Is the area of a circle ever an integer? Is the area of a circle ever an integer?


I was trying to answer someone else's question on yahoo answers today and I got thumbs down from people on my answer and have come here to get a thorough proof on it because now I just must know! :)

My assertion: Suppose we have two integers a and b. We can form a rational numbers out of each of these: $\frac{a}{1}$ and $\frac{b}{1}$. Now forming another rational by dividing them: $\frac{\frac{a}{1}}{\frac{b}{1}}$ = $\frac{a}{b}$. Now if we divide this by itself: $\frac{\frac{a}{b}}{\frac{a}{b}}$ = $\frac{ab}{ab}$ = 1. Therefore, we can conclude that given any rational number, we can always describe it using only integers thus always resulting in an integer number.

However, even if our radius is $\sqrt{\frac{1}{\pi}}$ then we will have $\pi\sqrt{\frac{1}{\pi}}^2$ = $\frac{\pi}{\pi}$, by definition of area = $\pi r^2$.

Now I think this the trickiest part. Does  $\frac{\pi}{\pi}$ = 1? Or can it only equal  $\frac{\pi}{\pi}$? How do we prove that  $\frac{\pi}{\pi} \neq$ 1? Or is my method sufficient at all? Thanks!
By definition, $\frac{1}{b}$ is the unique real number which, when multiplied by $b$, yields $1$.

By definition, $\frac{a}{b}$ is the product of $a$ and $\frac{1}{b}$.

Since $\frac{1}{\pi}$ is the unique real number that, when multiplied by $\pi$, yields $1$, then $\pi\left(\frac{1}{\pi}\right) = 1$. Hence, $\frac{\pi}{\pi}=1$.

If you allow any radius for a circle, then a circle has integer area if and only if its radius $r$ is the square root of an integer divided by $\pi$, that is, $r = \sqrt{a/\pi}$ for some nonnegative integer $a$. 

Your first paragraph, though, is completely irrelevant.
This is not a proof; Arturo has said all there is to it. But physical world thought experiments and arguments hint strongly that yes, there should be a circle with area 1. I have the following reasons for this. The general argument is:

"All sizes have been created equal." Consider the counter-question: Why should no circle have an integer-sized area? What, in your opinion, distinguishes a real-world area of one size from an area of another size? Have you seen Vi Hart's math lesson rant about parabolas? At 2:50 she says:


  All parabolas are exactly the same, just bigger or smaller or moved
  around. Your teacher could just as well just hand you parabolas
  already drawn and have you draw coordinate grids on parabolas rather
  than parabolas on coordinate grids.


This still makes my eyes water because this perfectly ordinary math lesson is so sad and Vi Hart is so smart and the rant is so funny. All at the same time.&mdash;

Back to the issue: What's true for the conic section called parabola is even truer (ha!) for the conic section called circle.


The numerical value of a real-world circle's area depends on the unit you measure it with! Obviously you have no doubt that the area of a circle can be $\pi$; well, you see the point &ndash; if you change your area measuring unit so that one unit is the area of $\pi$ units in the old system, voilà: circle of area 1!  This is not a trick argument; our common length- and associated area units are completely arbitrary. There simply are no area sizes which are special. (This is the real-world isomorphism of Arturo's mathematical argument.)
You can just shrink an image with a circle of area $\pi$ by a linear factor of $\sqrt{\pi}$. Voilà: Area 1. (This is another real-world isomorphism of Arturo's mathematical argument.)
Say, you roll dough out; you can give it any shape you like, square, circle, whatever. Obviously you can have a square of area 1; there is no reason why it should be more difficult to shape it into a circle than any other square. (No, this is not the unsolvable "square a circle" (or vice versa) problem. I do not claim that the circle's radius would be rational.) Perhaps it's easier to think in terms of volumes: If you put a water balloon in a cubic liter container and fill it with water, you have a cube with the volume 1 (if you measure in liters, see first bullet point). Now take the balloon out; if it's a perfect balloon and weightless (because you throw it &ndash; what else would you do with a water balloon ??), it will be a sphere with the volume 1, if we consider water incompressible. This applies analogously to areas.