# Is the diagonal of the unit square truly irrational? [closed]

I think the length of the diagonal of the unit square cannot be in the same class as transcendentals such as pi. I can make the length of the diagonal equal to exactly 2, for example, if I make the length of its sides equal to sqrt(2).

Further proof: A circle should be viewed as polygon with infinite number of sides. I can reduce those numbers of sides to 4 for a square and look at the diagonal like a proto-"diameter", then the square's diagonal cannot be a irrational, because this proto-diameter of the square cannot bear the same irrational and transcendental relationship as the diameter to a circle without violating the geometrical perfection of a circle.

What should be said about it then? The diagonal would be put into a new category of unknowable, rational numbers. It is unknowable because the relationship between diagonal and the length of its sides must retain indeterminate regardless of scale, yet non-transcendental because of the relationship noted above in reference to circles.

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## closed as not a real question by Asaf Karagila, Lord_Farin, Austin Mohr, Pedro Tamaroff, nullUserJun 5 '13 at 19:48

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Square root of 2 is indeed irrational. –  Sujaan Kunalan Jun 5 '13 at 19:47
"...a circle is defined as an infinitely-sided polygon, and therefore uncountable, a unit square is not." makes no sense. –  Pedro Tamaroff Jun 5 '13 at 19:50
A circle is the set of all points in the plane that are a certain fixed from a certain other point. The fixed distance is called the "radius" of the circle and the other point is called the "center" of the circle. –  MJD Jun 5 '13 at 20:08
That again makes no sense at all. –  Pedro Tamaroff Jun 6 '13 at 17:14

@MarkJ - Try disproving this elementary proof: $\sqrt{2} = \frac{m}{n}$, where $m$ and $n$ are the smallest possible. Then $2n^2 = m^2$ so $m$ must be even and $m = 2r$ for some integer $r$. Then $2n^2 = 4r^2$ and $n^2 = 2r^2$ implies $n$ is also even and $n = 2s$ for some integer $s$. So $\frac{m}{n} = \frac{2r}{2s} = \frac{r}{s}$ which is a contradiction since the initial fraction was the smallest. –  Legendre Jun 7 '13 at 14:06