- Yathirajsharma M.V. and Manjunatha M.R., "Partition of quadratic residues and non-residues in \(\mathbb{Z}_p^*\) for an odd prime \(p\)," Archiv der Mathematik, 2024 (https://doi.org/10.1007/s00013-023-01942-2).
Function: Center: Order: X Min: X Max: $$\textbf{Proof by Contradiction:}$$ 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 $$ \sqrt{2} = \frac{p}{q}. $$ Squ...
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