# How irrational quantities physically exist in nature?

We know that an irrational no has well defined decimal values upto infinite decimal places. These irrational quantities exist in nature in some kind of measurements. For an example, circumference of a circle is '2πr' , so if radius is rational then circumference will be irrational ,and this case is quite natural. But I am unable to understand , how can a physical measurement exist with infinite precision (due to precision of infinite decimal values in an irrational no) ?? Isn't it in contrary with the laws of Physics??

• Comments are not for extended discussion; this conversation has been moved to chat.
– Pedro
Sep 17, 2016 at 22:15
• I seem to have learned that my answer wasn't a great answer. What did you really mean to ask? Did you mean "Can we determine $\pi$ to arbitrary precision" using a Physics experiment? Would a better answer say something like "It's a mathematical constant so we can calculate a lot more digits than a simple experiment tells you, unlike the dimensionless fine structure constant which requires observations?" Did you define $\pi$ to be a different value for Earth because of curved space inside it and want to know whether $\pi$ for Earth can be measured that accurately? Nov 22, 2018 at 19:05
• Might be worth noting most numbers are irrational. There are countably many rationals and uncountably many irrationals. Pick a real number "at random" and its almost certainly irrational. Mar 13 at 15:01

I've got bad news for you: Even rational quantities don't exist in nature. Numbers, of whatever type, are just a way to describe our observations. But when we measure, we always have a measurement error, so we cannot really say "this measured quantity has the value $x$", all we can say is "this measured quantity is close to the value $x$". Well, we can do a bit more: Namely quantify how close it is. This is why measurement results are normally written in the form $x\pm y$. For example, a length might be measured to be $1.3\pm 0.5~\rm mm$.

Now we pretend that there is a "true value" of the measured quantity. For example, in the above example, we assume that there is a true value for the length, which is likely(!) less than $0.5~\rm mm$ away from the value $1.3~\rm mm$.

Exact numbers are therefore nothing we can measure, and nothing which we can even prove really exist in nature. They are abstract concepts we made up in our mind in order to describe what we observe. And they work very well for this purpose. But whether they really correspond to something "out there", nobody can really know.

But wait, you say, there are circles, and for those we can prove the circumference over diameter is $\pi$. So $\pi$ is part of our world, right? Wrong: You never have seen a circle. You've certainly seen lots of shapes that closely fit our concept of a circle. But likely if you look close enough, you'll find that it is not really a circle. And even if you cannot detect the difference, how do you know that if you had looked just a little bit closer, you'd have found a deviation?

And ultimately, those circle-lookalikes you've seen probably were all made of atoms anyway, which means they could not have been a circle, as they have volume, while the circle is an infinitely thin line.

• Comments are not for extended discussion; this conversation has been moved to chat.
– Pedro
Sep 17, 2016 at 22:14
• You seem to be mentioning only our inability of precision measuring the lengths which exist in nature; our inability of not measuring lengths exactly as they are, seems to not imply their non existence. Dec 20, 2018 at 13:39
• @ImmortalPlayer: “You seem to be mentioning only our inability of precision measuring the lengths which exist in nature” — no. I mention our inability to measure an exact length, period. The part “which exists in nature” is an assumption that cannot be derived from any experience. That's the whole point. Sure, it doesn't logically imply their non-existence. But in science you have to justify existence, not non-existence. And actually, the success of quantum mechanics strongly hints at there not being an actual “length which exists in nature”. Jan 27, 2019 at 9:36
• You might agree that lengths of one cm exist in nature? If we have a triangle of one cm height and one cm adjacent side, won't we be having hypotenuse of irrational length in reality? Feb 10, 2019 at 14:22
• @ImmortalPlayer: “You might agree that lengths of one cm exist in nature?” — No. When measuring with a not too precise measurement device, we will find that its length deviates from 1cm by less than our measurement precision. When using an extremely precise measurement device, we'll find that it doesn't have a fixed length at all. Feb 10, 2019 at 15:22

One odd thing about the question is the apparent belief that the decimal notation of a number is somehow fundamental to its meaning, and so an unending decimal expansion is a sign that the number itself is somehow unending or imprecise.

First, a simple factual point: the question claims:

We know that an irrational no has well defined decimal values upto infinite decimal places

I'm not exactly sure what this means, but I suspect it's not true$$^1$$. For example, we do have formulae that will tell us what the $$n^\mathrm{th}$$ decimal digit of $$\pi$$ is – in other words, we have a perfectly precise specification of any digit of $$\pi$$ you care to ask for. So there's no fundamental sense in which $$\pi$$ is less well-specified than, say $$4.6$$, which after all has an infinite decimal expansion as well, namely $$4.600000\dots$$

$$^1$$ edit: I originally interpreted "no has" as "doesn't have", but now I think "no" means "number", so I now guess the thing I'm quoting is true. I'll leave the rest of my answer unchanged, though.

Anyway, this fixation on decimal expansions is unnecessary. Here's how I choose to imagine the "true nature" of a real number: the essence of a real number is that it measures some continuous quantity – for example, each real number is the length of some idealized line segment.

Then decimal expansions are a way to understand these lengths by comparing them to fixed rational lengths that we know about, namely $$1$$, $$0.1$$, $$0.01$$, and etc. When we say $$\pi = 3.141\dots$$, we can view this as a series of comparisons, comparing $$\pi$$ to $$3$$ and saying it is larger, comparing it to $$3.15$$ and saying it is smaller, etc.

In this way, the decimal expansion isn't the "true nature" of a number, but simply as the view of a number you get when you look at it through the lens of powers of ten. This becomes more obvious when you realise that you can write the same number in different bases, view it through many different lenses, and indeed in some cases you get very different looking results, e.g. $$1/3$$ has an unending decimal expansion but a very simple ternary or nonary one.

Now ask again what it means for some real number, the length of an idealized line segment, to be a rational number. Well, to be rational is to be $$p/q$$ for some whole numbers $$p$$ and $$q$$, so a number is rational precisely if some multiple of it is a whole number (namely, $$q$$ times it is $$p$$). What that means is you can take your line segment and duplicate it $$q$$ times, you get a line segment that is exactly $$p$$ times a segment of length $$1$$.

Now the question becomes: why should it be rational? Why should that relationship ever hold exactly?

• "we do have formulae that will tell us what the nth decimal digit of π is " - I don't think this is possible in O(n)? (i.e. the only way is to compute all the digits up to the nth).
– M.M
Aug 22, 2016 at 23:48
• See Bailey–Borwein–Plouffe formula, although strictly speaking that's binary digits and not decimal digits. (In any case, computing all the digits up to the n$^\text{th}$ suffices for the point I was trying to make). Aug 23, 2016 at 6:39
• Binary digits is no help in finding the decimal digits (without computing all digits up to the point)
– M.M
Aug 23, 2016 at 7:06
• Yeah, so you do need to compute all the digits up to the nth, I guess. I think that's fine. Aug 23, 2016 at 8:07
• @M.M does it matter that its not in O(n)? We do have a formulae for the nth decimal digit. Maybe not an super efficient formulae but still. Aug 23, 2016 at 8:14

Philosophically, you could consider an "irrational quantity" as the limit of a sequences of "rational quantities", so e.g. 2pi is what the circumference of a unit circle approaches as you make it more and more perfect.

• :-) The completeness of it all. Aug 23, 2016 at 21:57
• More than philosophically, you can define real numbers as sets of sequences of rationals. Aug 24, 2016 at 1:32
• @YoTengoUnLCD right but here I'm talking about sequences of rational measurements in the physical world Aug 25, 2016 at 9:49

Relations between phenomena in Nature (φύσις, the origin of physics) existed probably before anyone thought about integers, rationals, and laws of physics. There is a legend about Kronecker saying integers were god-made, and the rest was a creation of mankind, but we leave that to the history of mathematics for the moment.

These relations therefore accommodate a variety of unit systems. The celerity of light, if constant, does not care about being measured in meter per second, or foot per lunar month.

Any instrument has finite precision, no phenomenon is "pure" enough to be measured precisely.

What matters, if laws are accurate enough, is to have sufficient precision. So, How Much Pi Do You Need?

NASA scientists keep the space station operational with only 15 or 16 significant digits of $\pi$, and the fundamental constants of the universe only require 32.

So, with "quadruple precision" you can almost work with "decimals", with respect to your floating point system. Which is much simple than with certain long-period rationals: $1/97$ has a period of 96 digits.

Finally, for $\pi$, even if you cannot work with infinite precision, you can still use indefinite precision, adding a decimal when you want, using the Bailey-Borwein-Plouffe formula, to directly calculate the value of any given digit (base 16) of $\pi$ without calculating the preceding digits:

$$\pi =\sum _{k=0}^{\infty }\left[{\frac {1}{16^{k}}}\left({\frac {4}{8k+1}}-{\frac {2}{8k+4}}-{\frac {1}{8k+5}}-{\frac {1}{8k+6}}\right)\right]\,.$$

An interesting question. It relates mathematical, i.e. mental concepts to "reality". Let us assume for the purpose of this discussion that there is a reality, and that it exists outside of our mind. Then the relation of the well-understood mental concept of irrational numbers and that reality depends on the properties of that reality (which, in all reality, are probably not completely known). I give three examples.

If the underlying structure of space and matter were completely continuous (i.e. there were no Planck constant, no atoms etc.), all properties of natural objects like length or weight were irrational: Because there are infinitely more irrational than rational numbers, it is virtually impossible to properly hit, say, 2 cm when cutting a piece of wood. It would always be 2cm + "random delta" (however small), and that delta will "always" be irrational. The same goes for the International Prototype Kilogram and all other artifacts and natural phenomena.

2. A blocky reality

If reality were like Minecraft, and all physical properties were "blocky" (albeit at an atomic scale), all properties would be rational: Everything is a multiple of block lengths, weights, times etc. and thus all properties can be expressed as fractions of integers.

3. Our quantum world

Our reality is a bit like Minecraft, except that things get fuzzy if you look too close. This means that no property of an object we want to measure can be measured exactly; on a fundamental level there is always a, well, uncertainty. (This is not a deficiency in our way of measuring -- those errors would come on top --, but a property of our universe.) Because exactness is limited one needs not be afraid of running out of decimal places.

Any interval produced by the amount of uncertainty has many rational numbers in them; so in a way you can probably get away without using irrational numbers in describing reality. (But you could also get away without ever using any rational numbers, mind you). Any existing circle can have both: A rational diameter and a rational circumference, as far as we can tell (+/- the Planck length).

Take a line segment of length 1. With compass and straight edge it is possible to construct a line segment starting from its endpoint that is perpendicular to it and also of length 1. Draw that line segment. Now draw the line connecting the two end points.

By the Pythagorean Theorem you have just drawn a line segment of length $\sqrt{2}$.

It is a fairly straightforward proof to show that this is an irrational number.

So it exists.

Now the question is - what law of physics you would believe this to violate. Sure - if you try to measure it, you won't be able to measure it to that accuracy (and okay, you probably didn't draw it to that accuracy), but assuming that you had the ability to exactly use a compass and straightedge then you would have drawn something with exactly length $\sqrt{2}$.

This is why the Pythagorean Theorem was a bit of a scandal. Up until this point it was assumed that for any two line segments we could find a smaller line segment that went into both of them evenly. This assumption made proofs about similar triangles quite easy. Mathematicians knew that $\sqrt{2}$ wasn't rational, but they could just argue no such length exists (much like $-1$ used to have no square root). The Pythagorean Theorem showed that $\sqrt{2}$ does too exist as a length.

Certain natural phenomena in probability theory such as Buffon's needle experiment have answers containing an irrational constant, namely $\pi$. Therefore such constants are arguably "in nature" (or any reasonable idealisation thereof) even if you are not particularly interested in the circle itself. The fact that there is an exact formula for the probability that involves an irrational number is precisely an argument against limiting precision in this fashion. Mathematicians have great respect for quantum theory but that theory involves a different type of idealisation of nature.

• The OP is arguing against infinite precision, and your example is exactly an infinite precision type of example. What happens if you limit your probability to something around the volume of the observable universe in Planck units. Using a uniform distribution, do $\pi$ and other irrational numbers ever come up then? Aug 27, 2016 at 19:05
• The fact that there is an exact formula for the probability that involves an irrational number is precisely an argument against limiting precision in this fashion. I have great respect for quantum theory and even published a paper in a quantum studies journal but you are confusing apples and oranges. Aug 28, 2016 at 11:11
• If you don't want to read my comment, don't expect me to answer yours properly. Have a nice day. Aug 28, 2016 at 11:55

What does it mean to say measure $$\pi$$ to arbitrary precision? Although it can be calculated entirely mathematically, it's possible that we could measure the circumference of a circle in a our universe to be a different number than $$\pi$$ times its diameter if our universe doesn't have Eucliden geometry. That can be considered measuring an incorrect value for $$\pi$$. In fact, general relativity predicts that our universe doesn't have Euclidean geometry but the tension from gravity in the measuring devices would have a greater effect on measurements than the geometry of the universe. I realize $$\pi$$ can't be physically measured to arbitrary precision because Planck's constant is not zero and the gravitational constant is not zero. It could have happened that the universe followed different laws where you can measure $$\pi$$ to arbitrary precision. The more precisely you want to measure it, the bigger the device you need.

I know Calculus is one possible explanation for the observations we make. They can be explained by real number induction on time. Here's my definition of real number induction. Take any subset $$S$$ of the real numbers that satisfies the following properties:

• $$0 \in S$$
• $$\forall x \in S\exists \epsilon \in \mathbb{R} (\epsilon > 0 \land \forall y \in \mathbb{R}((y > 0 \land y < \epsilon) \implies x + y \in S))$$
• $$\forall x \in \mathbb{R}(\exists \epsilon \in \mathbb{R} (\epsilon > 0 \land \forall y \in \mathbb{R}((y > x - \epsilon \land y < x) \implies y \in S))) \implies x \in S$$

It can be shown that $$S$$ includes all the nonnegative real numbers. If we lived in a universe with different laws where we can measure $$\pi$$ to arbitrary precision, it could also be explained the same way. I guess we could say all the real numbers physically exist because if you're missing one, real number induction doesn't work anymore. Despite that, I don't actually believe the fundamental laws of the universe have real number induction. Just because it's one possible explanation for our observations doesn't mean it's the correct explanation. Maybe the fundamental laws are actually a Conway's game of life that starts off a finite size whose grid keeps growing with time. How can it simulate a universe whose laws use real number induction? Maybe it turns out that in the simulation, it's consistent with observations that the length of timing between two precise events such as a wave function satisfying an exact mathematical property can be one length of time or another length of time that's an irrational number times that length but not all real numbers can possibly be the length of time between such events. We could then say that not all real numbers physically exist. Because the cause is what's going on at the fundamental level, we don't need real number induction to explain it so we don't need to include all real numbers so we can say that not all real numbers physically exist.

• How could you tell if $x$ is irrational or not, if you can only compute it to the billionth trillion decimal digit? Nov 21, 2018 at 23:32
• @AsafKaragila Just because we don't know what any of the later digits are doesn't mean we don't know a certain property of the set of all digits after that place, the property of not repeating any finite string of digits over and over. To prove that $\pi$ is irrational, it suffices to show that for any positive rational number less than 1, the $sin$ of it is not exactly $\frac{1}{2}$. Any positive rational less than 1 can be expressed as $\frac{p}{q}$ where the greatest common factor of p and q is 1 and p < q. The $sin$ of it is the limit of its Taylor series. The nth term must be a multiple Nov 22, 2018 at 1:46
• of $\frac{1}{n!q^n}$. Eventually you will get to the point where the absolute value of the sum of all the remaining terms is less than $\frac{1}{n!q^n}$. I believe I can prove that the sum of all the terms after the nth terms adds up to something that is not a multiple of $\frac{1}{n!q^n}$ but I'm not quite sure how to do that. Nov 22, 2018 at 3:21
• How can you prove that $\pi$ exists? As long as you can only measure finitely many digits, what makes you so sure that you're not just wrong? Or that your assumption in the proof that the value should be $\pi$ include non-physical assumptions (e.g. that space is continuous and not discrete)? Nov 22, 2018 at 7:54
• Timothy, that is exactly the point of the question. So you seem to have posted an answer that ignores the question entirely. Nov 22, 2018 at 18:52