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Is there any theorem saying that in below fraction we can't rationalize the denominator
$\frac{2}{\pi}$
I couldn't find any way and I don't think there is but I was wondering if this actually proved?
for example for $\frac{1}{\sqrt{2}}$ we say
$\frac{1}{\sqrt{2}}=\frac{1\times\sqrt{2}}{\sqrt{2}\times\sqrt{2}}=\frac{\sqrt{2}}{2}$
result have meaning if we look at fraction like Division

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    $\begingroup$ You can't do the same thing because $\pi$ is not algebraic. $\endgroup$ – mkspk Sep 17 '15 at 13:32
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Are you speaking of rationalizing the denominator and the numerator at the same time? If not, a trivial solution would be $\frac{2/\pi}{1}$, but that doesn't seem very interesting. So perhaps you are speaking about some sort of manipulation that creates $\frac{2}{\pi}=\frac{p}{q}$, where both $p$ and $q$ are rational. But, I would find that greatly disturbing, because then it would be the case that $\frac{p}{q}$ is also rational... in which case $\frac{2}{\pi}$ would be rational as well.

EDIT: Your example with $\sqrt 2$ is similar to the trivial solution above, but, in the case of a radical, might have some more use.

So, yes, you can rationalize the denominator, if you find such action useful. But, would you, in the case of $\frac{2}{\pi}$?

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  • $\begingroup$ i edited my question hope know it is clear what I meant $\endgroup$ – user148528 Sep 17 '15 at 13:08
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    $\begingroup$ +1 $\displaystyle\frac{2}{\pi}=\frac{p}{q}\implies \pi = \frac{2q}{p}$ $\endgroup$ – John Joy Sep 17 '15 at 13:19

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