9m + 8.

Problem 2. Show that the cube of any positive integer is of the form 4q, 4q + 1 or 4q + 3. Write a = 4k, 4k + 1, 4k + 2 or 4k + 3. (4k)³ = 64k³ = 4(16k³). (4k + 1)³ = 64k³ + 48k² + 12k + 1 = 4(16k³ + 12k² + 3k) + 1. (4k + 2)³ = 64k³ + 96k² + 48k + 8 = 4(16k³ + 24k² + 12k + 2). (4k + 3)³ = 64k³ + 144k² + 108k + 27 = 4(16k³ + 36k² + 27k + 6) + 3. So the cube is 4q, 4q + 1 or 4q + 3, and never 4q + 2.

Problem 3. Show that n² − 1 is divisible by 8 if n is an odd positive integer. Since n is odd, n = 4k + 1 or 4k + 3. If n = 4k + 1, n² − 1 = 16k² + 8k = 8(2k² + k). If n = 4k + 3, n² − 1 = 16k² + 24k + 8 = 8(2k² + 3k + 1). In both cases 8 divides n² − 1. An elegant alternative: n² − 1 = (n − 1)(n + 1), the product of two consecutive even numbers, one of which is a multiple of 4, so the product is a multiple of 8.

Problem 4. Show that one of every three consecutive positive integers is divisible by 3. Let the integers be n, n + 1, n + 2. Write n = 3q + r, r = 0, 1, 2. If r = 0, n is divisible by 3. If r = 1, n + 2 = 3q + 3 = 3(q + 1) is divisible by 3. If r = 2, n + 1 = 3q + 3 is divisible by 3. In each case exactly one of the three is a multiple of 3.

These problems are graded on structure: state the lemma, list the cases, expand each fully, factor out the modulus, and write a concluding sentence. Skipping a case loses the mark for that case; skipping the conclusion loses the final mark.

A neat family of optional problems asks you to show that the product of consecutive integers is always divisible by a fixed number. The principle behind all of them is that among any k consecutive integers, exactly one is divisible by k.

Problem 1. Show that the product of any two consecutive positive integers is divisible by 2. Let the integers be n and n + 1. By Euclid's lemma n = 2q or 2q + 1. If n = 2q, then n(n + 1) = 2q(2q + 1), divisible by 2. If n = 2q + 1, then n + 1 = 2q + 2 = 2(q + 1), so n(n + 1) = (2q + 1) × 2(q + 1), divisible by 2. Hence n(n + 1) is always even.

Problem 2. Show that the product of any three consecutive positive integers is divisible by 6. Let the product be P = n(n + 1)(n + 2). Since 6 = 2 × 3 with 2 and 3 coprime, it is enough to show P is divisible by 2 and by 3. Divisibility by 2 follows from Problem 1, because n(n + 1) is a factor of P. For divisibility by 3, write n = 3q, 3q + 1 or 3q + 2. If n = 3q, n is a multiple of 3. If n = 3q + 1, then n + 2 = 3q + 3 = 3(q + 1). If n = 3q + 2, then n + 1 = 3(q + 1). In each case one factor of P is a multiple of 3. Since P is divisible by both 2 and 3, it is divisible by 6.

Problem 3. Show that n³ − n is divisible by 6 for every positive integer n. Factorise: n³ − n = n(n² − 1) = (n − 1) n (n + 1), the product of three consecutive integers, which by Problem 2 is divisible by 6. This factorisation trick converts an algebraic statement into a consecutive-integer statement.

Problem 4. Show that for any positive integer n, n(n + 1)(2n + 1) is divisible by 6. Divisibility by 2 comes from n(n + 1). For 3: if n = 3q, done; if n = 3q + 1, then 2n + 1 = 6q + 3 = 3(2q + 1); if n = 3q + 2, then n + 1 = 3(q + 1). So the product is divisible by 6. This expression is six times the sum of the first n squares, which is why the sum 1² + 2² + ... + n² = n(n + 1)(2n + 1)/6 is always an integer.

Problem 5. Show that the product of four consecutive integers is divisible by 24. Among four consecutive integers, two are even and one of these is a multiple of 4, giving a factor 8; and one is a multiple of 3. Since 8 and 3 are coprime, the product is divisible by 24. For example 5 × 6 × 7 × 8 = 1680 = 24 × 70.

The common thread: to prove divisibility by a composite number, split it into coprime parts and prove each part separately; to prove divisibility by a prime p, classify the integer by its remainder on division by p.

The uniqueness part of the Fundamental Theorem of Arithmetic is used in two kinds of optional problem: deciding the last digit of a power, and proving that a given expression is composite.

Last digit problems. A number ends in 0 exactly when it is divisible by 10 = 2 × 5, that is, when both 2 and 5 appear in its prime factorisation. A number ends in 5 exactly when it is divisible by 5 but not by 2. A number ends in an even digit exactly when 2 appears in its factorisation.

Can 4ⁿ end with 0? 4ⁿ = 2²ⁿ; no factor 5, so never. Can 12ⁿ end with 0? 12ⁿ = 2²ⁿ × 3ⁿ; no factor 5, so never. Can 15ⁿ end with 0? 15ⁿ = 3ⁿ × 5ⁿ; no factor 2, so never; in fact 15ⁿ always ends in 5. Can 7ⁿ end with 5? 7ⁿ has no factor 5, so never. Can 20ⁿ end with 0? 20 = 2² × 5, so 20ⁿ has both 2 and 5 and ends in 0 for every n ≥ 1. The pattern of last digits of 7ⁿ is 7, 9, 3, 1, 7, 9, 3, 1, ..., repeating every 4, so 7ⁿ ends in 7, 9, 3 or 1 only.

Composite expressions. To prove that an expression is composite, exhibit it as a product of two integers each greater than 1. The usual route is to take out a common factor.

Show that 7 × 11 × 13 + 13 is composite: = 13(7 × 11 + 1) = 13 × 78. Since 13 and 78 are both greater than 1, the number is composite (it equals 1014). Show that 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is composite: = 5(7 × 6 × 4 × 3 × 2 × 1 + 1) = 5 × 1009, composite (it equals 5045). Show that 3 × 5 × 7 × 11 + 11 is composite: = 11(105 + 1) = 11 × 106 = 1166. Show that 17 × 5 × 11 × 3 × 2 + 2 × 11 is composite: = 2 × 11 × (17 × 5 × 3 + 1) = 22 × 256 = 5632.

Why does taking out a common factor prove compositeness? Because the Fundamental Theorem says the number has a unique factorisation, and we have found a factor other than 1 and itself. The remaining bracket must be greater than 1, which should be stated explicitly.

A related classic. Show that there is no natural number n for which 2ⁿ + 3ⁿ is divisible by 5 when n is a multiple of 4. Try n = 4: 16 + 81 = 97, not divisible by 5. Try n = 8: 256 + 6561 = 6817, not divisible by 5. The general argument uses the last-digit cycles of 2ⁿ (2, 4, 8, 6) and 3ⁿ (3, 9, 7, 1): when n is a multiple of 4, 2ⁿ ends in 6 and 3ⁿ ends in 1, so the sum ends in 7 and cannot be divisible by 5.

The optional exercise tests the finer points of HCF and LCM that the main exercise does not: three numbers, reconstructing a number from the HCF and LCM, and understanding when the product relation applies.

Three numbers. Find the HCF and LCM of 6, 72 and 120. 6 = 2 × 3, 72 = 2³ × 3², 120 = 2³ × 3 × 5. Common primes to all three: 2 and 3, smallest powers 2¹ and 3¹, so HCF = 6. All primes with greatest powers: 2³ × 3² × 5 = 360, so LCM = 360. Note that HCF × LCM = 2160 while 6 × 72 × 120 = 51840; they differ. The relation HCF × LCM = product holds only for two numbers.

Reconstructing a number. If HCF(a, 91) = 13 and LCM(a, 91) = 182, find a. From HCF × LCM = a × 91, a = 13 × 182 / 91 = 13 × 2 = 26. Check: 26 = 2 × 13, 91 = 7 × 13; HCF = 13, LCM = 2 × 7 × 13 = 182. Correct.

When the data are impossible. Can two numbers have HCF 18 and LCM 380? The HCF must always divide the LCM, because every common factor of the numbers divides each of them and hence divides any common multiple. Since 380 / 18 is not an integer (380 = 2² × 5 × 19 and 18 = 2 × 3², and 18 does not divide 380), no such pair exists. This test, HCF divides LCM, is a quick way to reject impossible questions.

Word problems with three numbers. Three tankers contain 403 litres, 434 litres and 465 litres of diesel. Find the maximum capacity of a container that can measure the diesel of each tanker an exact number of times. We need HCF(403, 434, 465). 403 = 13 × 31, 434 = 2 × 7 × 31, 465 = 3 × 5 × 31. HCF = 31 litres.

Find the least number which when divided by 12, 16, 24 and 36 leaves a remainder 7 in each case. First find LCM(12, 16, 24, 36) = 2⁴ × 3² = 144; the required number is 144 + 7 = 151. Check: 151 = 12 × 12 + 7, 16 × 9 + 7, 24 × 6 + 7, 36 × 4 + 7.

In a school, the number of students in each section must be the same and each section must contain students of one subject only. If there are 60 students of Mathematics, 84 of Science and 108 of English, find the minimum number of sections. HCF(60, 84, 108) = 12 students per section, so the sections needed are 60/12 + 84/12 + 108/12 = 5 + 7 + 9 = 21.

The sharpest fact. The HCF of two numbers is never larger than the smaller number, and the LCM is never smaller than the larger number; equality in both happens exactly when one number divides the other, as with 12 and 36, where HCF = 12 and LCM = 36.

The main exercise asked you to predict whether p/q terminates. The optional exercise asks the reverse: given a decimal, find the fraction and its denominator's prime factors, and confirm the prediction. It also asks for the actual expansion in some cases.

Terminating decimals. Write 43.123456789 as p/q. The decimal has 9 places, so it equals 43123456789 / 10⁹, and 10⁹ = 2⁹ × 5⁹. The denominator (before any cancelling) has only the primes 2 and 5, exactly as the theorem requires. After cancelling, the denominator can only lose factors, so it still has only 2s and 5s.

Repeating decimals. Write 0.120120120... as p/q. Let x = 0.120120... The repeating block has 3 digits, so multiply by 10³: 1000x = 120.120120... Subtracting, 999x = 120, so x = 120/999 = 40/333. Now 333 = 3² × 37, which has primes other than 2 and 5, consistent with the expansion being non-terminating repeating.

Write 43.123456789 with the digits 123456789 repeating as p/q. Let x be that number. Then 10⁹x = 43123456789.123456789... and subtracting x gives (10⁹ − 1)x = 43123456746, so x = 43123456746 / 999999999. The denominator 999999999 = 3⁴ × 37 × 333667 contains primes other than 2 and 5, again as expected.

Mixed decimals. Write 0.2333... as p/q. Let x = 0.2333... Then 10x = 2.333... and 100x = 23.333... Subtracting these, 90x = 21, so x = 21/90 = 7/30. Denominator 30 = 2 × 3 × 5; the factor 3 forces repetition. Write 0.5777... as p/q: 10x = 5.777..., 100x = 57.777..., 90x = 52, x = 52/90 = 26/45, and 45 = 3² × 5.

Producing the expansion. For 6/15 = 2/5 = 0.4 (terminating). For 35/50 = 7/10 = 0.7. For 1/7 the division gives 0.142857142857..., period 6. For 1/11 the expansion is 0.0909..., period 2. For 1/13 it is 0.076923076923..., period 6. For 1/37 it is 0.027027..., period 3, and indeed 37 × 27 = 999.

Why the period is less than the denominator. When dividing by q, each step produces a remainder between 0 and q − 1. If a remainder is 0, the expansion terminates. Otherwise there are only q − 1 possible non-zero remainders, so within q − 1 steps a remainder must repeat, and from that point the digits repeat. Hence the period of 1/q is at most q − 1. For q = 7 the period is exactly 6, the maximum; for q = 11 it is 2, much less.

A useful fraction to remember: 1/9 = 0.111..., 1/99 = 0.0101..., 1/999 = 0.001001..., which is why a block of k repeating digits d equals d/(10^k − 1).

Beyond √2 and 5 − √3, the optional exercise asks for proofs where the number involves two surds, a reciprocal, or where you must first decide whether the number is rational at all.

Problem 1. Prove that √2 + √3 is irrational. Assume √2 + √3 = a/b, rational. Then √3 = a/b − √2. Squaring: 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 (a quotient of integers with 2ab ≠ 0), so √2 would be rational, contradicting the known result. Hence √2 + √3 is irrational. The squaring step is essential; without it there is no way to isolate a single surd.

Problem 2. Prove that √3 − √5 is irrational. Same method: assume √3 − √5 = a/b; then √3 = a/b + √5; square: 3 = a²/b² + 2(a/b)√5 + 5; so √5 = (3 − 5 − a²/b²) × b/(2a) = (−2b² − a²)/(2ab), rational, contradiction. Note that a ≠ 0 here because √3 ≠ √5.

Problem 3. Prove that 1/√2 is irrational. Assume 1/√2 = a/b with a ≠ 0. Then √2 = b/a, which is rational, contradiction. Similarly 1/(2 + √3) is irrational: rationalising gives (2 − √3)/(4 − 3) = 2 − √3, and if 2 − √3 = a/b then √3 = 2 − a/b is rational, contradiction.

Problem 4. Prove that √2 + √5 is irrational and find its square. (√2 + √5)² = 2 + 2√10 + 5 = 7 + 2√10. If √2 + √5 were rational, its square would be rational, so 2√10 = (square) − 7 would be rational, so √10 would be rational. But 10 is not a perfect square, and the standard argument (if 10 = a²/b² in lowest terms then 2 divides a, then 2 divides b) shows √10 is irrational. Contradiction.

Problem 5. Is (√2 + √3)(√2 − √3) rational? The product is 2 − 3 = −1, which is rational. This shows that the product of two irrationals can be rational. Similarly (3 + √5) + (3 − √5) = 6 is rational. Optional questions often test whether you will blindly claim irrationality; always compute first.

Problem 6. Prove that √p + √q is irrational when p and q are distinct primes. Assume √p + √q = r rational. Then √p = r − √q, so p = r² − 2r√q + q, giving √q = (r² + q − p)/(2r), rational, contradicting the irrationality of √q. This is the general version of Problem 1.

In every proof the sentence that earns the crucial mark is the one asserting that the isolated expression is rational because it is formed from integers by addition, subtraction, multiplication and division with a non-zero divisor. Write that sentence explicitly.

Optional problems on logarithms go beyond log₂ 8 to bases that are fractions, surds or powers, and to arguments that are roots. The single technique that handles all of them is: write the argument as a power of the base, using the laws of exponents.

Fractional bases. log_1/2 8: 8 = 2³ = (1/2)⁻³, so the value is −3. log_1/3 81: 81 = 3⁴ = (1/3)⁻⁴, value −4. log_2/3 (8/27): 8/27 = (2/3)³, value 3. log_2/3 (27/8): 27/8 = (2/3)⁻³, value −3.

Surd bases. log_√2 4: 4 = 2² = (√2)⁴, value 4. log_√3 27: 27 = 3³ = (√3)⁶, value 6. log_√5 (1/25): 1/25 = 5⁻² = (√5)⁻⁴, value −4. In general, if the base is √a, write everything as a power of a and double the exponent.

Power bases. log₈ 2: 2 = 8^1/3, value 1/3. log₁₆ 64: 16 = 2⁴, 64 = 2⁶, so 64 = (2⁴)^6/4 = 16^3/2, value 3/2. log₂₅ 125: 25 = 5², 125 = 5³, value 3/2. log_{32 8: 32 = 2⁵, 8 = 2³, value 3/5. The rule: loga^m aⁿ = n/m.}

Root arguments. log₁₀ √1000 = log₁₀ 10^3/2 = 3/2. log₂ (cube root of 16) = log₂ 2^4/3 = 4/3. log₃ (1/√27) = log₃ 3^−3/2 = −3/2.

Mixed evaluation. Find log₂ 8 + log_1/2 8. The first is 3, the second is −3, sum 0. Find log_√2 4 × log₄ √2. The first is 4; for the second, √2 = 2^1/2 = 4^1/4, so it is 1/4; product 1. Find log₁₀ 0.0001 + log₁₀ 100: −4 + 2 = −2.

Solving for the base or argument. log_x 343 = 3 gives x³ = 343 = 7³, so x = 7. log_x (1/64) = −3 gives x⁻³ = 1/64, so x³ = 64, x = 4. log₅ x = −2 gives x = 1/25. log_√3 x = 4 gives x = (√3)⁴ = 9. log_x √2 = 1/4 gives x^1/4 = √2 = 2^1/2, so x = 2² = 4.

What must always hold. The base is positive and not 1; the argument is positive. If solving gives a negative base, reject it: from x² = 16, take x = 4 not −4. The value of a logarithm itself can be any real number, positive, negative or zero; it is only the base and argument that are restricted.

The signature optional problem on logarithms starts with an algebraic relation between x and y and asks you to prove a logarithmic identity. The method is: rewrite the relation as a perfect square, take logarithms of both sides, and apply the laws.

Problem 1. If x² + y² = 6xy, prove that 2 log (x + y) = log x + log y + 3 log 2. Add 2xy to both sides: x² + 2xy + y² = 8xy, that is, (x + y)² = 8xy. Take logarithms: log (x + y)² = log (8xy). Left side by the power law: 2 log (x + y). Right side by the product law: log 8 + log x + log y = log 2³ + log x + log y = 3 log 2 + log x + log y. Hence 2 log (x + y) = log x + log y + 3 log 2.

Problem 2. If x² + y² = 25xy, prove that 2 log (x + y) = 3 log 3 + log x + log y. Add 2xy: (x + y)² = 27xy. Taking logs: 2 log (x + y) = log 27 + log x + log y = 3 log 3 + log x + log y.

Problem 3. If log ((x + y)/3) = (1/2)(log x + log y), show that x² + y² = 7xy. The right side is (1/2) log (xy) = log √(xy) = log (xy)^1/2. So log ((x + y)/3) = log (xy)^1/2. Since the logarithm function takes each value only once, (x + y)/3 = √(xy). Squaring: (x + y)²/9 = xy, so x² + 2xy + y² = 9xy, giving x² + y² = 7xy. This problem runs the method in reverse: from logs to algebra.

Problem 4. If (2.3)^x = (0.23)^y = 1000, show that 1/x − 1/y = 1/3. Taking common logarithms, x log 2.3 = 3 and y log 0.23 = 3. So log 2.3 = 3/x and log 0.23 = 3/y. Now 2.3 = 0.23 × 10, so log 2.3 = log 0.23 + 1, giving 3/x = 3/y + 1, so 3/x − 3/y = 1, and dividing by 3, 1/x − 1/y = 1/3.

Problem 5. If 2^x+1 = 3^1−x, find x in terms of log 2 and log 3. Taking logs: (x + 1) log 2 = (1 − x) log 3. So x log 2 + log 2 = log 3 − x log 3, hence x (log 2 + log 3) = log 3 − log 2, and x = (log 3 − log 2)/(log 3 + log 2) = log (3/2)/log 6.

Problem 6. Show that log 2 + 2 log 5 − log 3 − 2 log 7 = log (50/147). Left side = log 2 + log 25 − log 3 − log 49 = log (2 × 25) − log (3 × 49) = log 50 − log 147 = log (50/147).

The three moves in every such problem are: complete a square or factor to get a product or power; take the logarithm of both sides (allowed because both sides are positive); and expand with the three laws. Conversely, when starting from logs, condense to a single log on each side and then drop the logs, which is valid because equal logarithms to the same base mean equal arguments.

The optional exercise contains expressions with several factors, powers and roots. Two systematic procedures, one for expanding and one for condensing, handle every case.

Expanding: outside in. Step 1: split the argument into numerator and denominator (quotient law). Step 2: split each into its factors (product law). Step 3: bring down every exponent, including fractional ones for roots (power law).

Expand log (x² y³ / z⁴): Step 1: log (x² y³) − log z⁴. Step 2: log x² + log y³ − log z⁴. Step 3: 2 log x + 3 log y − 4 log z.

Expand log (√(a³ b) / c²): = log √(a³ b) − log c² = (1/2) log (a³ b) − 2 log c = (1/2)(3 log a + log b) − 2 log c = (3/2) log a + (1/2) log b − 2 log c.

Expand log (128/625): 128 = 2⁷, 625 = 5⁴, so 7 log 2 − 4 log 5.

Expand log (343/125): 343 = 7³, 125 = 5³, so 3 log 7 − 3 log 5 = 3(log 7 − log 5).

Expand log (p² q³ r) − log (p q² r³): = (2 log p + 3 log q + log r) − (log p + 2 log q + 3 log r) = log p + log q − 2 log r.

Condensing: inside out. Step 1: move every coefficient up as an exponent (power law in reverse). Step 2: combine the plus terms into a product and the minus terms into a product. Step 3: write as a single quotient.

Condense 2 log 3 + 3 log 5 − 5 log 2: log 9 + log 125 − log 32 = log (1125/32).

Condense (1/2) log 25 − log 5 + log 1: (1/2) log 25 = log 5; log 1 = 0; so log 5 − log 5 + 0 = 0.

Condense 3 log x − (1/2) log y + 2 log z: log x³ − log √y + log z² = log (x³ z² / √y).

Condense log 2 + log 3 − log 4 + log 5 − log 6: numerator 2 × 3 × 5 = 30, denominator 4 × 6 = 24, so log (30/24) = log (5/4).

Evaluating after condensing. Find log₃ 54 + log₃ 9 − log₃ 6: = log₃ (54 × 9 / 6) = log₃ 81 = 4. Find log 4 + log 25 to base 10: log 100 = 2. Find 2 log 5 + log 8 − log 2: log 25 + log 8 − log 2 = log (200/2) = log 100 = 2. Find log₂ (1/8) + log₂ 64: −3 + 6 = 3.

Checking by substitution. When an expanded form has variables, test with x = y = z = 10 or with small powers of the base. For the first example with x = y = z = 10, left side is log 10 = 1 and right side is 2 + 3 − 4 = 1. A mismatch usually means a sign error on the denominator terms or a forgotten factor of 1/2 on a root.

Before calculators, tables gave only a few logarithms and everything else was computed from them by the laws. The optional exercise keeps this skill alive: given log 2 = 0.3010 and log 3 = 0.4771 (base 10), find as many other logarithms as possible.

Powers of 2 and 3. log 4 = 2 log 2 = 0.6020. log 8 = 3 log 2 = 0.9030. log 16 = 4 log 2 = 1.2040. log 9 = 2 log 3 = 0.9542. log 27 = 3 log 3 = 1.4313.

Products. log 6 = log 2 + log 3 = 0.7781. log 12 = log 4 + log 3 = 0.6020 + 0.4771 = 1.0791. log 18 = log 2 + 2 log 3 = 0.3010 + 0.9542 = 1.2552. log 24 = 3 log 2 + log 3 = 0.9030 + 0.4771 = 1.3801. log 36 = 2 log 2 + 2 log 3 = 1.5562. log 72 = 3 log 2 + 2 log 3 = 1.8572.

The trick for 5. Since 5 = 10/2, log 5 = log 10 − log 2 = 1 − 0.3010 = 0.6990. Then log 25 = 2 log 5 = 1.3980, log 15 = log 3 + log 5 = 1.1761, log 30 = log 3 + 1 = 1.4771, log 50 = log 5 + 1 = 1.6990, log 0.5 = log 5 − 1 = −0.3010 (equivalently −log 2), and log 2.5 = log 25 − 1 = 0.3980.

Quotients and decimals. log 1.5 = log 3 − log 2 = 0.1761. log 0.75 = log 3 − 2 log 2 = 0.4771 − 0.6020 = −0.1249. log 0.06 = log 6 − 2 = −1.2219. log 0.0016 = log 16 − 4 = 1.2040 − 4 = −2.7960. log 12.5 = log 100 − log 8 = 2 − 0.9030 = 1.0970.

Roots. log √2 = (1/2)(0.3010) = 0.1505. log √12 = (1/2)(1.0791) = 0.5396. log of the cube root of 2 = 0.1003. log √0.5 = −0.1505.

Estimating the number of digits. The common logarithm tells how many digits a number has: if log N = k.xxxx with k a non-negative integer, then N has k + 1 digits before the decimal point. So 2¹⁰⁰ has log 2¹⁰⁰ = 100 × 0.3010 = 30.10, hence 31 digits. And 3⁵⁰ has log = 50 × 0.4771 = 23.855, hence 24 digits. 6²⁰ has log = 20 × 0.7781 = 15.562, hence 16 digits. This is a favourite reasoning question because it shows the power of logarithms without any table beyond two values.

Comparing sizes. Which is bigger, 2³⁰⁰ or 3²⁰⁰? log 2³⁰⁰ = 90.30 and log 3²⁰⁰ = 95.42; the second is bigger. Since the logarithm is an increasing function, comparing logs compares the numbers.

Every value here is obtained from two given numbers and the number 1 = log 10 by the three laws. When answering, show the law used at each step; the numerical value alone does not earn the marks.

The optional exercise, and the harder examination questions modelled on it, do not announce which idea they need. This section is a decision guide.

If the problem says: show that a square, cube or product of consecutive integers has a certain form or is divisible by something. Use Euclid's division lemma. Write the integer as nk + r, list every remainder r, expand and factor. For divisibility by a composite number, split it into coprime parts. For n³ − n or n² − 1, factorise first into consecutive integers.

If the problem asks whether a power can end in a digit, or whether an expression is composite. Use the Fundamental Theorem of Arithmetic. For ending digits, look for 2 and 5 in the factorisation. For compositeness, take out a common factor and state that both factors exceed 1.

If the problem gives amounts and asks for the largest measure, the maximum number in equal groups, or the longest tape. Use HCF. If it asks for the least number, the next time events coincide, or the smallest amount that several quantities divide. Use LCM, and add the remainder if one is specified. If it gives HCF and LCM of two numbers and one of the numbers. Use HCF × LCM = product. If it gives an HCF that does not divide the LCM. Say the data are impossible.

If the problem gives a fraction and asks about its decimal. Reduce to lowest terms, factorise the denominator, look for primes other than 2 and 5. If it gives a repeating decimal and asks for a fraction. Multiply by 10^k where k is the period, subtract, solve, and then factorise the denominator to confirm.

If the problem asks to prove a number irrational. Assume it is a/b, isolate the surd, show the other side is rational, contradict. With two surds, square once. With a reciprocal, invert. With a product or sum of two surds, compute first, because the answer might be rational.

If the problem is a logarithm evaluation. Write the argument as a power of the base. For a fractional base, use negative exponents. For a surd base, double the exponent. For log_a^m aⁿ, the answer is n/m.

If the problem is a logarithm identity from an algebraic relation. Complete the square or factorise, then take logs of both sides and expand. In reverse, condense each side to one log and equate the arguments.

If the problem gives log 2 and log 3. Build the target from 2, 3 and 10 using log 5 = 1 − log 2 and log (N × 10^k) = log N + k.

Two final habits. First, check every answer against the definitions: a remainder below the divisor, an HCF dividing the LCM, a positive base and argument for every logarithm. Second, in a proof, write the concluding sentence that names the contradiction or the result. Students who master these two habits move from partial to full marks on this chapter.