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.