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If there are none why not?

Thanks in advance.

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    $\begingroup$ Nigh-trivially, if there were then the square root of two would be rational ($a^2+a^2=c^2$, so $2a^2=c^2$, so $2=\frac{c^2}{a^2}$). $\endgroup$ – Steven Stadnicki Jul 19 '13 at 21:58
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Note that this can be written as $$2=\left(\frac{c}a\right)^2$$

Is $\sqrt 2$ rational?

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