Question 708870
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Assume *[tex \LARGE \exists\,a,b\,\in\,\mathbb{Z}\,:\,\sqrt{3}\ =\ \frac{a}{b}\ \text{and }\frac{a}{b}\ \text{irreducible}]


Square both sides:  *[tex \LARGE\ \  3\ =\ \frac{a^2}{b^2}]


Then we can say:  *[tex \LARGE\ \  3b^2\ =\ a^2]


If *[tex \LARGE b] is odd, then *[tex \LARGE b^2] is odd because the product of any two odd numbers is odd.  Similarly, *[tex \LARGE 3b^2] must then also be odd.  Consequently, *[tex \LARGE a^2] is odd, and since by definition an odd number has no even factor, *[tex \LARGE a] must be odd.


If *[tex \LARGE b] is even, then *[tex \LARGE b^2], *[tex \LARGE a^2], and *[tex \LARGE a] must be even.   But if *[tex \LARGE a] and *[tex \LARGE b] are both even, then *[tex \LARGE \frac{a}{b}] is not irreducible.  Consequently, *[tex \LARGE a] and *[tex \LARGE b] must both be odd.


If *[tex \LARGE a] is the *[tex \LARGE m]th odd number, *[tex \LARGE a\ =\ 2m\ +\ 1].  Likewise, *[tex \LARGE b] can be represented as *[tex \LARGE 2n\ +\ 1]


Go back to *[tex \LARGE 3b^2\ =\ a^2] and substitute:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3(2n\ +\ 1)^2\ =\ (2m\ +\ 1)^2]


A little algebra (verification left as an exercise for the student) gives us:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6n^2\ +\ 6n\ +\ 1\ =\ 2(m^2\ +\ m)]


Since the LHS is clearly odd and the RHS is clearly even, this equation has no solutions over the integers.  Therefore, reductio ad absurdum, the square root of 3 cannot be expressed as a rational number, hence *[tex \LARGE \sqrt{3}] is irrational.   


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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