Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

closed as not a real question by Asaf Karagila, Lord_Farin, Austin Mohr, Pedro Tamaroff, nullUser Jun 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.

11  
Square root of 2 is indeed irrational. –  Sujaan Kunalan Jun 5 '13 at 19:47
5  
"...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
3  
2  
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
4  
That again makes no sense at all. –  Pedro Tamaroff Jun 6 '13 at 17:14
show 10 more comments

1 Answer 1

There are many proofs such as these.

Since it has been proven, the only possibility is that your proof is wrong. Perhaps you should focus on finding the error.

share|improve this answer
    
It has been proven, given a premise: Greek geometry and the natural numbers can be related in some way. My question really is about a definition. I think the diagonal can be represented by a countably infinite continued fraction, and should therefore simply be defined as a rational number. –  Mark J Jun 6 '13 at 4:59
    
Have you read and understood all 7 proofs on that Wikipedia page? –  Legendre Jun 6 '13 at 10:04
    
@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
    
I'm going to guess that you're equivocating with the word "is"/"are" and the symbol "=" (more specifically, your use of "are" in the first sentence). You can't always use them interchangeably. Question on definitions: Is 2i (of the complex domain) an "even" number? –  Mark J Jun 7 '13 at 17:33
    
I will tell you that I think the field has been conflating various domains that have only recently come to light. One of the main ones is the Platonic domain of geometry with the purely quantized realm of rationals. Think about how one can logicially and consistently interchange between a semi-discrete domain with a continuous domain of geometry and you'll see where the sqrt(2) lies at the boundary of both. I'm looking for collaborators to elucidate some of this. –  Mark J Jun 7 '13 at 18:01
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.