Real Numbers opens Class 10 Maths by looking closely at the integers and irrationals you met in Class IX. Its central result is the Fundamental Theorem of Arithmetic: every composite number is a product of primes in exactly one way, apart from order. From that one fact come HCF and LCM by prime factorisation, the rule HCF × LCM = product of two numbers, quick arguments such as why 6ⁿ never ends in 0, and proofs by contradiction that √2, √3 and √5 are irrational. Board papers test every one of these directly.
Key Concepts
1. Primes, Composites and the Fundamental Theorem of Arithmetic
A prime number is a natural number greater than 1 whose only factors are 1 and itself: 2, 3, 5, 7, 11, 13, 17, 19, 23 and so on. A composite number is a natural number greater than 1 that has at least one more factor: 4, 6, 8, 9, 10, 12 and so on. The number 1 is neither prime nor composite. The number 2 is the only even prime.
The NCERT chapter first works in the other direction. Multiply some primes, say 2, 3, 7, 11 and 23, repeating them as often as you like, and you get numbers such as 7 × 11 × 23 = 1771 and 23 × 3 × 73 = 8232.
There are infinitely many primes, so combining all of them in all possible ways gives an infinite collection of numbers. The question the chapter asks is whether every composite number can be produced this way. The answer is yes, and it comes with a second promise: each composite number is produced in only one way.
Breaking a number down with a factor tree
To factorise 32760, keep splitting off the smallest prime that divides what is left:
| Step | Divide by | What is left |
|---|---|---|
| 1 | 2 | 16380 |
| 2 | 2 | 8190 |
| 3 | 2 | 4095 |
| 4 | 3 | 1365 |
| 5 | 3 | 455 |
| 6 | 5 | 91 |
| 7 | 7 | 13 (prime, stop) |
So 32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 = 23 × 32 × 5 × 7 × 13. The NCERT text also gives a bigger case: 123456789 = 32 × 3803 × 3607, where 3803 and 3607 are both primes.
The theorem
Theorem 1.1 (Fundamental Theorem of Arithmetic): 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.
The theorem makes two separate claims:
- Existence: every composite number can be written as a product of primes.
- Uniqueness: there is only one such product. 2 × 3 × 5 × 7 and 3 × 5 × 7 × 2 count as the same factorisation, because only the order differs.
The same fact is often stated as: the prime factorisation of a natural number is unique, except for the order of its factors. If we agree to write the primes in ascending order, p1 ≤ p2 ≤ … ≤ pn, the factorisation becomes completely fixed. Grouping equal primes gives powers of primes, as in 23 × 32 × 5 × 7 × 13.
History: an equivalent version was probably first recorded as Proposition 14 of Book IX in Euclid’s Elements; the first correct proof was given by Carl Friedrich Gauss.
2. HCF and LCM by Prime Factorisation
The HCF (highest common factor) of two or more numbers is the largest number that divides each of them. The LCM (lowest common multiple) is the smallest positive number that each of them divides. Once the numbers are written as products of prime powers, both can be read off directly. This is called the prime factorisation method.
- HCF = product of the smallest power of each common prime factor in the numbers.
- LCM = product of the greatest power of each prime factor involved in the numbers.
NCERT Example 2 (6 and 20): 6 = 21 × 31 and 20 = 22 × 51. The only common prime is 2, and its smaller power is 21, so HCF(6, 20) = 2. The primes involved are 2, 3 and 5 with greatest powers 22, 31, 51, so LCM(6, 20) = 22 × 3 × 5 = 60.
Laying out the work in a table keeps you from missing a prime. Take 336 and 54: 336 = 24 × 3 × 7 and 54 = 2 × 33.
| Prime | Power in 336 | Power in 54 | For HCF (smaller) | For LCM (greater) |
|---|---|---|---|---|
| 2 | 4 | 1 | 21 | 24 |
| 3 | 1 | 3 | 31 | 33 |
| 7 | 1 | 0 | not common, left out | 71 |
HCF = 2 × 3 = 6 and LCM = 24 × 33 × 7 = 16 × 27 × 7 = 3024.
Three or more numbers work the same way; see Example 6 for 6, 72 and 120.
Two quick checks catch most errors: the HCF must divide every number, and every number must divide the LCM. Also, HCF ≤ smallest number and LCM ≥ largest number.
Coprime numbers. Two numbers are coprime when their HCF is 1, so they share no prime factor. For 17, 23 and 29, all primes, the HCF is 1 and the LCM is simply the product 17 × 23 × 29 = 11339.
3. The Relation HCF × LCM = Product of Two Numbers
For 6 and 20: HCF × LCM = 2 × 60 = 120, and 6 × 20 = 120. This is no accident. For any two positive integers a and b,
HCF(a, b) × LCM(a, b) = a × b
Why it works. For each prime, the HCF takes the smaller of the two powers and the LCM takes the greater. Together they use both powers, exactly as the product a × b does.
Using the relation. Once the HCF is known, the LCM follows without a second factorisation:
LCM(a, b) = (a × b) / HCF(a, b)
NCERT Example 3: 96 = 25 × 3 and 404 = 22 × 101, so HCF(96, 404) = 22 = 4 and LCM = (96 × 404)/4 = 96 × 101 = 9696.
It does not hold in general for three numbers. For 6, 72 and 120, the product is 6 × 72 × 120 = 51840, while HCF × LCM = 6 × 360 = 2160. The two are unequal. It can happen to hold for some triples (for 8, 9 and 25, HCF = 1 and LCM = 1800 = 8 × 9 × 25), so it is never a rule you can use for three numbers.
What does hold for three numbers. NCERT’s note to the reader gives these results for positive integers p, q and r:
- LCM(p, q, r) = p · q · r · HCF(p, q, r) / [HCF(p, q) · HCF(q, r) · HCF(p, r)]
- HCF(p, q, r) = p · q · r · LCM(p, q, r) / [LCM(p, q) · LCM(q, r) · LCM(p, r)]
Question 15 in the board questions below checks the LCM formula on 144, 180 and 192.
A consequence worth remembering: the HCF of two numbers always divides their LCM. So a pair with HCF 18 and LCM 380 cannot exist, because 380 ÷ 18 is not a whole number.
4. Applications of Unique Factorisation
The uniqueness part of the theorem turns many “is it possible” questions into a quick check of which primes are present.
(a) Can a power end in the digit 0?
A number ends in 0 exactly when it is divisible by 10 = 2 × 5. So its prime factorisation must contain both 2 and 5.
| Number | Prime factorisation | Contains 5? | Can end in 0? |
|---|---|---|---|
| 4n | 22n | No | Never |
| 6n | 2n × 3n | No | Never |
| 12n | 22n × 3n | No | Never |
| 15n | 3n × 5n | Yes, but no 2 | Never (it ends in 5) |
| 20n | 22n × 5n | Yes, with 2 | Always |
The written argument must mention uniqueness: “By the uniqueness of the Fundamental Theorem of Arithmetic, there are no other primes in the factorisation of 6n.”
(b) Showing a number is composite
Take out a common factor: 7 × 11 × 13 + 13 = 13 × 78, so it has a factor other than 1 and itself (see Example 8).
(c) Meeting again: an LCM word problem
When events repeat at fixed intervals and you want the first time they happen together again, find the LCM of the intervals. Bells, traffic lights, runners on a track and buses leaving a depot all fit this pattern.
(d) Largest equal groups: an HCF word problem
When you split several quantities into the largest possible equal groups, with nothing left over, find the HCF. Rows of students, stacks of sweets, lengths of rope cut into equal pieces and square tiles covering a floor fit this pattern.
| Key words in the question | Use |
|---|---|
| “meet again”, “ring together”, “least time”, “minimum distance”, “smallest number divisible by” | LCM |
| “maximum number of rows”, “largest size”, “greatest length”, “same number in each” | HCF |
5. Theorem 1.2: If a Prime Divides a2, It Divides a
Theorem 1.2: Let p be a prime number. If p divides a2, then p divides a, where a is a positive integer.
Proof idea (NCERT marks this proof as not from the examination point of view, but understanding it makes the irrationality proofs clear). Write a = p1 p2 … pn, a product of primes, not necessarily distinct. Then a2 = p12 p22 … pn2. If p divides a2, then p is one of the prime factors of a2. By uniqueness, the only prime factors of a2 are p1, p2, …, pn. So p is one of them, and therefore p divides a.
Illustration. 225 = 152 = 32 × 52. Both 3 and 5 divide 15: squaring repeats the same primes and brings in no new one.
The theorem needs p to be prime. 4 divides 36 = 62, and 4 does not divide 6. Because 4 is composite, the theorem says nothing about it. In proofs, always say “since 3 is prime” (or 2, or 5) when you use Theorem 1.2.
6. Proving that √2, √3 and √5 Are Irrational
A number s is called irrational if it cannot be written in the form p/q, where p and q are integers and q ≠ 0. Familiar examples are √2, √3, √15, π and 0.10110111011110…
In Class IX you used these numbers without a proof that they are irrational. This chapter proves it using proof by contradiction: assume the opposite of what you want, reason correctly, reach something impossible, and conclude that the assumption was wrong.
The proof that √2 is irrational (Theorem 1.3), step by step
- Assume the opposite. Suppose √2 is rational. Then √2 = r/s for integers r and s, s ≠ 0.
- Reduce to lowest terms. If r and s have a common factor other than 1, divide it out to get √2 = a/b, where a and b are coprime.
- Clear the root. b√2 = a. Squaring, 2b2 = a2. So 2 divides a2.
- Use Theorem 1.2. Since 2 is prime and 2 divides a2, 2 divides a. Write a = 2c for some integer c.
- Substitute back. 2b2 = 4c2, so b2 = 2c2. So 2 divides b2, and by Theorem 1.2, 2 divides b.
- Contradiction. Now 2 divides both a and b, which contradicts the fact that a and b are coprime.
- Conclude. The contradiction arose from the incorrect assumption that √2 is rational. So √2 is irrational.
The same template works for any prime p. Replace 2 by 3 and you get NCERT Example 5 (√3 is irrational): 3b2 = a2, so 3 divides a, a = 3c, 3b2 = 9c2, b2 = 3c2, so 3 divides b. Replace 2 by 5 and you get Exercise 1.2 Question 1. The chapter states the general result: √p is irrational whenever p is a prime.
| Step | For √p (p prime) |
|---|---|
| Assume | √p = a/b, a and b coprime, b ≠ 0 |
| Square | p b2 = a2 |
| Theorem 1.2 | p divides a, so a = pc |
| Substitute | p b2 = p2c2, so b2 = p c2 |
| Theorem 1.2 again | p divides b |
| Contradiction | p is a common factor of a and b |
7. Sums, Differences, Products and Quotients with Irrationals
The chapter uses these facts from Class IX:
- The sum or difference of a rational number and an irrational number is irrational.
- The product and quotient of a non-zero rational number and an irrational number is irrational (0 × √2 = 0 is rational).
To prove a particular case such as 5 − √3 or 3√2 irrational, you again argue by contradiction, and this time the contradiction is with a number already known to be irrational.
- Assume the number is rational, equal to a/b with a, b coprime and b ≠ 0.
- Rearrange to get the known irrational alone on one side.
- The other side is built from integers only, so it is rational.
- So the known irrational would be rational: a contradiction.
NCERT Example 6: if 5 − √3 = a/b, then √3 = 5 − a/b = (5b − a)/b, which is rational since a and b are integers. This contradicts the fact that √3 is irrational.
NCERT Example 7: if 3√2 = a/b, then √2 = a/(3b), which is rational since 3, a and b are integers. This contradicts the fact that √2 is irrational.
A caution. Two irrationals can combine to give a rational. √2 + (−√2) = 0 and √2 × √2 = 2. The rules above cover a rational mixed with an irrational, and nothing more.
Formula and Theorem Sheet
| Result | Statement | Remember |
|---|---|---|
| Theorem 1.1 (FTA) | Every composite number can be expressed as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur | Two parts: existence and uniqueness |
| HCF | Product of the smallest power of each common prime factor | Only primes common to all the numbers |
| LCM | Product of the greatest power of each prime factor involved | Every prime that appears anywhere |
| Two-number relation | HCF(a, b) × LCM(a, b) = a × b | Two positive integers only |
| Three numbers | HCF(p, q, r) × LCM(p, q, r) ≠ p × q × r in general | 6, 72, 120: 2160 vs 51840 |
| LCM of three numbers | p · q · r · HCF(p, q, r) / [HCF(p, q) · HCF(q, r) · HCF(p, r)] | From NCERT’s note to the reader |
| HCF of three numbers | p · q · r · LCM(p, q, r) / [LCM(p, q) · LCM(q, r) · LCM(p, r)] | From NCERT’s note to the reader |
| Theorem 1.2 | p prime, p divides a2 ⇒ p divides a (a a positive integer) | p must be prime |
| General result | √p is irrational for every prime p | √2, √3, √5, √7, … |
| Ends in 0 | Number divisible by 10 ⇔ factorisation contains both 2 and 5 | 4n, 6n never end in 0 |
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Important Definitions
- Prime number: a natural number greater than 1 whose only factors are 1 and the number itself.
- Composite number: a natural number greater than 1 that has at least one factor other than 1 and itself.
- Fundamental Theorem of Arithmetic: 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.
- HCF (Highest Common Factor): the greatest positive integer that divides each of the given numbers.
- LCM (Lowest Common Multiple): the smallest positive integer that is divisible by each of the given numbers.
- Coprime numbers: two integers whose only common factor is 1, that is, their HCF is 1.
- Rational number: a number that can be written as p/q, where p and q are integers and q ≠ 0.
- Irrational number: a number that cannot be written in the form p/q, where p and q are integers and q ≠ 0.
- Proof by contradiction: a method of proof in which you assume the statement is false, reason correctly to an impossible result, and conclude that the statement is true.
Solved Examples (NCERT-Based)
Example 1: Express each number as a product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429 (Exercise 1.1, Q1)
(i) 140 = 2 × 70 = 2 × 2 × 35 = 2 × 2 × 5 × 7 = 22 × 5 × 7
(ii) 156 = 2 × 78 = 2 × 2 × 39 = 2 × 2 × 3 × 13 = 22 × 3 × 13
(iii) 3825 ends in 5: 3825 = 5 × 765 = 5 × 5 × 153. The digits of 153 add to 9, so 153 = 3 × 51 = 3 × 3 × 17. So 3825 = 32 × 52 × 17.
(iv) 5005 = 5 × 1001 = 5 × 7 × 143 = 5 × 7 × 11 × 13. So 5005 = 5 × 7 × 11 × 13.
(v) 7429 is not divisible by 2, 3, 5, 7, 11 or 13. Trying 17: 7429 = 17 × 437. Then 437 = 19 × 23. So 7429 = 17 × 19 × 23.
Example 2: Consider the numbers 4n, where n is a natural number. Check whether there is any value of n for which 4n ends with the digit zero. (NCERT Example 1)
If 4n ended with the digit zero for some n, it would be divisible by 5. Then the prime factorisation of 4n would contain the prime 5.
But 4n = (22)n = 22n, so the only prime in its factorisation is 2.
By the uniqueness of the Fundamental Theorem of Arithmetic, there are no other primes in the factorisation of 4n. So 5 does not divide 4n.
∴ There is no natural number n for which 4n ends with the digit zero.
Example 3: Check whether 6n can end with the digit 0 for any natural number n. (Exercise 1.1, Q5)
For 6n to end with 0, it must be divisible by 10 = 2 × 5, so its prime factorisation must contain 5.
6n = (2 × 3)n = 2n × 3n. The only primes are 2 and 3.
By the uniqueness of the Fundamental Theorem of Arithmetic, 5 is not a prime factor of 6n.
∴ 6n cannot end with the digit 0 for any natural number n.
Example 4: Find the LCM and HCF of 510 and 92 and verify that LCM × HCF = product of the two numbers. (Exercise 1.1, Q2(ii))
510 = 2 × 3 × 5 × 17
92 = 22 × 23
The only common prime is 2; its smaller power is 21. HCF = 2.
Greatest powers of all primes involved: 22, 3, 5, 17, 23. LCM = 4 × 3 × 5 × 17 × 23 = 23460.
Verification: LCM × HCF = 23460 × 2 = 46920. Product = 510 × 92 = 46920. The two are equal, as required.
Example 5: Find the LCM and HCF of 12, 15 and 21 by applying the prime factorisation method. (Exercise 1.1, Q3(i))
12 = 22 × 3, 15 = 3 × 5, 21 = 3 × 7
The only prime common to all three is 3, with smallest power 31. HCF = 3.
Greatest powers of all primes involved: 22, 31, 51, 71. LCM = 4 × 3 × 5 × 7 = 420.
Check: 420 ÷ 12 = 35, 420 ÷ 15 = 28, 420 ÷ 21 = 20, all whole numbers.
Example 6: Find the HCF and LCM of 6, 72 and 120, using the prime factorisation method. (NCERT Example 4)
6 = 2 × 3, 72 = 23 × 32, 120 = 23 × 3 × 5
21 and 31 are the smallest powers of the common factors 2 and 3. So HCF(6, 72, 120) = 2 × 3 = 6.
23, 32 and 51 are the greatest powers of the primes involved. So LCM(6, 72, 120) = 8 × 9 × 5 = 360.
Remark: 6 × 72 × 120 = 51840, while HCF × LCM = 6 × 360 = 2160. The product of three numbers is not equal to the product of their HCF and LCM.
Example 7: Given that HCF (306, 657) = 9, find LCM (306, 657). (Exercise 1.1, Q4)
For two positive integers, HCF × LCM = product of the numbers.
LCM(306, 657) = (306 × 657) / 9
306 ÷ 9 = 34, so LCM = 34 × 657 = 22338.
Check with factorisations: 306 = 2 × 32 × 17 and 657 = 32 × 73. HCF = 32 = 9 (as given) and LCM = 2 × 32 × 17 × 73 = 18 × 1241 = 22338.
Example 8: Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers. (Exercise 1.1, Q6)
First number: 7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × (77 + 1) = 13 × 78 = 1014.
1014 has the factor 13 besides 1 and itself. So it is composite. In fact 1014 = 2 × 3 × 132.
Second number: 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1) = 5 × (1008 + 1) = 5 × 1009 = 5045.
5045 has the factor 5 besides 1 and itself. So it is composite.
Example 9: There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point? (Exercise 1.1, Q7)
Sonia is at the starting point after 18, 36, 54, … minutes (multiples of 18). Ravi is there after 12, 24, 36, … minutes (multiples of 12). They meet at the starting point at the first common multiple, which is the LCM.
18 = 2 × 32 and 12 = 22 × 3
LCM = 22 × 32 = 36
∴ They meet again at the starting point after 36 minutes. By then Sonia has done 2 rounds and Ravi 3 rounds.
Example 10: Prove that √5 is irrational. (Exercise 1.2, Q1)
Let us assume, to the contrary, that √5 is rational.
Then we can find integers a and b (b ≠ 0) such that √5 = a/b. If a and b have a common factor other than 1, we divide by it, so we may assume a and b are coprime.
So b√5 = a. Squaring both sides, 5b2 = a2. So 5 divides a2.
Since 5 is prime, by Theorem 1.2, 5 divides a. So a = 5c for some integer c.
Substituting, 5b2 = 25c2, that is, b2 = 5c2. So 5 divides b2, and by Theorem 1.2, 5 divides b.
So a and b have at least 5 as a common factor. This contradicts the fact that a and b are coprime.
This contradiction arose from our incorrect assumption that √5 is rational. ∴ √5 is irrational.
Example 11: Prove that 3 + 2√5 is irrational. (Exercise 1.2, Q2)
Let us assume, to the contrary, that 3 + 2√5 is rational.
Then we can find coprime integers a and b (b ≠ 0) such that 3 + 2√5 = a/b.
Rearranging: 2√5 = a/b − 3 = (a − 3b)/b, so √5 = (a − 3b)/(2b).
Since a and b are integers, a − 3b and 2b are integers and 2b ≠ 0. So (a − 3b)/(2b) is rational, and hence √5 is rational.
This contradicts the fact that √5 is irrational (Example 10).
∴ 3 + 2√5 is irrational.
Example 12: Prove that the following are irrationals: (i) 1/√2 (ii) 7√5 (iii) 6 + √2 (Exercise 1.2, Q3)
(i) Assume, to the contrary, that 1/√2 is rational, so 1/√2 = a/b with a, b coprime integers and b ≠ 0. Since 1/√2 ≠ 0, a ≠ 0 as well. Taking reciprocals, √2 = b/a. As a and b are integers with a ≠ 0, b/a is rational, so √2 is rational. This contradicts the fact that √2 is irrational. So 1/√2 is irrational.
(ii) Assume, to the contrary, that 7√5 = a/b with a, b coprime integers and b ≠ 0. Then √5 = a/(7b). Since 7, a and b are integers and 7b ≠ 0, a/(7b) is rational, so √5 is rational. This contradicts the fact that √5 is irrational. So 7√5 is irrational.
(iii) Assume, to the contrary, that 6 + √2 = a/b with a, b coprime integers and b ≠ 0. Then √2 = a/b − 6 = (a − 6b)/b. Since a and b are integers, (a − 6b)/b is rational, so √2 is rational. This contradicts the fact that √2 is irrational. So 6 + √2 is irrational.
Competency-Based Questions (with answers)
For every assertion-reason item below, choose one of these four options:
- (A) Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
- (B) Both Assertion (A) and Reason (R) are true, but R is not the correct explanation of A.
- (C) Assertion (A) is true, but Reason (R) is false.
- (D) Assertion (A) is false, but Reason (R) is true.
1. Case-based: The school bells
A school has three electronic bells. The first rings every 6 minutes, the second every 8 minutes and the third every 12 minutes. All three ring together at 8:00 a.m.
(a) Write the prime factorisations of 6, 8 and 12. 6 = 2 × 3, 8 = 23, 12 = 22 × 3.
(b) When will they next ring together? LCM = 23 × 3 = 24 minutes, so at 8:24 a.m.
(c) How many times will they ring together after 8:00 a.m., up to and including 10:00 a.m.? 120 minutes ÷ 24 = 5. They ring together at 8:24, 8:48, 9:12, 9:36 and 10:00, that is, 5 times.
2. Case-based: The marching contingent
An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns.
(a) Which idea gives the maximum number of columns: HCF or LCM? The number of columns must divide both 616 and 32, and we want the largest such number, so the HCF.
(b) Find it. 616 = 23 × 7 × 11 and 32 = 25. HCF = 23 = 8 columns.
(c) How many rows will each group then have? Contingent: 616 ÷ 8 = 77 rows. Band: 32 ÷ 8 = 4 rows.
3. Case-based: The sweet shop
A sweet seller has 420 kaju barfis and 130 badam barfis. She wants to stack them so that each stack has the same number of sweets and each stack holds only one kind, using the fewest stacks so the tray area used is least.
(a) How many barfis should go in each stack? The number per stack must divide 420 and 130, and should be as large as possible. 420 = 22 × 3 × 5 × 7 and 130 = 2 × 5 × 13. HCF = 2 × 5 = 10.
(b) How many stacks are there in all? 420 ÷ 10 + 130 ÷ 10 = 42 + 13 = 55 stacks.
4. Source-based: Reading a proof
Read this student’s proof: “Suppose √3 = a/b where a and b are coprime and b ≠ 0. Then 3b2 = a2. So 3 divides a2, and hence 3 divides a. Let a = 3c. Then b2 = 3c2, so 3 divides b. So 3 is a common factor of a and b, which is a contradiction. Hence √3 is irrational.”
(a) Which theorem justifies “3 divides a2, and hence 3 divides a”? Theorem 1.2: if a prime p divides a2, then p divides a. It applies because 3 is prime.
(b) Fill in the missing working between “Let a = 3c” and “b2 = 3c2”. Substituting a = 3c in 3b2 = a2 gives 3b2 = 9c2, and dividing by 3 gives b2 = 3c2.
(c) What exactly does the contradiction contradict? The starting assumption that a and b are coprime, that is, have no common factor other than 1.
5. Assertion-Reason
Assertion (A): HCF(6, 20) × LCM(6, 20) = 120.
Reason (R): For any two positive integers a and b, HCF(a, b) × LCM(a, b) = a × b.
Answer: (A). HCF = 2, LCM = 60 and 2 × 60 = 120 = 6 × 20, so A is true. R is the general result that gives A directly.
6. Assertion-Reason
Assertion (A): 6n never ends with the digit 0 for any natural number n.
Reason (R): The prime factorisation of 6n contains only the primes 2 and 3.
Answer: (A). Ending in 0 needs the prime 5. R says the only primes are 2 and 3, and uniqueness rules out a 5, which is exactly why A holds.
7. Assertion-Reason
Assertion (A): For the numbers 6, 72 and 120, HCF × LCM = 6 × 72 × 120.
Reason (R): For any two positive integers a and b, HCF(a, b) × LCM(a, b) = a × b.
Answer: (D). HCF × LCM = 6 × 360 = 2160, while the product is 51840, so A is false. R is true, but it is guaranteed only for two numbers.
8. Assertion-Reason
Assertion (A): 7 × 11 × 13 + 13 is a composite number.
Reason (R): The prime factorisation of a natural number is unique, except for the order of its factors.
Answer: (B). A is true because the number equals 13 × 78. R is true, but A follows from taking out the common factor 13; uniqueness of factorisation is not the reason it is composite.
9. Assertion-Reason
Assertion (A): 2 + √3 is an irrational number.
Reason (R): The product of any two irrational numbers is irrational.
Answer: (C). A is true: the sum of the rational 2 and the irrational √3 is irrational. R is false: √3 × √3 = 3, which is rational.
Important Questions for Board Exams
1-Mark Questions
Q1. If a = x3y2 and b = xy3, where x and y are prime numbers, find HCF(a, b) and LCM(a, b).
Answer: HCF takes the smaller powers: x1y2 = xy2. LCM takes the greater powers: x3y3.
Q2. The HCF of two numbers is 13 and their LCM is 182. If one number is 26, find the other.
Answer: Other number = (13 × 182)/26 = 2366/26 = 91.
Q3. Find the HCF of the smallest prime number and the smallest composite number.
Answer: Smallest prime = 2, smallest composite = 4. HCF(2, 4) = 2.
Q4. The product of a non-zero rational number and an irrational number is: (a) always rational (b) always irrational (c) rational or irrational (d) one
Answer: (b) always irrational.
Q5. Is 3 × 5 × 7 + 7 prime or composite? Give a reason.
Answer: 3 × 5 × 7 + 7 = 7 × (15 + 1) = 7 × 16 = 112. It has 7 as a factor besides 1 and itself, so it is composite.
2-Mark Questions
Q6. Find the LCM and HCF of 26 and 91 and verify that LCM × HCF = product of the two numbers.
Answer: 26 = 2 × 13 and 91 = 7 × 13. HCF = 13. LCM = 2 × 7 × 13 = 182. LCM × HCF = 182 × 13 = 2366, and 26 × 91 = 2366. Verified.
Q7. Check whether 12n can end with the digit 0 for any natural number n.
Answer: 12n = (22 × 3)n = 22n × 3n. To end in 0 it must have 5 as a prime factor. By the uniqueness of the Fundamental Theorem of Arithmetic, its only primes are 2 and 3. So 12n cannot end with 0.
Q8. Find the LCM and HCF of 8, 9 and 25 by applying the prime factorisation method.
Answer: 8 = 23, 9 = 32, 25 = 52. No prime is common, so HCF = 1. LCM = 23 × 32 × 52 = 8 × 9 × 25 = 1800.
Q9. Can two numbers have 18 as their HCF and 380 as their LCM? Give a reason.
Answer: The HCF of two numbers always divides their LCM. 380 ÷ 18 = 21.11…, which is not a whole number. So no such pair exists.
3-Mark Questions
Q10. Prove that √7 is irrational.
Answer: Assume, to the contrary, that √7 is rational. Then √7 = a/b for coprime integers a and b, b ≠ 0. So 7b2 = a2, and 7 divides a2. As 7 is prime, 7 divides a (Theorem 1.2). Let a = 7c. Then 7b2 = 49c2, so b2 = 7c2. So 7 divides b2 and hence 7 divides b. Now 7 is a common factor of a and b, which contradicts that a and b are coprime. Hence √7 is irrational.
Q11. Given that √5 is irrational, prove that 2 − 3√5 is irrational.
Answer: Assume, to the contrary, that 2 − 3√5 = a/b for coprime integers a and b, b ≠ 0. Then 3√5 = 2 − a/b = (2b − a)/b, so √5 = (2b − a)/(3b). Since a and b are integers and 3b ≠ 0, the right side is rational, so √5 would be rational. This contradicts the given fact. Hence 2 − 3√5 is irrational.
Q12. Find the LCM and HCF of 336 and 54 and verify that LCM × HCF = product of the two numbers.
Answer: 336 = 24 × 3 × 7 and 54 = 2 × 33. HCF = 2 × 3 = 6. LCM = 24 × 33 × 7 = 16 × 27 × 7 = 3024. LCM × HCF = 3024 × 6 = 18144. Product = 336 × 54 = 18144. Verified.
Q13. Three friends set off walking together. Their steps measure 40 cm, 42 cm and 45 cm. What is the minimum distance each should walk so that each covers it in a whole number of steps?
Answer: The distance must be a common multiple of all three step lengths, and the least one is the LCM. 40 = 23 × 5, 42 = 2 × 3 × 7, 45 = 32 × 5. LCM = 23 × 32 × 5 × 7 = 2520 cm = 25 m 20 cm.
5-Mark Questions
Q14. Prove that √2 is irrational. Hence show that 3 + 2√2 is irrational.
Answer: Part 1. Assume, to the contrary, that √2 is rational. Then √2 = a/b, where a and b are coprime integers and b ≠ 0. So b√2 = a, and squaring, 2b2 = a2. So 2 divides a2, and since 2 is prime, 2 divides a (Theorem 1.2). Write a = 2c. Then 2b2 = 4c2, so b2 = 2c2. So 2 divides b2, and hence 2 divides b. So 2 is a common factor of a and b, contradicting that they are coprime. Hence √2 is irrational.
Part 2. Assume, to the contrary, that 3 + 2√2 = p/q for coprime integers p and q, q ≠ 0. Then 2√2 = (p − 3q)/q, so √2 = (p − 3q)/(2q). Since p and q are integers and 2q ≠ 0, this is rational, so √2 would be rational. This contradicts Part 1. Hence 3 + 2√2 is irrational.
Q15. (i) Find the HCF and LCM of 144, 180 and 192 by the prime factorisation method. (ii) Show that the product of the three numbers is not equal to the product of their HCF and LCM. (iii) Verify your LCM using LCM(p, q, r) = p · q · r · HCF(p, q, r) / [HCF(p, q) · HCF(q, r) · HCF(p, r)].
Answer: (i) 144 = 24 × 32, 180 = 22 × 32 × 5, 192 = 26 × 3. HCF = 22 × 3 = 12. LCM = 26 × 32 × 5 = 64 × 45 = 2880.
(ii) Product = 144 × 180 × 192 = 25920 × 192 = 4976640. HCF × LCM = 12 × 2880 = 34560. These are unequal.
(iii) HCF(144, 180) = 22 × 32 = 36, HCF(180, 192) = 22 × 3 = 12, HCF(144, 192) = 24 × 3 = 48. Denominator = 36 × 12 × 48 = 20736. Numerator = 4976640 × 12 = 59719680. LCM = 59719680 ÷ 20736 = 2880, which matches part (i).
Common Mistakes and Examiner Tips
- Stopping a factor tree at a composite number. Writing 156 = 4 × 39 is not a prime factorisation. Keep splitting until every factor is prime: 22 × 3 × 13.
- Taking the greater power for HCF. HCF uses the smallest power of each common prime; LCM uses the greatest power of every prime. Say “smallest, common” and “greatest, all” as you work.
- Putting a non-common prime into the HCF. In HCF(336, 54), the prime 7 appears only in 336, so it is left out of the HCF but included in the LCM.
- Using HCF × LCM = product for three numbers. The rule is for two positive integers only. For 6, 72 and 120, 2160 ≠ 51840.
- Proving 4n or 6n cannot end in 0 by trying a few values of n. Checking n = 1, 2, 3 proves nothing about all n. Write the factorisation, show 5 is absent and quote the uniqueness of the Fundamental Theorem of Arithmetic.
- Leaving out “a and b are coprime” in an irrationality proof. Without it, finding a common factor is no contradiction. State it on the line where you write √p = a/b.
- Forgetting “b ≠ 0”. A rational number needs a non-zero denominator. Write it every time you assume a number equals a/b.
- Using Theorem 1.2 without saying the divisor is prime. Write “since 5 is prime and 5 divides a2, 5 divides a”. The theorem fails for composites: 4 divides 36 but not 6.
- Not naming the contradiction. End with a sentence that says what was contradicted (“this contradicts that a and b are coprime” or “this contradicts that √3 is irrational”) and then draw the conclusion.
- Claiming the answer “is rational” without a reason. In 3 + 2√5, write why (a − 3b)/(2b) is rational: a and b are integers, so the numerator and denominator are integers and 2b ≠ 0.
- Mixing up HCF and LCM in word problems. “Meet again” or “ring together” needs the LCM; “maximum number of columns” or “largest equal groups” needs the HCF. Answer with units: minutes, columns, cm.
Quick Revision Points
- A prime has exactly two factors, 1 and itself. 2 is the only even prime. 1 is neither prime nor composite.
- Fundamental Theorem of Arithmetic: every composite number can be expressed as a product of primes, uniquely apart from the order of the factors.
- 32760 = 23 × 32 × 5 × 7 × 13.
- HCF = product of the smallest power of each common prime factor.
- LCM = product of the greatest power of each prime factor involved.
- HCF(a, b) × LCM(a, b) = a × b for two positive integers.
- For three numbers, HCF × LCM is generally not equal to the product.
- HCF always divides LCM. HCF ≤ smallest number; LCM ≥ largest number.
- A number ends in 0 only if its factorisation contains both 2 and 5. So 4n, 6n and 12n never end in 0.
- Taking out a common factor shows a number is composite: 7 × 11 × 13 + 13 = 13 × 78.
- “Meet again” problems use LCM; “largest equal groups” problems use HCF.
- Theorem 1.2: if a prime p divides a2, then p divides a.
- Irrational: cannot be written as p/q with p, q integers and q ≠ 0.
- Proof by contradiction: assume rational, a/b coprime, b ≠ 0, reach a common factor, conclude irrational.
- √p is irrational for every prime p: √2, √3, √5, √7, …
- Rational ± irrational is irrational.
- Non-zero rational × irrational and non-zero rational ÷ irrational are irrational.
- Two irrationals can give a rational: √2 × √2 = 2.
- For 5 − √3, 3√2 and similar numbers, isolate the known irrational and show it would be rational.
Weightage in Board Exams
Check the current CBSE course structure for the marks given to this chapter. The chapter is assessed in a very consistent way.
| Question type | What is usually asked |
|---|---|
| MCQ and Assertion-Reason | HCF or LCM of expressions like x3y2 and xy3; a missing number from HCF and LCM; whether a number is rational or irrational; whether a power can end in 0 |
| Short answers | Prime factorisation; HCF and LCM with verification of HCF × LCM = product; showing a number is composite; 4n or 6n type reasoning |
| Proof questions | √2, √3, √5 or √p irrational; then a number such as 3 + 2√5 or 5 − √3 irrational |
| Case-based | Bells, traffic lights, runners or buses meeting again (LCM); columns, stacks or tiles (HCF) |
In the irrationality proof, write every line: the assumption with “coprime, b ≠ 0”, the squaring, each use of Theorem 1.2 with “since p is prime”, the named contradiction and the conclusion. Work through all of Exercises 1.1 and 1.2 and the NCERT examples first.
Class 10 Maths · Chapter 1 – swipe through all 10 cards to understand the whole chapter.
Primes and Composites
Primes are the building blocks: every other number greater than 1 is made from them.
1 is neither prime nor composite; 2 is the only even prime.
- Prime: exactly two factors, 1 and itself (2, 3, 5, 7, 11, …).
- Composite: more than two factors (4, 6, 8, 9, …).
- There are infinitely many primes.
Fundamental Theorem of Arithmetic
Every composite number is a product of primes, and that product is unique apart from order.
First correct proof by Gauss; an equivalent version is in Euclid’s Elements, Book IX.
- Two parts: the factorisation exists, and it is unique.
- 2 × 3 × 5 × 7 and 3 × 5 × 7 × 2 count as the same factorisation.
- Uniqueness tells you which primes can never appear.
Prime Factorisation Method
Read the HCF and LCM straight off the prime powers.
Check: HCF divides every number, every number divides the LCM.
- 6 = 2 × 3, 20 = 22 × 5 gives HCF 2, LCM 60.
- 336 = 24 × 3 × 7, 54 = 2 × 33 gives HCF 6, LCM 3024.
- A prime in only one number goes into the LCM, never the HCF.
HCF × LCM = Product
For two positive integers, HCF times LCM equals the product of the numbers.
Not true in general for three numbers: for 6, 72, 120, 2160 ≠ 51840.
- LCM(96, 404) = 96 × 404 ÷ 4 = 9696.
- Given HCF(306, 657) = 9, LCM = 22338.
- HCF must divide LCM, so HCF 18 with LCM 380 is impossible.
Can It End in Zero?
A number ends in 0 only if its prime factorisation has both 2 and 5.
Always quote the uniqueness of the Fundamental Theorem of Arithmetic.
- 4ⁿ = 22ⁿ has no 5, so it never ends in 0.
- 6ⁿ = 2ⁿ × 3ⁿ has no 5, so it never ends in 0.
- Trying a few values of n is not a proof.
LCM or HCF?
Meeting again means LCM; largest equal groups means HCF.
Always write the unit: minutes, columns, cm.
- Sonia 18 min, Ravi 12 min per round: they meet after LCM = 36 min.
- 616 soldiers and 32 band members: HCF = 8 columns.
- Steps of 40, 42 and 45 cm: LCM = 2520 cm.
Take Out the Common Factor
If every term shares a factor, factor it out to show the number is composite.
Write the factored form, then name a factor other than 1 and the number itself.
- 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 × 1009 = 5045.
- 3 × 5 × 7 + 7 = 7 × 16 = 112.
- A factor other than 1 and itself means composite.
Prime Divides a2, So Divides a
If a prime p divides a2, then p divides a, for a positive integer a.
NCERT marks its proof as not from the examination point of view.
- Squaring repeats the same primes and adds no new ones.
- 225 = 32 × 52: both 3 and 5 divide 15.
- Fails for composites: 4 divides 36 but not 6.
√p Is Irrational
Assume √p = a/b in lowest terms, then show p divides both a and b.
State ‘a and b are coprime, b ≠ 0’ or the contradiction has nothing to contradict.
- Works for √2, √3, √5, √7 and every prime p.
- Use Theorem 1.2 twice, saying ‘since p is prime’ each time.
- End by naming the contradiction, then conclude.
Rational With Irrational
Rational ± irrational is irrational; non-zero rational × or ÷ irrational is irrational.
Two irrationals can give a rational: √2 × √2 = 2.
- Isolate the known irrational on one side.
- 3 + 2√5 = a/b gives √5 = (a − 3b)/(2b), rational: contradiction.
- 3√2 = a/b gives √2 = a/(3b), rational: contradiction.
📝 Practice Real Numbers - 10 board questions
CBSE previous-year and competency-based · with answers & explanations
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Why A: 2 has exactly two factors, 1 and 2 itself, which is precisely the definition of a prime. It is the smallest prime and the only even prime.
Why not B: a composite number has more than two factors; 2 has only two.
Why not C: prime (exactly two factors) and composite (more than two factors) cannot both hold for the same number.
Why not D: the number that is neither prime nor composite is 1, which has only one factor.
Remember: 1 is neither, 2 is the only even prime, 4 is the smallest composite.
Why B: A is true: 36m² = 18m × 2m, so 18m divides 36m², and the HCF of a number and one of its multiples is the smaller number, 18m. R is true: a common factor must divide the smaller number, so it cannot exceed it. But R only gives an upper limit; it does not show why the HCF equals 18m, which comes from 18m dividing 36m².
Why not A: R is a true general bound, but it does not prove the HCF reaches that bound here.
Why not C: R is a true property of every HCF.
Why not D: A is true, because 18m is a factor of 36m².
Remember: if a divides b, HCF(a, b) = a; HCF ≤ smaller number is only a bound.
Why B: 6¹ = 6, 6² = 36, 6³ = 216; the units digit is always 6 because 6 × 6 = 36 ends in 6 again, so multiplying by 6 each time keeps the units digit at 6. Also 6ⁿ = 2ⁿ × 3ⁿ has no factor 5, so it can never end in 0.
Why not A: ending in 0 needs both 2 and 5 in the prime factorisation, and 6ⁿ = 2ⁿ × 3ⁿ has no 5.
Why not C: 3 is a prime factor of 6, but a prime factor is not the same as the last digit; 6ⁿ is even, so it cannot end in 3.
Why not D: 6ⁿ is even, but its units digit is fixed at 6, never 2.
Remember: numbers ending in 0, 1, 5 or 6 keep that last digit in every power.
Why C: A is true: if √3 + √5 = r were rational, then √5 = r − √3, and squaring gives 5 = r² − 2r√3 + 3, so √3 = (r² − 2)/(2r) would be rational, a contradiction. R is false: √2 and −√2 are both irrational, yet their sum is 0, which is rational.
Why not A: R is a false general rule, so it cannot be the correct explanation.
Why not B: this still claims R is true, but one counterexample (√2 + (−√2) = 0) breaks it.
Why not D: A is true; √3 + √5 is irrational, as the contradiction above shows.
Remember: irrational + irrational can be rational (√2 + (−√2) = 0), so each sum needs its own proof.
Why A: Step 1 (prime factorisation): 960 = 2⁶ × 3 × 5 and 432 = 2⁴ × 3³. Step 2 (HCF = smallest power of each common prime): the common primes are 2 and 3, with smallest powers 2⁴ and 3¹, so HCF = 16 × 3 = 48.
Why not B: 54 = 2 × 3³ uses the larger power of 3, and 54 does not even divide 960.
Why not C: 72 = 2³ × 3² does not divide 960 (960 ÷ 72 is not a whole number), so it cannot be a common factor.
Why not D: 36 = 2² × 3² does not divide 960, and it misses two of the four common 2s.
Remember: HCF takes each common prime at its SMALLEST power; then check that the answer divides both numbers.
Why A: Step 1 (divide out the smallest prime): 4004 = 2 × 2002 = 2² × 1001. Step 2 (continue until every factor is prime): 1001 = 7 × 11 × 13, so 4004 = 2² × 7¹ × 11¹ × 13¹. Step 3 (add the exponents): 2 + 1 + 1 + 1 = 5.
Why not B: 4 counts the number of distinct primes (2, 7, 11, 13) instead of adding their exponents.
Why not C: 3 comes from stopping at 4004 = 2² × 1001 and treating 1001 as prime.
Why not D: 2 is only the exponent of the prime 2; the primes 7, 11 and 13 also carry exponent 1 each.
Remember: 1001 = 7 × 11 × 13, and every prime written without a power has exponent 1.
Why C: (−1)⁸ = 1 because 8 is even, so the equation becomes (−1)ⁿ = −1. A power of −1 equals −1 exactly when the exponent is odd, so n must be odd.
Why not A: for an even positive integer such as n = 2, (−1)² + 1 = 2, not 0.
Why not B: the sign depends on whether n is odd or even, not on whether n is negative; n = −2 gives (−1)⁻² = 1 and the sum is 2.
Why not D: an even n makes (−1)ⁿ = 1, and 1 + 1 = 2.
Remember: (−1)ᵉᵛᵉⁿ = 1, (−1)ᵒᵈᵈ = −1.
Why D: √3 ≈ 1.732 and √5 ≈ 2.236. 1.857142 is a terminating decimal, so it is rational, and 1.732 < 1.857142 < 2.236.
Why not A: 1.4142… is less than √3 ≈ 1.732, so it is not between the two roots, and its non-repeating digits suggest it is not rational either.
Why not B: 2.3262626… repeats, so it is rational, but it is larger than √5 ≈ 2.236.
Why not C: π ≈ 3.14 is irrational and is also larger than √5.
Remember: terminating or repeating decimals are rational; always check both conditions, rational AND inside the interval.
Why A: Step 1 (HCF × LCM = a × b): LCM = 2520 × 6600 ÷ 40 = 2520 × 165 = 415800. Step 2 (solve for k): 252 × k = 415800, so k = 415800 ÷ 252 = 1650 (2520 = 252 × 10, so k = 10 × 165).
Why not B: 1600 gives 252 × 1600 = 403200, which is not 415800.
Why not C: 165 is 6600 ÷ 40; it forgets that 2520 is 10 × 252, so the factor 10 is lost.
Why not D: 1625 gives 252 × 1625 = 409500, not 415800.
Remember: divide by the HCF first, then match the given factor: 2520 = 252 × 10.
Why B: Step 1 (HCF × LCM = a × b): LCM = 65 × 104 ÷ 13 = 5 × 104 = 520. Step 2 (solve 40x = LCM): x = 520 ÷ 40 = 13.
Why not A: 5 is 65 ÷ 13, one of the co-factors, not the value of x.
Why not C: 40 is the coefficient in 40x; x = 40 would make the LCM 1600, but the LCM is 65 × 104 ÷ 13 = 520.
Why not D: 8 is 104 ÷ 13; 40 × 8 = 320, which is not a multiple of 65.
Remember: LCM = (a × b) ÷ HCF; here 65 = 13 × 5 and 104 = 13 × 8, so LCM = 13 × 5 × 8 = 520.
Chapter Navigation
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Frequently Asked Questions
It says every composite number can be expressed as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur. For example, 32760 = 2³ × 3² × 5 × 7 × 13, and no other set of primes multiplies to 32760. The uniqueness part is what lets you argue that a prime such as 5 can never appear in 4ⁿ or 6ⁿ.
Write each number as a product of prime powers. The HCF is the product of the smallest power of each prime common to all the numbers. The LCM is the product of the greatest power of every prime that appears in any of the numbers. For 6 = 2 × 3 and 20 = 2² × 5, HCF = 2 and LCM = 2² × 3 × 5 = 60.
Not in general. HCF(a, b) × LCM(a, b) = a × b is guaranteed only for two positive integers. For 6, 72 and 120, HCF × LCM = 6 × 360 = 2160, while the product is 51840. For three numbers NCERT gives a different result: LCM(p, q, r) = pqr × HCF(p, q, r) divided by HCF(p, q) × HCF(q, r) × HCF(p, r).
A number ends in 0 only if it is divisible by 10, so its prime factorisation must contain 5. But 6ⁿ = 2ⁿ × 3ⁿ, whose only primes are 2 and 3. By the uniqueness of the Fundamental Theorem of Arithmetic, no other prime can appear, so 5 is absent and 6ⁿ cannot end in 0 for any natural number n.
Theorem 1.2 says that if a prime p divides a², then p divides a, where a is a positive integer. It is the key step in proving √2, √3, √5 or √p irrational: from pb² = a² you get p divides a, write a = pc, and then show p divides b too. The divisor must be prime; 4 divides 36 but does not divide 6.
Assume √2 = a/b with a and b coprime integers and b ≠ 0. Squaring gives 2b² = a², so 2 divides a², and by Theorem 1.2 it divides a. Put a = 2c to get b² = 2c², so 2 divides b as well. Then 2 is a common factor of a and b, which contradicts coprimality. So √2 is irrational.
Assume 3 + 2√5 = a/b with a, b coprime integers and b ≠ 0. Rearranging gives √5 = (a − 3b)/(2b). Since a and b are integers and 2b ≠ 0, the right side is rational, so √5 would be rational. That contradicts the fact that √5 is irrational, so 3 + 2√5 must be irrational.
Use the LCM when events repeat and you need the first time they happen together again, such as bells ringing, runners meeting at the start of a track, or the minimum distance covered in whole steps. Use the HCF when you split quantities into the largest equal groups with nothing left over, such as the maximum number of columns or the largest stack size.
