GSEB Solutions Class 10 Maths Chapter 1 Real Numbers Ex 1.3

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.