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Let $x, y$ be real numbers, $x \gt 0, y \gt 0,$ and let $n$ be an integer $n > 0$. If $x < y$, then $\sqrt[n]{x} \lt \sqrt[n]{y}$. Proof. Recall that $(\sqrt[n]{x})^n = x$, $(\sqrt[n]{y})^n = y$. Thus \begin{align*} (\sqrt[n]{x})^n \lt (\sqrt[n]{y})^n\\ \iff (\sqrt[n]{y})^n - (\sqrt[n]{x})^n \gt 0\\ \iff (\sqrt[n]{y} - \sqrt[n]{x})\sum_{i=0}^{n} (\sqrt[n]{x})^i \cdot (\sqrt[n]{y})^{n-1-i} \gt 0\\ \iff (\sqrt[n]{y} - \sqrt[n]{x}) \gt 0\\ \iff \sqrt[n]{y} \gt \sqrt[n]{x}. \end{align*} QED.
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Mon, 07 Aug 2023 21:05 GMT