RS Aggarwal Solutions Chapter 1 Real Numbers Exercise  1D Class 10 Maths
Chapter Name  RS Aggarwal Chapter 1 Real Numbers Solutions 
Book Name  RS Aggarwal Mathematics for Class 10 
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Related Study  NCERT Solutions for Class 10 Maths 
Exercise 1D Solutions
1. Define
(i) rational numbers
(ii) irrational numbers
(iii) real numbers
Solution
(i) Rational numbers: The numbers of the form p/q where p, q are integers and q ≠ 0 are called rational numbers.
Example: 2/3
(ii) Irrational numbers: The numbers which when expressed in decimal form are expressible as nonterminating and nonrepeating decimals are called irrational numbers.
Example: √2
(iii) Real numbers: The numbers which are positive or negative, whole numbers or decimal numbers and rational numbers or irrational number are called real numbers.
Example: 2, 1/3, √2 , 3 etc.
2. Classify the following numbers as rational or irrational:
(i) 22/7
(ii) 3.1416
(iii) Ï€
(v) 5.636363
(vi) 2.040040004
(vii) 1.535335333
(viii) 3.121221222 3
(ix) √21
(x) √3
Solution
(i) 22/7 is a rational number because it is of the form of p/q, q ≠ 0.
(ii) 3.1416 is a rational number because it is a terminating decimal.
(iii) Ï€ is an irrational number because it is a nonrepeating and nonterminating decimal.
(iv)is a rational number because it is a repeating decimal.
(v) 5.636363… is a rational number because it is a nonterminating and nonrepeating decimal.
(vi) 2.040040004… is an irrational number because it is a nonterminating and nonrepeating decimal.
(vii) 1.535335333… is an irrational number because it is a nonterminating and non repeating decimal.
(viii) 3.121221222… is an irrational number because it is a nonterminating and non repeating decimal.
(ix) √21 = √3 × √7 is an irrational number because √3 and √7 are irrational and prime numbers.
(x)an irrational number because 3 is a prime number. So, √3 is an irrational number.
3. Prove that each of the following numbers is irrational:
(i) √6
(ii) 2√3
(iii) 3+√2
(iv) 2+√5
(v) 5+ 3√2
(vi) 3√7
(vii) 3/√5
(viii) 2 3√5
(ix) √3+ √5
Solution
(i) Let √6 = √2×√3 be rational.
Hence, √2, √3 are both rational.
This contradicts the fact that√2, √3 are irrational.
The contradiction arises by assuming √6 is rational.
Hence, √6 is irrational.
(ii) Let 2  √3 be rational.
Hence, 2 and 2  √3 are rational.
∴ (2  2 + √3)
= √3 = rational [∵ Difference of two rational is rational]
This contradicts the fact that √3 is irrational.
The contradiction arises by assuming 2  √3 is rational.
Hence, 2  √3 is irrational.
(iii) Let 3 + √2 be rational.
Hence, 3 and 3 + √2 are rational.
∴ 3 + √2 – 3
= √2 = rational [∵ Difference of two rational is rational]
This contradicts the fact that √2 is irrational.
The contradiction arises by assuming 3 + √2 is rational.
Hence, 3 + √2 is irrational.
(iv) Let 2 + √5 be rational.
Hence, 2 + √5 and √5 are rational.
∴ (2 + √5 ) – 2
= 2 + √5 – 2
= √5 = rational [∵ Difference of two rational is rational]
This contradicts the fact that √5 is irrational.
The contradiction arises by assuming 2  √5 is rational.
Hence, 2  √5 is irrational.
(v) Let, 5 + 3√2 be rational.
Hence, 5 and 5 + 3√2 are rational.
∴ (5 + 3√2 – 5)
= 3√2 = rational [∵ Difference of two rational is rational]
∴ 1/3 ×3√2
= √2 = rational [∵ Product of two rational is rational]
This contradicts the fact that √2 is irrational.
The contradiction arises by assuming 5 + 3√2 is rational.
Hence, 5 + 3√2 is irrational.
(vi) Let 3√7 be rational.
1/3 × 3√7
= √7 = rational [∵ Product of two rational is rational]
This contradicts the fact that √7 is irrational.
The contradiction arises by assuming 3√7 is rational.
Hence, 3√7 is irrational.
(vii) Let 3/√5 be rational.
∴ 1/3 × 3/√5
= 1/√5 [∵ Product of two rational is rational]
This contradicts the fact that is 1/√5 irrational.
∴ (1×√5)/(√5×√5)
= 1/5.√5
So, if 1/√5 is irrational, then 1/5.√5 is rational
∴ 5 × 1/5.√5
= √5 rational [∵ Product of two rational is rational]
Hence, 1/√5 is irrational
The contradiction arises by assuming 3/√5 is rational.
Hence, 3/√5 is irrational.
(viii) Let 2  3√5 be rational.
Hence 2 and 2  3√5 are rational.
∴ 2 – (2  3√5)
= 2 – 2 + 3√5
= 3√5 = rational [∵ Difference of two rational is rational]
∴ 1/3 × 3√5
= √5 = rational [∵ Product of two rational is rational]
This contradicts the fact that √5 is irrational.
The contradiction arises by assuming 2  3√5 is rational.
Hence, 2  3√5 is irrational.
(ix) Let √3 + √5 be rational.
∴ √3 + √5 = a, where a is rational.
∴ √3 = a  √5 ...(1)
On squaring both sides of equation (1), we get
3 = (a  √5)^{2 }
⇒ 3 = a^{2 }+ 5  2√5a
⇒ √5 = (a^{2} + 2)/2a
This is impossible because righthand side is rational, whereas the lefthand side is irrational.
This is a contradiction.
Hence, √3 + √5 is irrational.
4. Prove that 1/√3 is irrational.
Solution
Let 1/√3 be rational.
∴ 1/√3 = a/b, where a, b are positive integers having no common factor other than 1
∴ √3 = b/a ...(1)
Since a, b are nonzero integers, b/a is rational.
Thus, equation (1) shows that √3 is rational.
This contradicts the fact that √3 is rational.
The contradiction arises by assuming √3 is rational.
Hence, 1/√3 is irrational.
5. (i) Give an example of two irrationals whose sum is rational.
(ii) Give an example of two irrationals whose product is rational.
Solution
(i) Let (2 + √3), (2  √3) be two irrationals.
∴ (2 + √3) + (2  √3) = 4 = rational number
(ii) Let 2√3, 3√3 be two irrationals.
∴ 2√3 × 3√3 = 18 = rational number.
6. State whether the given statement is true or false:
(i) The sum of two rationals is always rational
(ii) The product of two rationals is always rational
(iii) The sum of two irrationals is an irrational
(iv) The product of two irrationals is an irrational
(v) The sum of a rational and an irrational is irrational
(vi) The product of a rational and an irrational is irrational
Solution
(i) The sum of two rationals is always rational  True
(ii) The product of two rationals is always rational  True
(iii) The sum of two irrationals is an irrational  False
Counter example:
2 + √3 and 2  √3 are two irrational numbers. But their sum is 4, which is a rational number.
(iv) The product of two irrationals is an irrational – False
Counter example:
2 √3 and 4 √3 are two irrational numbers. But their product is 24, which is a rational number.
(v) The sum of a rational and an irrational is irrational  True
(vi) The product of a rational and an irrational is irrational  True
7. Prove that (2 √3 – 1) is irrational.
Solution
Let x = 2 √3 – 1 be a rational number.
x = 2√3 – 1
⇒ x^{2 }= (2√3 – 1)^{2}
⇒ x^{2 }= (2√3 )^{2 }+ (1)^{2 }– 2(2√3)(1)
⇒ x^{2 }= 12 + 1  4√3
⇒ x^{2 }– 13 =  4√3
⇒ (13 – x^{2})/4 = √3
Since x is rational number, x^{2 }is also a rational number.
⇒ 13  x^{2 }is a rational number
⇒ (13 – x^{2})4 is a rational number
⇒ √3 is a rational number
But √3 is an irrational number, which is a contradiction.
Hence, our assumption is wrong.
Thus, (2 √3 – 1) is an irrational number.
8. Prove that (4  5√2) is irrational.
Solution
Let x = 4  5√2 be a rational number.
x = 4  5√2
⇒ x^{2 }= (4  5√2)^{2}
⇒ x^{2 }= 4^{2 }+ (5√2)^{2 }– 2(4) (5√2)
⇒ x^{2 }= 16 + 50 – 40√2
⇒ x^{2 }– 66 = – 40√2
⇒ (66 – x^{2})/40 = √2
Since x is a rational number, x^{2 }is also a rational number.
⇒ 66  x^{2 }is a rational number
⇒ (66  x^{2})/40 is a rational number
⇒ √2 is a rational number
But √2 is an irrational number, which is a contradiction.
Hence, our assumption is wrong.
Thus, (4  5√2) is an irrational number.
9. Show that (5  2√3) is irrational.
Solution
Let x = 5  2√3 be a rational number.
x = 5  2√3
⇒ x^{2 }= (5  2√3)^{2}
⇒ x^{2 }= 5^{2 }+ (2√3)^{2 }– 2(5) (2√3)
⇒ x^{2 }= 25 + 12 – 20√3
⇒ x^{2 }– 37 = – 20√3
⇒ (37 – x^{2})/20 = √3
Since x is a rational number, x^{2 }is also a rational number.
⇒ 37  x^{2 }is a rational number
⇒ (37 – x^{2})/20 is a rational number
⇒ √3 is a rational number
But √3 is an irrational number, which is a contradiction.
Hence, our assumption is wrong.
Thus, (5  2√3) is an irrational number.
10. Prove that 5√2 is irrational.
Solution
Let 5√2 is a rational number.
∴ 5√2 = p/q, where p and q are some integers and HCF (p, q) = 1 …(1)
⇒ 5√2q = p
⇒ (5√2q)^{2 }= p^{2}
⇒ 2(25q^{2}) = p^{2}
⇒ p^{2 }is divisible by 2
⇒ p is divisible by 2 ….(2)
Let p = 2m, where m is some integer.
∴ 5√2q = 2m
⇒ (5√2q)^{2 }= (2m)^{2}
⇒2(25q^{2}) = 4m^{2}
⇒25q^{2 }= 2m^{2}
⇒ q^{2 }is divisible by 2
⇒ q is divisible by 2 ….(3)
From (2) and (3) is a common factor of both p and q, which contradicts (1).
Hence, our assumption is wrong.
Thus, 5√2 is irrational.
11. Show that 2/√7 is irrational.
Solution
2/√7 = 2√7 × √7/√7 = 2/7√7
Let 2/7.√7 is a rational number.
∴ 2/7.√7 = p/q, where p and q are some integers and HCF (p, q) = 1 ...(1)
⇒ 2√7q = 7p
⇒ (2√7q) ^{2 }= (7p)^{2}
⇒ 7(4q^{2}) = 49p^{2}
⇒ 4q^{2 }= 7p^{2}
⇒ q^{2 }is divisible by 7
⇒ q is divisible by 7 ...(2)
Let q = 7m, where m is some integer.
∴ 2√7q = 7p
⇒ [2√7(7m)]^{2 }= (7p)^{2}
⇒ 343(4m^{2}) = 49p^{2}
⇒ 7(4m^{2}) = p^{2}
⇒ p^{2 }is divisible by 7
⇒ p is divisible by 7 ….(3)
From (2) and (3), 7 is a common factor of both p and q, which contradicts (1).
Hence, our assumption is wrong.
Thus, 2/√7 is irrational.