ML Aggarwal Rational and Irrational Number Exe-1.2 Class 9 ICSE Maths Solutions. We Provide Step by Step Answer of Exe-Exe-1.2 Questions for Rational and Irrational Number as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-9.
ML Aggarwal Rational and Irrational Number Exe-1.2 Class 9 ICSE Maths Solutions
Board | ICSE |
Subject | Maths |
Class | 9th |
Chapter-1 | Rational and Irrational |
Topics | Solution of Exe-1.2 Questions |
Edition | 2024-2025 |
Rational and Irrational Number Exe-1.2
ML Aggarwal Class 9 ICSE Maths Solutions
Question 1. Prove that, √5 is an irrational number.
Ans: Let us consider √5 be a rational number, then
√5 = p/q, where ‘p’ and ‘q’ are integers, q ≠ 0 and p, q have no common factors (except 1).
So,
5 = p2 / q2
p2 = 5q2 …. (1)
As we know, ‘5’ divides 5q2, so ‘5’ divides p2 as well. Hence, ‘5’ is prime.
So 5 divides p
Now, let p = 5k, where ‘k’ is an integer
Square on both sides, we get
p2 = 25k2
5q2 = 25k2 [Since, p2 = 5q2, from equation (1)]
q2 = 5k2
As we know, ‘5’ divides 5k2, so ‘5’ divides q2 as well. But ‘5’ is prime.
So 5 divides q
Thus, p and q have a common factor 5. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).
We can say that, √5 is not a rational number.
√5 is an irrational number.
Hence proved.
Question 2. Prove that √7 is an irrational number.
Ans : Let us consider √7 be a rational number, then
√7 = p/q, where ‘p’ and ‘q’ are integers, q ≠ 0 and p, q have no common factors (except 1).
So,
7 = p2 / q2
p2 = 7q2 …. (1)
As we know, ‘7’ divides 7q2, so ‘7’ divides p2 as well. Hence, ‘7’ is prime.
So 7 divides p
Now, let p = 7k, where ‘k’ is an integer
Square on both sides, we get
p2 = 49k2
7q2 = 49k2 [Since, p2 = 7q2, from equation (1)]
q2 = 7k2
As we know, ‘7’ divides 7k2, so ‘7’ divides q2 as well. But ‘7’ is prime.
So 7 divides q
Thus, p and q have a common factor 7. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).
We can say that, √7 is not a rational number.
√7 is an irrational number.
Hence proved.
Question 3. Prove that √6 is an irrational number.
Ans: Let us consider √6 be a rational number, then
√6 = p/q, where ‘p’ and ‘q’ are integers, q ≠ 0 and p, q have no common factors (except 1).
So,
6 = p2 / q2
p2 = 6q2 …. (1)
As we know, ‘2’ divides 6q2, so ‘2’ divides p2 as well. Hence, ‘2’ is prime.
So 2 divides p
Now, let p = 2k, where ‘k’ is an integer
Square on both sides, we get
p2 = 4k2
6q2 = 4k2 [Since, p2 = 6q2, from equation (1)]
3q2 = 2k2
As we know, ‘2’ divides 2k2, so ‘2’ divides 3q2 as well.
‘2’ should either divide 3 or divide q2.
But ‘2’ does not divide 3. ‘2’ divides q2 so ‘2’ is prime.
So 2 divides q
Thus, p and q have a common factor 2. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).
We can say that, √6 is not a rational number.
√6 is an irrational number.
Hence proved.
Question 4. Prove that 1/√11 is an irrational number.
Ans : Let us consider 1/√11 be a rational number, then
1/√11 = p/q, where ‘p’ and ‘q’ are integers, q ≠ 0 and p, q have no common factors (except 1).
So,
1/11 = p2 / q2
q2 = 11p2 …. (1)
As we know, ‘11’ divides 11p2, so ‘11’ divides q2 as well. Hence, ‘11’ is prime.
So 11 divides q
Now, let q = 11k, where ‘k’ is an integer
Square on both sides, we get
q2 = 121k2
11p2 = 121k2 [Since, q2 = 11p2, from equation (1)]
p2 = 11k2
As we know, ‘11’ divides 11k2, so ‘11’ divides p2 as well. But ‘11’ is prime.
So 11 divides p
Thus, p and q have a common factor 11. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).
We can say that, 1/√11 is not a rational number.
1/√11 is an irrational number.
Hence proved.
Question 5. Prove that √2 is an irrational number. Hence show that 3 — √2 is an irrational.
Ans : – Let us consider √2 be a rational number, then
√2 = p/q, where ‘p’ and ‘q’ are integers, q ≠ 0 and p, q have no common factors (except 1).
So,
2 = p2 / q2
p2 = 2q2 …. (1)
As we know, ‘2’ divides 2q2, so ‘2’ divides p2 as well. Hence, ‘2’ is prime.
So 2 divides p
Now, let p = 2k, where ‘k’ is an integer
Square on both sides, we get
p2 = 4k2
2q2 = 4k2 [Since, p2 = 2q2, from equation (1)]
q2 = 2k2
As we know, ‘2’ divides 2k2, so ‘2’ divides q2 as well. But ‘2’ is prime.
So 2 divides q
Thus, p and q have a common factor 2. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).
We can say that, √2 is not a rational number.
√2 is an irrational number.
Now, let us assume 3 – √2 be a rational number, ‘r’
So, 3 – √2 = r
3 – r = √2
We know that, ‘r’ is rational, ‘3- r’ is rational, so ‘√2’ is also rational.
This contradicts the statement that √2 is irrational.
So, 3- √2 is irrational number.
Hence proved.
Question 6. Prove that, √3 is an irrational number. Hence, show that 2/5×√3 is an irrational number.
Ans : – Let us consider √3 be a rational number, then
√3 = p/q, where ‘p’ and ‘q’ are integers, q ≠ 0 and p, q have no common factors (except 1).
So,
3 = p2 / q2
p2 = 3q2 …. (1)
As we know, ‘3’ divides 3q2, so ‘3’ divides p2 as well. Hence, ‘3’ is prime.
So 3 divides p
Now, let p = 3k, where ‘k’ is an integer
Square on both sides, we get
p2 = 9k2
3q2 = 9k2 [Since, p2 = 3q2, from equation (1)]
q2 = 3k2
As we know, ‘3’ divides 3k2, so ‘3’ divides q2 as well. But ‘3’ is prime.
So 3 divides q
Thus, p and q have a common factor 3. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).
We can say that, √3 is not a rational number.
√3 is an irrational number.
Now, let us assume (2/5)√3 be a rational number, ‘r’
So, (2/5)√3 = r
5r/2 = √3
We know that, ‘r’ is rational, ‘5r/2’ is rational, so ‘√3’ is also rational.
This contradicts the statement that √3 is irrational.
So, (2/5)√3 is irrational number.
Hence proved.
Question 7. Prove that √5 is an irrational number. Hence, show that -3 + 2√5 is an irrational number.
Ans : – Let us consider √5 be a rational number, then
√5 = p/q, where ‘p’ and ‘q’ are integers, q ≠ 0 and p, q have no common factors (except 1).
So,
5 = p2 / q2
p2 = 5q2 …. (1)
As we know, ‘5’ divides 5q2, so ‘5’ divides p2 as well. Hence, ‘5’ is prime.
So 5 divides p
Now, let p = 5k, where ‘k’ is an integer
Square on both sides, we get
p2 = 25k2
5q2 = 25k2 [Since, p2 = 5q2, from equation (1)]
q2 = 5k2
As we know, ‘5’ divides 5k2, so ‘5’ divides q2 as well. But ‘5’ is prime.
So 5 divides q
Thus, p and q have a common factor 5. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).
We can say that, √5 is not a rational number.
√5 is an irrational number.
Now, let us assume -3 + 2√5 be a rational number, ‘r’
So, -3 + 2√5 = r
-3 – r = 2√5
(-3 – r)/2 = √5
We know that, ‘r’ is rational, ‘(-3 – r)/2’ is rational, so ‘√5’ is also rational.
This contradicts the statement that √5 is irrational.
So, -3 + 2√5 is irrational number.
Hence proved.
Question 8. Prove that the following numbers are irrational:
(i) 5 +√2
(ii) 3 – 5√3
(iii) 2√3 – 7
(iv) √2 +√5
Ans : –
(i) 5 +√2
Now, let us assume 5 + √2 be a rational number, ‘r’
So, 5 + √2 = r
r – 5 = √2
We know that, ‘r’ is rational, ‘r – 5’ is rational, so ‘√2’ is also rational.
This contradicts the statement that √2 is irrational.
So, 5 + √2 is irrational number.
(ii) 3 – 5√3
Now, let us assume 3 – 5√3 be a rational number, ‘r’
So, 3 – 5√3 = r
3 – r = 5√3
(3 – r)/5 = √3
We know that, ‘r’ is rational, ‘(3 – r)/5’ is rational, so ‘√3’ is also rational.
This contradicts the statement that √3 is irrational.
So, 3 – 5√3 is irrational number.
(iii) 2√3 – 7
Now, let us assume 2√3 – 7 be a rational number, ‘r’
So, 2√3 – 7 = r
2√3 = r + 7
√3 = (r + 7)/2
We know that, ‘r’ is rational, ‘(r + 7)/2’ is rational, so ‘√3’ is also rational.
This contradicts the statement that √3 is irrational.
So, 2√3 – 7 is irrational number.
(iv) √2 +√5
Now, let us assume √2 +√5 be a rational number, ‘r’
So, √2 +√5 = r
√5 = r – √2
Square on both sides,
(√5)2 = (r – √2)2
5 = r2 + (√2)2 – 2r√2
5 = r2 + 2 – 2√2r
5 – 2 = r2 – 2√2r
r2 – 3 = 2√2r
(r2 – 3)/2r = √2
We know that, ‘r’ is rational, ‘(r2 – 3)/2r’ is rational, so ‘√2’ is also rational.
This contradicts the statement that √2 is irrational.
So, √2 +√5 is irrational number.
— : End of ML Aggarwal Rational and Irrational Number Exe-1.2 Class 9 ICSE Maths Solutions :–
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In question 8 in all parts root no. are not rational no.