Q.Verify by the method of contradiction.
p: $\sqrt{7}$ is irrational

Claim: $\sqrt7$ is irrational.

Proof by contradiction: Suppose, for contradiction, that $\sqrt7$ is rational. Then we can write $\sqrt7 = \dfrac{p}{q}$ where $p, q$ are integers with $q \neq 0$ and $\gcd(p, q) = 1$ (the fraction is in lowest terms).

Squaring both sides: $7 = \dfrac{p^2}{q^2}$, so $p^2 = 7q^2$.

This means $7$ divides $p^2$. Since 7 is prime, it follows that $7$ divides $p$. Write $p = 7k$ for some integer $k$.

Substituting: $(7k)^2 = 7q^2 \Rightarrow 49k^2 = 7q^2 \Rightarrow q^2 = 7k^2$.

This means $7$ divides $q^2$, and since 7 is prime, $7$ divides $q$ as well.

But now both $p$ and $q$ are divisible by 7, which contradicts our assumption that $\gcd(p, q) = 1$.

This contradiction shows our assumption was false. Hence $\sqrt7$ cannot be written as a ratio of two integers, i.e., $\sqrt7$ is irrational.

Assume the negation (√7 is rational, expressible as p/q in lowest terms), manipulate algebraically to show 7 must divide both p and q, contradicting the lowest-terms assumption — this contradiction proves the original statement true.

• Forgetting to first reduce p/q to lowest terms (gcd = 1) — without this, no contradiction arises.
• Skipping the step of showing 7 | p² ⇒ 7 | p (this uses that 7 is prime).