Natural numbers
Get to grips with the natural numbers for Year 11 Specialist Mathematics in Queensland (QCAA). The counting numbers one, two, three and onwards are the foundation of number theory, where you meet factors, multiples, and prime and composite numbers.
You will learn to build prime factorisations, find the HCF and LCM, and reason about closure and the well-ordering principle using examples and counterexamples — the groundwork for proof later in the course.
Theory
The natural numbers \(\mathbb{N}=\{1,2,3,\dots\}\) are the counting numbers and the starting point for number theory in Year 11 Specialist Mathematics (QCAA, Queensland). This page covers factors and multiples, primes and composites, divisibility, HCF and LCM, and the ideas of closure and the well-ordering principle — with full worked examples.
The natural numbers are the counting numbers \(\mathbb{N}=\{1,2,3,4,\dots\}\). In the QCAA convention these are the positive integers \(\mathbb{Z}^{+}\); every whole number you use to count belongs to \(\mathbb{N}\).
A factor (or divisor) of a natural number \(n\) is a natural number that divides \(n\) exactly, with no remainder. A multiple of \(n\) is any number in the list \(n, 2n, 3n, \dots\). A number is even if \(2\) is a factor and odd if it is not.
A prime number has exactly two factors, \(1\) and itself (so \(2,3,5,7,11,\dots\)). A composite number has more than two factors. The number \(1\) is neither prime nor composite. Every composite splits into a unique product of primes — its prime factorisation.
The highest common factor (HCF) of two numbers is the largest factor they share; the lowest common multiple (LCM) is the smallest multiple they share. \(\mathbb{N}\) is closed under addition and multiplication (a sum or product of naturals is always a natural), but not under subtraction or division. The well-ordering principle says every non-empty subset of \(\mathbb{N}\) has a least element.
If a natural number has prime factorisation \(n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}\), the number of factors is found by adding \(1\) to each index and multiplying:
For any two natural numbers \(a\) and \(b\), the HCF and LCM satisfy:
How to find the HCF and LCM
- Factorise each number into primes (a factor tree helps), and write it in index form.
- HCF: multiply the primes common to both, each taken to the lower power that appears.
- LCM: multiply every prime that appears in either number, each taken to the higher power.
- Check: confirm \(\mathrm{HCF}\times\mathrm{LCM}=a\times b\).
Pair the factors as products that give \(48\), then count them:
| \(48\) | \(=\) | \(1 \times 48\) |
| \(=\) | \(2 \times 24\) | |
| \(=\) | \(3 \times 16\) | |
| \(=\) | \(4 \times 12\) | |
| \(=\) | \(6 \times 8\) | |
| \(\text{factors}\) | \(=\) | \(1,2,3,4,6,8,12,16,24,48\) |
| \(\text{count}\) | \(=\) | \(10\) |
\(48\) has \(10\) factors.
Split \(90\) with a factor tree until every leaf is prime:
| \(90\) | \(=\) | \(9 \times 10\) |
| \(9\) | \(=\) | \(3 \times 3 = 3^{2}\) |
| \(10\) | \(=\) | \(2 \times 5\) |
| \(90\) | \(=\) | \(2 \times 3^{2} \times 5\) |
Add \(1\) to each index and multiply:
| \(\tau(90)\) | \(=\) | \((1+1)(2+1)(1+1)\) |
| \(=\) | \(2 \times 3 \times 2\) | |
| \(=\) | \(12\) |
\(90=2\times3^{2}\times5\), which has \(12\) factors.
Factorise both numbers into index form:
| \(36\) | \(=\) | \(2^{2} \times 3^{2}\) |
| \(48\) | \(=\) | \(2^{4} \times 3\) |
HCF: common primes to the lower power; LCM: every prime to the higher power:
| \(\mathrm{HCF}\) | \(=\) | \(2^{2} \times 3\) |
| \(=\) | \(12\) | |
| \(\mathrm{LCM}\) | \(=\) | \(2^{4} \times 3^{2}\) |
| \(=\) | \(144\) |
Check the product identity:
| \(\mathrm{HCF}\times\mathrm{LCM}\) | \(=\) | \(12 \times 144 = 1728\) |
| \(36 \times 48\) | \(=\) | \(1728\) |
\(\mathrm{HCF}=12\), \(\mathrm{LCM}=144\); both give \(1728\), so the identity holds.
Closure would mean the result of the operation stays inside \(\mathbb{N}\). Pick two naturals whose difference escapes:
| \(4\) | \(\in\) | \(\mathbb{N}\) |
| \(7\) | \(\in\) | \(\mathbb{N}\) |
| \(4 - 7\) | \(=\) | \(-3\) |
| \(-3\) | \(\notin\) | \(\mathbb{N}\) |
Because one example leaves \(\mathbb{N}\), closure fails:
| \(\mathbb{N}\) | \(\text{ is not closed under}\) | \(\text{subtraction}\) |
\(4-7=-3\) is not a natural number, so \(\mathbb{N}\) is not closed under subtraction.
Common pitfalls
Frequently asked questions
Is 1 a prime number?
No. A prime number has exactly two different factors, \(1\) and itself. Since \(1\) has only one factor, it is neither prime nor composite. The smallest prime is \(2\).
What is the difference between a factor and a multiple?
A factor of \(n\) divides \(n\) exactly (for example \(4\) is a factor of \(12\)). A multiple of \(n\) is \(n\) times a natural number (for example \(36\) is a multiple of \(12\)).
How do I find the HCF and LCM using prime factorisation?
Write each number in index form. For the HCF, take each common prime to the lower power and multiply. For the LCM, take every prime that appears to the higher power and multiply.
Why are the natural numbers not closed under subtraction?
Closure means the result always stays in the set. One counterexample breaks it: \(3-5=-2\), and \(-2\) is not a natural number, so \(\mathbb{N}\) is not closed under subtraction.
What is the well-ordering principle?
It states that every non-empty subset of the natural numbers has a least (smallest) element. For example, the multiples of \(3\) greater than \(17\) have a least element, \(18\).
How many factors does a number have?
Write it as \(p_1^{a_1}\cdots p_k^{a_k}\), add \(1\) to each index, then multiply. For example \(90=2\times3^{2}\times5\) has \((1+1)(2+1)(1+1)=12\) factors.