Gujarat Board GSEB Solutions Class 10 Maths Chapter 1 Real Numbers Ex 1.3 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 10 Maths Chapter 1 Real Numbers Ex 1.3

Question 1.
Prove that (sqrt { 5 } ) is irrational.
Solution:
Let (sqrt { 5 } ) be a rational number.
Therefore we can find two integers a, b (b ≠ 0) such that (frac{a}{b}) = (sqrt { 5 } )
Let a and b have a common factor other than 1 then we divide both by common factor.
Then, we get (frac{p}{q}) = (sqrt { 5 } ) …(1)
[Where p and q are co-prime]
Squaring both sides
(frac{p^{2}}{q^{2}}) = 5
p² = 5q² …(2)
therefore p² is divisible by 5, then p will be divisible by 5.
Let p = 5r Putting value of p in eqn (2) we get
25r² = 5q²
5r² = q²
This means q² is divisible by 5 then q will be divisible by 5.
Which means p and q have common factor 5. We reach at the contradiction as our supposition that p and q are co-prime is wrong.
Hence (sqrt { 5 } ) is irrational.

Question 2.
Prove that 3 + 2 (sqrt { 5 } ) is irrational.(CBSE)
Solution:
Let us suppose 3 + 2(sqrt { 5 } ) is a rational.
Then (frac{a}{b}) = 3 + 2(sqrt { 5 } )
[Where a and b are integers]
⇒ (frac{a}{b}) – 3 = 2(sqrt { 5 } )
(frac{1}{2})[(frac{a}{b}) – 3] = (sqrt { 3 } )
(frac{1}{2})[(frac{a}{b}) – 3] is rational as a and b are integers therefore (sqrt { 3 } ) should be rational. This contradicts the fact that (sqrt { 3 } ) is an irrational.
Hence 3 + 2(sqrt { 5 } ) is an irrational.

Question 3.
Prove that the following are irrationals:

1. (frac { 1 }{ sqrt { 2 } } )
2. 7(sqrt { 5 } )
3. 6 + (sqrt { 2 } )

Solution:
1. Let (frac { 1 }{ sqrt { 2 } } ) be a rational.
Therefore we can find two integers a, b (b ≠ 0)
Such that (frac { 1 }{ sqrt { 2 } } ) = (frac{a}{b})
(sqrt { 2 } ) = (frac{b}{a})
(frac{b}{a}) is a rational. Therefore (sqrt { 2 } ) will also be rational which contradicts to the fact that (sqrt { 2 } ) is irrational.
Hence our supposition is wrong and (frac { 1 }{ sqrt { 2 } } ) is irrational.

2. Let 7(sqrt { 5 } ) be a rational
Therefore 7(sqrt { 5 } ) = (frac{a}{b})
[Where a and b are integers]
(sqrt { 5 } ) = (frac{a}{7b})
(frac{a}{7b}) is rational as a and b are integers
Therefore (sqrt { 5 } ) should be rational.
This contradicts the facts that is (sqrt { 5 } ) irrational therefore our supposition is wrong. Hence 7(sqrt { 5 } ) is irrational.

3. Let 6 + (sqrt { 2 } ) be rational
Therefore 6 + (sqrt { 2 } ) = (frac{a}{b})
[where a and b are integers]
(sqrt { 2 } ) = (frac{a}{b}) – 6
(frac{a}{b}) – 6 is rational as a and b are integers therefore, (sqrt { 2 } ) should be rational.
This contradicts the fact that (sqrt { 2 } ) is irrational, therefore, our supposition is wrong.
Hence 6 + (sqrt { 2 } ) is irrational.