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![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Jul 2008 21:40:02 IST
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![\text{ Let cube root of 2 is rational}\\\\Then\ \sqrt[3]{2}=\frac{p}{q}\ \text{ (where (p,q)=1 and q is not equal to 0) }\\\\\text{Cubing both sides}\\\\2=\frac{p^3}{q^3}\\\\\Rightarrow p^3=2q^3\\\\let\ p=2k\\\\8k^3=2q^3\\\\Or\ q^3=2(2k^3)\\\\\text{ So 2 is a common factor for both p and q}\\\\\text{ Hence our assumption is wrong as (p,q)=2 and not (p,q)=1}\\\\\text{Hence}\ \sqrt[3]{2}\ \text{is irrational}](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/4/4/9/44962636a861a42cf151308094565057aa69eb60.gif)
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God does not care about our mathematical difficulties. He integrates empirically. ~~~Albert Einstein (1879-1955)~~~~
To divide a cube into two other cubes, a fourth power or in general any power whatever into two powers of the same denomination above the second is impossible, and I have assuredly found an admirable proof of this, but the margin is too narrow to contain it.~~~Pierre de Fermat (1601-1665)~~~
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