In an effort to develop better number sense (and to create my own journey from fish to infinity), I have been going through Khan Academy's math material from the very beginning. So far, I have been able to develop strong intuitions and metaphors for most elementary mathematical ideas. I now see that arithmetic, division, negative numbers, fractions, decimals, factors, multiples, etc. all have clear grounding in reality; I have a much stronger grasp of the material.

So far, the only place my intuition has broken down is irrational numbers. It was surprising to me because it happened so suddenly, like the ground giving way. It was as if we saw this gaping hole in the ground, gingerly walked around it, and pretended it wasn't there. This bothers me.

Is there any straightforward metaphor for irrational numbers? Could you explain it to a child?

  • 2,025
  • 2
  • 19
  • 32
  • 7
    What do you not understand about them? – Euler_Salter Dec 20 '16 at 15:05
  • 2
    @Euler_Salter I think he wants to know the real world meaning of irrational numbers – Razin Dec 20 '16 at 15:06
  • @Razin as in $\pi$ in a circle? – Euler_Salter Dec 20 '16 at 15:07
  • 2
    Not easy : there are not enough rational numbers to "count" all the possible way you can cut a line. – Mauro ALLEGRANZA Dec 20 '16 at 15:09
  • 1
    If you're looking for "real life contexts" where they would show up: think of objects you can easily describe with rational numbers, but where an irrational one pops up; e.g. you want to diagonally walk from one corner on a football field to the opposite corner in a straight line. If the football field is 100 m long and 65 m wide, what distance will you have walked? Or with simpler numbers: you want to tile your kitchen with square tiles with surface area 2 dm², what is the required length of a side? You can then show that this isn't possible with rational numbers. – StackTD Dec 20 '16 at 15:10
  • @Euler_Salter, yes. For example, if I were to teach fractions to a child, I might talk about pizza, cutting them into different equally sized numbers of pieces, and so on. For division of fractions, I might say, "If you have 3/4 of a pizza and want to give each of your friends 1/4 of a pizza, how many friends can you feed?" This stuff helps me (maybe it matters less to people here). I am looking for something similar with irrational numbers—I understand the mathematical definition. – jds Dec 20 '16 at 15:11
  • @gwg are you looking for a metaphor that would work for any irrational number or to specific metaphors for specific irrational numbers? – Euler_Salter Dec 20 '16 at 15:14
  • 9
    @gwg In your pizza example, you get fractions if you split the pizza into equal slices. What happens if you just cut a wedge from the pizza? It is almost never the case the wedge you will cut can than be repeated exactly around the pizza to perfectly split the pizza up. In this case the size of your wedge to the original pizza is irrational. – Leon Sot Dec 20 '16 at 15:18
  • 2
    A [Lissajous curve](https://en.wikipedia.org/wiki/Lissajous_curve) is closed iff the frequencies are commensurable, otherwise it turns into a space-filling curve. – dxiv Dec 20 '16 at 15:50
  • 4
    The Clinton-Trump election results seemed like irrational numbers. – Michael Dec 20 '16 at 17:09
  • 2
    "It was as if we saw this gaping whole in the ground, gingerly walked around it, and pretended it wasn't there." What is the hole in this metaphor, and in what way are we pretending that it isn't there? – Tanner Swett Dec 20 '16 at 18:34
  • 2
    https://arxiv.org/abs/math/0411418 – Count Iblis Dec 20 '16 at 18:56
  • 1
    @TannerSwett, Sal Khan devotes extraordinary energy to explaining simple concepts from multiple angles in order to really drive home the point. Look at this explanation of fractions: https://www.khanacademy.org/math/arithmetic-home/arith-review-fractions. By contrast, he spent two videos on irrational numbers. It seems to better to either ignore them for the time being or introduce them properly. – jds Dec 20 '16 at 19:00
  • Se my answers [here](http://math.stackexchange.com/questions/1216125/why-is-the-construction-of-the-real-numbers-important/1216154#1216154) and [here](http://math.stackexchange.com/questions/645009/teaching-irrational-numbers/649333#649333). – Jack M Dec 20 '16 at 19:02
  • 2
    Wait till you get to imaginary numbers... – MercyBeaucou Dec 20 '16 at 19:34
  • In my experience the most valuable uses for irrational numbers (outside of the pure math topics like infinities) is in the form of negative proofs. As many have shown here, it's hard to find a proof that you *can* do something because irrational numbers exist, but it is easier to show you *can't* do something (the jogging along the diagonal example seems most obvious to me). Irrational numbers show the limit of what you can do with ratios. – Cort Ammon Dec 20 '16 at 20:05
  • Of course, the real world application of them is limited as our tools lack the precision to actually construct objects to demonstrate the limits of rational numbers. Irrational numbers merely show that, as our tools get better and better, there are some places they can simply never go. – Cort Ammon Dec 20 '16 at 20:07
  • 39
    IMO, trying to understand mathematics through metaphor is the exact opposite of the path that leads to actually understanding mathematics. –  Dec 21 '16 at 00:05
  • That's normal you have difficulties to get an intuition of real numbers. It took quite a lot of time for mathematicians to understand them, from their discovery by Pythagoreans to the modern understanding of them as the metric completion of the rational numbers. More important than their "intuitive" meaning (not everything can be intuitive), what is important to understand at early stage is that since they are by a non-property (they are not "something"), many proofs concerning irrational numbers will use induction. – Taladris Dec 21 '16 at 09:42
  • 1
    Rationnals represent finite presition while real numbers represent infinite precision. In the real world, finite precision is all we have so there's no real reason to use real numbers but if you start doing maths, to speak about infinite precision, you need sequences of rationnal numbers (that get closer and closer to whatever you're approximating). But sequences are annoying to work with. Fortunately, we can "pack" those sequences in some technical way and call the result real numbers, and most of our intuition still works on those new numbers (and we can still do stuff like $+$, $\times$..). – xavierm02 Dec 21 '16 at 12:50
  • 8
    @Hurkyl: I have to disagree. Mathematical discovery almost *requires* the ability to think intuitively/metaphorically/symbolically/pictorially about mathematical objects. How else will new discoveries emerge? – Lee Mosher Dec 21 '16 at 20:18
  • 7
    @LeeMosher This particular question is about _real-world_ metaphors. I think trying to make things too "real" can be an impediment to doing math. But I think you're correct to question an overly sweeping generalization about the use of metaphor in mathematics. – David K Dec 22 '16 at 17:43
  • 1
    Picture the current U.S. president. Now picture the U.S. president-elect. One is rational, the other is irrational. I'll let you figure out which is which. – user541686 Dec 23 '16 at 02:38
  • @Hurkyl, serious question: how would you teach a small child counting or arithmetic? – jds Dec 23 '16 at 22:39
  • @gwg: I'm not in the business of teaching small children *anything*. I don't see how your question is relevant though; e.g. I would not call the application of numbers to counting a "metaphor". –  Dec 23 '16 at 22:44
  • @Hurkyl, I did not suggest that you were in that business. I asked how you would do it. I suspect you would, in the process, use a metaphor. – jds Dec 23 '16 at 22:56
  • @gwg: I've thought some about it, and the only teaching method I could come up with to teach counting is by example and repetition. How would *you* teach a small child counting or arithmetic via metaphor? I can't even imagine what you could use as a metaphor for counting for teaching purposes. I suspect you would use something that I would not consider a metaphor. –  Dec 24 '16 at 00:27
  • @Hurkyl, metaphor: "A thing regarded as representative or symbolic of something else, especially something abstract." [Here is Sesame Street introducing counting](https://www.youtube.com/watch?v=hgZwSRpfouQ). Or [a professor introducing fractals](http://yaledailynews.com/blog/2011/11/08/curious-geometries/). Or von Neumann: "I think that it is a relatively good approximation to truth... that mathematical ideas originate in empirics." Often an abstract idea is introduced by relating it to something concrete. Mathematical depth doesn't end in metaphor, but I am not so sure it cannot begin there. – jds Dec 24 '16 at 03:23
  • @Hurkyl Exactly. Maths is abstraction to the extreme about the actual, analog world we live in. Just like the formal approaches for lingusitics and AI have failed to accurately model human behaviors, while the seemingly much more "chaotic" neural network approach has seen massive success, if one tries to link formalisms too tightly with real world processes it won't be a good match imo. – xji Dec 28 '16 at 19:32

24 Answers24


Here is a physical metaphor:

Draw a circle, and prepare two sticks:

  • One with the length of the circle's diameter (2r)
  • One with the length of the circle's circumference (2πr)

You cannot cut both sticks into pieces of the same length, no matter what length you choose.

In other words, you cannot measure both sticks using the same measurement-unit.

You can try meters, feet, inches, miles, or even invent your own measurement-unit.

You will never be able to accomplish this, because the ratio between the sticks is irrational.

barak manos
  • 42,243
  • 8
  • 51
  • 127
  • 27
    Though I admit that it's kinda oxymoronic to say something like "ir**ratio**nal ratio"... – barak manos Dec 20 '16 at 15:37
  • 118
    If you want to use sticks, perhaps you should prefer to use one as the side of a square and the other to the diagonal of the same square. Then they can both be straight. Circular sticks are unusual. – MJD Dec 20 '16 at 16:07
  • 1
    @MJD: Haha, that's actually quite a good idea. This way, I could omit the part of the instructions saying "prepare sticks with length...", since those sticks would already be "prepared" :) – barak manos Dec 20 '16 at 16:09
  • @MJD: But please note that I essentially meant to prepare a stick with the length of the circumference, not to actually take the circumference. This would indeed be a very unusual stick... – barak manos Dec 20 '16 at 16:12
  • 16
    (Your first comment) A usual historical word for this is two say that the two lengths are _incommensurable_. A use for irational numbers is being able to "calculate" with such incommensurable geometric quantities. – Jeppe Stig Nielsen Dec 20 '16 at 17:27
  • If I have a circle with diameter of 6, the circumference is about 18.85. Why can't I cut my 18.85 stick to be 6? What am I not realizing?? – BruceWayne Dec 20 '16 at 18:10
  • 2
    @BruceWayne The idea is to try to cut the sticks so that _all_ the pieces of the cut-up sticks are the same size. Sure, you can cut $6$ from a stick of length $6\pi$, but now you have pieces of length $6$ and length $6\pi-6$, and they are not the same size. – David K Dec 20 '16 at 18:25
  • @DavidK - Ohh, okay I get it now, thanks for that clarification! – BruceWayne Dec 20 '16 at 18:27
  • @ConfusinglyCuriousTheThird: Sorry, I don't understand your comment. What does "lining copies" mean? – barak manos Dec 21 '16 at 06:17
  • @barakmanos it might be better to point out that, lining up copies of stick 1, end to end, would never equal the same length as copies of stick 2 lined up in the same manner. – anon01 Dec 21 '16 at 06:19
  • @ConfusinglyCuriousTheThird: Why would it be better to say it this way? – barak manos Dec 21 '16 at 06:37
  • 3
    @barakmanos Ah, I was confused by your first "cut sticks into pieces of the same length", I understand your statement now. The measurement bit though: you *never* have objects that are commensurate with a (predetermined) unit of measurement... and that doesn't keep you from performing a measurement. You just can't write down the exact value. Physicist here :) – anon01 Dec 21 '16 at 06:45
  • @ConfusinglyCuriousTheThird Has a very good point; that occurred to me a while ago. The thing is that if you cut a stick or whatever, there is always some inaccuracy, some tolerance, some random delta (Planck lengths be ignored for the argument). Since there are so many more irrational than rational numbers, it is all but impossible to accidentally hit a rational length. In reality we effectively *always deal with irrational, incommensurable quantities in our daily lives.* We just don't really care, and don't have to. – Peter - Reinstate Monica Dec 21 '16 at 14:06
  • 2
    @barakmanos But in real life you can't prepare a stick whose length is the circumference, because, as we all know, in real life you can only use a compass and a straightedge. – JiK Dec 22 '16 at 13:28
  • 1
    @JiK: Then attach a sewing thread (or a really thin shoe-lace) on the tip of that straightedge as you draw the circle, and when you're done, take it out and stretch it into a straight line. – barak manos Dec 22 '16 at 13:34
  • That example is actually stronger than was requested, since π is not merely irrational but transcendental. It might be better to use a square whose side is a stick of length 1, and whose diagonal is therefore a stick of length √2. – Mike Scott Dec 24 '16 at 17:35
  • @MikeScott: Thank you. However, I don't see any difference between algebraic irrational and transcendental irrational **in the context of this question**. If you think that there is any such difference (again - in the context of this question), then can you please point it out? – barak manos Dec 24 '16 at 19:10
  • @barakmanos The issue is that someone might say "Yes, I can see that your example gives me an intuitive feeling for transcendental numbers -- but other irrational numbers might not behave like that, and so it doesn't help me with irrational numbers that are not transcendental." – Mike Scott Dec 24 '16 at 19:30
  • @MikeScott: I agree, but in the context of this question I find it hardly relevant. The question says "Is there any straightforward metaphor for irrational numbers? Could you explain it to a child?". The terms *algebraic* and *transcendental* are relatively advanced topics, not something that a child is taught. And if you **do** direct this answer towards someone who is already familiar with these terms, then surely that person will not mistaken the description that I have provided as something which is true only for transcendental numbers. – barak manos Dec 24 '16 at 19:50

I am a historian of science and I am intrigued by this question, because I am a non-specialist in mathematics but for many years I taught classes dealing with the Pythagoreans. One thing you might consider in making irrational numbers "real" is to imagine why they were deemed so revolutionary and important by the ancient Greeks. In other words, instead of looking for an illustration of their practical reality, look instead to why they were so darned interesting. One answer that hasn't yet been addressed is the contention that shapes are a better representation of reality than numbers. An irrational number is the only way, in the language of numbers, to represent any "real" distance" that cannot be expressed as a relationship between two whole numbers. And yet we all know that such distances are in fact real, i.e. the hypotenuse of a right triangle. That reality, by contrast, is quite easily expressed visually through geometry, which is one reason Euclid spent so much time laying out the rules of geometric shapes. Remember that he was fundamentally a philosopher. One could argue that geometric shapes (to which Platonic philosophers believed all reality could be reduced, i.e. the "forms") are a better representation of reality than numbers are.

By bringing up metaphors, you've put your finger on the fascinating problem of reality raised by irrational numbers. The "whoa" moment regarding irrational numbers usually comes when we rethink what's behind the theorem we all had to memorize in school. Irrational has a literal meaning (no ratio possible), but also it suggests the cognitive dissonance between numbers and shapes, and helps to explain why forms were perceived as fundamental among some of the ancient Greeks.

  • 529
  • 3
  • 2

Here a metaphor based on the irrationality of $\sqrt{2}$:

Every day two friends $A$ and $B$ enter a square field at one of its vertices, jog for a while, and leave by the same vertex. $A$ jogs along the boundary of the field (always in the same direction) whilst $B$ jogs along the diagonal joining the entry vertex and its opposite (changing the direction only when he/she reaches a vertex).

$A$ and $B$ never jog the same distance.

  • I don't understand the meaning of the last sentence. Suppose every day $A$ jogs along the long side of a rectangle that has a length twice its width, while $B$ jogs along the width. Again, $A$ and $B$ never jog the same distance, but no irrational numbers are required. Don't get me wrong, I'm sure there's a good metaphor in your answer, it's just not clear to me what that metaphor is. – Todd Wilcox Dec 20 '16 at 22:52
  • 3
    @ToddWilcox It is a _square_ field. $A$ jogs $4sk$ for some natural $k$ and $s$ the length of any size of the field. $B$ jogs $2n\sqrt{2}s$ for some natural $n$. Can they be equal? –  Dec 20 '16 at 22:58
  • 5
    In the body of your answer, you didn't specify that neither $A$ nor $B$ ever stop in the middle of a segment while jogging. It also wasn't clear whether they each jogged their lengths exactly once or twice or an arbitrary number of times. Perhaps putting the formulae from your comment into words would make it clear. The comment makes perfect sense. The answer proper less so. – Todd Wilcox Dec 20 '16 at 23:09
  • 1
    @ToddWilcox Well, it is implicitly implied that both joggers leave the field after some time and following its path... –  Dec 20 '16 at 23:27
  • It could be useful to indicate that $A$ and $B$ complete an integer quantity $a$ and $b$ of edges respectively, or something or the ilk. Otherwise, they would surely run the same distance given the same speed and differing distances given differing speeds! – zahbaz Dec 21 '16 at 01:45
  • @zahbaz **It is a real-world metaphor**, so each jogger leaves the field at some time, so of course they complete an integer number of edges/diagonals. I feel like you understood the metaphor but are being deliberately ridiculous about its accuracy. I don't think any more explanation could be useful, so I am not editing the answer. –  Dec 21 '16 at 02:14
  • A better metaphor would not rely on the reader's understanding of irrational numbers to fill in the gaps. – zahbaz Dec 21 '16 at 02:52
  • 9
    And, looking at this discussion, we see the problem of trying to understand math by metaphor. – Martin Argerami Dec 21 '16 at 04:55
  • 5
    Your problem would be pithier if you did not require them to do the jog "every day". If $A$ does his jog some number of times, and if $B$ does her jog some number of times, then no matter what those numbers are, $A$ and $B$ do not jog the same distance. – Lee Mosher Dec 21 '16 at 20:14
  • @LeeMosher Indeed!! The 'never' in the last sentence is actually meaning "all time records". –  Dec 21 '16 at 20:27
  • 1
    Hmmm.... in the context of the problem, the evident meaning of "never" is "on no day". – Lee Mosher Dec 21 '16 at 20:54
  • 1
    @LeeMosher Forgive my English, I am still learning. I mean that 'never' in the last sense can also be understood that way; that the impreciseness is deliberated. Clearly "in finite time" the 'all times record' interpretation is intuitively stronger, but anyway mathematically the assertion 'They never yog the same distance' have the same implications in either case. –  Dec 21 '16 at 21:07
  • 4
    All the effort you spent defending your original answer could have instead been spent making it more clear. – Austin Mohr Dec 22 '16 at 09:01
  • It is possible both runners take the same number of strides. And if both runners strides are the same, then it's possible the both have traveled the same distance. Basically it boils down to your unit of measure of distance, of which there is always some error involved. – gogators Dec 22 '16 at 20:51
  • 1
    @gogators Maybe it is my English but I don't understand your point. –  Dec 22 '16 at 21:55
  • @mathbeing Most answers here essentially are comparing a measured quantity (e.g. distance) to an irrational number. The point is no matter how accurately you measure something, it will never exactly equal some irrational number because all measurements are rational numbers. – gogators Dec 22 '16 at 23:04
  • @gogators Well that is obvious. There is no bigger math than that going on, but the point of the answers in this thread is not about math, but about the inavoidability of irrational numbers in our real-world experience, about how they arise naturally even if we don't know about them. –  Dec 23 '16 at 02:22
  • @gogators And if the square is very small, I can cover the diagonal with one stride and the edge with one stride. But the strides do not have the same length, so I have still not covered the same distance. –  Dec 24 '16 at 17:05

Grab a sheet of grid paper and draw two axes.

enter image description here

Draw a line — neither horizontal nor vertical — passing through the origin. If this line passes through the corner of one of the little squares (the origin excluded), then its slope is rational. If not, then its slope is irrational.

In other words, the existence of irrational numbers means that we can "stand" at the origin and point a laser pointer in some special directions and not "hit" any corner of any "little square".

Two lines passing through the origin are depicted below. The blue line has slope $2$ and the pink line has slope $\pi$.

enter image description here

The blue line passes through $(1,2)$ and $(2,4)$, which are corners of "little squares". The laser beam "hits a target". This is no coincidence. After all, to draw the blue line, we simply connected $(0,0)$ and $(1,2)$. We hit the target intentionally.

If we do not want the laser beam to "hit" anything, we can make the line pass through $(1,\pi)$. Since $\pi$ is irrational, the laser beam will travel forever, always missing the corners. Irrationality tells us that there is an escape from the "tyranny" of the grid.

More rigorously, we have an equivalence relation on $\mathbb Z \times \mathbb Z \setminus \{0\}$

$$(a,b) \equiv (c, d) \iff \frac{a}{b} = \frac{c}{d}$$

The blue line depicts the equivalence class

$$\cdots \equiv (-3,-6) \equiv (-2,-4) \equiv (-1,-2) \equiv (1,2) \equiv (2,4) \equiv (3,6) \equiv \cdots$$

which corresponds to the rational number $\frac 12$.

Rodrigo de Azevedo
  • 18,977
  • 5
  • 36
  • 95
  • 2
    Related: [What fraction of the integer lattice can be seen from the origin?](http://mathoverflow.net/q/151706/91764) – Rodrigo de Azevedo Dec 21 '16 at 08:53
  • A nice answer, but mainly so if one is already "sold" on the notion that there are indeed such things as irrational numbers. As a geometric object, the pink line, which has slope $\pi$ and goes through the $(0,0)$ and $(1,\pi)$, cannot 'prove' that it will *never* go through any pair of integer-valued numbers (other than $(0,0)$, of course). How might you revise your answer to provide some intuition that the assertion is valid? – Mico Dec 23 '16 at 14:47
  • 1
    @Mico I should have chosen slope $\sqrt{2}$. – Rodrigo de Azevedo Dec 23 '16 at 14:54
  • 3
    As a non-mathematician, I've read through most of the answers to this question, and I think this one explains the concept the best. @Mico, the question wasn't about proving irrational numbers, it was about understanding them and I feel that this answer explains it better than the others. – esteuart Dec 23 '16 at 21:13
  • @esteuart - Thanks. I guess I had interpreted the OP's final two sentences -- "Is there any straightforward metaphor for irrational numbers? Could you explain it to a child?" -- as asking for intuition regarding the very meaning and "logic" of irrational numbers, rather than "just" for some of their properties (such as a line through the origin with slope $\pi$ (or $\sqrt{2}$) never going through any other pairs of integers). However, I will readily concede that I may well have misinterpreted the OP's objective! – Mico Dec 23 '16 at 21:21
  • @Mico My point was that if irrational numbers exist, then the laser beam won't "hit" any points of the lattice $\mathbb Z^2$. – Rodrigo de Azevedo Dec 23 '16 at 21:24
  • This doesn't sit well with me. I completely understand your point and agree with it from a mathematical perspective. But in real life, if one pointed a laser from the origin in any direction it would eventually hit a vertex. This relies on how we define a vertex. In mathematics it is assumed to be an infinitesimal point but such a thing does not exist in the physical world. With a large enough grid where the laser does not "hit" any vertex it will still come so close to a vertex that we begin to delve into Planck lengths and for all intents and purposes the laser has hit the vertex. – Darren H Dec 24 '16 at 18:47
  • Your proposal does indeed hold for very large grids, but not infinite grids in the real world. This makes it more confusing because humans generally find infinities and their effects difficult to comprehend. – Darren H Dec 24 '16 at 18:49

Suppose we have a square whose side length is $1$, then according to Pythagoras theorem, its diagonal length is a number whose square is $1^2+1^2=2$.


It would indeed be catastrophic for geometry if there were no actual number that could describe the length of the diagonal of a square.

The Pythagoreans tried to make do with a notion of ‘actual number’ that could be described simply in terms of ratios of whole numbers; i.e. find a number $\frac{a}{b}$ such that $\left(\frac{a}{b} \right)^2=2$. But it can be proved that $\left(\frac{a}{b} \right)^2=2$ has no solution for integers $a$ and $b$, where we can take these integers to be positive. Thus, there is no such rational number that squares to be $2$.

Therefore, there should be other kinds of numbers that are not rationals and hence, irrational numbers were discovered.

  • 663
  • 1
  • 5
  • 25

An interesting question.

You're not the first to fall into a hole when faced with the idea of irrational numbers. Here's part of the myth about the Greek discoverer:


Irrational numbers may be the first thing you find in your mathematical journey that has no good counterpart in "ordinary life". You never need them to measure things as accurately as you wish, or to cut up pizza. They come up only when you want to pursue pure mathematics for its own sake. That effort started with the Greeks. Book X of Euclid's Elements begins with a definition of "incommensurable" (explained too in @BarackManos 's answer). Then in Proposition 10 Euclid proves that the side and the diagonal of a square are incommensurable. (The notes at that site say this proposition may not really belong to the original Elements.)

When you start to learn about infinite sets you'll discover that there are (in a sense that can be made precise) many more irrational numbers than rational ones.

Enjoy the journey.

Ethan Bolker
  • 80,490
  • 6
  • 95
  • 173
  • 1
    "They come up only when you want to pursue pure mathematics for its own sake." So, there are no know practical applications which require the use of irrational numbers? – Anonymous Coward Dec 20 '16 at 19:38
  • 2
    @JoseAntonioDuraOlmos None that I know of directly. Indirectly, there's lots of important applied mathematics that couldn't really be developed until the pure part of mathematics was powerful enough to support the idea of irrational numbers. – Ethan Bolker Dec 20 '16 at 19:53
  • How about exponential growth as a practical application? Euler's constant lies at its heart. – Austin Mohr Dec 21 '16 at 06:07
  • 2
    @AustinMohr Indeed e is irrational (even transcendental), but you don't have to understand that to understand exponential growth. The OP asked for a real world metaphor; I don't think exponential growth provides an answer. – Ethan Bolker Dec 21 '16 at 13:43
  • I was responding to your claim that irrational numbers come up only in pure mathematics. – Austin Mohr Dec 21 '16 at 14:00
  • 4
    @AustinMohr Do you think that your calculator uses an exact representation of e when you press that button? ;-). Obviously, in all practical calculations we use (rational) approximations. – Peter - Reinstate Monica Dec 21 '16 at 14:13
  • Re "no good counterpart in "ordinary life" ": I think that is simply not true. As I said in another comment, in fact reality presents us almost exclusively with irrational numbers because it is so inexact. We just don't (have to) care. Like, define the famous coast length of England, or tell me *exactly* how long that stick is, or how much the kilogram prototype weighs (the force). – Peter - Reinstate Monica Dec 21 '16 at 14:16
  • Of course not, but I don't think it's entirely accurate to say that irrational numbers don't "come up" merely because we use rational approximations in practice. – Austin Mohr Dec 21 '16 at 14:17
  • An aesthetic and elegant attempt to draw a boundary between a pure math and applied one. – kaka Jan 05 '17 at 07:12
  • @AustinMohr Exponential growth can occur in integers, can it not? 2, 4, 8, 16, 32... the parable of the rice and the chess board comes to mind. – Michael Jan 11 '17 at 02:29

If the "gap in the ground" of the irrational numbers is more noticeable and intimidating, then that's because you are understanding the situation accurately.

Let's go over definitions, first. You know what integers are, and how they fit on the number line. If you try to divide by integers, you get rational numbers; these too fit on the number line. You might think, then, that the rational numbers "fill up" the whole number line. Certainly, between any two rational numbers, you'll be able to find another rational number. Nevertheless, I intend to convince you that there are gaps between rational numbers, which we will fill with irrational numbers. If we want the number line to be complete, without any gaps, we have to add something.

The usual example is $\sqrt{2}$. By definition, $\sqrt{2}$ is this number $x$ that, when multiplied by itself, yields $2$. That is $x^2 = x\cdot x = 2$. Notably, there is no rational number that behaves this way (often, this is proven by infinite descent).

Let's look at this another way: pretend for the moment that all numbers are rational. We can divide the positive half of the number line into two pieces. One piece (call it $S$) is the set of all positive $x$'s for which $x^2 < 2$, and another is the set of all positive $x$'s for which $x^2 > 2$ (call it $T$). That is, we've created a Dedekind cut.

Clearly, every rational number is either in $S$ or in $T$, and any number in $T$ is bigger than every number from $S$. Intuitively, this means that these two sets cut the positive number line into two pieces. However, we haven't touched any rational numbers with our cut; there is no rational number on the border between $S$ and $T$. Our cut through the number line (which was supposed to be made completely out of rational numbers) has missed; it didn't hit any numbers.

What we can do, however, is look at the location of the cut on the number line, and declare that at this boundary is a number, which we call $\sqrt{2}$. If we do so with every gap in the number line, then we get the real numbers: the full, gap-less number line.

So, what are the irrational numbers? They're the numbers that fit between the rational numbers. Their defining property is that, like all other numbers, we can put them in order. For example, I could tell you that $1.41 < \sqrt{2} < 1.42$, or that $1.414 < \sqrt{2} < 1.415$, but $\sqrt{2}$ is not itself a rational number.

Ben Grossmann
  • 203,051
  • 12
  • 142
  • 283
  • 1
    Most rational numbers are not whole numbers, and most real numbers are irrational. In fact, most real numbers are transcendental (not algebraic.) Moreover, most real numbers are not even *computable.* This is truly bizarre to contemplate when you understand it, but on the flip side (philosophically) you can consider that computable numbers are Order, in contrast to the Chaos of non-computable numbers. – Wildcard Dec 21 '16 at 03:00
  • NB: None of the example irrational numbers you specify in your answer are transcendental, much less non-computable. All are algebraic. (It's probably impossible to specify a non-computable number, by definition.) – Wildcard Dec 21 '16 at 03:04
  • 1
    @Wildcard see [Chaitin's constant](https://en.wikipedia.org/wiki/Chaitin's_constant) and also [this post](http://math.stackexchange.com/questions/462790/are-there-any-examples-of-non-computable-real-numbers) for examples of non-computable numbers. In any case, I didn't think the notion of algebraic/transcendental was appropriate for the scope of the question. – Ben Grossmann Dec 21 '16 at 03:33
  • I agree with you re the question scope (level), I just thought it should be noted in the comments to expand on this quite excellent answer. :) Thanks for the links re non-computable numbers. – Wildcard Dec 21 '16 at 04:00

Here is an idea for a metaphor:

First, I wish to point out that what I find difficult about the irrational numbers is that they are uncountable. And so something like the rational numbers, $\mathbb{Q}$, do not seem unmanageable to me even if they are unioned with, say, all the various roots, e.g., $\sqrt{2}$ or $(5/6)^{3/4}$ (algebraic numbers) or some of the numbers that have special names, e.g., $\pi$ or $e$ (specific transcendental numbers).

Bearing all that in mind, how about as the metaphor for rationals: Colors that you can name; thus, as the metaphor for irrationals: colors that go nameless.

Individual colors can be named, but try as you might - and even given an infinite amount of time and an infinite amount of paint - you could not possibly list names for all of the different colors.

I like this metaphor because it allows you to recognize the rationals (say, by viewing them as a ratio of one color to another) and even to pick other, individual examples (simply by pointing to a given mixture and giving it a name). But, again, you cannot name all the colors that arise through [a suitably continuous approach to] mixing them together.

This metaphor is far from perfect, and I am sure that there are ways to pick it apart. But you can also push through some related notions (e.g., around density) that might make it worthwhile.

Benjamin Dickman
  • 13,259
  • 2
  • 39
  • 82

I hear this once, not sure its origin:

The rationals viewed as points of light on the real number line are as the stars in the night sky.

Rene Schipperus
  • 38,213
  • 2
  • 28
  • 74
  • 10
    The stars are dense in the night sky? I don't think so. – celtschk Dec 20 '16 at 17:32
  • 2
    @celtschk I think what Rene is saying is that "everywhere we look, we see a star (if our vision is sharp enough)". Imagine projecting the entire sky to the plane of our eyeballs. Then - assuming the universe is infinitely large, and matter isn't distributed *too* un-randomly - the points on this plane corresponding to stars are indeed dense in that plane. Now, we will not actually *see* them all, since most of them are [too far away (hence redshifted)](https://en.wikipedia.org/wiki/Olbers'_paradox#The_mainstream_explanation) to be visible, but the idealized picture still holds. – Noah Schweber Dec 20 '16 at 19:12
  • No, the point being that there are enormous gaps between those points. An excellent, simple, real-world metaphor. Also, I believe the OP was not looking for math to be explained by yet more math, as many of the answers do. –  Dec 20 '16 at 19:17
  • Even with perfect vision, you'll not see a star everywhere, see [Olbers' paradox.](https://en.wikipedia.org/wiki/Olbers'_paradox) Of course you could think of an abstract sky where you'd see stars dense in each direction, but that's not the sky we actually see, neither with our eyes, nor with arbitrarily advanced instruments. Moreover since stars are spheres, not points (and for the difference between rationals and reals that *does* matter!), the stars in that infinite universe would cover *every* point, without any gaps whatsoever, not even irrational ones. – celtschk Dec 20 '16 at 19:26
  • 6
    @mickeyf: There are *no* enormous gaps between the rationals. That's exactly the point of my first comment. And thus it's *not* an apt metaphor for the rationals. All the gaps between the rationals are infinitely small. – celtschk Dec 20 '16 at 19:28
  • 6
    @celtschk: I like this answer a lot in view of the [Hubble Deep Field](https://en.wikipedia.org/wiki/Hubble_Deep_Field) image. No matter how far you zoom into the gap between stars, you find more stars between them. Yet, all the stars together are not enough to fill up the night sky. Of course these facts are not literally true for stars, and as you've argued the stars are not literally dense in the night sky, but it's still an apt metaphor I think. –  Dec 20 '16 at 22:01
  • 2
    I don't think it's unreasonable to claim that the stars could be dense. If you suppose that the universe is infinite and all that jazz, and let enough time pass, you will have a star shining at any point you choose. – neptun Dec 20 '16 at 22:23
  • 1
    flawed as it may be, probably the only answer that is a metaphor instead of an example. Kudos for answering the question! – anon01 Dec 21 '16 at 06:18

To expand on my comment above, which as stated was somewhat incorrect, the usual idea as fractions as wedges of a circle can be expanded upon to include irrational numbers.

Now, to reexamine the rational case: normally when you explain $1/7$ to a child you picture a circle divided into seven equal wedges and $1/7$ corresponds to one wedge while $2/7$ corresponds to two wedges. A better way to think about this, in some respects atleast, is to notice that $1/7$ is the wedge that if you repeat it seven times it goes around the circle once. While $2/7$ is the wedge that if you repeat it seven times it goes around the circle twice. From this we can then say any rational number $r/s$ corresponds to the wedge that if you repeat it $s$ times goes around the circle $r$ times.

What happens then if you pick any arbitrary wedge? Well almost always no matter how many times you repeat it, it will never exactly go around the circle an integer many times. These wedges correspond to the irrationals.

Edit: The fact that each wedge we can choose almost always corresponds to an irrational is due to the cardinality of the rationals vs the irrationals. However, this doesn't provide any intuition of why we should expect this.

If we draw the wedges on the circle, by drawing lines from the centre to the circumference. We may eventually draw a line that seems to overlap a previous line. However, what we are drawing are really representation of an "infinitely thin" line. If I drew a line that was infinitely thin and then asked you to trace it with your own infinitely thin line, do you think you could do so? I doubt it, the more likely outcome is that if we zoom in close enough, eventually we will see that your line and my line are distinct. In some ways this intuition carries over to why almost all wedges will never exactly overlap. It would be just as difficult to draw a wedge that when repeated goes around exactly an integer many times.

Leon Sot
  • 1,253
  • 6
  • 10
  • It depends how sharp your pencil is, doesn't it? Unless it's infinitely sharp, it can't draw an infinitely thin line. ;) (This is where the idea of "real" numbers breaks down as mirroring our understanding of what *real* is.) But very good answer; I like it. Incidentally, the idea of an infinitely sharp knife is part of Phillip Pullman's series ["His Dark Materials,"](https://en.wikipedia.org/wiki/His_Dark_Materials) best known by the title of the first book, "The Golden Compass." – Wildcard Dec 21 '16 at 04:04

The reason irrational numbers are so hard to grasp is that the rational numbers are already dense: In an arbitrary small region around any rational number you'll find infinitely many other rational numbers. In particular, if you have a rational number, there is no "next higher" rational number. Therefore it intuitively seems there's no gap left in between for even more numbers.

But it turns out that there are such gaps, although you cannot label them with a single pair of rational numbers. This is unlike with the integers where you can clearly say "there's a gap between 0 and 1, where more numbers can be filled in."

To specify the gaps between the rational numbers, you have to specify not just one pair, but a whole sequence of pairs.

For example, consider the square root of two. It can be proven that there does not exist a rational number whose square is 2, and you surely already know that proof. But is this a fundamental problem (like the fact that there is no number $x$ such that $0x=1$) or just because we are missing some numbers?

One thing we can find for sure is a rational number whose square is less than $2$; for example, $1$ is such a number, as $1^2 = 1 < 2$. We can also find another rational number whose square is larger than $2$, for example, $\left(\frac32\right)^2 = \frac94 > 2$. Since for positive numbers, $x>y$ implies $x^2>y^2$, so if we assume that this should hold also for those "extra numbers" (which we term "irrational" because, well, they are not rational), we get for the (for now, hypothetical) number $x$ with $x^2=2$ that $1 < x < \frac32$.

But we can find a larger number than $1$ whose square is less than $2$. For example, $1 < (5/4)^2 = 25/16 < 2$. And equivalently, we can find another, smaller rational less than $\frac32$ whose square is larger than $1$.

Indeed, we can find a whole growing sequence of rational numbers whose squares are all smaller than $2$, but which eventually will grow above any rational number whose square is less than $2$. That is, any rational number that is above all the numbers in that sequence must have a square that is at least $2$. Similarly, we can find a falling sequence of rational numbers with squares larger than $2$ that eventually falls below any rational with square larger than $2$. Now it is obvious that there is no rational number that is larger than any number in the first sequence, and at the same time smaller than each number in the second sequence, as the square of such a rational number would be neither greater nor less than $2$, and we already know that a rational number with square exactly $2$ does not exist. Thus those two sequences describe a gap in the rational numbers; a gap tjhat is above all the number of the first sequence, but below the numbers in the second sequence.

As a cross check, we can do the same with $4$ in the place of $2$. Here we get a growing sequence of rational numbers with square below $4$ and a second sequence of rational numbers with square above $4$. Of course the first sequence consists of rational numbers below $2$, and the second sequence consists of numbers above $2$. And there is exactly one rational number enclosed by those sequences, namely $2$. And indeed, $2^2=4$.

Now if you look at the two sequences, you see that as you progress, the terms of those sequences get arbitrary close to each other, but the first always stays below the second. And one can show that whenever this condition is met, then there can be at most one rational number in between, but there might be none.

Now where there is none in between, those sequences identify a gap in the rational numbers; and we can now just fill in the gaps by declaring that each of the sequence pairs has exactly one number enclosed. In the case where this is no rational number, it is an irrational number (which just means, not rational).

Now of course one has to prove that this gives a consistent definition for those numbers, but one can do that, and indeed, this is one of the standard definitions of real numbers (though usually the two sequences are described as end points of nested intervals, but that is technical details).

So in short, the irrational numbers are the gaps in between the rational numbers, but because the rational numbers are already dense, you cannot specify that gap with a single pair of rational numbers, but you need to use two sequences of numbers "closing up" to each other in order to identify such a gap.

  • 41,315
  • 8
  • 68
  • 125

Imagine you can build chronometers. They all share a common mechanism such that one round takes the same unit of time. To fix ideas, say it is one minute. Now, you can draw any number of ticks (indices?), regularly marked around the dial's edge. It could be $60$ (marking seconds), but you can choose $17$, or any integer.

An irrational is a duration (expressed in your unit of time, here one minute) such that, no matter the number of rounds around the clock, you will never be able to design one chronometer (with regularly spaced ticks) for which the hand would stop exactly on one of the ticks, for this duration.

Laurent Duval
  • 6,164
  • 1
  • 18
  • 47

As you want a metaphor:

Consider you two numbers, say 6 and 8, and you want to related them by lining them up end to end to see where they meet up. Starting at 0, you lay out the six and it goes to 6. You lay out the eight and it goes to 8 so they don't match. You lay out a second six and it goes 12 (no match). You lay out the eight to 16 (no match). You lay out a third six to 18 (no match). You lay out a third eight to 24 (no match). You lat out a fourth six to 24 and you have a match.

4 sixes = 3 eights. six to eight is a ratio of three to four or 3/4. Such a nice ratio of whole numbers is called a rational number.

Try it with something not so nice. $5 \frac {16}{23}$ and $52 \frac {10}{21}$ for instance will be pretty tedious but they will eventually match up if we lay out 582958 of the $5 \frac {16}{23}$ and 2751 of the $52 \frac {10}{21}$ creating a ratio of 2751 to 582958 or the rational number $\frac{2751}{582958}$.

Now if we pick two arbitrary values, there is absolutely no reason on earth to assume that they ever will match up in any whole number of steps to attempt to cram them together.

If we try to match $1$ with $\sqrt{2}$ or $1$ with $\pi$ we'll never be able to. So the ratios of $\sqrt{2}$ to $1$ or $\pi$ to $1$ will never have any nice whole number ratios.

It's important to note that we can get as close to cramming to within any close but not complete margin as we like. It's also important to note any value we can express in terms of fractions has already been divided neatly into even parts so any two fractions (or whole numbers, or terminating decimals, are combinations thereof) will match up eventually. But if we pick values abtrarily by tossing an infinitely precise dart at a mile long yard stick there is no reason to believe the resulting value is a fraction of anything.

The question is how do we "really get" what such a value is if we have no method of expressing it. That is where it feels the "floor is pulled out from under us". But we shouldn't feel as the the floor was removed. We should feel instead that we had been building a floor out of cards flimsily relating whole numbers to whole numbers when there is no reason to believe all values relate to whole numbers. We should feel instead that we built a flimsy but very wide net upon which to walk, only to discover there is a solid granite bedrock foundation beneath it.

It should make us feel good. But we're humans so instead we feel as though the wind was knocked out of us.

  • 1
  • 5
  • 39
  • 125

There are two words involved here: ratio and reason or reasoning.

It seems that the word "irrational" in current mathematics is related just to "ratio" rather than "reason". A rational number is a number that can be written as a ratio of whole numbers while an irrational is just a number that it is not rational.

However, the first irrational that has been known is the length of the diagonal of a side $1$ square. This happened for the first time when non-rational numbers were unknown (even the negative integers were still at their birth aurora). And in such a circumstance it would seem that $\sqrt 2$ was something contrary to reason, i.e. “irrational”.

It is well known that there are many more irrational than rational numbers which would indicate that in politics often what counts most for the meaning of "irrational" is the old sense of $\sqrt2$. Can you get of this a real-word metaphor for irrational numbers?

  • 25,368
  • 3
  • 24
  • 47

What metaphor do you have for "number"? (specifically, "real number"; e.g. positive real numbers are the kinds of numbers you use to measure distance in the Euclidean plane)

That's what an irrational number is. Nearly everything that is a number is an irrational number.

There are some rare numbers that happen to be ratios of integers and we call those special numbers "rational" numbers. The adjective "irrational" simply means that the described number doesn't happen to be one of those special ones.


Imagine that you have a sifter like this one, and that you are sifting through sand or soil. Let's call the internal lengths of the sifter $x$ and $y$, and let's assume that every particle which falls over the sifter represents a number from $0$ to $1$. The number $0.657$ for example, would be represented by a particle that falls a distance (along the x-axis) of $0.657*x$ from one rim, and a distance of $0.657*y$ (along the y-axis) from the other rim.

Each way of dividing the space $x*y$ into a grid pattern represents a certain numeral system, utilizing a limited number of decimals.

If a particle falls exactly on the grid, it represents a value that can be written using the numeral system and number of decimals represented by the grid.

Now, imagine that the spaces in the grid pattern are evenly distributed, and there are as many spaces along the $x$ axis as along the $y$ axis.

If for example there are $n$ spaces along each axis, the first perpendicular thread along the x-axis will appear at a distance of $\frac{x*1}{n}$ from the rim, the second at a distance of $\frac{x*2}{n}$ amd so on. enter image description here
Now, only those particles that fall exactly on the grid will stuck, the rest will fall through. Let's say there are $5^3 - 1$ threads along each axis in the grid, dividing each axis into $5^3$ evendly distributed spaces.

This means that any particle which falls a distance from the rim (along the x-axis), consisting of a multiple of $\frac{x}{5^3}$ will get stuck on the grid.

Now, if you have a particle that falls a distance (along the x-axis) of $\frac{x*10}{5^3} + \frac{x}{5^4}$ from the rim of the sifter, it will fall through, unless you increase the number of spaces by $5$, so that you have $5^4$ spaces, and $5^4-1$ threads along each axis.

Now, imagine a particle that falls such a length from the rim of the sifter, along each axis, that it wont get stuck unless you have a grid pattern which is infinitely small. Such a particle represents an irrational number.

  • 314
  • 1
  • 7

The "high school" definition is usually something like

A number is irrational if its decimal expansion neither terminates nor repeats

In other words, whenever a number is rational, you can express it with a decimal that either eventually stops (as in $1.234$) or repeats (as in $1.234234234234... = 1.\overline{234}$). However, we could come up with numbers where the decimal does something else. For example, $\pi = 3.14159...$ neither ends nor repeats, and $0.10100100010000100000...$ follows a non-repeating pattern.

These non-repeating decimals are numbers; you can easily order them in between rational numbers. For example, it's clear from comparing the decimal expansions that $3 < \pi < 3.\overline{3}$. However, we can't write these numbers as a fraction of integers.

It is troubling, however, that one must take it on faith that these objects are really "numbers" in the same sense as rational numbers.

Ben Grossmann
  • 203,051
  • 12
  • 142
  • 283
  • I have absolutely no difficulty accepting that the square root of two must be a number, despite being easily proven irrational. The alternative leads to having to argue that it is impossible to construct a right-angled triangle with both short sides exactly one unit long, or that Pythagoras' theorem is false! – nigel222 Dec 20 '16 at 18:03
  • @nigel222 to be sure, there are alternate perspectives (such as [PRA](https://en.wikipedia.org/wiki/Primitive_recursive_arithmetic)). For example, if we suppose that "the only true numbers are rational numbers", one might say that although one cannot construct a perfect isosceles right triangle, one can come "arbitrarily close" to doing so. – Ben Grossmann Dec 20 '16 at 18:10
  • @Omnomnomnom Can you clarify the relevance of PRA here? I believe that PRA can prove the irrationality of $\sqrt{2}$ without trouble; it's Robinson arithmetic that can't do this, if my recollection is correct. – Noah Schweber Dec 20 '16 at 18:38
  • I'm more of a physics - comp.sci person than a Mathematician. I'd not have commented were it not for the framing of the question trying to ground the concepts in "reality" (to which I mentally added "physical"). – nigel222 Dec 20 '16 at 18:41
  • @Noah I could be wrong about which brand of finitism applies here. I may have been thinking Robinson. – Ben Grossmann Dec 20 '16 at 19:04
  • 2
    @Omnomnomnom Well PRA is frequently believed to correspond to finitistic reasoning, but (unless I'm missing something) it can prove e.g. "There are no numbers $p, q$ such that $2q^2=p^2$," which is the natural finitization of the statement "The square root of $2$ is irrational." Similarly, basic statements about $\pi$ and $e$ can be framed appropriately and proved in PRA. In general I don't think the finitism of PRA gets in the way of developing a basic theory irrational numbers. – Noah Schweber Dec 20 '16 at 19:06

Irrational numbers do not exist in the "real world". For example, You cannot physically construct an exact square or a perfect circle. These are purely mathematical concepts. Take for example the corners of one face of a bcc diamond crystal. Although they form a "square", they never form a "perfect" square due to thermal fluctuations and quantum effects. So to say that the length of the diagonal is $\sqrt 2$ times the length of its side is never exactly true. Even if you take a statistical average, you will never reach a value of $\sqrt 2$. However, your average will always be an rational number. So in that sense, an irrational number is the limit toward which your statistical average (a rational number) approaches, but never reach, as you increase the number of observations of a measurement.

  • 207
  • 1
  • 4
  • 3
    I'm surprised at this example of micro-measuring. Wouldn't this argument imply that rational numbers are equally unreal? When measured to a degree of precision that includes quantum effects _nothing_ is exactly one inch long -- in fact, at quantum precision no human-scale object from our "real world" has a knowable length at all that we can be accurate about on repeated measure. Interesting to contrast this approach with answers such as by @TrueVoice, whose claims that irrational numbers often seem *better* at reflecting the real world than rational ones. – JeremyDouglass Dec 20 '16 at 22:15
  • To complete the thought -- I'm suspicious of arguing that an imprecise diamond hypotenuse illustrates irrationals "do not exist" because it also implies that rational numbers "do not exist" according to the same standard, and if there is no distinction then it may not be a useful explanation. You could illustrate this with the sides from the same bcc diamond: it is a square, but the sides aren't equal lengths, therefor units don't equal themselves, therefor "there are no rationals." – JeremyDouglass Dec 20 '16 at 22:25
  • 1
    Precisely. But irrational numbers get the rap while "exactly one" get off unscathed. I think it's more that in the real world exactness doesn't exist and without exact numbers irrational and rationals are not distinguishable. All numbers whithin 23 decimal places are more or less equal. – fleablood Dec 21 '16 at 00:14
  • 2
    Rational number exists in the real world. For example, "there are 10 people in this room", or "0.7 of the people in this room are wearing pants". They convey exact information because they are based on counting. I can't think of any physical example, either intensive or extensive, that is exactly described by an irrational number. Another example: you could conceivably divide a certain pizza exactly into two parts consisting of the same numbers of protons, neutrons, and electrons. But you cannot divide any pizza into 1/pi and (1 - 1/pi) pieces. – gogators Dec 22 '16 at 20:42

Two circus performers are riding unicycles. The tires on the two unicycles are not the same size (different diameters). They are riding along, side by side, when they run over a freshly painted line, and so now they each make a dot on the ground at regular intervals. When they crossed the line, the painted parts of their tires were both on the ground at the same. But as long as they both stay abreast of each other (their distances since crossing the line are both the same), they will NEVER EVER have their paint spots on the ground at the same time again. Enough paint to fill the entire universe wouldn't keep those spots wet long enough to reach another moment of unity like they experienced back at the line. Surely, such absurdity is preposterous! How could you ever explain such a thing? Well, it turns out that when the smaller rider (having the smaller tire) cuts across the tire and perfectly flattens it out, it happens to be the EXACT height (diameter) of the other tire!


What does it mean for the ratio of the lengths of the sides of a rectangle to be rational, say $\frac{5}{3}$? It means that if the long side of the rectangle is divided into five equal parts, and if one counts out three of those parts, then the length of the resulting line segment equals the length of the short side of the rectangle. The short side can now be divided into three parts all equal to the parts of the long side. Hence the rectangle can be tiled as a $3\times5$ array of squares.

3x5 square tiling

Conversely an $m\times n$ array of squares forms a rectangle whose ratio of sides is the rational number $\frac{n}{m}$. So to say that the ratio of sides of a rectangle is rational is the same as to say that the rectangle can be tiled as an array of squares. The question now is, can every rectangle be tiled as an array of squares? The answer isn't obvious. One might imagine that as long as we make the squares small enough, it can always be done.

Before answering this question, let's imagine that someone has told you that a rectangle can be tiled as an array of squares but hasn't told you what size square to use. How would you go about finding the size of square? To approach this, notice that if you manage to find a square that tiles a given rectangle, the same square will tile the shorter rectangle you get by chopping off a square section from the long end of the rectangle.

Chop square

On the other hand, if the rectangle is, in fact, a square, then the rectangle is a square tiling of itself (using only one tile). Because of these two properties, we can find the square we want by chopping square sections off of the rectangle until the remainder rectangle is itself a square.

iterated chop square

By reversing the process, this remainder square will tile the original rectangle, as the following image should make clear.

iterated add squares

At this point, the idea that every rectangle can be tiled as an array of squares may seem much less plausible than it did previously. In order for a tiling with squares to exist, the remainder rectangle in the chopping-off process must eventually be a square. But it is not clear that this always has to happen. Why couldn't it be the case that, for certain rectangles, the chopping-off process continues forever, never resulting in a square? After all, for the remainder rectangle to be a square, its two sides have to be precisely equal. That the two sides should always be at least slightly unequal seems much more probable, from a random starting rectangle, than that they should ever be exactly the same.

These misgivings are all well-founded, but the true situation can actually be even worse than this. For certain starting rectangles, the sides of the remainder rectangle never even get close to approximate equality, much less exact equality. This happens, for example, when the sequence of remainder rectangles falls into a repeating pattern, which occurs when a remainder rectangle is geometrically similar to an earlier remainder rectangle in the sequence.

The simplest example of such a rectangle is the golden rectangle, which is defined by the property that chopping a square section off of the long end of the rectangle results in a rectangle that is similar to the starting rectangle. The ratio of the long side to the short side is known as the golden ratio and has decimal expansion $1.61803\ldots$. All remainder rectangles in the sequence have side lengths in this ratio, and hence none is ever close to being square. As a consequence, the golden rectangle cannot be tiled as a rectangular array of squares, and therefore the ratio of its side lengths is not rational.

golden rectangle iteration

The defining property of the golden ratio implies that its value is $(1+\sqrt{5})/2$. It turns out that similar numbers involving square roots, that is, numbers of the form $r+s\sqrt{d}$ where $d$ is a natural number that is not a perfect square and $r$ and $s$ are rational numbers, always fall into a repeating, but generally more complicated, pattern of remainder rectangles. None of these numbers are rational.

These examples are but the simplest way to see the phenomenon of irrationality. An interesting irrational number has decimal expansion $1.433127\ldots$. If the chopping-off process is carried out on a rectangle whose horizontal side has this length and whose vertical side has length $1$ then, after chopping a square off the long side, the vertical side becomes the long side; after chopping two squares off the long side, the horizontal side again becomes the long side; after chopping three squares off the long side, the vertical side becomes the long side; after chopping four squares off the long side, the horizontal side becomes the long side; and so on. Hence among the remainder rectangles are rectangles that become progressively longer and longer relative to their width. Assuming that this continues, it follows that the remainder rectangle is never a square and hence that that original rectangle does not have a whole-number ratio of sides. Actually proving that this pattern continues for this particular ratio (which is $I_0(2)/I_1(2)$, where the functions $I_n(z)$ are things called modified Bessel functions of the first kind) is considerably more work than in the case of the golden ratio.

A more familiar number that exhibits a similar, but more complicated pattern is $e$, which is therefore also irrational. The chopping-off pattern in this case is $[2,1,2,1,1,4,1,1,6,1,1,8,\ldots]$, where the the numbers represent how many squares get chopped off with each change in orientation of the long edge. Unlike the examples we have looked at so far, however, most irrational numbers exhibit a rather unpredictable chopping-off pattern. For example, $\pi$ has the pattern $[3,7,15,1,292,1,1,1,2,\ldots]$. In any case, it is only when the chopping-off process terminates that the side lengths have a whole-number ratio.

Incidentally, the chopping-off process is usually called the Euclidean algorithm, and the sequences of numbers representing squares chopped off with each change in orientation of the long side are called the coefficients of the continued fraction. Examples of continued fraction coefficients for various numbers, some of which were discussed in this post, are listed below. $$ \begin{aligned} 5/3&=[1,1,2]\\ 22/9&=[2,2,4]\\ 99/34&=[2,1,10,3]\\ (1+\sqrt{5})/2&=[1,1,1,\ldots]\\ \sqrt{2}&=[1,2,2,2,\ldots]\\ (6+\sqrt{10})/4&=[2,3,2,3,1,3,2,3,1,3,2,3,1,\ldots]\\ I_0(2)/I_1(2)&=[1,2,3,4,5,6,\ldots]\\ e&=[2,1,2,1,1,4,1,1,6,1,1,8,1,1,\ldots]\\ \sqrt[3]{2}&=[1,3,1,5,1,1,4,1,1,8,1,14,1,10,2,1,\ldots]\\ \pi&=[3,7,15,1,292,1,1,1,2,1,3,1,14,2\ldots] \end{aligned} $$

Will Orrick
  • 18,856
  • 1
  • 44
  • 78

Irrational numbers entering the 'real world' of point masses and weightless strings: the harmonic oscillator.

Hang a point mass on a (infinitesimally thin yet non-stretchable) string and observe the period of oscillation in the limit of small amplitudes.

Now create a string of the same material and of exactly twice the length. Attach a point mass also to this string and observe the period of oscillations in the limit of small amplitudes. You notice the period of this pendulum is about $1.4$ times longer than the period of the half-length pendulum.

Now let's determine the exact ratio between the two periods. You start both oscillators in sync by letting go both weights from an elevated position at exactly the same moment, and wait till both masses are again simultaneously at their highest point.

When the short oscillator has completed 7 periods, the long oscillator is a bit shy of 5 periods. So you wait a bit longer. And longer. And again longer... You never observe both oscillator to be in sync.

  • 1,062
  • 1
  • 6
  • 18

Here's a simpler metaphor. Imagine an infinite flat plane, a precise sign saying "here", a dart point down some distance away. You measure the distance in miles. It isnt an exact number of miles. You measure it in 1/2 miles. It isn't an exact number of 1/2 miles. You measure it in 1/3, 1/4, 1/5, 1/6, 1/7 miles it isn't an exact measure of any of those either. You keep doing smaller increments, 1/2,897,854 miles, 1/2,897,855 miles. Is there any reason to believe this point will eventually measure exactly into any of those.

Well, a typical student will say well there must eventually be some fuzzy edge where you can't measure any finer and how do you mean the dart is "there"? The edge of the tip? The center of the tip? Where do the quantuum particles that make the dart begin and where does the "not dart" end? Try not to slap or browbeat or bully the poor student but gently point out that this is math and in math we have infinite precision.

Is there any reason to believe this point is any whole number measure of $1/n$ miles for some whole number $n$?

The answer is no, there is no reason to assume that and if it isn't so for that point, that distance is an irrational number.

An irrational value is a value that can not be divided into a distinct whole number of fraction parts with whole number denominators.

Yes, it feels like the floor is being pulled out because it never occurred to the student that any values wouldn't be divisible like that, but when one thinks about it, there was never any reason to think that they should.

  • 103
  • 2
  • 1
  • 5
  • 39
  • 125

There seem to be a lot of answers that dance around this issue so I'll get directly to the point. The $\sqrt(2)$ examples and the circle example, and ranting about stars are trying to find a simple way to explain a profound mathematical concept, density of numbers. While barak manos' circle example seems to be the most liked for it's intuition, it's incomplete.

I would ask a very simple follow up question. Why is it not possible to construct the prescribed lengths? The answer is, in some sense, because the rationals are too coarse. If you were only allowed to measure things up to a $\frac{1}{16}$th (like a typical school ruler) there would be lots of things you couldn't measure (width of a human hair for example).

You might then say, "Ok, I'll get a better ruler, one that is finer". But then the question becomes how fine can you go? As it turns out, the answer is infinitely fine. But even if you had an infinitely fine ruler, you would still have irrationals between the infinitely fine rational points. This is what is meant by "dense". In the stars analogy, the irrationals are the empty space that fills in the gap between stars.

If you consider Omnomnomnom decimal expansion approach, you can get a feel for how big the "space" is. Consider 0.5 and 0.51, and ask how many decimals are there between those two numbers. Since there is no limit to the number of digits you can tack the back of the expansion (e.g. 0.502, 0.5002, etc...), there are infinitely many. Some of these are rational (each time you tack on just one more digit, it's still rational because the decimal terminates). But for each rational you add, an infinite number of irrationals come along for the ride.

Why does this happen? Consider adding .501. It terminates, and thus is rational. Now ask how many decimals are there that start with .501 and have more digits past the 1? The answer is again infinitely many. You might stop and say well about .5011, it's in that list and is rational? But we can turn the question around again and ask about decimals that start with .5011. This process never ends but each time you add 1 rational, an infinite number of irrationals come with it.

This gets to the other point that is missed, the notion of "next". If I told you an natural number $\mathcal{(N)}$ for example 7, and asked you what the next number was, the answer would be 7+1=8. For natural numbers next makes sense, but If I gave you an irrational number and asked the same question, the choice is not obvious (7.1, 7.01,...?). Now you may ask, what is the next rational for a given rational. It turns out this can be defined but it's not obvious how (look up Cantor counting).


Noah Schweber points out that the cantor ordering is not natural ordering of $\mathcal{Q}$, and this is correct. This starts to dig into a bigger problem of asking how do you tell when two sets of numbers have the same amount of numbers (called the cardnality) in them. But to resolve this, we'll need more mathematical machinery, specifically bijective maps between sets.

There is some insight to be gained in the idea that the density and the ordering of a set of numbers end up being closely related. The bigger picture that I was aiming for, is that when you reach the point where you are trying to discern how the rationals are different from the irrationals, you've actually found the surface of a very complex problem. Gaining some intuition about this is actually a very good step on your journey.

James S.
  • 208
  • 1
  • 5
  • The last sentence is extremely misleading. There is no next rational in the usual ordering. This should be made explicit if you're going to mention Cantor's well-ordering of $\mathbb{Q}$. – Noah Schweber Dec 21 '16 at 20:32

There is no real world metaphor

$\mathbb{R}$eal numbers are the union of rational an irrational numbers. The irrational numbers are the step to abstract world of unrealistically continuous space.

They are a contradiction to $\mathbb{Q}$uantum world and if we believe in it, then the irrational numbers indeed are a hole in the non $\mathbb{R}$eal world. But I like to think it more as a pixel grid, so that there is no holes.

There is concepts of the uncountably infinite sets and even more uncountable to that. Irrational numbers are kind of the stepping point between abstract mathematics and the real world mathematics. Too hard to explain for the children, without watering the concept. The real world has no perfect shapes.

  • 163
  • 7