Rational and Irrational Numbers Class 9 RS Aggarwal Exe-1C Goyal Brothers ICSE Foundation Mathmatics Solutions. In this article you will learn how to do mathematical operationa with irrational numbers . Visit official Website **CISCE** for detail information about ICSE Board Class-9 Mathematics.

## Rational and Irrational Numbers Class 9 RS Aggarwal Exe-1C Goyal Brothers ICSE Foundation Mathmatics Solutions.

Board | ICSE |

Publications | Goyal brothers Prakshan |

Subject | Maths |

Class | 9th |

Chapter-1 | Rational and Irrational Numbers |

Writer | RS Aggrawal |

Book Name | Foundation |

Topics | Solution of Exe-1 C |

Academic Session | 2024-2025 |

### How to do Mathematical Operationa with Irrational Numbers

Addition, subtraction, multiplication and division of between **two irrational number** is **may or may not be irrational.**

Addition, subtraction, multiplication and division** between rational and irrational** number is **always irrational.**

**Page- 15,16**

**Exercise- 1C**

Rational and Irrational Numbers Class 9 RS Aggarwal Exe-1C Goyal Brothers ICSE Foundation Mathmatics Solutions

**Que-1: Classify the rational and irrational numbers from the following :**

**(i) 5 (ii) 9/14 (iii) √3 (iv) π (v) 3.1416 (vi) ****√4 (vii) –****√5 (viii) ∛**

**8 (ix)**

**∛****3 (x) 2**

**√6 (xi) 0.36 (xii) 0.202202220 …. (xiii) 2/√3 (xiv) 22/7**

**Solution- **Any fraction with non-zero denominators is a rational number.

Any real number which cannot be expressed in the form of p/q, where p and q are integers, can be called irrational number.

So, √3, π, -√5, ∛3, 2√6, 0.202202220…., 2/√3 are irrational numbers.

5, 9/14, 3.1416, √4, ∛8, 0.36, 22/7 are rational numbers.

**Que-2: Separate the rationals and irrationals from among the following numbers:**

**(i) -8 (ii)** **√25 (iii) -3/5 (iv) ****√8 (v) 0 (vi) π (vii) ****∛5 (viii) 2.4 (ix) –****√3**

**Solution- **(i) −8: This is a rational number because it can be expressed as −8/1.

(ii) √25: This simplifies to 5, which is a rational number because it can be expressed as 5/1.

(iii) −3/5: This is a rational number because it is already in the form of a fraction.

(iv) √8: This is an irrational number because its decimal expansion is non-repeating and non-terminating (≈2.828427….).

(v) 0: This is a rational number because it can be expressed as 0/1.

(vi) π: This is an irrational number because it cannot be expressed as a fraction and its decimal expansion is non-repeating and non-terminating.

(vii)∛5: This is an irrational number because the cube root of 5 is not a rational number.

(viii) 2.4: This is a rational number because it can be expressed as 24/10 or 12/5.

(ix) −√3: This is an irrational number because √3 is irrational, and multiplying by −1 does not change that.

Rational number are : -8, √25, -3/5, 0, 2.4.

Irrational number are : √8, π, ∛5, -√3.

**Que-3: Represent each of the following on the real line :**

**(i)** **√3 (ii) ****√5 (iii) ****√6 (iv) ****√10**

**Solution-**

**Que-4: Write the values of :**

**(i) (2****√3****)² (ii) (3/2 ****√2****)² (iii) (5+****√3)² (iv) (****√6-3****)² (v) (3+2****√5****)² (vi) (****√5+****√6****)² (vii) (3/2****√2)² (viii) (5-6****√3****)²**

**Solution- **(i) (2√3)² = 4 x 3 = 12

(ii) (3/2 √2)² = (9 x 2)/4 = 9/2

(iii) (5 + √3)² = 5² + √3² + 2x5x√3

= 25 + 3 + 10√3 = 28 + 10√3.

(iv) (√6-3)² = √6² + 3² – 2x3x√6

= 6 + 9 – 6√6 = 15 – 6√6.

(v) (3+2√5)² = 3² + (2√5)² + 2x3x2√5

= 9 + 20 + 12√5 = 29 + 12√5.

(vi) (√5+√6)² = √5² + √6² + 2√5√6

= 5 + 6 + 2√30 = 11 + 2√30

(vii) (3/2√2)² = 9/4×2 = 9/8

(viii) (5-6√3)² = 5² + (6√3)² – 2x5x6√3

= 25 + 108 – 60√3 = 133 – 60√3.

**Que-5: State, giving reason, whether the given number is rational or irrational :**

**(i) (3+****√5****) (ii) (-1+****√3****) (iii) 5****√6 (iv) –****√7 (v) ****√6/4 (vi) 3/****√2 (vii) (3+****√3****)****(3-****√3****)**

**Solution- **(i) Let us assume that 3 + √5 is a rational number.

Now,

3 + √5 = a/b

[Here a and b are co-prime numbers, where b ≠ 0]

√5 = a/b – 3

√5 = (a – 3b)/b

Here, {(a – 3b)/b} is a rational number.

But we know that √5 is an irrational number.

So, {(a – 3b)/b} should also be an irrational number.

Hence, it is a contradiction to our assumption.

Thus, 3+√5 is an irrational number.

(ii) Let us assume that the square root of the prime number p is rational.

Hence we can write √p = a/b where a and b are coprime numbers.

Then p = a²/b² and so pb² = a².

Hence, p divides a², so p divides a.

Substitute a by pk. Find out that p divides b.

Hence, this is a contradiction as they should be relatively prime i.e., H.C.F. (a,b) = 1.

∴ √3 is irrational.

⇒ -1+√3 is an irrational number.

(iii) First, recall that a number is irrational if it cannot be expressed as a ratio of two integers.

The square root of √6, 6, is known to be irrational. This is because 6 is not a perfect square and cannot be expressed as a ratio of two integers. Therefore, √6 is an irrational number.

Next, consider the product of a rational number and an irrational number. If a is a non-zero rational number and b is an irrational number, then the product ab is always irrational.

In this case:

5 is a rational number (since it can be expressed as 5/1).

√6 is an irrational number.

The product of 5 and √6:

5√6

Since 5 is rational and √6 is irrational, the product 5√6 is irrational.

Thus, 5√6 is an irrational number.

(iv) √7: The square root of 7 is a number that, when multiplied by itself, equals 7. Since 7 is not a perfect square, √7 cannot be expressed as a fraction of two integers. Its decimal expansion is non-repeating and non-terminating, which makes √7 an irrational number.

Multiplying an irrational number by −1 (or any non-zero rational number) does not change its irrationality. Therefore, −√7 is also an irrational number.

(v) √6 is an irrational number. This is because 6 is not a perfect square, and its square root cannot be expressed as a fraction of two integers. Its decimal expansion is non-repeating and non-terminating.

When you divide an irrational number by a rational number (in this case, 4, which is rational because it can be written as 4/1), the result is still an irrational number.

Therefore, √6/4 is an irrational number.

(vi) First, observe that √2 is an irrational number. Rational numbers cannot be expressed as a fraction of two integers when the denominator is an irrational number. However, to make the process clear, let’s simplify the expression:

3/√2 = 3/√2 × √2/√2 = 3√2/2

Now, we have 3√2/2.

In this form, 3√2 is still irrational because √2 is irrational, and multiplying an irrational number by a rational number (in this case, 3) still results in an irrational number. Dividing this product by 2 (a rational number) does not change its irrationality. Thus, 3√2/2 remains an irrational number.

Therefore, the given number 3/√2 is irrational number.

(vii) We use the difference of squares formula:

(a+b)(a−b) = a² − b²

Here, a = 3 and b = √3. Applying the formula, we get:

(3+√3)(3−√3) = 3² − (√3)²

Now, calculate each term separately:

3² = 9

(√3)² = 3

Substitute these values back into the expression:

9−3 = 6

The result is 6, which is a rational number because it can be expressed as a fraction (6/1).

Thus, (3+√3)(3−√3) is a rational number.

**Que-6: Show that each of the following is irrational :**

**(i) (2+****√5****)² (ii) (3-****√3****)² (iii) (****√5****+√3****)² (iv) 6/****√3**

**Solution- **(i) Let’s first compute the expression:

(2+√5)² = (2+√5)(2+√5)

Using the distributive property (also known as the FOIL method for binomials):

(2+√5)(2+√5) = 2⋅2 + 2⋅√5 + √5⋅2 + √5⋅√5

Simplify each term:

= 4 + 2√5 + 2√5 + (√5)²

= 4 + 2√5 + 2√5 + 5

= 4 + 5 + 4√5

= 9 + 4√5

Now, we need to show that 9+4√5 is irrational.

It can be expressed as a fraction of two integers:

9 + 4√5 = p/q

4√5 = p/q − 9

Multiply both sides by q:

4𝑞√5 = 𝑝−9𝑞

Thus,

√5 = [𝑝−9𝑞/4𝑞]

Since p and q are integers, [𝑝−9𝑞/4𝑞] is a rational number.

Hence, 9+4√5 is irrational. Therefore, (2+√5)² is irrational.

(ii) (3 – √3)²:

Expanding the expression using the binomial formula, we have:

(3 – √3)² = 3² – 2(3)(√3) + (√3)²

= 9 – 6√3 + 3

= 12 – 6√3

Assume that 12 – 6√3 is rational. Then it can be expressed as the quotient of two integers:

12 – 6√3 = p/q, where p and q are integers with q ≠ 0.

Rearranging the equation, we have:

6√3 = 12 – p/q

Since the left side (√3) is irrational and the right side (12 – p/q) is rational, we have a contradiction. Therefore, 12 – 6√3 is irrational.

(iii) (√5 + √3)²:

Expanding the expression using the binomial formula, we have:

(√5 + √3)² = (√5)² + 2(√5)(√3) + (√3)²

= 5 + 2√15 + 3

= 8 + 2√15

Assume that 8 + 2√15 is rational. Then it can be expressed as the quotient of two integers:

8 + 2√15 = p/q, where p and q are integers with q ≠ 0.

Rearranging the equation, we have:

2√15 = p/q – 8

Since the left side (√15) is irrational and the right side (p/q – 8) is rational, we have a contradiction. Therefore, 8 + 2√15 is irrational.

(iv) 6/√3:

To show that 6/√3 is irrational, we assume the opposite and suppose that 6/√3 is rational. Then it can be expressed as the quotient of two integers:

6/√3 = p/q, where p and q are integers with q ≠ 0.

Rearranging the equation, we have:

√3 = 6q/p

Squaring both sides, we get:

3 = 36q²/p²

This implies that 3p² = 36q², which simplifies to p² = 12q².

From this equation, we can see that the left side is divisible by 3, while the right side is not divisible by 3. This contradicts the assumption that p and q are integers. Therefore, 6/√3 is irrational.

**Que-7: Prove that ****√5 is an irrational number.**

**Solution- **If √5 is rational, that means it can be written in the form of a/b, where a and b that have no common factor other than 1 and b ≠ 0. i.e., a and b are

√5/1 = a/b

√5b = a

Squaring both sides,

5b^{2} = a^{2} … (1)

This means 5 divides a^{2}.

From this, 5 also divides a.

Then a = 5c, for some ‘c’.

On squaring, we get

a^{2} = 25c^{2
}Put the value of a^{2} in equation (1).

5b^{2} = 25c^{2
}b^{2} = 5c^{2
So, √5 is an irrational number.}

**Que-8: Write down the examples of 4 distinct irrational number.**

**Solution- **Sure, here are four distinct irrational numbers:

$√2 , √$3, √5, √6Each of these numbers cannot be expressed as a fraction of two integers and has non-repeating, non-terminating decimal expansions.

**Que-9: Prove that (√3+√7) is irrational.**

**Solution- **So, we have:

√3+√7 = a/b

Squaring both sides:

(√3+√7)² = (a/b)²

3+2√21+7 = a²/b²

10+2√21 = a²/b²

2√21= a²/b²−10

√21 = a²/2b²−5

This implies that √21 is rational, which contradicts our previous proof that √21 is irrational.

Therefore, our initial assumption that √3+√7 is rational must be false. Hence, √3+√7 is irrational.

**Que-10: Prove that ****(√2+√3) is irrational.**

**Solution- **Let us assume that √2+√3 is a rational number.

So it can be written in the form 𝑎𝑏

√2+√3 = 𝑎/𝑏

Here 𝑎 and 𝑏 are coprime numbers and 𝑏≠0

√2+√3 = 𝑎/𝑏

√2 = 𝑎/𝑏 – √3

On squaring both the sides we get,

(√2)² = [𝑎/𝑏 -3]²

We know that

(𝑎–𝑏)² = 𝑎² + 𝑏² – 2𝑎𝑏

So the equation [𝑎/𝑏 – 3]² can be written as

[𝑎/𝑏 – 3]² = 𝑎²/𝑏² + 3–2 (𝑎/𝑏) √3

Substitute in the equation we get,

2 = 𝑎²/𝑏² + 3–2 𝑎/𝑏 √3

Rearranging the equation we get,

𝑎²/𝑏² + 3-2 = 2(𝑎/𝑏) √3

𝑎²/𝑏² + 1 = 2 (√3) (𝑎/𝑏)

(𝑎²+𝑏²)/𝑏² × 𝑏/2𝑎 = 3

(𝑎²+𝑏²)/2𝑎𝑏 = 3

Since, a, b are integers, (𝑎²+𝑏²)/2𝑎𝑏 is a rational number.

√3 is a rational number.

It contradicts to our assumption that is √3 irrational.

Therefore, our assumption is wrong

Thus, √2+√3 is irrational.

**Que-11: Write two irrational numbers between √14 and √19.**

**Solution- **We first need to approximate the values of these square roots:

√14 ≈ 3.742

√19 ≈ 4.359

Here are two examples:

√15:

√15 ≈ 3.873

√17:

√17 ≈ 4.123

Since 3.873 < 4.123 is a two irrational number between √14 and √19.

Thus, two irrational numbers between √14 and √19 are:

√15 and √17.

**Que-12: ****Write three irrational numbers between √2 and √7.**

**Solution-** We first need to approximate the values of these square roots:

√2 ≈ 1.4142

√7 ≈ 2.6467

Here are three examples:

√3:

√3 ≈ 1.7323 ≈ 1.732

√5:

√5 ≈ 2.2365 ≈ 2.236

√6:

√6 ≈ 2.4496 ≈ 2.449

Since 1.414 < 2.449 < 2.646 √6 is a third irrational number between √2 and √7.

Thus, three irrational numbers between √2 and √7 are:

√3, √5, and √6.

**Que-13: State in each case, whether true or false :**

**(i) The sum of two rationals is a rational.**

**(ii) The sum of two irrationals is a irrational.**

**(iii) The products of two rationals is a rational.**

**(iv) The product of two irrationals is an irrational.**

**(v) The sum of a rational and an irrational is an irrational.**

**(vi) The product of a rational and an irrational is a rational.**

**Solution- **(i) True

(ii) False

(iii) True

(iv) False

(v) True

(vi) False

**Que-14: What are rational numbers? Give ten examples.**

**Solution- **Any fraction with non-zero denominators is a rational number. Some of the examples of rational numbers are 1/2, 1/5, 3/4, 2/6, 8/6, 7/5, 2/8, 4/7, 6/5, 7/2 and so on.

**Que-15: ****What are irrational numbers? Give ten examples.**

**Solution-**Any real number which cannot be expressed in the form of p/q, where p and q are integers, can be called irrational number.

Examples of irrational number are : √2, √3, √5, √7, √11, √13, √17, √19, √23, √29 and so on.

-: End of Rational and Irrational Numbers Class 9 RS Aggarwal Exe-1C Goyal Brothers ICSE Solutions : –

Return to :- **RS Aggarwal Solutions for ICSE Class-9 Mathematics (Goyal Brother Prakashan)**

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