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

(N/A) Let us assume,to the contrary,that $\sqrt{3}$ is rational.
That is,we can find integers $a$ and $b$ $(b \neq 0)$ such that $\sqrt{3} = \frac{a}{b}$.
Suppose $a$ and $b$ have a common factor other than $1$. Then we can divide by the common factor and assume that $a$ and $b$ are coprime.
So,$b\sqrt{3} = a$.
Squaring on both sides,we get $3b^2 = a^2$.
This implies that $a^2$ is divisible by $3$,and by the fundamental theorem of arithmetic,$a$ is also divisible by $3$.
So,we can write $a = 3c$ for some integer $c$.
Substituting $a = 3c$ in $3b^2 = a^2$,we get $3b^2 = (3c)^2 = 9c^2$,which simplifies to $b^2 = 3c^2$.
This means that $b^2$ is divisible by $3$,and consequently,$b$ is also divisible by $3$.
Therefore,$a$ and $b$ have at least $3$ as a common factor.
But this contradicts the fact that $a$ and $b$ are coprime.
This contradiction has arisen because of our incorrect assumption that $\sqrt{3}$ is rational. Hence,we conclude that $\sqrt{3}$ is irrational.