This method of proof by contradiction works identically for any prime root, such as $\sqrt{2}$, $\sqrt{5}$, or $\sqrt{7}$.
Memorizing this logical structure is extremely helpful as it is a highly expected question in board exams!
Step 1: Start the contradiction the usual way. Suppose, for the sake of argument, that $\sqrt3$ is rational. Then we can write $\sqrt3=\dfrac{p}{q}$ for coprime integers $p,q$ with $q\neq0$. Squaring gives $p^2=3q^2$. Step 2: Track the prime factor 3 through both numbers. Since $p^2=3q^2$, the number 3 clearly divides $p^2$, and since 3 is prime, it must divide $p$ itself. So write $p=3m$ for some integer $m$. Substituting back: $9m^2=3q^2$, which simplifies to $q^2=3m^2$. By the same reasoning, 3 must also divide $q$. Step 3: Spot the contradiction and conclude. We've now shown 3 divides both $p$ and $q$, but we assumed $p$ and $q$ share no common factor besides 1. This is a contradiction, so our starting assumption must be wrong. Therefore, $\sqrt3$ cannot be written as a ratio of two integers, and it is proved to be an irrational number.
