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anonymous
 4 years ago
How can I use this theorem
"Let \(\alpha\) be a root of the polynomial \(x^n+c_{n1}x^{n1}+...+c_1x+c_0\), where the coefficients \(c_0, c_1, ..., c_n\) are integers. Then \(\alpha\) is either an integer or an irrational number."
to show that \(\sqrt[3]{5}\) is irrational?
anonymous
 4 years ago
How can I use this theorem "Let \(\alpha\) be a root of the polynomial \(x^n+c_{n1}x^{n1}+...+c_1x+c_0\), where the coefficients \(c_0, c_1, ..., c_n\) are integers. Then \(\alpha\) is either an integer or an irrational number." to show that \(\sqrt[3]{5}\) is irrational?

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JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1The solutions of \( x^3 5 \) must therefore be integral or irrational, and it's not hard to show they are not integers.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ah! I see. I will try that now. Thank you.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Proof: The solutions of \(x^35\) must be either integral or irrational. Thus, we are looking for an integer \(x\) such that \(x^3=5\). However, there is no such integer since \(1^3=1<5\) and \(2^3=8>5\). Therefore, \(x\) must be irrational, and \(\sqrt[3]{5}\) is irrational. \(\blacksquare\) Is this okay?
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