Prove that the number $\sqrt{21}$ is irrational.

(N/A) Assume,to the contrary,that $\sqrt{21}$ is a rational number.
Then,there exist two coprime positive integers $a$ and $b$ $(b \neq 0)$ such that $\sqrt{21} = \frac{a}{b}$.
Squaring both sides,we get $21 = \frac{a^2}{b^2}$,which implies $a^2 = 21b^2$.
This means $a^2$ is divisible by $21$,so $a$ must also be divisible by $21$ (since $21$ is a product of two distinct primes $3$ and $7$).
Let $a = 21k$ for some integer $k$.
Substituting this into the equation,we get $(21k)^2 = 21b^2$,which simplifies to $441k^2 = 21b^2$,or $b^2 = 21k^2$.
This implies $b^2$ is divisible by $21$,so $b$ must also be divisible by $21$.
Thus,$a$ and $b$ have a common factor of $21$,which contradicts our assumption that $a$ and $b$ are coprime.
Therefore,our assumption is false,and $\sqrt{21}$ is an irrational number.