Theorem 1.1 (Fundamental Theorem of Arithmetic): 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.
(The prime factorisation of a natural number is unique, except for the order of its factors.)
Theorem 1.2: Let p be a prime number. If p divides a², then p divides a, where a is a positive integer.
(Requires proof in the exam; proof uses FTA.)
Theorem 1.3: √2 is irrational.
(Requires proof in the exam; proof by contradiction using Theorem 1.2.)
Standard results (proved in text): - √3 is irrational (same method as Theorem 1.3). - If a rational number is subtracted from or added to an irrational, the result is irrational (e.g., 5 – √3 is irrational). - Product of a non-zero rational and an irrational is irrational (e.g., 3√2 is irrational).
No definitions of new terms beyond the statements above; all results are derived from FTA.
| Expression | Meaning | Symbols |
|---|---|---|
| HCF(a, b) | Product of smallest powers of common prime factors of a and b | a, b positive integers |
| LCM(a, b) | Product of greatest powers of all prime factors involved in a and b | a, b positive integers |
| HCF(a, b) × LCM(a, b) | Equals a × b (valid only for two numbers) | — |
| a = p₁ p₂ … pₙ | Prime factorisation of composite a (unique up to order) | pᵢ primes |
Method: Repeatedly divide by smallest prime factors until the quotient is 1; write as product of powers of primes in ascending order.
Type 2: Find HCF and LCM of two or three numbers by prime factorisation
Method: Factorise each number; HCF = product of lowest powers of common primes; LCM = product of highest powers of all primes appearing. Verify HCF × LCM = product only when exactly two numbers are involved.
Type 3: Use HCF to find LCM (or vice versa)
Method: Compute one using prime factorisation; apply relation LCM = (a × b) / HCF for two numbers.
Type 4: Check whether a number of the form aⁿ ends with digit 0
Method: Assume it ends with 0 ⇒ divisible by 5; check whether 5 appears in prime factorisation of aⁿ; use uniqueness from FTA to reach contradiction if it does not.
Type 5: Prove √p (p prime) is irrational
Method: Assume √p = a/b in lowest terms (a, b coprime, b ≠ 0); square to get p b² = a²; apply Theorem 1.2 to show p divides both a and b, contradicting coprimeness; conclude irrational.
Type 6: Prove an expression involving rational and irrational is irrational
Method: Assume the expression is rational; rearrange algebraically to isolate the irrational part; show this implies the irrational is rational, leading to contradiction.
A study aid reviewed by GFIS faculty — always verify with your textbook and teacher.