Question 243988
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Prove *[tex \LARGE \sqrt[3]{6}] is irrational.


Assume *[tex \LARGE \sqrt[3]{6}] is rational.  That means *[tex \LARGE \exists a,\,b\ \in\ Z\ |\ \frac{a}{b}\ =\ \sqrt[3]{6}] where *[tex \LARGE \frac{a}{b}] is reduced to lowest terms, that is *[tex \LARGE a] and *[tex \LARGE b] have no common integer factors.


It follows then that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{a^3}{b^3}\ =\ 6]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^3\ =\ 6b^3]


Since at least one factor of the RHS is even, the entire RHS must be even.  Since the RHS is even, the LHS must therefore also be even.  Since the product of two odd numbers is always odd, it follows that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^3\, \text{mod}\,2 =\ 0\ \Rightarrow\ a^2\, \text{mod}\, 2\ =\ 0\ \Rightarrow\ a\, \text{mod}\, 2\ =\ 0]


Since *[tex \LARGE a] is even, it follows that *[tex \LARGE \exists\ c\ |\ c\ =\ \frac{a}{2}\ \Rightarrow\ 2c\ =\ a]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (2c)^3\ =\ 6b^3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2^2c^3\ =\ 3b^3]


Now, since at least one factor of the LHS is even, the LHS must be even.  Therefore the RHS must be even and *[tex \LARGE b^3] must be even.  But if *[tex \LARGE b^3] is even, then *[tex \LARGE b] must be even.


Since the the original assumption leads to the conclusion that both *[tex \LARGE  a] and *[tex \LARGE b] are even, contradicting the part of the original assumption that *[tex \LARGE a] and *[tex \LARGE b] have no common integer factors, the assumption that *[tex \LARGE \sqrt[3]{6}] is rational must be false.


Therefore *[tex \LARGE \sqrt[3]{6}] is irrational.  QED.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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