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I have learned that $\pi$ is an irrational quantity and a product of an irrational number with a rational number is always irrational. Does This imply that area of a circle with radius $r$, which is $\pi$.r$^2$ is always an irrational quantity? 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?

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    $\begingroup$ Consider $r=1 / \sqrt{\pi}$ for example. $\endgroup$ – dxiv Nov 8 '16 at 4:43
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    $\begingroup$ This is because you maybe taking $/pi$ as 22/7 which it is not. $\endgroup$ – deepa kapoor Nov 8 '16 at 5:22
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    $\begingroup$ "when I take r=7, the answer come out to be 154" No, it doesn't.... $\pi r^2$ = 49\pi \ne 154$. Why on earth did you say the answer was 154. $\endgroup$ – fleablood Nov 8 '16 at 5:36
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    $\begingroup$ "Does this mean I can never calculate the "exact" area of a circle" !!!NO!!!!! The If $r = 7$ then the area is $49\pi$. That IS a calculation and it !!!!IS!!!! and exact calculation. Just because something is expressed an irrational number does !!!!NOT!!!! mean it neither calculated or not exact. That is a naive misconception with no justification. $\endgroup$ – fleablood Nov 8 '16 at 5:43
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    $\begingroup$ @fleablood Kindly use *markdown formatting* in the future. Otherwise it looks like you're screaming. $\endgroup$ – grooveplex Nov 8 '16 at 11:30
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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.

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    $\begingroup$ Isn't it easier to say $a,b \in \mathbb{N}$ with $b \neq 0$ $\endgroup$ – Mindwin Nov 8 '16 at 11:43
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    $\begingroup$ @Mindwin Perhaps, but if it is true then why would you avoid saying it was true for integers? It's a little subjective to say it is 'easier' in any case ;) $\endgroup$ – Sam Walls Nov 8 '16 at 14:57
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    $\begingroup$ Since 0 isn't positive, the $b \neq 0$ is redundant. Probably the simplest way to write this is as "$a,b$ to be any positive integers". If you want symbols, you could go with $a,b \in \mathbb{Z}^{+}$. $\endgroup$ – jpmc26 Nov 9 '16 at 0:01
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    $\begingroup$ Are you aware that you cannot take a irrational radius? how do you measure it? (joking, but not so much). $\endgroup$ – GameDeveloper Nov 10 '16 at 12:19
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    $\begingroup$ @DarioOO: show us a real circle first, then we'll start thinking about how to measure it... $\endgroup$ – RemcoGerlich Nov 10 '16 at 12:27
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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.

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    $\begingroup$ does this mean that for a circle with radius r such that r is rational the area of a circle is not exact but for an irrational r it is? $\endgroup$ – deepa kapoor Nov 8 '16 at 4:49
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    $\begingroup$ "the square of an irrational number is also a rational"? $\endgroup$ – Joel Reyes Noche Nov 8 '16 at 4:50
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    $\begingroup$ @deepa Were you saying "yes" to my question? Are you saying $\pi^2$ is rational? $\endgroup$ – Joel Reyes Noche Nov 8 '16 at 4:53
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    $\begingroup$ The square of an irrational is not always irrational: $\sqrt{2} \cdot \sqrt{2}=2$, which is certainly not irrational. $\endgroup$ – mathematics2x2life Nov 8 '16 at 4:59
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    $\begingroup$ Pi is not merely irrational, but transcendental, so raising it to an integer power will never produce an integer. For what it's worth, although Pi is probably the best know of them, most real (and complex) numbers are transcendental--to be more specific, transcendental numbers are uncountable, while algebraic numbers (e.g., roots of integers) are countable. $\endgroup$ – Jerry Coffin Nov 8 '16 at 16:59
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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.

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    $\begingroup$ You have confused irrational with integer $\endgroup$ – polfosol Nov 8 '16 at 10:34
  • $\begingroup$ ...I mean, the definition of a rational number is a ratio of two integers. $\endgroup$ – LegionMammal978 Nov 8 '16 at 11:22
  • $\begingroup$ +1. This aspect of technically constructing a circle with radius $1\mathrm m$ (which I'd typeset upright as \mathrm m) appears related to Irrational numbers in reality. $\endgroup$ – MvG Nov 8 '16 at 11:45
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This is a sneaky blind spot that a lot of people get. When asked if there are any solutions to

$$ A = \pi r^2$$

where $A$ is rational, they think "what can I pick for $r$?"

But that is exactly the wrong way to approach this problem! The real trick is to think "what can I pick for $A$?" Once you decide to pick, say, $A=1$, it's a trivial matter to find the corresponding $r$.

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You need to keep the following three postulates about rational and irrational numbers in mind :

  1. Product of a non-zero rational number with an irrational number is ALWAYS irrational.

  2. Product of two irrational numbers is SOMETIMES rational.

  3. Product of two rational numbers is ALWAYS rational.

Now for a given radius $r$ of a circle we can have following cases:

Case 1: $r$ is a rational number(>0). In this case the area of a circle will ALWAYS be irrational, since $\pi$ is irrational and $r^2$ always rational.

Case 2: $r$ is an irrational. In this case $r^2$ can either be rational (e.g. r= $\sqrt{2}$, then r$^2$=$\sqrt{2}$.$\sqrt{2}$ = $2$) giving the area of a circle ALWAYS as irrational, Or, r$^2$ can be irrational giving the area of a circle SOMETIMES as rational (e.g, r= $\sqrt{1/\pi}$, then r$^2$= $1/\pi$) and SOMETIMES as irrational.

As to your other question about the accuracy of the area of a circle, then it is always accurate to express it in terms of the number $\pi$. The integer representation of the area of a circle may sometimes lead to a rational approximation of the otherwise irrational quantity at times depending upon $r$.

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More generally, note that the function

$$f:\mathbb{R}_{+} \to \mathbb{R}_{+}$$

given by $f(r) = \pi r^2$ is a bijective map. Consequently, not only can the area of a circle be rational, but there exists a unique number $r_c$ such that for any positive number $c$ there exists a circle of radius $r_c$ and area $c$.

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  • $\begingroup$ Yeah, for the positive quantities involved, we can write $r=\sqrt{\frac{A}{\pi}}$ which is really a formula for what you call $f^{-1}$ and shows how to find the sought for $r = r_c$ for any desired area $A=c$. $\endgroup$ – Jeppe Stig Nielsen Nov 9 '16 at 9:35
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Only for circles with a rational radius. You can work the formula backwards, define A as "10", and there will be some wacky, non-terminating radius length out there that gives you a nice, neat area value.

You're not far off, but the conclusion would have to be worded "Either the area OR the radius will always be incalculable" in order to be true. Either one can be a normal number- just not both.

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  • $\begingroup$ This is the best answer, in my opinion. Straight to the point and without the unnecessary use of equations. $\endgroup$ – Spencer Müller Diniz Jan 27 '18 at 9:59
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The short summary is that it's trivial to define a circle with a rational area, but if you do so you end up with a diameter/radius that's irrational.

Pi defines the ratio of the diameter to circumference, and it's irrational. Moreover, it's not just rational, but also transcendental. That means (among other things) that raising it to any integer power still results in an irrational result.

Therefore, the same basic idea applies to things like the surface area or volume of a sphere: you can define a sphere with an irrational diameter/radius that has a rational volume or surface area, or you can define it with a rational radius/diameter, in which case it'll have an irrational volume/surface area.

Moving to an object in more dimensions doesn't change that either. The volume (or surface area) of an n-dimensional sphere can be slightly more complex to compute, but the ratio of radius/diameter to area/volume still involves Pi (in a way that doesn't cancel with another Pi, or anything like that), raised to an integer power (directly proportional to the number of dimensions) so we still get the basic fact that if the radius/diameter is rational, the surface area/volume will be irrational, and if the area/volume is rational, the radius/diameter will be irrational (though there's one corner case that might exist where both the radius and area are rational: with a radius of zero, you obviously get an area of zero, but it may be open to argument whether this qualifies as a circle, or just a point).

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You already got plenty of answers about rationals, irrationals, and examples for circles with rational area.

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?

Let me expand the topic a bit: You have to clearly separate between mathematics and reality. As noted in other answers, irrational numbers are just as exact as any other number, speaking in mathematical terms.

But mathematics is pure theory. It exists only in our minds, it is a construct of logic. Mathematical concepts do not exist in reality. You can never ever measure any length in reality and get any mathematically precise answer.

If I tell you to give me the exact, precise measure of "2" (no matter what unit - let's call it "2 meters") - in reality, I'll have to tell you that I can't. Nobody will be able to measure anything (not length, nor time, nor temperature etc.) down to mathematical precision. There is always some inaccuracy.

For length, this is easy to see. Even if we could measure a large quantity like 2m down to the accuracy of individual atoms, you are still not "there". Atoms are not billiard balls that have a defined volume, so you cannot say "my stick ends exactly at the outer border of this atom". You cannot even say "the outer border of this atom is where that electron is currently" because the location of an electron is not working like that (=> Heisenberg). You don't even find consensus between physicists what it means to say that two atoms are touching or whether atoms actually are ever touching in any intuitive sense (they very likely are not).

Even then, atoms always move around, even at a hypothetical 0°K temperature.

And then let's not get into how - if the thing to be measured is inherently undefined - you would define the thing that does the measuring...

So,

  • yes, you can trivially calculate the exact area of a circle for any kind of number whatsoever, and
  • no, it does not make sense to use the term "exact" with any number, even the most trivial ones, as soon as you are not only thinking about theoretical mathematical objects which, per se, never have an actual represenation in the "real" reality.
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