# Is it possible for a physical object to have a irrational length?

Suppose I have a caliper that is infinitely precise. Also suppose that this caliper returns not a number, but rather whether the precise length is rational or irrational.

If I were to use this caliper to measure any small object, would the caliper ever return an irrational number, or would the true dimensions of physical objects be constrained to rational numbers?

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This is much more a physics question than a math question, as it is about how space behaves at very small distances. – Lieven Jan 27 at 0:04
Can one have such calipers? (Heisenberg.) – Brian M. Scott Jan 27 at 0:04
We are starting by assuming something that isn't so. And there is no such thing as the true dimension. – André Nicolas Jan 27 at 0:08
What units are you using for length? It's possible that the units in regular use in my country are an irrational multiple of the units used in yours --- then what will you do? – Gerry Myerson Jan 27 at 0:14
The problem I see with your question is that you have defined a hypothetical caliper and are complaining when people are applying it to hypothetical objects. You can only measure physical objects with physical calipers, and hypothetical objects with hypothetical calipers. – Rahul Narain Jan 27 at 0:15

## migrated from math.stackexchange.comJan 27 at 0:28

The set of irrational numbers densely fills the number line. Even assuming that quantum mechanics doesn't disable the preimse of your question, the probability that you will randomly pick an irrational number out of a hat of all numbers is roughly $1 - \frac{1}{\infty} \approx 1$.

So the question should be "is it possible to have an object with rational length?

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This is the appropriate question. – KDN Jan 27 at 2:33
All of what you have said makes sense, but isn't $1-1/\infty$ = 1 because $1/\infty$ = 0? – Nick Anderegg Jan 27 at 4:26
But wait, that doesn't actually make sense. There are an infinite number of rational numbers as well. The irrational number may be "densely" packed at whatever precision you choose, but at infinite precision, there would have to be an even distribution of both rational and irrational numbers. – Nick Anderegg Jan 27 at 4:32
@NickAnderegg: yes, there are an infinity of rational numbers. But there is a bigger infinity of irrational numbers. Namely, the number of rational number is countably infinite, while the number of irrational numbers is uncountably infinite. – Jerry Schirmer Jan 27 at 10:45
Didn't you mean that the probability of randomly picking an irrational number is $1$? @NickAnderegg Even though the set of rational number is dense in $\mathbb{R}$, it's measure is zero. This means that if we take any set and remove all rational numbers from it, we cannot tell the difference by measuring it. – Petr Pudlák Jan 27 at 16:26

Is it possible for a physical object to have an irrational length?

It's a bit of a philosophical question, but one could say this:

Just for fun, assume you have a perfect 45-degree right triangular piece of metal whose base and height is rational. Then it's hypotenuse is irrational because its length is the base times $\sqrt{2}$.

So it is possible to have a physical object of irrational length IF you can have a physical object of rational length.

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This seems to be where my whole premise falls apart and I'm not able to communicate my thinking clearly. Basically, what I'm asking is it possible for that hypotenuse to exist. Perhaps the base and height cannot both be equal because then the hypotenuse would be irrational. But otherwise, this make sense. – Nick Anderegg Jan 27 at 4:30
you also need to be able to assume you can have an object which has a perfect right angle. – RoundTower Jan 27 at 14:48

On very short distances, the position of particles isn't deterministic, but random. The electrons on the very outer edge of this object won't necessarily be the exact same distance from neighboring electrons, but occupy positions according to a wave function that obeys a number of laws and is based on neighboring electrons. So imagine your calipers are exactly flat and have infinite precision, then if the outermost electrons on each end can occupy some non-discrete interval then it would have to be able to occupy an irrational distance.

I'm not however, a physicist, and even the current state of physics isn't sure about how things act on very short distances. Below one Planck length it doesn't really make sense to measurably distinguish between the location of two points, and since there is a rational number arbitrarily close to any real number, we will never be able to actually experimentally determine the truth of this statement. So since this is really a physical question and not a mathematical one, and all current physical knowledge points in the direction that this question is unanswerable, I'm going to rule it as such.

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physical objects do not have well-defined lengths (there is this thing called quantum mechanics conceived in its entirety upon this concept). A more interesting question is if dimensionless numbers in physics can be irrational, for instance, the ratio between the mass of the electron and the proton.

Theoretically, we will need a numerical expansion and some limiting argument to tell to what domain of the reals the limit belongs (irrational, transcendental, rational). Experimentally this can never be asserted, as naturally all experimental numbers are known with a finite number of digits of precision

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Suppose your infinitely precise caliper gives the answer $2.00000000000000\dots$ How would you know whether this is $2$ exactly, or if somewhere past the trillionth decimal it starts to deviate from $2$? How would you read your infinitely precise caliper?

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Well, that's just cheating the question. These are clearly not any sort of calipers in existence. I've modified the question to accommodate this response. – Nick Anderegg Jan 27 at 0:09
You're still assuming, without justification, that there is such a thing as a "precise length" of a physical object. – Gerry Myerson Jan 27 at 0:13
But matter is quantized: atoms/quarks/...strings? Even strings are quantized. If everything is quantized we don't have infinite precision or infinite decimal places. – Raindrop Jan 28 at 2:21

If we assume that the universe is continuous, and say fix everything at a certain time frame. Then everything has an irrational length, regardless to how well we can measure it. Simply because we can define a unit of measure whose result would be irrational.

For example, measure my foot. Now define the unit of measure $1\ \small\bf Karf$ to be the square root of twice the length. Then my foot would be exactly $\sqrt\frac12\ \small\bf Karf$ long. As we know $\sqrt\frac12$ is irrational.

But this requires the assumption that the universe is continuous and that we can freeze time and measure with infinite precision. If the universe is discrete, or if we cannot measure accurately, then we can't really say too much. Not to mention that everything changes all the time (cells falling off, atoms released, etc. etc.) so there's no constant length to anything large enough.

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 I think, comming up with a scale like that isn't what the question asks for: Take the right Isosceles triangle in the example. It assumes that you meassure the side-lengths with length 1 and thus, the hypothenuse has to be $/sqrt{2}$. The question essentially is: Given you use an infinitely accurate scale in which one of the sides comes out rational, would, on a physical level, all sides be rational (two of them of miniscully different length) or could two of them possibly be exactly the same, making the third side irrational? (or the third side could be rational and the other two irrational.) – kram1032 Jan 27 at 23:32

If you are talking about real, physical objects, then your question collapses completely, because such objects are composed of particles which have no definite positions and momenta according to Heisenberg's uncertainty principle.

So lets stick to a stick in classical mechanics, then your caliper can return irrational numbers.

But a mathematical line-segment doesnt even have to have rational or irrational length, it could have an even 'finer' scale, a so called non-standard number.

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From the point of view of measure theory, the probability of measuring a rational length is actually zero.

Consider, without loss of generality, the interval $[0,1]$. Using the standard Lebesgue measure, the measure of this set (its length) is 1. If we consider the subset which consists of all the rational numbers from this set, its measure is actually 0. This starts to make sense if one considers how miniscule the size of the rational numbers is compared with all the other real numbers. In fact, it turns out that the only subsets of our interval with non-zero measure are continuous ones (eg $[a, b]$, where $a<b$ and the measure is $b-a$) and ones that contain so-called normal numbers. Only the normal numbers are said to 'take up any space' on the real number line. That is, virtually all the real numbers are actually normal numbers (which can never be written down on paper), and so the probability of measuring anything that's not a normal number is 0.

http://en.wikipedia.org/wiki/Normal_number

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Let's take the smallest possible case of such a triangle. It would be made of three atoms of equal size, linked together in a L-shape with a 90° angle in between.

If you have an arrangement like that, and something similar might be chemically possible, the centers of mass of the more distant two atoms would be apart [exactly][1] $\sqrt{2}\times$the distance between the directly touching ones.

Presumably, if you take a more rigorous and accurate approach, if you look at the bonding structure of water (which, of course, won't feature a right angle but the situation is equivalent), the centers of mass of the two Hydrogen atoms would also be an irrational distance appart compared the the distances of the centers of mass of each Hydrogen to the Oxygen. No matter what scale you use, at least one of the two distances will always be irrational.

If you can somehow limit the set of all possible distances to a countable infinity, I'd suspect this set not to be the rationals but rather the algebraic numbers. (or at least the subset of them that are positive)

[1]: modulo Heisenberg but I didn't use proper orbitals either. Let's, for the sake of the argument, define a distance on quantum level by the distances of expected values of the corresponding probability clouds.

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One can give an argument based on measure theory and the like, but one must not forget that physics is about measurement. The question whether the length can be rational or irrational would need an infinitely precise measurement, which is not possible (measurements bear an error). Hence this question cannot be answered from the physics viewpoint. Any answer will be just speculation.

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The hypotenuse of a right angled triangle with legs 1 is irrational.

Alternatively, consider a pyramid. As you take measurements of the 'base length' towards the apex, you get a continuous sets of values. One of these must be irrational.

Of course, you can then start an argument about what 'physical' object is, and if length is truly continuous, or it has to be discrete because it is constructed by atoms.

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 Well, but then you have to find a right angled triangle, and you have to have the legs equal exactly $1$ of something rational... – Asaf Karagila Jan 27 at 0:09 A theoretic triangle is not a physical object. – Nick Anderegg Jan 27 at 0:10 @NickAnderegg How about your set square? – Calvin Lin Jan 27 at 0:10 The question is asking more about precision. It's more along the lines of "Can I have a physical triangle with a hypotenuse of $\sqrt[]{2}$. Perhaps it wouldn't be possible to construct a triangle with legs that are exactly 1 unit. Perhaps one leg is so slightly shorter in a way as to allow a hypotenuse near $\sqrt[]{2}$. – Nick Anderegg Jan 27 at 0:14

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