If a prime number \( p \) divides \( a^2 \), then \( p \) divides \( a \) — that is Theorem 1.2 of NCERT Class 10 Mathematics, Chapter 1 Real Numbers, printed on page 6 of the textbook.
This page gives you the book’s own statement, a step-by-step proof you can follow in an exam, the footnote that marks this proof ‘Not from the examination point of view’, and the two places on page 7 where the chapter leans on the theorem.
The theorem exactly as NCERT states it
Theorem 1.2 (NCERT Class 10 Mathematics, page 6) Let \(p\) be a prime number. If \(p\) divides \(a^2\), then \(p\) divides \(a\), where \(a\) is a positive integer.
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.
Three conditions carry the weight of this statement. p must be a prime number — composite values can fail, as the counterexample below shows.
a must be a positive integer, because the proof runs on the Fundamental Theorem of Arithmetic, which is stated for natural numbers. The conclusion is driven by the hypothesis that p divides the square \( a^2 \) — without that input there is nothing to conclude.
What the theorem says in plain words
To say that \( p \) divides \( a^2 \) means \( a^2 = p \times k \) for some integer \( k \) — the division leaves no remainder. The theorem says that when \( p \) is prime, you can strip the square: the same \( p \) must already divide \( a \).
Squaring a number writes its prime factorisation twice. Every exponent doubles — \( 60^2 \) carries \( 2^4 \), not \( 2^2 \) — but no new prime appears. The primes inside \( a^2 \) are exactly the primes inside \( a \), only with doubled exponents. Keep the mnemonic: squaring doubles exponents but adds no new primes.
Run the argument on a real number. Take \( a = 60 = 2^2 \times 3 \times 5 \). Then \[ 60^2 = 3600 = 2^4 \times 3^2 \times 5^2. \]
The primes that divide 3600 are exactly 2, 3 and 5 — each already divides 60. If a prime \( p \) divides 3600, the theorem forces \( p \) to be one of these three.
The condition ‘p is prime’ is doing real work. Check \( p = 4 \), \( a = 6 \): \( 4 \) divides \( 6^2 = 36 \) because \( 36 = 4 \times 9 \), yet \( 4 \) does not divide \( 6 \). The theorem is false for composite \( p \).
| Part of the statement | Why the proof needs it | What breaks without it |
|---|---|---|
| \( p \) is a prime number | Only a prime is guaranteed to appear as a named factor in a prime factorisation | \( 4 \mid 6^2 = 36 \) but \( 4 \nmid 6 \) |
| \( a \) is a positive integer | The Fundamental Theorem of Arithmetic gives \( a \) a prime factorisation to work with | The statement, as printed, covers only positive integers |
| \( p \) divides \( a^2 \) | This hypothesis starts the argument | Without it, there is nothing to prove |
The figure NCERT gives for this theorem
The book prints no diagram for Theorem 1.2. This is a purely algebraic proof — it works with symbolic prime factorisations, not shapes — so there is no figure to draw in an exam.
The ‘mental figure’ is the factorisation chain itself: \( a = p_1 p_2 \dots p_n \), its square \( a^2 = p_1^2 p_2^2 \dots p_n^2 \), and the single prime \( p \) forced to be one of \( p_1, p_2, \dots, p_n \).
Given, to prove, and the opening move
- Given: \( p \) is a prime number; \( a \) is a positive integer; \( p \) divides \( a^2 \) (NCERT, p. 6).
- To prove: \( p \) divides \( a \).
- Opening move: no geometric construction here — the proof begins by writing the prime factorisation \( a = p_1 p_2 \dots p_n \), where \( p_1, p_2, \dots, p_n \) are primes, not necessarily distinct (NCERT, p. 6).
That factorisation exists by the Fundamental Theorem of Arithmetic (Theorem 1.1, NCERT, p. 3): every composite number factorises as a product of primes, uniquely apart from order. The proof of Theorem 1.2 uses each half of that theorem separately — existence in one step, uniqueness in another.
Proof of Theorem 1.2, step by step
The book gives the full proof on printed page 6 (NCERT, p. 6). Here it is expanded into six numbered steps, each with the reason behind it.
- Factorise \( a \). Since \( a \) is a positive integer, the existence part of the Fundamental Theorem of Arithmetic lets us write \( a = p_1 p_2 \dots p_n \), where each factor is prime and the primes are not necessarily distinct — the same prime may repeat, as 2 repeats in \( 12 = 2 \times 2 \times 3 \).
- Square the factorisation. Writing the product twice gives \( a^2 = (p_1 p_2 \dots p_n)(p_1 p_2 \dots p_n) = p_1^2 p_2^2 \dots p_n^2 \), which is a prime factorisation of \( a^2 \).
- Read the hypothesis. We are given \( p \mid a^2 \), so the prime \( p \) divides the product \( p_1^2 p_2^2 \dots p_n^2 \). By the existence part of the Fundamental Theorem of Arithmetic, \( p \) is one of the prime factors of \( a^2 \).
- Apply uniqueness. The uniqueness part of the Fundamental Theorem of Arithmetic says \( a^2 \) has only one prime factorisation, apart from order. The factorisation found in step 2 is one such factorisation — so the prime \( p \) from step 3 must be among the factors listed there. Hence \( p \) is one of \( p_1, p_2, \dots, p_n \).
- Return to \( a \). Since \( a = p_1 p_2 \dots p_n \), the factor \( p \) appears in the product that makes \( a \).
- Conclude. A factor of a product divides that product. Therefore \( p \mid a \). Hence proved.
The two appeals to the Fundamental Theorem of Arithmetic do separate jobs. Step 1 uses its existence half — a factorisation exists. Step 4 uses its uniqueness half — there is only one, so \( p \) is forced into the list. Confusing the two halves is the usual source of errors in this proof.
One footnote settles how to revise this proof. An asterisk beside the word ‘Proof’ on printed page 6 points to the line: ‘Not from the examination point of view.’ The proof is taught, but the book itself marks it as not examined; the statement, by contrast, does the key work on page 7, four times over.
Try it yourself: a live practice figure
Where the chapter uses Theorem 1.2 next
Section 1.3, ‘Revisiting Irrational Numbers’, opens on printed page 6 and states its programme: prove that \( \sqrt{2}, \sqrt{3}, \sqrt{5} \) and, in general, \( \sqrt{p} \) for a prime \( p \) are irrational (NCERT, p. 6). Theorem 1.2 is introduced as the tool needed before the first of those proofs.
On printed page 7 the theorem is applied four times, always in the same rhythm: from ‘\( p \) divides the square’ to ‘\( p \) divides the number’ (NCERT, p. 7).
| Where it appears (p. 7) | The step that triggers it | How Theorem 1.2 enters |
|---|---|---|
| Proof of Theorem 1.3 (\( \sqrt{2} \) irrational), first use | \( 2 \mid a^2 \) | By Theorem 1.2 with \( p = 2 \), \( 2 \mid a \) |
| Proof of Theorem 1.3, second use | \( 2 \mid b^2 \) | Again by Theorem 1.2, \( 2 \mid b \) |
| Example 5 (\( \sqrt{3} \) irrational), first use | \( 3 \mid a^2 \) | By Theorem 1.2 with \( p = 3 \), \( 3 \mid a \) |
| Example 5, second use | \( 3 \mid b^2 \) | Again by Theorem 1.2, \( 3 \mid b \) |
The full \( \sqrt{2} \) argument — a proof by contradiction that calls on Theorem 1.2 twice — lives on its own page: Theorem 1.3: root 2 is irrational. Example 5 on the same page runs the identical pattern for \( \sqrt{3} \).
The converse and the theorems around it
The converse is immediate. If \( p \mid a \), then \( p \mid a^2 \), because \( a^2 = a \times a \) is a multiple of \( a \). So for a prime \( p \), \( p \mid a \) if and only if \( p \mid a^2 \) — the theorem supplies the difficult direction, the converse is trivial.
The engine of the proof is the Fundamental Theorem of Arithmetic (Theorem 1.1, NCERT, p. 3), which the chapter’s Real Numbers page introduces. Its immediate successor, Theorem 1.3, root 2 is irrational, is the first place the chapter puts the statement to work.
The book adds a historical aside on printed page 3, right after stating the Fundamental Theorem of Arithmetic: an equivalent version was probably first recorded as Proposition 14 of Book IX of Euclid’s Elements, and the first correct proof was given by Carl Friedrich Gauss in his Disquisitiones Arithmeticae (NCERT, p. 3).
Common mistakes when proving Theorem 1.2
Five slips appear more often than any others in this proof, and each has a quick correction.
| Mistake | Correct rule | How to check your answer |
|---|---|---|
| Applying the theorem to a composite \( p \), e.g. \( 4 \mid 6^2 = 36 \) | \( p \) must be prime; prime factorisation names primes individually | Re-test with \( p = 4 \), \( a = 6 \): the conclusion fails, so your statement must keep ‘prime’ |
| Squaring the factorisation wrongly — writing \( a^2 = p_1 p_2 \dots p_n \) | Each factor appears twice, so \( a^2 = p_1^2 p_2^2 \dots p_n^2 \) | Try \( a = 6 = 2 \times 3 \): the square is \( 36 = 2^2 \times 3^2 \) |
| Using only existence and skipping uniqueness | Existence shows \( p \) is a prime factor of \( a^2 \); uniqueness shows its only prime factors are \( p_1, \dots, p_n \) | If your proof never says ‘unique’, add the uniqueness step before ‘p is one of the factors’ |
| Declaring \( p \mid a \) too early, before \( p \) is matched to a factor | The conclusion comes last, after \( p \) equals one of \( p_1, \dots, p_n \) | Check that the line ‘\( p \mid a \)’ follows the factor list, not precedes it |
| Reading the exam note as covering the whole theorem | The asterisk exempts the proof, not the statement | You meet the statement in the \( \sqrt{2} \) and \( \sqrt{3} \) proofs, so learn it regardless |
Questions students search about this theorem
Why does p dividing a2 force p to divide a when p is a prime?
Because \( a^2 \) is the factorisation of \( a \) multiplied by itself: \( a^2 = (p_1 p_2 \dots p_n)(p_1 p_2 \dots p_n) = p_1^2 p_2^2 \dots p_n^2 \).
The uniqueness part of the Fundamental Theorem of Arithmetic says \( a^2 \) has no other prime factors, so a prime that divides \( a^2 \) must be one of \( p_1, \dots, p_n \) — and those all divide \( a \).
Is the proof of Theorem 1.2 really not from the examination point of view?
Yes. The asterisk by ‘Proof’ on printed page 6 refers to the footnote: ‘Not from the examination point of view.’ Textbook contents and the examinable syllabus are not always identical — check the current official syllabus to see what your board actually asks.
Where is Theorem 1.2 used later in the Real Numbers chapter?
On printed page 7: twice in the proof that \( \sqrt{2} \) is irrational (Theorem 1.3) and twice in Example 5, which proves \( \sqrt{3} \) irrational. Each use is the same move — from ‘\( p \) divides \( a^2 \)’ to ‘\( p \) divides \( a \)’ — with \( p = 2 \) or \( p = 3 \).
Does Theorem 1.2 work when p is not a prime number?
No. \( 4 \) divides \( 6^2 = 36 \), but \( 4 \) does not divide \( 6 \). The prime condition is what guarantees \( p \) appears as a named factor in the prime factorisation.
Do I need to memorise the proof of Theorem 1.2 for the exam?
The book marks this proof ‘Not from the examination point of view’, so the proof itself is not examined as the book states it. Memorise the statement and how to apply it — it does the key work in the \( \sqrt{2} \) and \( \sqrt{3} \) irrationality proofs. Textbook contents and the examinable syllabus are not always identical — check the current official syllabus.
Related resources and sources
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.
Revise the theorem with its neighbours: the proof that root 2 is irrational, the Class 10 Mathematics hub, and the Class 10 hub.
If you want to check the statement against the printed book, the official PDF of Chapter 1, Real Numbers, is published on ncert.nic.in — open the NCERT chapter PDF and go to page 6, where Theorem 1.2 and its proof sit together.
This page follows the current official NCERT Class 10 Mathematics textbook, Chapter 1 Real Numbers, and its printed page numbers. NCERT settles textbook editions and official PDFs; CBSE settles the examinable syllabus and the examinations.
Reference: NCERT Class 10 Mathematics textbook, chapter 1, Theorem 1.2, official edition on ncert.nic.in.
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