Real Numbers

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CLASS X Mathematics ~6 marks/year Ch 1 of 14
Real Numbers

Class 10 · Mathematics · NCERT chapter notes · Akanksha Classes

Snapshot
  • Every composite number is built from prime building blocks in exactly one way — the Fundamental Theorem of Arithmetic (FTA).
  • Prime factorisation gives HCF (smallest powers of common primes) and LCM (greatest powers of all primes).
  • For any two numbers: $\text{HCF}(a,b)\times\text{LCM}(a,b)=a\times b$ — get one instantly from the other.
  • FTA lets us prove $\sqrt2,\ \sqrt3,\ \sqrt5$ (and $\sqrt p$ for any prime $p$) are irrational, using proof by contradiction.
  • Board weightage: ~6 marks/year — usually one HCF/LCM word problem (2–3 marks) and one "prove it is irrational" proof (3 marks).
Detailed notes

1. Where this chapter sits

In Class 9 you built the full number system. Quick recap, from inside out:

  • Natural numbers $\mathbb{N}=\{1,2,3,\dots\}$ — the counting numbers.
  • Whole numbers $\mathbb{W}=\{0,1,2,3,\dots\}$ — naturals plus zero.
  • Integers $\mathbb{Z}=\{\dots,-2,-1,0,1,2,\dots\}$ — include negatives.
  • Rational numbers $\mathbb{Q}$ — anything of the form $\dfrac{p}{q}$ with $q\neq0$ (e.g. $\tfrac34,\ -5,\ 0.25,\ 0.\overline{3}$). Their decimals either terminate or repeat.
  • Irrational numbers — cannot be written as $\dfrac{p}{q}$ (e.g. $\sqrt2,\ \pi$). Their decimals are non-terminating, non-repeating.
  • Real numbers $\mathbb{R}$ = rationals + irrationals (everything on the number line).

This chapter zooms into the positive integers and answers two questions that look small but run deep:

  • Building up: can every number be made by multiplying primes? (Yes — the FTA.)
  • Breaking down: what does that tell us about HCF, LCM and which square roots are irrational?

2. Primes and composites — the raw material

A prime number has exactly two factors: $1$ and itself ($2,3,5,7,11,13,\dots$). A composite number has more than two factors ($4,6,8,9,12,\dots$). Note: $1$ is neither prime nor composite, and $2$ is the only even prime.

You already know any number can be written as a product of its prime factors — e.g. $2=2,\ 4=2\times2,\ 253=11\times23$. Now look the other way: take a few primes, say $2,3,7,11,23$, and multiply them in any combination, repeating as you like:

$$7\times11\times23=1771,\qquad 2^{2}\times3\times7\times11\times23=21252$$

Since there are infinitely many primes, combining them all possible ways produces an enormous collection of numbers. The natural question: does this collection give us every composite number, or is some composite "missed"? The answer is the Fundamental Theorem.

3. The Fundamental Theorem of Arithmetic (FTA)

Theorem 1.1 (FTA): 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.

Two ideas are packed in here:

  • Existence: every composite can be broken into primes.
  • Uniqueness: there is only one such set of primes (order doesn't matter). So $2\times3\times5\times7$ and $7\times5\times3\times2$ are the same factorisation.

In general, a composite $x$ can be written $x=p_{1}p_{2}\cdots p_{n}$, with primes in increasing order $p_{1}\le p_{2}\le\dots\le p_{n}$. Grouping repeats gives powers of primes — once we fix the order as ascending, the factorisation is unique. (Historically this was hinted at in Euclid's Elements; the first correct proof was given by Carl Friedrich Gauss.)

Think of it like LEGO: primes are the bricks, and every number is a tower built from those bricks in exactly one way. This single fact powers everything else in the chapter.

4. Factor trees — building the factorisation

To factorise a number, keep splitting it into (a prime) × (the rest) until only primes remain. Take $32760$:

$$32760=2\times16380=2\times2\times8190=\dots=2^{3}\times3^{2}\times5\times7\times13$$

Reading off the leaves of the tree: $32760=2\times2\times2\times3\times3\times5\times7\times13$, which we tidy as $2^{3}\times3^{2}\times5\times7\times13$. Even a scary-looking number like $123456789=3^{2}\times3803\times3607$ obeys the same rule (here $3803$ and $3607$ happen to be primes).

NCERT Example 1 — can $4^{n}$ end in the digit 0?

A number ends in $0$ only if it is divisible by $10=2\times5$, i.e. its factorisation must contain a $5$. Now $4^{n}=(2^{2})^{n}=2^{2n}$ — the only prime here is $2$. By the uniqueness of FTA, no $5$ can appear. So $4^{n}$ never ends with $0$, for any natural number $n$.

5. HCF and LCM by prime factorisation

HCF (Highest Common Factor) is the largest number that divides all the given numbers. LCM (Lowest Common Multiple) is the smallest number divisible by all of them. Once each number is in prime-power form, it becomes a "pick the powers" game:

  • HCF = product of the smallest power of each common prime.
  • LCM = product of the greatest power of every prime that appears.
NCERT Example 2 — HCF & LCM of 6 and 20

$6=2^{1}\times3^{1}$ and $20=2^{2}\times5^{1}$.

Common prime is $2$; its smallest power is $2^{1}$, so $\text{HCF}(6,20)=2$.

All primes at greatest powers: $2^{2}\times3^{1}\times5^{1}=60$, so $\text{LCM}(6,20)=60$.

6. The HCF × LCM = product rule

From Example 2, notice $\text{HCF}\times\text{LCM}=2\times60=120=6\times20$. This always holds for two positive integers:

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

So once you have the HCF, the LCM drops out instantly — no need to factorise twice. Rearranged: $\text{LCM}=\dfrac{a\times b}{\text{HCF}}$.

NCERT Example 3 — HCF and LCM of 96 and 404

$96=2^{5}\times3$ and $404=2^{2}\times101$. Common prime $2$, smallest power $2^{2}$, so $\text{HCF}=4$.

Then $\text{LCM}=\dfrac{96\times404}{4}=\dfrac{38784}{4}=9696.$

7. Three or more numbers

The "pick the powers" method extends to three numbers — but the product shortcut does not.

NCERT Example 4 — HCF and LCM of 6, 72, 120

$6=2\times3,\quad 72=2^{3}\times3^{2},\quad 120=2^{3}\times3\times5.$

Common primes $2,3$ at smallest powers: $\text{HCF}=2^{1}\times3^{1}=6$.

All primes at greatest powers: $\text{LCM}=2^{3}\times3^{2}\times5=360.$

Important warning: for three numbers, $\text{HCF}\times\text{LCM}\neq$ product. Here $6\times72\times120=51840$ but $\text{HCF}\times\text{LCM}=6\times360=2160$. The correct three-number identities (good to know) are:

$$\text{LCM}(p,q,r)=\dfrac{p\,q\,r\cdot\text{HCF}(p,q,r)}{\text{HCF}(p,q)\,\text{HCF}(q,r)\,\text{HCF}(p,r)}$$

8. "Ending digit" and "is it composite?" tricks

These are favourite 1–2 mark questions. The logic is always FTA + uniqueness.

  • Ending in 0: needs both $2$ and $5$ as factors. $6^{n}=2^{n}3^{n}$ has no $5$, so $6^{n}$ never ends in $0$.
  • Is a sum composite? Factor out a common term. $7\times11\times13+13=13(7\times11+1)=13\times78$ — has a factor $13$, so it is composite.
  • Likewise $7\times6\times5\times4\times3\times2\times1+5=5(1008+1)=5\times1009$ — composite.

9. Irrational numbers — the idea

A number $s$ is irrational if it cannot be written as $\dfrac{p}{q}$ with integers $p,q$ and $q\neq0$. Familiar examples: $\sqrt2,\ \sqrt3,\ \sqrt{15},\ \pi,\ 0.10110111011110\dots$ In Class 9 you placed these on the number line but never proved they were irrational. We do that now, and the key tool is — again — the FTA.

10. Helper theorem: $p\mid a^{2}\Rightarrow p\mid a$

Theorem 1.2: Let $p$ be a prime. If $p$ divides $a^{2}$, then $p$ divides $a$ (for a positive integer $a$).

Why it's true: write $a=p_{1}p_{2}\cdots p_{n}$ in primes. Then $a^{2}=p_{1}^{2}p_{2}^{2}\cdots p_{n}^{2}$ — the primes of $a^{2}$ are exactly the primes of $a$, each appearing twice. If a prime $p$ divides $a^{2}$, it must be one of those $p_{i}$ (by uniqueness), and hence it already divides $a$.

This little theorem is the engine behind every irrationality proof below.

11. Proving $\sqrt2$ is irrational

This uses proof by contradiction: assume the opposite, then derive something impossible.

NCERT Theorem 1.3 — full proof

1. Assume, to the contrary, that $\sqrt2$ is rational. Then $\sqrt2=\dfrac{a}{b}$ where $a,b$ are coprime (share no common factor) and $b\neq0$.

2. So $b\sqrt2=a$. Squaring: $2b^{2}=a^{2}$. Hence $2$ divides $a^{2}$, and by Theorem 1.2, $2$ divides $a$. Write $a=2c$.

3. Substitute: $2b^{2}=(2c)^{2}=4c^{2}\Rightarrow b^{2}=2c^{2}$. So $2$ divides $b^{2}$, and again by Theorem 1.2, $2$ divides $b$.

4. Now $2$ divides both $a$ and $b$ — contradicting that they are coprime. The only faulty step was our assumption. Therefore $\sqrt2$ is irrational. $\blacksquare$

12. Proving $\sqrt3,\ \sqrt5$ and $\sqrt p$ irrational

The same three-step pattern works — just swap the prime.

NCERT Example 5 — $\sqrt3$ is irrational

Assume $\sqrt3=\dfrac{a}{b}$ (coprime). Then $3b^{2}=a^{2}$, so $3\mid a^{2}\Rightarrow 3\mid a$; write $a=3c$. Then $3b^{2}=9c^{2}\Rightarrow b^{2}=3c^{2}$, so $3\mid b$. Both share $3$ — contradiction. Hence $\sqrt3$ is irrational.

Replacing $3$ by $5$ proves $\sqrt5$ irrational; in general $\sqrt p$ is irrational for every prime $p$. (The argument needs $p$ to be prime — that's exactly where Theorem 1.2 is used.)

13. Sums and products with irrationals

Two facts (stated in Class 9, used constantly here):

  • (rational) $\pm$ (irrational) = irrational.
  • (non-zero rational) $\times$ or $\div$ (irrational) = irrational.
NCERT Example 6 — $5-\sqrt3$ is irrational

Assume $5-\sqrt3=\dfrac{a}{b}$ (rational). Then $\sqrt3=5-\dfrac{a}{b}=\dfrac{5b-a}{b}$. The right side is rational (integers only) — but $\sqrt3$ is irrational. Contradiction. So $5-\sqrt3$ is irrational.

NCERT Example 7 — $3\sqrt2$ is irrational

Assume $3\sqrt2=\dfrac{a}{b}$. Then $\sqrt2=\dfrac{a}{3b}$, which is rational — contradicting that $\sqrt2$ is irrational. So $3\sqrt2$ is irrational.

14. NCERT Exercise 1.1 — fully solved

Q1. Express as a product of primes.

  • (i) $140=2^{2}\times5\times7$
  • (ii) $156=2^{2}\times3\times13$
  • (iii) $3825=3^{2}\times5^{2}\times17$
  • (iv) $5005=5\times7\times11\times13$
  • (v) $7429=17\times19\times23$

Q2. Find LCM & HCF and verify LCM × HCF = product.

  • (i) $26=2\times13,\ 91=7\times13\Rightarrow\text{HCF}=13,\ \text{LCM}=182.$ Check $13\times182=2366=26\times91.$
  • (ii) $510=2\times3\times5\times17,\ 92=2^{2}\times23\Rightarrow\text{HCF}=2,\ \text{LCM}=23460.$ Check $2\times23460=46920=510\times92.$
  • (iii) $336=2^{4}\times3\times7,\ 54=2\times3^{3}\Rightarrow\text{HCF}=6,\ \text{LCM}=3024.$ Check $6\times3024=18144=336\times54.$

Q3. Find LCM & HCF (three numbers).

  • (i) $12=2^{2}\times3,\ 15=3\times5,\ 21=3\times7\Rightarrow\text{HCF}=3,\ \text{LCM}=420.$
  • (ii) $17,23,29$ are all prime $\Rightarrow\text{HCF}=1,\ \text{LCM}=17\times23\times29=11339.$
  • (iii) $8=2^{3},\ 9=3^{2},\ 25=5^{2}\Rightarrow\text{HCF}=1,\ \text{LCM}=1800.$

Q4. Given $\text{HCF}(306,657)=9$, find LCM. $\text{LCM}=\dfrac{306\times657}{9}=22338.$

Q5. Can $6^{n}$ end in $0$? $6^{n}=2^{n}3^{n}$ has no factor $5$, so no — it can never end in $0$.

Q6. $7\times11\times13+13=13(77+1)=13\times78$ and $7\times6\times5\times4\times3\times2\times1+5=5\times1009$ — both have a factor other than $1$ and themselves, so both are composite.

Q7. Circular track. Sonia takes $18$ min, Ravi $12$ min; they meet at the start after $\text{LCM}(18,12)$. $18=2\times3^{2},\ 12=2^{2}\times3\Rightarrow\text{LCM}=2^{2}\times3^{2}=36$ minutes.

15. NCERT Exercise 1.2 — fully solved

Q1. Prove $\sqrt5$ is irrational. Assume $\sqrt5=\dfrac{a}{b}$ (coprime) $\Rightarrow 5b^{2}=a^{2}\Rightarrow 5\mid a\Rightarrow a=5c\Rightarrow b^{2}=5c^{2}\Rightarrow 5\mid b$. Both share $5$ — contradiction. So $\sqrt5$ is irrational.

Q2. Prove $3+2\sqrt5$ is irrational. Assume $3+2\sqrt5=\dfrac{a}{b}$ (rational). Then $\sqrt5=\dfrac{1}{2}\!\left(\dfrac{a}{b}-3\right)=\dfrac{a-3b}{2b}$, which is rational — contradicting that $\sqrt5$ is irrational. Hence $3+2\sqrt5$ is irrational.

Q3. Prove these are irrational:

  • (i) $\dfrac{1}{\sqrt2}$: if it were rational $r$, then $\sqrt2=\dfrac{1}{r}$ would be rational — contradiction. So $\dfrac{1}{\sqrt2}$ is irrational.
  • (ii) $7\sqrt5$: if rational $=\dfrac{a}{b}$, then $\sqrt5=\dfrac{a}{7b}$ rational — contradiction. Irrational.
  • (iii) $6+\sqrt2$: if rational $=\dfrac{a}{b}$, then $\sqrt2=\dfrac{a}{b}-6$ rational — contradiction. Irrational.

16. Common mistakes to avoid

  • Using $\text{HCF}\times\text{LCM}=$ product for three numbers — it only works for two.
  • Calling $1$ a prime — it is neither prime nor composite.
  • In an irrationality proof, forgetting to state $a,b$ are coprime — the contradiction depends on it.
  • Writing $\text{HCF}$ using the greatest powers (that's LCM) and vice-versa — keep "HCF = smallest, LCM = greatest".
  • Saying a number "ends in 0" without checking for a factor $5$.

17. Quick revision checklist

  • FTA: every composite = unique product of primes.
  • Factorise into prime powers first, then HCF = smallest common powers, LCM = greatest powers.
  • $\text{HCF}\times\text{LCM}=a\times b$ — two numbers only.
  • To end in $0$ a number needs both $2$ and $5$.
  • Irrationality proof: assume rational with coprime $a,b$ → both share a prime factor → contradiction.
  • Helper: $p\mid a^{2}\Rightarrow p\mid a$ (prime $p$).
Practice MCQs
1. The HCF of $2^{3}\times3^{2}$ and $2^{2}\times3^{3}$ is:
  1. $2^{3}\times3^{3}$
  2. $2^{2}\times3^{2}$
  3. $2^{2}\times3^{3}$
  4. $6$
Answer: (B) $2^{2}\times3^{2}=36$ — smallest powers of common primes.
2. If $\text{HCF}(a,b)=12$ and $a\times b=1800$, then $\text{LCM}(a,b)=$
  1. $150$
  2. $120$
  3. $1800$
  4. $15$
Answer: (A) $\text{LCM}=\dfrac{1800}{12}=150.$
3. $6^{n}$ (natural $n$) can end with the digit:
  1. $0$
  2. $5$
  3. $6$
  4. $0$ or $5$
Answer: (C) $6^{n}=2^{n}3^{n}$ has no $5$, so it never ends in $0$ or $5$.
4. $7\times11\times13+13$ is:
  1. Prime
  2. Composite
  3. Equal to $1$
  4. Irrational
Answer: (B) $=13(77+1)=13\times78$ — composite.
5. Which of these is irrational?
  1. $\sqrt{16}$
  2. $0.\overline{3}$
  3. $\sqrt5$
  4. $\dfrac{22}{7}$
Answer: (C) $\sqrt5$ — non-terminating, non-recurring decimal.
6. The prime factorisation of $3825$ is:
  1. $3^{2}\times5^{2}\times17$
  2. $3\times5^{3}\times17$
  3. $3^{2}\times5\times85$
  4. $5^{2}\times153$
Answer: (A) $3825=3^{2}\times5^{2}\times17.$
7. The LCM of the smallest prime and the smallest composite number is:
  1. $2$
  2. $4$
  3. $6$
  4. $8$
Answer: (B) smallest prime $=2$, smallest composite $=4$; $\text{LCM}(2,4)=4.$
8. If $n=2^{3}\times3^{4}\times5$, the number of consecutive zeros at the end of $n$ is:
  1. $3$
  2. $2$
  3. $1$
  4. $0$
Answer: (C) zeros = number of $10=2\times5$ pairs = $\min(3,1)=1.$
9. HCF of two coprime numbers is:
  1. their product
  2. the larger one
  3. $1$
  4. $0$
Answer: (C) coprime means no common prime factor, so HCF $=1$.
10. The decimal expansion of a rational number is always:
  1. terminating
  2. non-terminating non-recurring
  3. terminating or non-terminating recurring
  4. irrational
Answer: (C) rationals either terminate or repeat; only irrationals are non-terminating non-recurring.
Assertion–Reason
A: $\sqrt2$ is irrational.   R: If a prime $p$ divides $a^{2}$, then $p$ divides $a$.
Answer: Both A and R are true, and R is the correct explanation of A — Theorem 1.2 is exactly the step used in the $\sqrt2$ proof.
A: $\text{HCF}(a,b)\times\text{LCM}(a,b)=a\times b$ for all positive integers.   R: The same product rule holds for three numbers.
Answer: A is true, R is false — the product rule fails for three numbers (see §7).
Previous-year questions
Q1. Prove that $\sqrt5$ is irrational. (CBSE, 3 marks)
Outline: assume $\sqrt5=\dfrac{a}{b}$ coprime $\Rightarrow 5b^{2}=a^{2}\Rightarrow 5\mid a\Rightarrow a=5c\Rightarrow b^{2}=5c^{2}\Rightarrow 5\mid b$ — contradiction. Hence irrational.
Q2. Two tankers hold $850$ L and $680$ L of diesel. Find the maximum capacity of a container that measures both exactly. (CBSE, 3 marks)
Answer: $\text{HCF}(850,680)$. $850=2\times5^{2}\times17,\ 680=2^{3}\times5\times17\Rightarrow\text{HCF}=2\times5\times17=170$ litres.
Q3. Find the largest number that divides $245$ and $1029$ leaving remainder $5$ in each case. (CBSE, 3 marks)
Answer: Subtract the remainder: needs $\text{HCF}(240,1024)$. $240=2^{4}\times3\times5,\ 1024=2^{10}\Rightarrow\text{HCF}=2^{4}=16.$
Q4. Prove that $3+2\sqrt5$ is irrational. (CBSE, 3 marks)
Outline: if rational, then $\sqrt5=\dfrac{a-3b}{2b}$ is rational — contradiction. Hence $3+2\sqrt5$ is irrational.
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