Is Area of a circle always irrational

Solution 1:

Just because $A=\pi r^2$ has a $\pi$, which is irrational, does not mean that $A$ has to be irrational. This is because $r^2$ could be irrational. For example, take $a,b \in \mathbb{Z}$ to be any positive integers with $b \neq 0$. We can form the fraction $a/b$. Then taking $r=\sqrt{\dfrac{a}{b\pi}}$, we have $$A=\pi r^2=\pi \sqrt{\dfrac{a}{b \pi}}^2=\dfrac{a\pi}{b\pi}=\dfrac{a}{b}$$ which is certainly rational.

Essentially, this results from the fact that product of two irrational numbers need not be irrational. We take the irrational number $\pi$ and multiply by the irrational number $r^2$ and get a rational. Another example would be $\sqrt{2} \cdot \sqrt{2}=2$. However, it is certain that if $r \neq 0$ were rational then $r^2$ would be rational and then the area $A$ would be irrational.

Solution 2:

No, the square of an irrational number may be irrational, but the product of two irrational numbers can be rational.

As @dxiv said, consider $r=\frac{1}{\sqrt{\pi}}$.
$\pi r^2 = 1$ in this case.

$1$ is certainly rational.

Solution 3:

There are 2 questions here:

First:

Is Area of a circle always irrational?

Answer: No.

Make $r= \frac{1}{ \sqrt{ \pi}}$ like in E. Nusinovich’s answer or more generally mathematics2x2life answer.

Second:

Does this mean I can never calculate the "exact" area of a circle but always just an approximate which always has some "room" for more accuracy?

Answer: No.

One should not confuse irrational with inaccurate. $ \pi$ is an irrational and precise number.

If $r=1$, $A= \pi$ (or using $r=7$, $A=49 \pi$ from comments above) this calculation is precise. $ \pi$ is the symbolic representation of the entire chain of that infinite sequence of numbers. It is as accurate as the symbol $ \infty$ is (add to the list $i$ and $e$).

So, if one could technically construct a circle with radius of $1 \mathrm m$, then its area would be precisely $ \pi \mathrm m^2$. Now you may dispute on how one could technically construct a circle with exact radius of $1 \mathrm m$. But this is an entirely different problem.

If you ask any trained experimentalist he would tell you that any measurement contain an associated error, but that don’t make our calculation inaccurate, since by this same argument, not even the area of a square would be possible to measure precisely since one cannot precisely measure the side (length) of a square with absolute accuracy.