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.