Prove that the following number is irrational: $\sqrt{11}$

(N/A) Assume,to the contrary,that $\sqrt{11}$ is a rational number.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $\sqrt{11} = \frac{a}{b}$.
Squaring both sides,we get $11 = \frac{a^2}{b^2}$,which implies $a^2 = 11b^2$.
This means $a^2$ is divisible by $11$. Since $11$ is a prime number,$a$ must also be divisible by $11$.
Let $a = 11k$ for some integer $k$.
Substituting this into the equation,we get $(11k)^2 = 11b^2$,which simplifies to $121k^2 = 11b^2$,or $b^2 = 11k^2$.
This implies $b^2$ is divisible by $11$,and consequently,$b$ is divisible by $11$.
Since both $a$ and $b$ are divisible by $11$,they have a common factor $11$,which contradicts our assumption that $a$ and $b$ are coprime.
Therefore,our assumption is false,and $\sqrt{11}$ is an irrational number.