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February 29th, 2016, 04:45 PM   #61
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Quote:
 Originally Posted by topsquark If that's supposed to be the square root of 2 it is not rational. There is no "n" that makes $\displaystyle \sqrt{2} \cdot 10^n$ an integer. Are we sidetracked? I don't see how this applies to prime numbers? -Dan
Here is why prime is implied
Prime numbers and irrational numbers

Here is the detailed explanation.
PengKuan on Maths: Prime numbers and irrational numbers

February 29th, 2016, 04:46 PM   #62
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Quote:
 Originally Posted by Pengkuan 1.4142135623730950488… is decimal number, it is rational. It makes no difference how many decimals it has is. End of story.
Between Pythagoras' proof that $\sqrt2$ is irrational and Euclid's proof of the infinitude of primes, the ancient Greeks had already dealt your work the fatal blows more than 2000 years before you were born.

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