NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.3 || CBSE Class 10 maths chapter 1 Ex 1.3

Let's assume that \( \sqrt{5 } \) is a rational number.
Hence, \( \sqrt{ 5} \) can be written in the form \( \frac{a }{ b} \) [where \( a \) and \( b(b \neq 0) \) are co-prime (i.e. no common factor other than 1)]
\(\therefore \sqrt{5 } =\frac{\mathrm{a} }{ \mathrm{b}}\)
\(\Rightarrow \sqrt{ 5} \mathrm{~b}=\mathrm{a}\)
Squaring both sides,
\(\Rightarrow(\sqrt{ 5} b)^{2}=a^{2}\)
\(\Rightarrow 5 b^{2}=a^{2}\)
\(\Rightarrow \frac{a^{2} }{ 5}=b^{2}\)
Hence, 5 divides \( \mathrm{a}^{2} \)
By theorem, if p is a prime number and p divides \( \mathrm{a}^{2} \), then p divides a, where \( a \) is a positive number
So, 5 divides a too
Hence, we can say \( \frac{\mathrm{a} }{ 5}=\mathrm{c} \) where, c is some integer
So, \( \mathrm{a}=5 \mathrm{c} \)
Now we know that,
\(5 b^{2}=a^{2}\)
Putting \( \mathrm{a}=5 \mathrm{c} \),
\(\Rightarrow 5 b^{2}=(5 c)^{2}\)
\(\Rightarrow 5 b^{2}=25 c^{2}\)
\(\Rightarrow b^{2}=5 c^{2}\)
\(\therefore \frac{\mathrm{b}^{2} }{ 5}=\mathrm{c}^{2}\)
Hence, 5 divides \( \mathrm{b}^{2} \)
By theorem, if p is a prime number and p divides \( \mathrm{a}^{2} \), then p divides a, where a is a positive number
So, 5 divides b too
By earlier deductions, 5 divides both a and b
Hence, 5 is a factor of \( a \) and \( b \)
\( \therefore \mathrm{a} \) and b are not co-prime.
Hence, the assumption is wrong.
\( \therefore \) By contradiction,
\( \therefore \sqrt{5} \) is irrational

To Prove: \( 3+2 \sqrt{5 } \) is irrational
Proof: Let \( 3+2 \sqrt{5} \) is rational
A number is said to be rational if it can be expressed in the form \( \frac{\mathrm{p} }{ \mathrm{q}} \) where \( q \neq 0 \)
Therefore,
We can find two integers \( p \ \& \ q \) where, \( (q \neq 0) \) such that
\(3+2 \sqrt{5}=\frac{p}{q}\)
\(2 \sqrt{5}=\frac{p}{q}-3\)
Since p and q are integers, \( \frac{1}{2}\left(\frac{p}{q}-3\right) \) will also be rational and therefore, \( \sqrt{5} \) is rational.
We know that \( \sqrt{5 } \) is irrational but according to above statement it has to be rational
So, both the comments are contradictory, Hence, the number should have been irrational to make the statement correct.
Therefore, \( 3+2 \sqrt{5} \) is irrational.