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Zak wrote:So we have g>1, and we haven't sqr(2)/2, sqr(2)/3, thus we can't say that we have integer value in denominator of fraction. And all the squares of all integer values are integer, but we have an irrational sqr(2), why?
x3mEn wrote:Of course, you always can build a cuboid having 3 integer lengths, but why did you decide that the space diagonal of such cuboid should be integer?
Zak wrote:Why not?
Zak wrote:it is quite possible that it has two sides as the legs of the smallest Pythagorean triple: a = 3, b = 4, and the third side crushes together with the spatial diagonal to infinity, and we will rather have not a brick, but a long beam, a straight line.
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