ICSE Solutions for Selina Concise Chapter 1 Rational and Irrational Numbers Class 9 Maths
Exercise 1(A)
1. Is zero a rational number ? Can it be written in the form P/q, where p and q are integers and q≠0 ?Answer
Yes, zero is a rational number.
As it can be written in the form of , where p and q are integers and q≠0 ?
⇒ 0 = 0/1
(i) Every whole number is a natural number.
(ii) Every whole number is a rational number.
(iii) Every integer is a rational number.
(iv) Every rational number is a whole number.
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(i) False, zero is a whole number but not a natural number.
(ii) True, Every whole can be written in the form of, where p and q are integers and q≠0.
(iii) True, Every integer can be written in the form of , where p and q are integers and q≠0.
(iv) False.
Example: is a rational number, but not a whole number.
Also, find the difference between the largest and smallest of these rational numbers. Express this difference as a decimal fraction correct to one decimal place.
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Also, find the sum of the lowest and largest of these fractions. Express the result obtained as a decimal fraction correct to two decimal places.
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Given number is 7/16
Since 16 = 2×2× 2×2 = 24 = 24 × 50
I.e. 16 can be expressed as 2m × 5n
∴ 7/16 is convertible into the terminating decimal.
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Given number is 23/125
Since 125 = 5×5×5 = 53 = 20 × 53
i.e. 125 can be expressed as 2m×5n
∴ 231/25 is convertible into the terminating decimal.
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Given number is 9/14
Since 14 = 2×7 = 21× 71
i.e. 14 cannot be expressed as 2m × 5n
∴ 9/14 is not convertible into the terminating decimal.
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Given number is 32/45
Since, 45 = 3×3×5 = 32 × 51
i.e. 45 cannot be expressed as 2m × 5n
∴ 32/45 is not convertible into the terminating decimal.
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Given number is 46/50
Since 50 = 2×5×5 = 21 × 52
i.e. 50 can be expressed as 2m × 5n
∴ 43/50 is convertible into the terminating decimal.
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Given number is 17/40
Since 40 = 2×2 × 2 × 5 = 23 × 51
i.e. 40 can be expressed as 2m × 5n
∴ 17/40 is convertible into the terminating decimal.
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Given number is 61/75
Since, 75 = 3×5×5 = 31 × 52
i.e. 75 cannot be expressed as 2m × 5n
∴ 61/75 is not convertible into the terminating decimal.
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Given number is 123/250
Since 250 = 2 ×5×5×5 = 21 × 53
i.e. 250 can be expressed as 2m × 5n.1
∴ 123/250 is convertible into the terminating decimal.
Exercise – 1(B)
1.1. State, whether the following numbers is rational or not : (2 + √2)2
Answer
(2 + √2)2 = 22 + 2 (2) (√2) + (√2)2
= 4 + 4√2 + 2
= 6 + 4√2
Irrational number.
Answer
(3 – √3)2 = 32 + 2(3) (√3) + (√3)2
= 9 – 6√3 + 3
= 12 – 6√3 = 6 (2 – √3)
Irrational number.
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(5 + √5)(5 – √5) = (5)2 – (√5)2
= 25 – 5 = 20
Rational Number
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(√3 – √2)2 = (√3)2 – 2(√3)(√2) + (√2)2
= 3 – 2√6 + 2
= 5 – 2√6
Irrational Number
(32√2)²
Answer
(√76√2)
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(√3 + √2)2 = (√3)2 + 2(√3)(√2) + (√2)2
= 3 + 2√6 + 2
= 5 + 2√6
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(√5 – 2)2 = (√5)2 – 2(√5)(2) + (2)2
= 5 – 4√5 + 4
= 9 – 4√5
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(3 + 2√5)2 = 32 + 2(3)(2√5) + (2√5)2
= 9 + 12√5 + 20
= 29 + 12√5
√2 + √3 = √5
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False
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2√4 + 2 = 2×2 + 2 = 4 + 2 = 6 which is True.
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3√7 – 2√7 = √7
True.
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False Because 2/7= 0.28571427 = 0.285714
which is recurring and non-terminating and hence it is rational.
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True, because 5/11= 0.45 which is recurring and non-terminating
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True
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False
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True
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Given Universal set is
From the given set, find : Set of irrational numbers
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Given Universal set is
{-6, -5 3/4, -√4, -3/5, -3/8, 0, 4/5, 1, 1 2/3, √8, 3.01, π, 8.47}
We need to find the set of integers.
Set of integers consists of zero, the natural numbers and their additive inverses.
The set of integers is Z.
Z = {…-3, -2, -1, 0, 1, 2, 3,….}
Here, the set of integers is U ∩ Z = {-6, √4, 0, 1}
From the given set, find : Set of non - negative integers
AnswerGiven universal set = { −6, −5(3/4), −√4 , −3/5, −3/8, 0 , 4/5 , 1, 1(2/3), √8 , 3.01, 𝛑 , 8.47}
From the given set, find : Set of non - negative integers
We need to find the set of non-negative integers.
Set of non-negative integers consists of zero and the natural numbers.
The set of non-negative integers is Z+ and Z+ = { 0, 1, 2, 3,……}
Here, the set of integers is U ∩ Z+ = {0, 1}
Answer
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3 + √2
Let √3 + √2 be a rational number.
⇒ √3 + √2 = x
Squaring on both the sides, we get
(√3 + √2)2 = x2
⇒ 3 + 2 + 2 x √3 x √2 = x2
⇒ x2 is an irrational number.
⇒ x is an irrational number.
But we have assume that x is a rational number.
∴ we arrive at a contradiction.
So, our assumption that √3 + √2 is a rational number is wrong.
∴ √3 + √2 is an irrational number.
Answer
3 – √2
Let 3 – √2 be a rational number.
⇒ 3 – √2 = x
Squaring on both the sides, we get
(3 – √2)2 = x2
⇒ 9 + 2 – 2 x 3 x √2 = x2
⇒ 11 – x2 = 6√2
⇒ 11 – x2 is an irrational number.
⇒ x2 is an irrational number.
⇒ x is an irrational number.
But we have assume that x is a rational number.
∴ we arrive at a contradiction.
So, our assumption that 3 – √2 is a rational number is wrong.
∴ 3 – √2 is an irrational number.
Answer
√5 – 2
Let √5 – 2 be a rational number.
⇒ √5 – 2 = x
Squaring on both the sides, we get
(√5−2)2 = x2(√5-2)2 = x2
⇒ 5 + 4 – 2x2 ×√5 = x2
⇒ 9 – x2 = 4√5
⇒ x2 = 9 - 4√5
Here, x is a rational number.
⇒ 9 – x2 is an irrational number.
⇒ x2 is an irrational number.
⇒ x is an irrational number.
But we have assume that x is a rational number.
∴ we arrive at a contradiction.
So, our assumption that √5 – 2 is a rational number is wrong.
∴ √5 – 2 is an irrational number.
7. Write a pair of irrational numbers whose sum is irrational.
Answer
√3 + 5 and √5 – 3 are irrational numbers whose sum is irrational.
(√3 + 5) + (√5 – 3) = √3 + √5 + 5 – 3 = √3 + √5 + 2 which is irrational.
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√3 + 5 and 4 – √3 are two irrational numbers whose sum is rational.
(√3 + 5) + (4 – √3) = √3 + 5 + 4 – √3 = 9
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√3 + 2 and √2 – 3 are two irrational numbers whose difference is irrational.
(√3 + 2) – (√2 – 3) = √3 – √2 + 2 + 3 = √3 – √2 + 5 which is irrational.
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√5 – 3 and √5 + 3 are irrational numbers whose difference is rational.
( √5 – 3 ) – ( √5 + 3 ) = √5 – 3 – √5 – 3 = -6 which is rational.
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Consider two irrational numbers (5 + √2) and (√5 – 2)
Thus, the product, (5 + √2) × (√5 – 2) = 5√5 – 10 + √10 – 2√2 is irrational.
Consider √2 as an irrational number.
√2×√2= √4= 2 which is a rational number.
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14.2. Write in descending order: 7√3 and 3√7Answer
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We know that 5 = √25 and 6 = √36
Thus consider the numbers.
√25 < √26 < √27 < √28 < √29 < √30 < √31 < √32 < √33 < √34 < √35 < √36.
Therefore, any two irrational numbers between 5 and 6 is √27 and √28.
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We know that 2√5 = √4×54×5 = √20 and 3√3 = √20
Thus, We have, √20 < √21 < √22 < √23 < √24 < √25 < √26 < √27.
So, any five irrational numbers between 2√5 and 3√3 are :
√21, √22, √23, √24, and √26.
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(3 + √2)(4 + √7)
= 3×4 + 3×√7 + 4×√2 + √2×√7
= 12 + 3√7 + 4√2 + √14
Answer
(√3 – √2 )2
= ( √3 )2 + ( √2 )2 – 2×√3×√2
= 3 + 2 – 2√6
= 5 – 2√6
Exercise 1(C)
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√180 = √2×2×5×3×32×2×5×3×3 = 6√5 which is irrational.
∴ √180 is a surds.
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3√25×3√40
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√π not a surds as π is irrational.
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√3+√2 is not a surds because 3 + √2 is irrational.
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5√2×5√2 = 5×2 = 10 which is rational.
∴ lowest rationalizing factor is √2
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(√5 – 3)(√5 + 3) = (√5)2 – (3)2 = 5 – 9 = -4
∴ lowest rationalizing factor is (√5 + 3)
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7 – √7
(7 – √7)(7 + √7) = 49 – 7 = 42
Therefore, lowest rationalizing factor is (7 + √7).
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∴ lowest rationalizing factor is √2
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√5 – √2
(√5 – √2)(√5 + √2) = (√5)2 – (√2)2 = 3
Therefore lowest rationalizing factor is √5 + √2
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(√13 + 3)(√13 – 3) = (√13)2 – 32 = 13 – 9 = 4
Its lowest rationalizing factor is √13 – 3.
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3√2 + 2√3
= ( 3√2 + 2√3 )( 3√2 – 2√3 )
= ( 3√2)2 – (2√3)2
= 9×2 – 4×3
= 18 – 12
= 6
Its lowest rationalizing factor is 3√2 – 2√3.
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3√5×√5√5 = 3√5/5
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3.4. Rationalise the denominators of : 3/√5+√2
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x2 + y2 + xy
= 161 – 72√5 + 161 +72√5 + 1
= 322 + 1 = 323
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10. If x = 5 - 2√6, find x2 + 1/x2
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12. Rationalise the denominator of : 1/(√3−√2+1)
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13.2. If √2 = 1.4 and √3 = 1.7, find the value of : 1/(3+2√2)
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13.3. Simplify: 2−√3/√3
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