Q.Prove by the method of contradiction that $\sqrt{3}$ is irrational.

Assume the opposite (that $\sqrt3$ is rational) and show it forces a contradiction — both the numerator and denominator of the lowest-terms fraction end up divisible by 3.

To prove: $\sqrt3$ is irrational.

Proof (by contradiction):

Suppose, if possible, $\sqrt3$ is rational. Then $\sqrt3=\dfrac pq$ for integers $p,q$ with $q\ne0$ and $\gcd(p,q)=1$ (the fraction is in lowest terms).

Squaring: $3=\dfrac{p^2}{q^2} \implies p^2=3q^2$. $\quad(*)$

So $3$ divides $p^2$. Since $3$ is prime, $3$ must divide $p$. Write $p=3m$ for some integer $m$.

Substitute into $(*)$: $(3m)^2=3q^2 \implies 9m^2=3q^2 \implies q^2=3m^2$.

So $3$ divides $q^2$, and since $3$ is prime, $3$ divides $q$ too.

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