Real Numbers
Why This Matters
Every number you’ll ever meet is built out of a tiny set of “building-block” numbers called primes — and it’s built in exactly one way. That single fact is quietly doing the work behind a surprising amount of everyday life: it’s why your online banking and UPI payments stay secure, why two flashing lights or two buses on a route line up again after a fixed time, and why a fraction either gives a clean terminating decimal or a repeating one.
This chapter is about understanding numbers from the inside. You’ll learn how to break any number down to its prime building blocks, how that instantly gives you the HCF and LCM of two numbers, and finally how to prove — not just believe — that numbers like √2 can never be written as a neat fraction.
That last part matters more than it looks. Most of your earlier maths was about calculating an answer. Here you’ll do something a mathematician does: assume the opposite of what you want to show, follow it logically until it collapses into a contradiction, and conclude you were right all along. It’s a way of thinking you’ll use for the rest of the subject.
The Big Idea
Every composite number is a product of primes, and — apart from the order — there is only one such product. Primes are the atoms of arithmetic. Once you know a number’s prime “recipe”, almost everything about it (its HCF and LCM with another number, whether it can end in a 0, whether a root of it is irrational) falls out immediately.
Let’s Break It Down
Every number is built from primes
A prime number has exactly two factors: 1 and itself (2, 3, 5, 7, 11, 13, …). A composite number has more (4, 6, 8, 9, …). The number 1 is neither.
The Fundamental Theorem of Arithmetic says something that feels obvious but is incredibly powerful:
Every composite number can be written as a product of primes, and this factorisation is unique, except for the order in which the primes are written.
So 12 = 2 × 2 × 3, and there is no other set of primes that multiplies to 12. To find the primes, keep splitting a number into factors until everything left is prime — a factor tree.
We usually write the primes in increasing order and group repeats as powers: 3825 = 3² × 5² × 17.
The theorem really has two halves. Existence — that some prime factorisation exists — is easy to believe: a factor tree always ends, because every split makes the numbers smaller until only primes are left. Uniqueness — that there’s only one such set of primes — is the deep and powerful half. (Its full proof is beyond Class 10, but it’s exactly the half we lean on below to prove the next theorem and the irrationality of roots.)
Express 156 as a product of its prime factors.
- 156 is even, so pull out 2: 156 = 2 × 78.
- 78 is even too: 78 = 2 × 39. So far 156 = 2 × 2 × 39.
- 39 = 3 × 13, and 13 is prime, so we stop.
- Collecting the primes: 156 = 2² × 3 × 13.
Reading off the HCF and LCM
Once you have two numbers in prime-power form, their HCF (highest common factor) and LCM (lowest common multiple) require no guessing:
- HCF = product of the smallest power of each prime that appears in both numbers (the shared part).
- LCM = product of the greatest power of every prime that appears in either number (the combined part).
| HCF (common factor) | LCM (common multiple) | |
|---|---|---|
| Which primes | only primes in BOTH | every prime in EITHER |
| Which power | the smallest | the greatest |
| Size | ≤ both numbers | ≥ both numbers |
| Think of it as | the largest tile that fits both | when two cycles next align |
Find the HCF and LCM of 96 and 404.
- Factorise each: 96 = 2⁵ × 3, and 404 = 2² × 101.
- HCF: the only prime in both is 2; its smallest power is 2². So HCF = 2² = 4.
- LCM: take every prime at its greatest power — 2⁵, 3, and 101. So LCM = 2⁵ × 3 × 101 = 32 × 303 = 9696.
- So HCF(96, 404) = 4 and LCM(96, 404) = 9696.
For two numbers there’s a lovely shortcut: the part you put in the HCF and the part you put in the LCM together use up every prime exactly once, so
HCF(a, b) × LCM(a, b) = a × b
This lets you find the LCM the instant you know the HCF (or check your work). Notice the check above: 4 × 9696 = 38784 = 96 × 404. ✓
Given that HCF(306, 657) = 9, find LCM(306, 657).
- For two numbers, HCF × LCM = product, so LCM = (306 × 657) / HCF.
- 306 × 657 = 201042. Divide by the HCF, 9: LCM = 201042 / 9.
- 201042 / 9 = 22338, so LCM(306, 657) = 22338.
Warning: the HCF × LCM = product rule works for two numbers only. For three numbers, HCF(a, b, c) × LCM(a, b, c) is not generally a × b × c.
Using uniqueness to settle a question
Because a number’s prime recipe is the only one it has, you can rule things out completely. A number ends in 0 only if it’s divisible by 10 = 2 × 5 — that is, only if both 2 and 5 appear in its factorisation.
Is there any natural number n for which 6ⁿ ends in the digit 0?
- To end in 0, the number must be divisible by 10 = 2 × 5, so 5 must appear in its prime factorisation.
- But 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ. The only primes here are 2 and 3 — never 5.
- By the uniqueness of the Fundamental Theorem of Arithmetic, 6ⁿ has no other prime factors, so 5 can never appear.
- So 6ⁿ can never end in the digit 0, for any natural number n.
Why some numbers can’t be written as fractions
A rational number can be written as p/q with p, q integers and q ≠ 0. An irrational number cannot. Irrational numbers are not strange or “unreal” — they’re genuine points on the number line. √2, for instance, is simply the length of the diagonal of a 1 × 1 square (by Pythagoras, 1² + 1² = 2, so the diagonal is √2). Swing that diagonal down with a compass and it lands at a definite spot between 1 and 2:
To prove a number is irrational we use proof by contradiction: assume it is rational, then show that assumption forces an impossibility.
We need one small fact first — and, since it’s a theorem, here is its proof (it leans on the uniqueness half of the Fundamental Theorem of Arithmetic):
Theorem. If a prime p divides a², then p divides a (where a is a positive integer).
Show that whenever a prime p divides a², it must also divide a.
- Write a in its prime factorisation: a = p₁ × p₂ × … × pₙ (its prime building blocks, not necessarily distinct).
- Square it: a² = (p₁ × p₂ × … × pₙ)² = p₁² × p₂² × … × pₙ². So the primes appearing in a² are exactly p₁, p₂, …, pₙ — the same primes as a.
- We’re told the prime p divides a². By the uniqueness of prime factorisation, the only primes inside a² are p₁, …, pₙ, so p must be one of them.
- But p₁, …, pₙ are precisely the primes of a — so p divides a. ∎ (This little theorem is the engine inside every “root is irrational” proof.)
Prove that √2 cannot be written as a fraction.
- Assume the opposite: √2 = a/b, where a and b are integers with no common factor (the fraction is in lowest terms) and b ≠ 0.
- Then b√2 = a. Square both sides: 2b² = a². So 2 divides a², and by the fact above, 2 divides a. Write a = 2c.
- Substitute: 2b² = (2c)² = 4c², so b² = 2c². Then 2 divides b², so 2 divides b as well.
- But now 2 divides both a and b — they share the factor 2. That contradicts “no common factor”. The only thing we assumed was that √2 is rational, so that must be false: √2 is irrational.
The exact same argument (with 3 in place of 2, or 5, or any prime) proves √3, √5, … are irrational. And once you know √3 is irrational, you can show combinations like 5 − √3 or 3√2 are irrational too — by assuming they’re rational and rearranging until a known irrational is forced to equal a fraction.
Prove that 5 − √3 is irrational, given that √3 is irrational.
Assume the opposite: 5 − √3 is rational, say 5 − √3 = a/b with integers a, b (b ≠ 0).
Rearrange to isolate the root: √3 = 5 − a/b = (5b − a)/b.
The right-hand side is a difference and quotient of integers, so it is rational. That would make √3 rational — but we are told √3 is irrational. Contradiction.
So our assumption was wrong: 5 − √3 is irrational.
Common Mistakes
The number 1 is a prime number (its factors are 1 and itself).
Primes are described as 'divisible only by 1 and themselves', and 1 fits that phrase loosely — 1 is divisible by 1 and by 1.
A prime must have exactly TWO different factors. 1 has only one factor (itself), so 1 is neither prime nor composite. (If 1 were prime, prime factorisation would no longer be unique — you could stick on extra 1s.)
For the LCM you take the smallest powers, and for the HCF the greatest powers.
Both rules are about 'powers of primes', and it's easy to attach the wrong word to each since they're mirror images of each other.
It's the reverse: HCF takes the SMALLEST power of each shared prime (a common factor can't be bigger than what both have); LCM takes the GREATEST power of every prime (a common multiple must contain all of each).
HCF × LCM = product of the numbers works for any set of numbers.
It's a clean, memorable rule and it really does hold for two numbers, so it feels natural to extend it to three.
It holds for TWO numbers only. For three numbers, HCF(a,b,c) × LCM(a,b,c) is generally not a × b × c — check with 6, 72, 120: HCF = 6, LCM = 360, and 6 × 360 = 2160, but 6 × 72 × 120 is far larger.
To prove √2 is irrational you just compute √2 = 1.41421356… and note the decimal never repeats.
A non-repeating decimal does signal irrationality, and a calculator's display looks convincingly endless.
A calculator only shows a finite chunk — you can never SEE that a decimal never repeats. Irrationality must be proved by contradiction (assume √2 = a/b in lowest terms, derive that a and b share a factor, contradiction).
Quick Check
What does the Fundamental Theorem of Arithmetic guarantee about a composite number?
If 90 = 2 × 3² × 5 and 24 = 2³ × 3, what is HCF(90, 24)?
For two numbers, HCF = 9 and LCM = 90. What is the product of the two numbers?
Why is 7 × 11 × 13 + 13 a composite number?
Practice Problems
Easy
Express 5005 as a product of its prime factors.
5005 ends in 5, so divide by 5: 5005 = 5 × 1001.
1001 = 7 × 143, and 143 = 11 × 13.
So 5005 = 5 × 7 × 11 × 13 (all four are prime).
Find the HCF and LCM of 26 and 91, and verify HCF × LCM = product.
26 = 2 × 13 and 91 = 7 × 13.
HCF: the only shared prime is 13, so HCF = 13.
LCM: every prime at its greatest power = 2 × 7 × 13 = 182.
Check: HCF × LCM = 13 × 182 = 2366, and 26 × 91 = 2366. ✓
Medium
There is a circular track. Sonia takes 18 minutes for one round and Ravi takes 12 minutes. They start together from the same point. After how many minutes do they next meet at the starting point?
They meet at the start again after a time that is a whole number of both their lap times — the LCM of 18 and 12.
18 = 2 × 3² and 12 = 2² × 3.
LCM = greatest powers = 2² × 3² = 4 × 9 = 36 minutes.
(After 36 minutes Sonia has done 36/18 = 2 laps and Ravi 36/12 = 3 laps, both back at the start.)
Explain why 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is a composite number.
Both terms share a factor of 5. The first term, 7 × 6 × 5 × 4 × 3 × 2 × 1, clearly contains 5, and the second term is 5.
Take out the 5: 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1) = 5 × (1008 + 1) = 5 × 1009.
It’s a product of two integers each greater than 1, so it is composite.
Challenge
Prove that 3 + 2√5 is irrational, given that √5 is irrational.
Assume the opposite: 3 + 2√5 is rational, say 3 + 2√5 = a/b with integers a, b (b ≠ 0).
Isolate the root: 2√5 = a/b − 3 = (a − 3b)/b, so √5 = (a − 3b)/(2b).
The right-hand side is built only from integers using subtraction and division, so it is rational. That makes √5 rational — contradicting the given fact that √5 is irrational.
So the assumption fails: 3 + 2√5 is irrational.
Summary
You should now be able to explain:
- A prime has exactly two factors; 1 is neither prime nor composite.
- The Fundamental Theorem of Arithmetic: every composite number is a product of primes, and that factorisation is unique apart from order.
- To find HCF, multiply the smallest power of each shared prime; for LCM, the greatest power of every prime.
- For two numbers, HCF × LCM = product — but not for three or more.
- Uniqueness of factorisation lets you rule out possibilities (e.g. 6ⁿ can never end in 0, because 5 is never one of its primes).
- A number is irrational if it can’t be written as p/q. We prove √2, √3, √5 irrational by contradiction, using “if a prime divides a², it divides a”.
What’s Next
Next, in Polynomials, we move from numbers to expressions — things like x² − 5x + 6. You’ll see that polynomials have “building blocks” too (their zeroes, the values that make them 0), and a relationship between those zeroes and the coefficients that echoes the structure you just met here.