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 = p^{2} / q^{2}

p^{2} = 5q^{2} …. (1)

As we know, ‘5’ divides 5q^{2}, so ‘5’ divides p^{2} 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

p^{2} = 25k^{2}

5q^{2} = 25k^{2} [Since, p^{2} = 5q^{2}, from equation (1)]

q^{2} = 5k^{2}

As we know, ‘5’ divides 5k^{2}, so ‘5’ divides q^{2} 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 = p^{2} / q^{2}

p^{2} = 7q^{2} …. (1)

As we know, ‘7’ divides 7q^{2}, so ‘7’ divides p^{2} 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

p^{2} = 49k^{2}

7q^{2} = 49k^{2} [Since, p^{2} = 7q^{2}, from equation (1)]

q^{2} = 7k^{2}

As we know, ‘7’ divides 7k^{2}, so ‘7’ divides q^{2} 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 = p^{2} / q^{2}

p^{2} = 6q^{2} …. (1)

As we know, ‘2’ divides 6q^{2}, so ‘2’ divides p^{2} 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

p^{2} = 4k^{2}

6q^{2} = 4k^{2} [Since, p^{2} = 6q^{2}, from equation (1)]

3q^{2} = 2k^{2}

As we know, ‘2’ divides 2k^{2}, so ‘2’ divides 3q^{2} as well.

‘2’ should either divide 3 or divide q^{2}.

But ‘2’ does not divide 3. ‘2’ divides q^{2} 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 = p^{2} / q^{2}

q^{2} = 11p^{2} …. (1)

As we know, ‘11’ divides 11p^{2}, so ‘11’ divides q^{2} 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

q^{2} = 121k^{2}

11p^{2} = 121k^{2} [Since, q^{2} = 11p^{2}, from equation (1)]

p^{2} = 11k^{2}

As we know, ‘11’ divides 11k^{2}, so ‘11’ divides p^{2} 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 = p^{2} / q^{2}

p^{2} = 2q^{2} …. (1)

As we know, ‘2’ divides 2q^{2}, so ‘2’ divides p^{2} 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

p^{2} = 4k^{2}

2q^{2} = 4k^{2} [Since, p^{2} = 2q^{2}, from equation (1)]

q^{2} = 2k^{2}

As we know, ‘2’ divides 2k^{2}, so ‘2’ divides q^{2} 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 = p^{2} / q^{2}

p^{2} = 3q^{2} …. (1)

As we know, ‘3’ divides 3q^{2}, so ‘3’ divides p^{2} 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

p^{2} = 9k^{2}

3q^{2} = 9k^{2} [Since, p^{2} = 3q^{2}, from equation (1)]

q^{2} = 3k^{2}

As we know, ‘3’ divides 3k^{2}, so ‘3’ divides q^{2} 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 = p^{2} / q^{2}

p^{2} = 5q^{2} …. (1)

As we know, ‘5’ divides 5q^{2}, so ‘5’ divides p^{2} 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

p^{2} = 25k^{2}

5q^{2} = 25k^{2} [Since, p^{2} = 5q^{2}, from equation (1)]

q^{2} = 5k^{2}

As we know, ‘5’ divides 5k^{2}, so ‘5’ divides q^{2} 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 = r^{2} + (√2)^{2} – 2r√2

5 = r^{2} + 2 – 2√2r

5 – 2 = r^{2} – 2√2r

r^{2} – 3 = 2√2r

(r^{2} – 3)/2r = √2

We know that, ‘r’ is rational, ‘(r^{2} – 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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