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LESSON 1
Build numbers from primes
Break a whole number into primes and explain why the final factors are unique.
A factor divides a number exactly. A prime has exactly two positive factors: 1 and itself. The number 1 is not prime.
Start with a product
Write 60 as 6 × 10. Both branches multiply back to 60, so the total has not changed.
Keep splitting composite factors
The number 6 becomes 2 × 3, and 10 becomes 2 × 5. Each remaining factor is prime, so stop splitting.
Collect repeated primes
The leaves are 2, 3, 2 and 5. Rearrange them: 60 = 2 × 2 × 3 × 5 = 2² × 3 × 5. Multiplication order does not change the product.
Explain what stays the same
Starting with 60 = 3 × 20 gives the same prime factors. For each whole number greater than 1, its prime-factor list is unique apart from order. This is the Fundamental Theorem of Arithmetic.
Worked example: 84 = 7 × 12 = 7 × 3 × 4 = 7 × 3 × 2 × 2 = 2² × 3 × 7. Check: 4 × 3 × 7 = 84.
Watch for this: Stopping at 6 × 10 is a factorisation, but it is not a prime factorisation: 6 and 10 are composite.
Discuss: A friend writes 90 = 9 × 10 and says the job is finished. Explain what must still be split and why.
LESSON 2
Choose HCF or LCM
Use prime factors to solve equal-group and repeating-event problems.
A common factor divides both numbers. A common multiple is divisible by both. First decide whether the situation is about dividing into groups or waiting for repeats.
Factor both numbers
For 18 and 30: 18 = 2 × 3 × 3; 30 = 2 × 3 × 5. Keep repeated factors because they affect the result.
Use the shared primes for HCF
Both lists contain one 2 and one 3. Their product is 6. No second 3 is shared, so HCF(18, 30) = 6. With powers, take the smaller exponent of each common prime.
Cover both lists for LCM
Use one 2, two 3s and one 5: 2 × 3 × 3 × 5 = 90. This contains enough prime factors to be divisible by both 18 and 30. Take the greater exponent for every prime that appears.
Check the answer and its meaning
6 divides both numbers; 90 is divisible by both. For two positive integers, HCF × LCM equals their product: 6 × 90 = 18 × 30 = 540. Do not assume this two-number identity works for three numbers.
Worked example: A box has 24 red beads and 36 blue beads. To make the greatest number of identical bags with no beads left, use HCF(24, 36) = 12 bags. Each bag gets 2 red and 3 blue beads. If lights flash every 8 and 12 seconds, their next shared flash is after LCM(8, 12) = 24 seconds.
Watch for this: HCF is for the largest equal grouping or size. LCM is for the earliest shared repeat. The word “largest” or “smallest” alone is not enough: identify what must divide what.
Discuss: Create an equal-group story and a repeating-event story using 12 and 20. Explain why the answers are different.
LESSON 3
Explain why a root is irrational
Build a contradiction proof for √2, √3 or √5 and explain the role of lowest terms.
A rational number can be written as a/b, with integers a and b and b ≠ 0. An irrational number cannot. Coprime integers share no positive factor other than 1.
Assume a fraction in lowest terms
To test √2, suppose √2 = a/b with positive coprime integers a and b. Using lowest terms is essential: it gives us something precise to contradict.
Square and use prime divisibility
Squaring gives a² = 2b², so 2 divides a². If a prime divides a square, it divides the original integer: the prime factors of a² are just those of a repeated. Thus 2 divides a; write a = 2k.
Substitute to test the denominator
Now (2k)² = 2b², so 4k² = 2b² and b² = 2k². Therefore 2 divides b² and hence b. Both numerator and denominator have a factor 2.
Find the contradiction
We assumed a and b were coprime, but both are divisible by 2. The assumption that √2 is rational is impossible, so √2 is irrational. The same argument uses the prime 3 or 5 for √3 or √5.
Worked example: Suppose 4 + √3 were rational, say r. Then √3 = r − 4 would be rational because subtracting rational numbers gives a rational number. This contradicts the irrationality of √3. Therefore 4 + √3 is irrational.
Watch for this: A calculator showing many decimal digits does not prove irrationality. Also, a square root is not always irrational: √9 = 3. The divisibility argument uses a prime.
Discuss: Write the proof for √5 in your own words. Identify the exact line that contradicts the starting assumption. A teacher or partner should check your reasoning.
