Theory:

Let us prove the theorem by assuming its contrary.
Example:
1. Prove that \(\sqrt{7}\) is an irrational number.
 
Proof: Let us assume that \(\sqrt{7}\) is a rational number.
 
Therefore, it can be written as \(\sqrt{7} = \frac{p}{q}\) where \(p, q\) are integers and \(q \neq 0\).
 
Here, \(p\) and \(q\) are coprime(no common factors between \(p\) and \(q\)).
 
Thus, \(\sqrt{7} = \frac{p}{q}\)
 
\(q \sqrt{7} = p\)
 
Squaring on both sides, we have:
 
\(7q^2 = p^2\) ---- (\(1\))
 
Therefore, \(p^2\) is divisible by \(7\).
 
And, \(p\) is also divisible by \(7\) [If \(p\) be a prime number. If \(p\) divides \(a^2\), then \(p\) divides \(a\), where \(a\) is a positive integer.]
 
So, we can write \(p = 7a\) for some integer \(a\).
 
Substituting the value of \(p\) in equation (\(1\)), we have:
 
\(7q^2 = 49a^2\)
 
\(q^2 = 7a^2\)
 
This implies that \(q^2\) is divisible by \(7\) and \(q\) is also divisible by \(7\). [If \(p\) be a prime number. If \(p\) divides \(a^2\), then \(p\) divides \(a\), where \(a\) is a positive integer.]
 
Therefore, \(p\) and \(q\) have atleast \(7\) as a common factor.
 
This contradicts the fact that \(p\) and \(q\) are coprime.
 
Therefore, our assumption is wrong.
 
Thus, \(\sqrt{7}\) is an irrational number.
 
Hence, we proved.
 
 
2. Prove that \(8 - \sqrt{3}\) is an irrational number.
 
Proof: Let us assume that \(8 - \sqrt{3}\) is a rational number.
 
Therefore, it can be written as \(8 - \sqrt{3} = \frac{p}{q}\) where \(p, q\) are integers and \(q \neq 0\).
 
Here, \(p\) and \(q\) are coprime(no common factors between \(p\) and \(q\)).
 
Thus, \(8 - \sqrt{3} = \frac{p}{q}\)
 
\(8 - \frac{p}{q} = \sqrt{3}\)
 
\(\frac{8q - p}{q} = \sqrt{3}\)
 
Since \(8\), \(p\) and \(q\) are integers, then \(\frac{8q - p}{q}\) is rational. This implies that \(\sqrt{3}\) is a rational number.
 
But, we know that \(\sqrt{3}\) is an irrational number.
 
Therefore, our assumption is wrong.
 
Thus, \(8 - \sqrt{3}\) is an irrational number.
 
Hence, we proved.
Important!
In grade \(9\), we have learnt that
 
1. The sum or difference of a rational and an irrational number is irrational.
 
2. The product and quotient of a non-zero rational and irrational number is irrational.
Example:
Show that \(\frac{3}{\sqrt{5}}\) is an irrational number.
 
Proof:
 
Let us assume that \(\frac{3}{\sqrt{5}}\) is a rational number.
 
Therefore, it can be written as \(\frac{3}{\sqrt{5}} = \frac{p}{q}\) where \(p, q\) are integers and \(q \neq 0\).
 
Here, \(p\) and \(q\) are coprime(no common factors between \(p\) and \(q\)).
 
Thus, we have:
 
\(\frac{3}{\sqrt{5}} = \frac{p}{q}\)
 
\(3q = \sqrt{5}p\)
 
\(\frac{3q}{p} = \sqrt{5}\)
 
Since we know that \(3\), \(q\) and \(p\) are integers and \(\frac{3q}{p}\) is a rational number. This implies that \(\sqrt{5}\) is a rational number.
 
But, we know that \(\sqrt{5}\) is an irrational number.
 
Therefore, our assumption is wrong.
 
Thus, \(\frac{3}{\sqrt{5}}\) is an irrational number.
 
Hence, we proved.