Prove that $\sqrt{3}$ is an irrational number.

(N/A) Assume,to the contrary,that $\sqrt{3}$ is a rational number.
Then,there exist coprime positive integers $a$ and $b$ such that $\sqrt{3} = \frac{a}{b}$,where $gcd(a, b) = 1$.
Squaring both sides,we get $3 = \frac{a^2}{b^2}$,which implies $a^2 = 3b^2$ .......... $(1)$.
Since $3$ divides $a^2$,it follows that $3$ must also divide $a$ (by the Fundamental Theorem of Arithmetic).
Let $a = 3k$ for some integer $k$.
Substituting this into equation $(1)$,we get $(3k)^2 = 3b^2$,which simplifies to $9k^2 = 3b^2$,or $b^2 = 3k^2$.
This implies that $3$ divides $b^2$,and consequently,$3$ must divide $b$.
Thus,$3$ is a common factor of both $a$ and $b$,which contradicts our initial assumption that $gcd(a, b) = 1$.
Therefore,our assumption that $\sqrt{3}$ is rational is false.
Hence,$\sqrt{3}$ is an irrational number.