Let us assume, to the contrary, that, \(\sqrt{7}\) is a rational number. Then, there exist coprime positive integers a and \(b\) such that
$$
\sqrt{7}=\frac{a}{b}, \quad b \neq 0
$$
So, \(\quad a=\sqrt{7} b\)
Squaring both sides, we have
$$
a^{2}=7 b^{2}
$$
\(\Rightarrow 7\) divides \(a^{2} \Rightarrow 7\) divides \(a\)
So, we can write
$$
a=7 c \text {, }
$$
(where \(c\) is any integer)
Putting the value of \(a=7 c\) in \((i)\), we have
$$
49 c^{2}=7 b^{2} \Rightarrow 7 c^{2}=b^{2}
$$
It means 7 divides \(b^{2}\) and so 7 divides \(b\).
So, 7 is a common factor of both \(a\) and \(b\) which is a contradiction. So, our assumption that \(\sqrt{7}\) is rational is wrong.
Hence, we conclude that \(\sqrt{7}\) is an irrational number.
