$$\textbf{Proof by Contradiction:}$$
Assume, for contradiction, that $\sqrt{2}$ is rational. Then there exist coprime integers \( p \) and $q$ (with $q \neq 0$) such that
$$\sqrt{2} = \frac{p}{q}.$$
\[\sqrt{2} = \frac{p}{q}.\]
Squaring both sides gives:
$$
2 = \frac{p^2}{q^2} \quad \Longrightarrow \quad 2\,q^2 = p^2.
$$
Since $p^2$ is a perfect square, every prime factor of $p^2$ appears with an even exponent. However, because 2 is square-free and not a perfect square, at least one prime factor of 2 appears only once. This mismatch in the exponents (odd versus even) leads to a contradiction.
Therefore, the assumption that $\sqrt{2}$ is rational must be false. Hence, $\sqrt{2}$ is irrational.
$$
\textbf{Proof by Contradiction:}
$$
Assume, for contradiction, that $\sqrt{2}$ is rational. Then there exist coprime integers $p$ and $q$ (with $q \neq 0$) such that
$$
\sqrt{2} = \frac{p}{q}.
$$
Squaring both sides gives:
$$
2 = \frac{p^2}{q^2} \quad \Longrightarrow \quad 2\,q^2 = p^2.
$$
Since $p^2$ is a perfect square, every prime factor of $p^2$ appears with an even exponent. However, because 2 is square-free and not a perfect square, at least one prime factor of 2 appears only once. This mismatch in the exponents (odd versus even) leads to a contradiction.
Therefore, the assumption that $\sqrt{2}$ is rational must be false. Hence, $\sqrt{2}$ is irrational.
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