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

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 Maths Solutions

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

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,
5b2 = a2 … (1)
This means 5 divides a2.
From this, 5 also divides a.
Then a = 5c, for some ‘c’.
On squaring, we get
a2 = 25c2
Put the value of a2 in equation (1).
5b2 = 25c2
b2 = 5c2
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:
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 : –

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