Real Numbers

Real numbers are the rationals and the irrationals together. This chapter takes what you already know about numbers and makes two things precise: that every composite number breaks into primes in exactly one way, and that some numbers cannot be written as fractions at all.

What CBSE examines here (2026-27). Only two things: the Fundamental Theorem of Arithmetic (and the HCF/LCM work that rests on it), and proofs of irrationality for √2, √3 and √5. Euclid's division lemma and decimal expansions of rationals are no longer part of this chapter — see the appendix at the end if you want them as background.

Unit I (Number Systems) carries 6 marks in the 80-mark paper, and Real Numbers is the whole unit.

1. The Fundamental Theorem of Arithmetic

Every composite number can be written as a product of primes, and that factorisation is unique apart from the order in which you write the primes.

The second half is the part that does the work. Uniqueness is what lets you argue about a number by arguing about its primes — if a prime is not in the factorisation, it is not a factor, and no amount of rewriting will produce it.

HCF and LCM from the primes

Once both numbers are factorised:

• HCF — take each prime the two numbers share, at the lowest power present.
• LCM — take every prime that appears in either, at the highest power present.

The reason is worth holding on to. The HCF has to divide both numbers, so it can only use what both actually have. The LCM has to be divisible by both, so it must carry enough of every prime to cover the greedier of the two.

Worked example — 96 and 404.

$96=2^5\cdot 3,404=2^2\cdot 101.$

The only shared prime is 2, and the lower power is $2^2$:

$\text{HCF}=2^2=4,\text{LCM}=2^5\cdot 3\cdot 101=9696.$

Check: $4\times 9696=38,784$, and $96\times 404=38,784$. ✓

The identity — and its one restriction

$\text{HCF}(a,b)\times \text{LCM}(a,b)=a\times b$

This holds for two numbers only. For three it fails, and examiners like that trap: for 12, 15 and 21 the HCF is 3 and the LCM is 420, so $3\times 420=1260$, while $12\times 15\times 21=3780$. With three numbers, go through the primes.

Worked example — given HCF(306, 657) = 9, find the LCM.

$9\times \text{LCM}=306\times 657=201,042\Rightarrow \text{LCM}=22,338.$

Using uniqueness to rule things out

Can $6^n$ ever end in 0? To end in 0 a number must be divisible by $10=2\times 5$, so it needs a factor of 5. But $6^n=(2\cdot 3{)}^n=2^n\cdot 3^n$, which contains only 2s and 3s. Because prime factorisation is unique, there is no hidden 5 anywhere. So $6^n$ never ends in 0 — for any $n$.

Notice this argument settles all $n$ at once. Checking 6, 36 and 216 would only show it has not happened yet.

Why is $7\times 11\times 13+13$ composite? Take out the common 13: it equals $13\times (77+1)=13\times 78$. Displaying one factor other than 1 and itself is all "composite" requires — you never need the full factorisation.

2. Irrational Numbers

A number is irrational if it cannot be written as $p/q$ with $p,q$ integers and $q\ne 0$.

The proof that $\sqrt{5}$ is irrational

Every proof in this section is the same argument with a different number, so learn the shape once.

Suppose $\sqrt{5}=p/q$ where $p$ and $q$ are integers with no common factor — cancel first if they have one. Then $p^2=5q^2$, so 5 divides $p^2$.

Here is the step that needs the Fundamental Theorem: since 5 is prime, if it divides $p^2$ it must appear in the factorisation of $p$ itself. So write $p=5k$. Substituting, $25k^2=5q^2$, that is $q^2=5k^2$ — and by the same reasoning 5 divides $q$.

Now 5 divides both $p$ and $q$, contradicting the "no common factor" we arranged at the start. The only assumption made was that $\sqrt{5}$ is rational, so that is what is false.

The identical argument proves $\sqrt{2}$, $\sqrt{3}$, or $\sqrt{p}$ for any prime $p$.

Building on it

Once $\sqrt{5}$ is known irrational, numbers like $3+2\sqrt{5}$ do not need the argument again. Assume $3+2\sqrt{5}=a/b$, then rearrange until the surd is alone:

$\sqrt{5}=\frac{a-3b}{2b}.$

The right side is one integer over another, so it is rational — which contradicts what we just proved. Any expression of the form (rational) + (non-zero rational) × $\sqrt{5}$ falls the same way.

Key results to memorise

• Every composite number has a unique prime factorisation.
• HCF = common primes at lowest powers; LCM = all primes at highest powers.
• $\text{HCF}(a,b)\times \text{LCM}(a,b)=a\times b$ — two numbers only.
• If a prime $p$ divides $a^2$, then $p$ divides $a$. (This is the engine of every irrationality proof.)

Practice (try before checking)

1. Find the HCF and LCM of 196 and 38220 by prime factorisation.
2. Show that $5-2\sqrt{3}$ is irrational.
3. The HCF of two numbers is 27 and their LCM is 162. One number is 54 — find the other.
4. Find the smallest number that leaves remainder 5 when divided by 12, 15 and 18.
5. Explain why $7\times 6\times 5\times 4\times 3\times 2\times 1+5$ is composite.

Answers

1. $196=2^2\cdot 7^2$ and $38220=2^2\cdot 3\cdot 5\cdot 7^2\cdot 13$. Common at lowest powers: $2^2\cdot 7^2=196$, so HCF = 196 (196 divides 38220). LCM $=2^2\cdot 3\cdot 5\cdot 7^2\cdot 13=38,220$.
2. Suppose $5-2\sqrt{3}=r$ is rational. Then $\sqrt{3}=(5-r)/2$, a rational number — contradicting the irrationality of $\sqrt{3}$.
3. $54\times b=27\times 162\Rightarrow b=81$.
4. LCM(12, 15, 18): $12=2^2\cdot 3$, $15=3\cdot 5$, $18=2\cdot 3^2$, so LCM $=2^2\cdot 3^2\cdot 5=180$. Required number $=185$.
5. Both terms are divisible by 5, so it equals $5\times (1008+1)=5\times 1009=5045$ — a product of two factors greater than 1.

Common mistakes

1. Swapping the HCF and LCM rules. Take lowest powers for HCF, highest for LCM. Recover it from meaning rather than memory: the HCF must divide both, so it cannot use more than both have.
2. Using HCF × LCM = product for three numbers. It holds for two only.
3. Dropping "no common factor" from an irrationality proof. Without it there is no contradiction at the end, and the proof scores almost nothing.
4. Proving a number composite by full factorisation. One factor other than 1 and itself is enough — look for the common term before multiplying anything out.

CBSE exam focus

The irrationality proof is the most reliably repeated item in this chapter. It is worth being able to write it out cleanly from memory.

Quick self-test

1. Find the HCF and LCM of 12, 15 and 21. (HCF 3, LCM 420 — and note the product identity does not apply here.)
2. Show that $3\sqrt{2}$ is irrational. (If $3\sqrt{2}=p/q$ then $\sqrt{2}=p/3q$, which is rational — contradiction.)
3. Prove $\sqrt{5}$ is irrational. (Assume $\sqrt{5}=p/q$ coprime; $p^2=5q^2$ ⇒ 5 | p ⇒ p = 5k ⇒ $q^2=5k^2$ ⇒ 5 | q. Both divisible by 5 — contradiction.)
4. Two people jog a circular track in 18 and 12 minutes. When do they next meet at the start? (LCM(18, 12) = 36 minutes.)
5. Express 7429 as a product of primes. (17 × 19 × 23.)

Appendix — beyond the current syllabus

Not examinable in CBSE 2026-27. Both topics below were removed from this chapter when the syllabus was rationalised. They are kept here because they are genuinely useful — Euclid's algorithm is standard in computer science, and decimal expansions explain something you see constantly — but do not spend board-exam preparation time on them, and do not use Euclid's lemma as your method in an answer that asks for the HCF.

Euclid's division lemma and algorithm

Given positive integers $a$ and $b$, there are unique integers $q$ and $r$ with

$a=bq+r,0\le r<b.$

To find the HCF, apply it repeatedly, each time replacing the pair by (divisor, remainder). The last non-zero remainder is the HCF.

HCF(135, 225): $225=135\times 1+90$; $135=90\times 1+45$; $90=45\times 2+0$. So the HCF is 45 — the same answer prime factorisation gives.

It also drives "show that every odd integer is of the form $8k+1$ when squared" style arguments: any odd integer is $4q+1$ or $4q+3$, and squaring either gives $8k+1$.

Decimal expansions of rational numbers

A rational $p/q$ in lowest terms has a terminating decimal expansion exactly when $q=2^n{5}^m$; otherwise the expansion is non-terminating recurring.

Simplify the fraction before checking the denominator: 77/210 looks bad until it reduces to 11/30.

Once Real Numbers is solid, Polynomials carries the same idea into algebra — factorising expressions instead of integers, and reading the zeroes of a polynomial off its coefficients.