If you need to prove that root 2 is irrational for Class 10, here is the exact NCERT Theorem 1.3 statement and the proof as the book builds it on printed pages 6–7 of Chapter 1, Real Numbers.
Every step below carries its reason, the supporting theorem is named, and the same template is then run on a new prime so you can reuse it in the exam.
The Exact Statement: Root 2 Is Irrational (Theorem 1.3)
Theorem 1.3 (NCERT Class 10 Mathematics, page 6) \(\sqrt{2}\) is irrational.
New: practise this theorem. Drag the book’s figure and watch the numbers change, try to break the theorem on purpose, then solve fresh questions with hints.
The theorem is Theorem 1.3 of Chapter 1, Real Numbers, printed on page 6 at the start of section 1.3, “Revisiting Irrational Numbers”. The section opens by announcing what it will prove: \( \sqrt{2} \), \( \sqrt{3} \), \( \sqrt{5} \) and, in general, \( \sqrt{p} \) for any prime \(p\) (NCERT, p. 6).
What “Root 2 Is Irrational” Means in Plain Words
The book’s definition of irrational, printed on page 6: a number \(s\) is irrational if it cannot be written in the form \( \frac{p}{q} \), where \(p\) and \(q\) are integers and \(q \neq 0\). A rational number, by contrast, can always be written that way.
So the claim “\( \sqrt{2} \) is irrational” is about this one number: no matter which integers you try for \(p\) and \(q\), you can never make \( \frac{p}{q} \) equal to \( \sqrt{2} \). The book’s own examples of irrational numbers (NCERT, p. 6) are \( \sqrt{2}, \sqrt{3}, \sqrt{15}, \pi, -\frac{\sqrt{2}}{\sqrt{3}} \) and \( 0.10110111011110 \dots \).
- Rational: can be written as \( \frac{p}{q} \) with integers \(p, q\), \( q \neq 0 \) — for example \( \frac{3}{4} \), \( -2 \), \( 0.5 \).
- Irrational: cannot be written that way — no fraction equals it.
- Why the proof matters: in Class IX the book introduced irrationals and located them on the number line, but did not prove they were irrational; section 1.3 supplies those proofs (NCERT, p. 6).
No Figure Is Needed for This Proof
Unlike geometry theorems, this theorem has no diagram in the book. The proof on printed pages 6–7 is purely algebraic — a proof by contradiction built from squaring and divisibility — so a figure would add nothing to it.
The only visual a student might recall, locating irrationals on the number line, is something the book mentions only in prose on page 6 as a Class IX result; it plays no part in the proof of Theorem 1.3.
Given, To Prove, and the Setup of the Proof
| Element | What the book sets up |
|---|---|
| Given | \( \sqrt{2} \) is a real number, and a number is irrational if it cannot be written as \( \frac{p}{q} \) with integers \(p, q\) and \( q \neq 0 \) (NCERT, p. 6). |
| To prove | \( \sqrt{2} \) is irrational. |
| Method | Proof by contradiction — the book names this technique on page 6 and points to Appendix 1 for a fuller discussion. |
The two moves that set the proof up (NCERT, p. 7) act like the “construction” in a geometry proof. First, assume \( \sqrt{2} = \frac{r}{s} \) with integers \(r\) and \(s\) (\( s \neq 0 \)).
Second, if \(r\) and \(s\) share a common factor other than 1, divide it out so that \( \sqrt{2} = \frac{a}{b} \) with \(a\) and \(b\) coprime — meaning their HCF is exactly 1.
This reduction is the hinge of the whole proof. The contradiction at the end is precisely that \(a\) and \(b\) turn out to share factor 2, even though we first removed every common factor. Without the coprime step, the ending would have nothing to contradict.
Proof of Theorem 1.3: Root 2 Is Irrational, Step by Step
The book’s own proof runs in seven moves, on printed pages 6–7. Work through each move in order, giving the reason for every step.
Step 1 — Assume the contrary: suppose \( \sqrt{2} \) is rational.
The book names this technique — proof by contradiction — on page 6 and notes it is discussed in Appendix 1.
The plan is to show this assumption forces a contradiction, so its opposite must be true.
Step 2 — Write it as a fraction in lowest terms: since \( \sqrt{2} \) is assumed rational, \( \sqrt{2} = \frac{r}{s} \) for integers \(r\) and \(s\), \( s \neq 0 \).
If \(r\) and \(s\) share a common factor other than 1, divide it out to get \( \sqrt{2} = \frac{a}{b} \), where \(a\) and \(b\) are coprime.
This is the step the final contradiction depends on.
Step 3 — Square both sides: from \( b\sqrt{2} = a \), squaring and rearranging gives \[ 2b^2 = a^2. \]
So \( a^2 = 2b^2 \), which means 2 divides \( a^2 \) (NCERT, p. 7).
Step 4 — Apply Theorem 1.2 once: 2 is a prime number, so Theorem 1.2 says that if a prime divides a square, it divides the number itself.
Hence 2 divides \(a\), and we can write \( a = 2c \) for some integer \(c\).
The step is valid only because 2 is prime — a point the common-mistakes section below makes concrete.
Step 5 — Substitute and apply Theorem 1.2 again: putting \( a = 2c \) into \( 2b^2 = a^2 \) gives \[ 2b^2 = 4c^2 \Rightarrow b^2 = 2c^2. \]
So 2 divides \( b^2 \), and by Theorem 1.2 again (with \( p = 2 \)), 2 divides \(b\).
Step 6 — State the contradiction: 2 divides \(a\) and 2 divides \(b\), so \(a\) and \(b\) have at least 2 as a common factor.
This directly contradicts the fact, fixed in Step 2, that \(a\) and \(b\) have no common factor other than 1.
Step 7 — Conclude: the contradiction arose because of the assumption that \( \sqrt{2} \) is rational.
That assumption is therefore false, and \( \sqrt{2} \) is irrational.
Which proof carries the asterisk? On printed page 6 the book prints “* Not from the examination point of view” under the proof of Theorem 1.2, not under Theorem 1.3. So the root 2 proof itself is not the part the book marks that way — learn it in full, including the coprime setup and the two applications of Theorem 1.2.
Applying the Same Template: Proving Root 7 Is Irrational
This example is this page’s own working, modelled move for move on the book’s proof. The book itself proves \( \sqrt{2} \) (Theorem 1.3) and \( \sqrt{3} \) (Example 5, p. 7), and leaves \( \sqrt{5} \) as a closing question on p. 9. Here is the same seven steps with the prime replaced by 7:
- Step 1: Assume \( \sqrt{7} \) is rational.
- Step 2: \( \sqrt{7} = \frac{a}{b} \), with \(a\) and \(b\) coprime after dividing out any common factor.
- Step 3: From \( b\sqrt{7} = a \), squaring gives
\[ 7b^2 = a^2, \]
so 7 divides \( a^2 \).
- Step 1: Since 7 is prime, Theorem 1.2 gives that 7 divides \(a\) — write \( a = 7c \).
- Step 2: Substitute: \( 7b^2 = (7c)^2 = 49c^2 \), so \( b^2 = 7c^2 \), and 7 divides \( b^2 \); by Theorem 1.2 again, 7 divides \(b\).
- Step 3: Now \(a\) and \(b\) share factor 7, contradicting that they are coprime.
- Step 4: The assumption was false, so \( \sqrt{7} \) is irrational.
The proof in six moves, with a memory aid. This is the entire exam-ready skeleton:
- Assume \( \sqrt{2} \) is rational.
- Write \( \sqrt{2} = \frac{a}{b} \) with \(a, b\) coprime.
- Square: \( 2b^2 = a^2 \), so 2 divides \( a^2 \).
- Theorem 1.2: 2 divides \(a\); write \( a = 2c \).
- Substitute: \( b^2 = 2c^2 \), so 2 divides \(b\).
- Contradiction with coprimality — hence \( \sqrt{2} \) is irrational.
Memory aid: Assume, Reduce, Square, Divide, Substitute, Contradict.
Try it yourself: a live practice figure
Where the Book Uses This Proof Next
Everything after Theorem 1.3 in section 1.3 either repeats its template or leans on its conclusion. Every item below is named, never solved.
| Number | Where the book treats it | Printed page | What the book does |
|---|---|---|---|
| \( \sqrt{2} \) | Theorem 1.3 | p. 6–7 | Proves irrational in full — the proof reproduced above. |
| \( \sqrt{3} \) | Example 5 | p. 7 | Solves it with the identical template, replacing 2 by 3. |
| \( \sqrt{5} \) | First closing question of section 1.3 | p. 9 | Sets it as a question for the student to prove. |
- Example 6 (p. 8): shows \( 5 – \sqrt{3} \) is irrational. Its method differs from the squaring template: it assumes the whole expression rational, rearranges to isolate \( \sqrt{3} \), and calls the result a contradiction to Example 5’s conclusion.
- Class IX facts recalled on p. 8: the sum or difference of a rational and an irrational is irrational; the product and quotient of a non-zero rational and an irrational is irrational. These two statements are the levers Examples 6 and 7 pull.
- Example 7 (p. 8–9): shows \( 3\sqrt{2} \) is irrational by isolating \( \sqrt{2} \) and using Theorem 1.3’s conclusion.
- Closing questions of section 1.3 (p. 9): Q1 proves \( \sqrt{5} \) irrational; Q2 proves \( 3 + 2\sqrt{5} \) irrational; Q3 covers \( \frac{1}{\sqrt{2}} \), \( 7\sqrt{5} \) and \( 6 + \sqrt{2} \).
A note on how this is usually tested. Irrationality appears in Class 10 board-style papers as a “prove that” question — either \( \sqrt{p} \) for a prime \(p\), or a rational-combination variant like the examples above. A full answer must contain the coprime reduction, both applications of Theorem 1.2, and the closing contradiction sentence.
One caveat applies to all of it: textbook contents and the examinable syllabus are not always identical — check the current official CBSE syllabus before deciding what to prepare.
The Supporting Theorem 1.2, and the Question of a Converse
The proof applies Theorem 1.2 twice — if a prime \(p\) divides \( a^2 \), then \(p\) divides \(a\) — the theorem proved on printed page 6, whose own proof is based on the Fundamental Theorem of Arithmetic. Read its statement, proof and page reference on the Theorem 1.2 page.
The chain of reasoning rests on the Fundamental Theorem of Arithmetic, which is Theorem 1.1 of this chapter and is restated in the chapter summary on page 9. When naming it, always use the current number, Theorem 1.1.
Is there a converse to Theorem 1.3? Not a meaningful one — “\( \sqrt{2} \) is irrational” is a statement about a single number, so there is no second claim to reverse. The direction students actually meet next is the book’s announcement on page 6 that \( \sqrt{p} \) is irrational for every prime \(p\), which the same seven-step template proves.
Common Mistakes in This Proof, With Corrections
Six mistakes lose full credit more often than the algebra does. Each row gives the mistake, the correct rule, and a one-line check.
| Mistake | Correct rule | How to check your answer |
|---|---|---|
| Skipping the reduce-to-coprime step. | Always end the setup with \( \sqrt{2} = \frac{a}{b} \), \(a\) and \(b\) coprime (HCF 1). | Your final line must say “contradicts that \(a\) and \(b\) are coprime” — without the reduction, that sentence has nothing to contradict. |
| Using “\(p\) divides \( a^2 \) so \(p\) divides \(a\)” for a non-prime. | The implication is valid only for a prime \(p\) — that is exactly the hypothesis of Theorem 1.2. | Test the counterexample: 4 divides \( 6^2 = 36 \), but 4 does not divide 6. The step fails for 4, which is why the proof works because 2 is prime. |
| Squaring \( b\sqrt{2} = a \) wrongly. | The square is \( (b\sqrt{2})^2 = a^2 \), i.e. \( 2b^2 = a^2 \). | Check with \( b = 1 \): \( \sqrt{2} = a \) should give \( a^2 = 2 \), which \( 2b^2 = a^2 \) reproduces. |
| Stopping after showing 2 divides \(a\). | The chain must run twice: 2 divides \(a\) and 2 divides \(b\), so both share factor 2. | If your proof ends at “\( a = 2c \)”, there is no contradiction yet — \(b\) is still unrestricted. |
| Ending without the contradiction sentence. | Close with: “This contradicts that \(a\) and \(b\) are coprime; the assumption was false; hence \( \sqrt{2} \) is irrational.” | Read the last two lines: they must name the contradiction and the conclusion. |
| In variants like \( 5 – \sqrt{3} \), not isolating the known irrational. | Rearrange the assumed rational equality until the irrational you already proved stands alone on one side. | Before writing “contradiction”, one side should be exactly \( \sqrt{3} \), \( \sqrt{2} \) or \( \sqrt{5} \), and the other side visibly rational. |
Questions Students Ask About Root 2 and Irrationality
Why is root 2 irrational? Can it be explained in one simple line?
If \( \sqrt{2} \) were rational, its fraction in lowest terms \( \frac{a}{b} \) would force both \(a\) and \(b\) to be even — and a fraction in lowest terms cannot have both numerator and denominator even. So \( \sqrt{2} \) cannot be rational.
Why do we start the proof by assuming root 2 is rational?
Because this is a proof by contradiction, the technique the book names on page 6. You assume the opposite of what you want to prove, show that the assumption leads to an impossible conclusion, and conclude the assumption was false.
What does “coprime” mean, and why does the whole proof depend on it?
Two integers are coprime when their HCF is exactly 1 — they share no common factor other than 1. The proof’s ending is a contradiction because Steps 4 and 5 show \(a\) and \(b\) both have factor 2, which can only be a contradiction because we deliberately removed every common factor at the start.
Why does the same argument fail for root 4?
Because 4 is not a prime. From \( 4b^2 = a^2 \) we can say 4 divides \( a^2 \), but Theorem 1.2 only lets a prime divisor of \( a^2 \) force the same divisor of \(a\). The counterexample is 4 dividing \( 6^2 = 36 \) while 4 does not divide 6. And in fact \( \sqrt{4} = 2 \) is rational, so any proof that it is irrational must fail somewhere.
Is proving that root 2 is irrational asked in the Class 10 board exam?
Irrationality proofs are standard “prove that” material in this chapter, but textbook contents and the examinable syllabus are not always identical — check the current official CBSE syllabus before deciding what to prepare. When the proof is asked, the full version needs the coprime setup, both applications of Theorem 1.2, and the closing contradiction.
How do you prove that 2 + 3 root 5 is irrational?
Assume the opposite and isolate the known irrational. This is this page’s own working on the pattern of Examples 6–7, with original numbers. Suppose \( 2 + 3\sqrt{5} = \frac{a}{b} \) for integers \(a, b \neq 0 \). Rearranging, \[ 3\sqrt{5} = \frac{a}{b} – 2 \Rightarrow \sqrt{5} = \frac{a – 2b}{3b}. \]
The right side is a quotient of integers, so it is rational. That forces \( \sqrt{5} \) to be rational, contradicting the fact that \( \sqrt{5} \) is irrational (announced on page 6 and proved as the first closing question of section 1.3). The assumption fails, so \( 2 + 3\sqrt{5} \) is irrational.
Related Resources and the Source
This listing is maintained for the 2026-27 academic session using the NCERT textbook information available to us. NCERT remains the authority for confirming the latest edition.
Keep the whole chapter in view with the NCERT book page for Real Numbers, and use the Class 10 theorems hub to jump to any theorem of the year.
For wider revision, the Class 10 mathematics notes and the Class 10 notes hub group the working set of pages for the term, and you can browse every subject and class from the notes home. A sibling theorem page on the same site, the Basic Proportionality Theorem, shows the same statement-proof-exam structure applied to geometry.
The statement and proof steps on this page follow the official edition of the NCERT Class 10 Mathematics textbook on ncert.nic.in — open it to see Theorem 1.3 and its proof on printed pages 6–7 inside Chapter 1, Real Numbers.
For the syllabus side, CBSE’s official website carries the current Class 10 syllabus — textbook contents and the examinable syllabus are not always identical.
For parents and teachers: this page covers NCERT Class 10 Mathematics, Chapter 1 (Real Numbers), Theorem 1.3, printed on pages 6–7 of the official edition. NCERT settles the textbook and its editions; CBSE settles the syllabus and the examinations.
Reference: NCERT Class 10 Mathematics textbook, chapter 1, Theorem 1.3, official edition on ncert.nic.in.
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