Real Numbers

A prime number has exactly two factors, 1 and itself (2, 3, 5, 7, 11, 13, ...). A composite number has more than two factors. The Fundamental Theorem of Arithmetic states that every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

Worked example: Express 3825 as a product of primes. We divide step by step by the smallest possible prime each time: 3825 = 3 × 1275 = 3 × 3 × 425 = 3 × 3 × 5 × 85 = 3 × 3 × 5 × 5 × 17. So 3825 = 3² × 5² × 17.

Because the factorisation is unique, the set of primes that appear and their powers are fixed for a given number. This uniqueness lets us answer questions without computing directly. For example, can 6ⁿ end with the digit 0 for any natural number n? A number ending in 0 is divisible by 10 = 2 × 5, so the prime 5 must appear in its factorisation. But 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ contains only the primes 2 and 3. By uniqueness, 5 can never occur, so 6ⁿ can never end with 0.

The same idea shows numbers like 7 × 11 × 13 + 13 are composite: 7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × 78, which clearly has 13 as a factor besides 1 and itself.

Once we have prime factorisations, finding HCF (Highest Common Factor) and LCM (Lowest Common Multiple) is straightforward:

• HCF = product of the smallest power of each common prime factor.
• LCM = product of the greatest power of each prime factor that appears in either number.

Worked example: Find HCF and LCM of 96 and 404. First, 96 = 2⁵ × 3 and 404 = 2² × 101. The only common prime is 2, and its smallest power is 2², so HCF = 2² = 4. For the LCM, take the greatest power of every prime: LCM = 2⁵ × 3 × 101 = 32 × 303 = 9696.

Verification of the relation: HCF × LCM = 4 × 9696 = 38784, and 96 × 404 = 38784. They match, confirming that for any two positive integers a and b, HCF(a, b) × LCM(a, b) = a × b. This relation is very useful: if you know three of these four quantities, you can find the fourth.

Note: This product relation holds for exactly two numbers. For three numbers the simple product HCF × LCM = a × b × c is not valid, so compute HCF and LCM separately from the factorisations.

A rational number can be written as p/q where p and q are integers and q ≠ 0. An irrational number cannot be written in this form. The proofs rely on one key result: if a prime p divides a², then p divides a.

Proof that √2 is irrational (by contradiction): Suppose √2 is rational. Then √2 = a/b for some integers a, b (b ≠ 0) with a and b having no common factor other than 1 (in lowest terms). Squaring: 2 = a²/b², so a² = 2b². Then 2 divides a², so 2 divides a. Write a = 2c. Substituting: (2c)² = 2b², i.e. 4c² = 2b², so b² = 2c². Then 2 divides b², so 2 divides b. But now 2 divides both a and b, contradicting that they have no common factor. Hence our assumption is wrong and √2 is irrational.

The proofs for √3 and √5 are identical in structure, replacing 2 by 3 or 5 (using that if 3 | a² then 3 | a, and if 5 | a² then 5 | a).

Using these results: Once we know √5 is irrational, we can prove numbers like 3 + 2√5 are irrational. Suppose 3 + 2√5 = a/b (rational). Then 2√5 = a/b − 3 = (a − 3b)/b, so √5 = (a − 3b)/(2b), which is rational. This contradicts √5 being irrational. Hence 3 + 2√5 is irrational. In general, the sum or difference of a rational and an irrational number is irrational, and a non-zero rational times an irrational is irrational.

Every rational number has a decimal expansion that is either terminating (ends after finitely many digits) or non-terminating recurring (a block of digits repeats forever). The type is decided entirely by the denominator.

Rule: Let p/q be a rational number in its lowest terms (p and q have no common factor, q ≠ 0). Then p/q has a terminating decimal expansion if and only if q can be written as 2^m × 5ⁿ for some non-negative integers m and n. If q has any prime factor other than 2 or 5, the expansion is non-terminating recurring.

Worked examples:

• 17/8 = 17/2³. The denominator is only powers of 2, so it terminates: 17/8 = 2.125.
• 13/3125 = 13/5⁵. Only powers of 5, so it terminates. Multiplying top and bottom by 2⁵ = 32: (13 × 32)/10⁵ = 416/100000 = 0.00416.
• 1/7: the denominator is 7, not of the form 2^m × 5ⁿ, so it is non-terminating recurring: 0.142857142857... .
• 64/455: here 455 = 5 × 7 × 13. The primes 7 and 13 are present, so the expansion is non-terminating recurring.

Counting decimal places: For a terminating expansion, the number of decimal places equals the larger of m and n in q = 2^m × 5ⁿ. For example, 23/(2³ × 5²) = 23/200 = 0.115, which terminates after max(3, 2) = 3 places.

Many exam problems are word problems that hide an HCF or LCM. A quick guide: if you want the largest / greatest / maximum quantity that fits or divides several numbers, use HCF. If you want the next time / smallest / minimum value at which repeating events coincide, use LCM.

HCF example (with remainders): Find the largest number that divides 245 and 1029 leaving a remainder of 5 in each case. First subtract the remainder: the required number must exactly divide 245 − 5 = 240 and 1029 − 5 = 1024. Now 240 = 2⁴ × 3 × 5 and 1024 = 2¹⁰. The only common prime is 2, so HCF = 2⁴ = 16. Hence the largest such number is 16.

HCF example (largest measure): Two tankers hold 850 L and 680 L of petrol. The maximum capacity of a container that can measure each exactly is HCF(850, 680). Since 850 = 2 × 5² × 17 and 680 = 2³ × 5 × 17, HCF = 2 × 5 × 17 = 170 L.

LCM example (events together): Three bells ring at intervals of 9, 12 and 15 minutes. When do they next ring together? Take the LCM: 9 = 3², 12 = 2² × 3, 15 = 3 × 5, so LCM = 2² × 3² × 5 = 180 minutes = 3 hours.