# 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.

If I look at the diagonal like a proto-diameter akin to a circle, then the diagonal cannot be irrational, because this proto-diameter cannot bear the same irrational relationship as the diameter to a circle without violating a principle of geometry (i.e. a circle as a polygon with infinite sides).

If true, then the diagonal would be either something like an "unknowable rational" and non-transcendental (something like inf/inf).

-
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
You don't need to worry about $\sqrt2$. You need to grapple with Zeno's paradox of the arrow first. –  MJD Jun 5 '13 at 20:20
That again makes no sense at all. –  Pedro Tamaroff Jun 6 '13 at 17:14

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

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

@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