REAL NUMBERS
1. Important Concepts/ Result
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.
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.
Theorem 1.2: Let p be a prime number. If p divides a2, then p divides a, whereas is a positive integer.
Proof: Let the prime factorisation of 𝑎 be as follows:
𝑎 = 𝑝1 , 𝑝2 , . . . , 𝑝𝑛 , where 𝑝1 , 𝑝2 , . . . , 𝑝𝑛 are primes, not necessarily distinct. Therefore,
𝑎2 = ( 𝑝1𝑝2 . . . 𝑝𝑛)( 𝑝1𝑝2 . . . 𝑝𝑛) = 𝑝2 1𝑝2 2 . . . 𝑝2 𝑛. now, we are given that p divides a2.
Therefore, from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a2.
However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only
prime factors of a2 are 𝑝1 , 𝑝2 , . . . , 𝑝𝑛 . So p is one of 𝑝1 , 𝑝2 , . . . , 𝑝𝑛 .
Now, since 𝑎 = 𝑝1 , 𝑝2 , . . . , 𝑝𝑛 , and 𝑝 divides a.
We are now ready to give a proof that 2 is irrational. The proof is based on a technique called „proof by
contradiction‟.
Theorem 1.3: 2 is irrational.
Proof: Let us assume, to the contrary, that 2 is rational.
r
So, we can find integers r and s (≠ 0) such that 2 = s . Suppose r and s have a common factor other than 1.
a
Then, we divide by the common factor to get 2 , where a and b are co-prime.
b
So, b 2 = a.
Squaring on both sides and rearranging, we get 2b2 = a2 . Therefore, 2 divides a2.
Now, by Theorem 1.3, it follows that 2 divides a.
So, we can write a = 2c for some integer c.
Substituting for a, we get 2b2 = 4c2 , that is, b2 = 2c2.
This means that 2 divides b2, and so 2 divides b (again using Theorem 1.3 with p = 2). Therefore, a and b
have at least 2 as a common factor.
But this contradicts the fact that a and b have no common factors other than 1. This contradiction has arisen
because of our incorrect assumption that 2is rational. So, we conclude that 2 is irrational.
2. Some Illustrations/ Examples
i) MCQs
1. What is the HCF of the least prime number and the least composite number
(a) 1 (b) 2 (c) 3 (d) 4 Solution: (b) Least prime number = 2,
Least composite number = 4, HCF = 2
2. If two positive integers p and q are written as p = x5y2, q = x3y3, where x, y are prime no‟s then HCF (p, q)
is : (a) xy (b) x2y2 (c) x3y2 (d) x5y3
5 2 3 3
Solution: (c) p = x y , q=x y, HCF = x3y2
3. The values of r and s in the given figure is
4
3
s
r 7
(a) r = 10, s = 14 (b) r = 21, s = 84 (c) r = 21, s = 25 (d) r = 10, s = 40
Solution: (b) r = 3 x 7 = 21, s = 4 x r = 4 x 21 = 84,
2
, 4. The sum of exponents of prime factors in the prime factorisation of 196 is
(a) 3 (b) 4 (c) 5 (d) 2
Solution: (b) 4, 196 = 22 x 72, Sum = 2 + 2 = 4
5. The prime factorisation of 96 is:
(a) 25 × 3 (b) 26 (c) 24 × 3 (d) 24 × 32
Solution: (a) 25 × 3, The prime factorisation of 96 is: 96 = 2 × 2 × 2 × 2 × 2 × 3 = 2 5 × 3
ii) Short answer type questions
1. Show that 6- 2 is irrational
𝑎
Solution: Let 6- 2 is rational. Then, 6- 2 = 𝑏
(where a, b are co-prime integers and b ≠ 0)
𝑎 6𝑏−𝑎
6- 𝑏
= 2=> 𝑏
= 2
𝑎
Since a and b are integers, we get 6 - 𝑏
is rational and so 2 is rational.
But this contradicts the fact that 2is irrational. Hence out assumption 6- 2 is rational is wrong. So 6- 2is
irrational.
2. Find the HCF and LCM of 12, 14 and 16 using prime factorisation method.
Solution: 12 = 2 x 2 x 3 = 22 x 31, 14 = 2 x 7 = 21 x 71, 16 = 2 x 2 x 2 x 2 = 24
H.C.F (12, 14, 16) = 21 = 2, LCM (12, 14, 16) = 24 x 31 x 71 = 16 x 21= 336
3. If the HCF of 65 and 117 is expressible in the form 65m – 117, then the value of m is
Solution: 117 > 65, 117 = 65 × 1 + 52, 65 = 52 × 1 + 13, 52 = 13 × 4 + 0, HCF(65,117) = 13
According to the given, 65𝑚 – 117 = 13, 65m = 117 + 13, 65m = 130, m = 130/65 = 2
3. Questions for Practice:
i) MCQs
1. The LCM of two numbers is 2079 and their HCF is 27. If one of the number is 297.The other number is
(a) 190 (b) 189 (c) 163 (d) 199
2. The values of x and y in the given below figure are:
(a) 𝑥 = 25 , 𝑦 = 75 (b) 𝑥 = 20 , 𝑦 = 80 (c) 𝑥 = 21 , 𝑦 = 84 (d) 𝑥 = 23 , 𝑦 = 92
3. 5050 as product of its prime factors is.
(a) 2 × 52 × 101 (b) 2 × 5 × 5 x 5 × 101 (c)2 × 53 × 100 (d) 2 × 52
4. The expression of 2658 as a product of its prime factors.
(a) 2 × 3 × 443 (b) 2 × 4 × 443 (c)2 × 3 × 333 (d) 2 × 2 × 443
5. The ratio between the LCM and HCF of 5, 15 and 20.
(a) 1:12 (b) 12 : 11 (c)14 : 1 (d) 12 : 1
6. The HCF of two numbers 𝑎 𝑎𝑛𝑑 𝑏 𝑖𝑠 5 and their 𝐿𝐶𝑀 𝑖𝑠 200. The product 𝑎𝑏 is:
(a)1001 (b) 1000 (c)100 (d) 2000
7. The product of two numbers is 1050 and their HCF is 25. Their LCM is:
(a) 24 (b) 42 (c) 44 (d) 40
8. The LCM of two numbers is 182 and their HCF is 13. If one of the numbers is 26, the other number is.
(a)84 (b) 90 (c) 81 (d) 91
9. The least number that is divisible by all the numbers from 1 to 5 is:
(a) 70 (b) 60 (c) 80 (d) 90
10. ASSERTION: 5 is an example of a rational number.
REASON: The square root of all positive integers is irrational numbers.
(a) Both assertion (A) and reason (R) are true and assertion reason R is the correct explanation of
assertion A.
(b) Both assertion A and reason R are true but reason R is not the correct explanation of assertion A
(c) Assertion A is true but reason R is false. (d) Assertion A is false but reason R is true.
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1. Important Concepts/ Result
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.
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.
Theorem 1.2: Let p be a prime number. If p divides a2, then p divides a, whereas is a positive integer.
Proof: Let the prime factorisation of 𝑎 be as follows:
𝑎 = 𝑝1 , 𝑝2 , . . . , 𝑝𝑛 , where 𝑝1 , 𝑝2 , . . . , 𝑝𝑛 are primes, not necessarily distinct. Therefore,
𝑎2 = ( 𝑝1𝑝2 . . . 𝑝𝑛)( 𝑝1𝑝2 . . . 𝑝𝑛) = 𝑝2 1𝑝2 2 . . . 𝑝2 𝑛. now, we are given that p divides a2.
Therefore, from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a2.
However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only
prime factors of a2 are 𝑝1 , 𝑝2 , . . . , 𝑝𝑛 . So p is one of 𝑝1 , 𝑝2 , . . . , 𝑝𝑛 .
Now, since 𝑎 = 𝑝1 , 𝑝2 , . . . , 𝑝𝑛 , and 𝑝 divides a.
We are now ready to give a proof that 2 is irrational. The proof is based on a technique called „proof by
contradiction‟.
Theorem 1.3: 2 is irrational.
Proof: Let us assume, to the contrary, that 2 is rational.
r
So, we can find integers r and s (≠ 0) such that 2 = s . Suppose r and s have a common factor other than 1.
a
Then, we divide by the common factor to get 2 , where a and b are co-prime.
b
So, b 2 = a.
Squaring on both sides and rearranging, we get 2b2 = a2 . Therefore, 2 divides a2.
Now, by Theorem 1.3, it follows that 2 divides a.
So, we can write a = 2c for some integer c.
Substituting for a, we get 2b2 = 4c2 , that is, b2 = 2c2.
This means that 2 divides b2, and so 2 divides b (again using Theorem 1.3 with p = 2). Therefore, a and b
have at least 2 as a common factor.
But this contradicts the fact that a and b have no common factors other than 1. This contradiction has arisen
because of our incorrect assumption that 2is rational. So, we conclude that 2 is irrational.
2. Some Illustrations/ Examples
i) MCQs
1. What is the HCF of the least prime number and the least composite number
(a) 1 (b) 2 (c) 3 (d) 4 Solution: (b) Least prime number = 2,
Least composite number = 4, HCF = 2
2. If two positive integers p and q are written as p = x5y2, q = x3y3, where x, y are prime no‟s then HCF (p, q)
is : (a) xy (b) x2y2 (c) x3y2 (d) x5y3
5 2 3 3
Solution: (c) p = x y , q=x y, HCF = x3y2
3. The values of r and s in the given figure is
4
3
s
r 7
(a) r = 10, s = 14 (b) r = 21, s = 84 (c) r = 21, s = 25 (d) r = 10, s = 40
Solution: (b) r = 3 x 7 = 21, s = 4 x r = 4 x 21 = 84,
2
, 4. The sum of exponents of prime factors in the prime factorisation of 196 is
(a) 3 (b) 4 (c) 5 (d) 2
Solution: (b) 4, 196 = 22 x 72, Sum = 2 + 2 = 4
5. The prime factorisation of 96 is:
(a) 25 × 3 (b) 26 (c) 24 × 3 (d) 24 × 32
Solution: (a) 25 × 3, The prime factorisation of 96 is: 96 = 2 × 2 × 2 × 2 × 2 × 3 = 2 5 × 3
ii) Short answer type questions
1. Show that 6- 2 is irrational
𝑎
Solution: Let 6- 2 is rational. Then, 6- 2 = 𝑏
(where a, b are co-prime integers and b ≠ 0)
𝑎 6𝑏−𝑎
6- 𝑏
= 2=> 𝑏
= 2
𝑎
Since a and b are integers, we get 6 - 𝑏
is rational and so 2 is rational.
But this contradicts the fact that 2is irrational. Hence out assumption 6- 2 is rational is wrong. So 6- 2is
irrational.
2. Find the HCF and LCM of 12, 14 and 16 using prime factorisation method.
Solution: 12 = 2 x 2 x 3 = 22 x 31, 14 = 2 x 7 = 21 x 71, 16 = 2 x 2 x 2 x 2 = 24
H.C.F (12, 14, 16) = 21 = 2, LCM (12, 14, 16) = 24 x 31 x 71 = 16 x 21= 336
3. If the HCF of 65 and 117 is expressible in the form 65m – 117, then the value of m is
Solution: 117 > 65, 117 = 65 × 1 + 52, 65 = 52 × 1 + 13, 52 = 13 × 4 + 0, HCF(65,117) = 13
According to the given, 65𝑚 – 117 = 13, 65m = 117 + 13, 65m = 130, m = 130/65 = 2
3. Questions for Practice:
i) MCQs
1. The LCM of two numbers is 2079 and their HCF is 27. If one of the number is 297.The other number is
(a) 190 (b) 189 (c) 163 (d) 199
2. The values of x and y in the given below figure are:
(a) 𝑥 = 25 , 𝑦 = 75 (b) 𝑥 = 20 , 𝑦 = 80 (c) 𝑥 = 21 , 𝑦 = 84 (d) 𝑥 = 23 , 𝑦 = 92
3. 5050 as product of its prime factors is.
(a) 2 × 52 × 101 (b) 2 × 5 × 5 x 5 × 101 (c)2 × 53 × 100 (d) 2 × 52
4. The expression of 2658 as a product of its prime factors.
(a) 2 × 3 × 443 (b) 2 × 4 × 443 (c)2 × 3 × 333 (d) 2 × 2 × 443
5. The ratio between the LCM and HCF of 5, 15 and 20.
(a) 1:12 (b) 12 : 11 (c)14 : 1 (d) 12 : 1
6. The HCF of two numbers 𝑎 𝑎𝑛𝑑 𝑏 𝑖𝑠 5 and their 𝐿𝐶𝑀 𝑖𝑠 200. The product 𝑎𝑏 is:
(a)1001 (b) 1000 (c)100 (d) 2000
7. The product of two numbers is 1050 and their HCF is 25. Their LCM is:
(a) 24 (b) 42 (c) 44 (d) 40
8. The LCM of two numbers is 182 and their HCF is 13. If one of the numbers is 26, the other number is.
(a)84 (b) 90 (c) 81 (d) 91
9. The least number that is divisible by all the numbers from 1 to 5 is:
(a) 70 (b) 60 (c) 80 (d) 90
10. ASSERTION: 5 is an example of a rational number.
REASON: The square root of all positive integers is irrational numbers.
(a) Both assertion (A) and reason (R) are true and assertion reason R is the correct explanation of
assertion A.
(b) Both assertion A and reason R are true but reason R is not the correct explanation of assertion A
(c) Assertion A is true but reason R is false. (d) Assertion A is false but reason R is true.
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