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In the book "Zero: The Biography of a Dangerous Idea", author Charles Seife claims that a dart thrown at the real number line would never hit a rational number. He doesn't say that it's only "unlikely" or that the probability approaches zero or anything like that. He says that it will never happen because the irrationals take up all the space on the number line and the rationals take up no space. This idea almost makes sense to me, but I can't wrap my head around why it should be impossible to get really lucky and hit, say, 0, dead on. Presumably we're talking about a magic super sharp dart that makes contact with the number line in exactly one point. Why couldn't that point be a rational? A point takes up no space, but it almost sounds like he's saying the points don't even exist somehow. Does anybody else buy this? I found one academic paper online which ridiculed the comment, but offered no explanation. Here's the original quote:

"How big are the rational numbers? They take up no space at all. It's a tough concept to swallow, but it's true. Even though there are rational numbers everywhere on the number line, they take up no space at all. If we were to throw a dart at the number line, it would never hit a rational number. Never. And though the rationals are tiny, the irrationals aren't, since we can't make a seating chart and cover them one by one; there will always be uncovered irrationals left over. Kronecker hated the irrationals, but they take up all the space in the number line. The infinity of the rationals is nothing more than a zero."

regularmike
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    If the probability that an event occurs is 0, that (counterintuitively) does not mean that it is impossible. This Wikipedia article may help. – Zev Chonoles Jun 07 '12 at 14:34
  • Of course it is possible. It just has probability $0$. – André Nicolas Jun 07 '12 at 14:36
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    food for thought : if you take a random real number between 0 and 1, the probability to get that number was 0. and yet you got it ! that should explain intuitively why we need to give zero probability to some events that can actually happen – Albert Jun 07 '12 at 14:43
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    I don't need help understanding why the probability is 0 or why it should still be theoretically possible. I'm just trying to confirm that he's wrong to say "never". – regularmike Jun 07 '12 at 14:44
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    Hitting any individual number has probability zero. So with this logic we will never hit any point when throwing a dart. Can you see the flaw in this? Eventhough rationals cumulate zero mass on the real line, this doesn't mean that they don't exist. – T. Eskin Jun 07 '12 at 14:45
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    Ok, so what happends if you throw uncountable many darts? – TROLLHUNTER Jun 07 '12 at 14:53
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    I think it is important that the author says it will not happen, not that it cannot happen. So, it is not impossible, it just is not going to happen... I think this is a nice way to explain probability zero, although not mathematically exact. – Ansgar Esztermann Jun 07 '12 at 15:02
  • @AnsgarEsztermann Yes, that's an important distinction. But exactly how sure can we be? How many years would it take infinite monkeys throwing darts to hit a rational? – regularmike Jun 07 '12 at 15:04
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    I think a valid answer to all those questions is : measure theory is just a mathematical theory, it's not perfect and while it's useful, you can't pretend to have an acurate description of physical reality. actually, the mere concept of "is going to happen" or "cannot happen" is already a heavy philosphical assumption on the nature of the future and the possible... – Albert Jun 07 '12 at 15:17
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    @ZevChonoles and others: Don't you agree that this is purely a matter of interpretation? In the real world, we will never be able to see the difference, and in mathematics people don't really distinguish between, say, the measure on $[0,1]$ with density $1$ and the measure on $\mathbb{R}$ with density $\chi_{[0,1]}$. My interpretation of "probability zero" has always been "impossible", and I'm not convinced at all that this is wrong or misguided. – Stefan Jun 07 '12 at 18:28
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    @StefanWalter: It really depends on how you define the word impossible. If impossible means that it can not happen, then clearly this does not coincide with the concept of probability zero and it would indeed be wrong/misguided to say that it would. – T. Eskin Jun 07 '12 at 19:56
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    @ThomasE.: That's what I mean by impossible. Extreme example: Consider the experiment of throwing a coin: Let's model it by a set with three elements $A,B,C$. $A$ stands for "head", $B$ stands for "tail" and $C$ stands for some event that is absolutely unthinkable. Let $p_A=p_B=0.5$ and $p_C=0$. Though redundant, this model accurately describes a fair coin flip (maybe that's where you disagree, but why?). But including $C$ in the mathematical description surely cannot change the fact that $C$ is impossible. – Stefan Jun 07 '12 at 20:45
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    @Xnyyrznaa obviously you would ruin a perfectly good number line-themed dart board – Phillip Schmidt Jun 07 '12 at 20:47
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    If the dart has a finite width at the tip, you will simultaneously hit infinitely many rational and irrational numbers... – zzzzBov Jun 07 '12 at 21:11
  • @Xnyyrznaa It's still possible that you don't hit a rational, since a set of uncountable darts have a bijective correspondence to the set of irrational numbers. In fact, since the set of irrationals within any sub-interval of [0,1] can be put in bijective correspondence with an uncountable number of darts, you need not hit all the irrationals either. Now, the probability of one dart hitting a rational number is exactly zero, and therefore the probability of an uncountably infinite set of darts hitting a rational number can be proved by transfinite induction. (I believe...?) – mboratko Jun 08 '12 at 03:32
  • @regularmike We can even include any countable set of irrational numbers - say we take the union of the rationals and all the square roots of all the rationals, some of which will be irrational. This set has measure zero, so we hit one of the elements of this set with probability zero. We can expand this even countably many times - say we take the union of all the rationals and all $n$-th roots of the rationals, where $n$ is any integer. You could continue appending countable sets of irrational numbers, and since a countable set of countable sets is countable the result would be the same. – mboratko Jun 08 '12 at 03:44
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    @StefanWalter: This doesn't really change anything. Either $C$ is an event of neither tails nor heads (e.g. wind blew the coin away) in which case, again, if $C$ is zero measurable it does not mean that it can not happen. It only says that tails or heads are gotten almost surely. On the hand, if you consider no other options than tail or heads, then $C$ is empty. For the r.v. that takes two values, any set not containing them has empty preimage. In fair coin tossing there are no non-empty zero measurable sets separating $A$ and $B$. – T. Eskin Jun 08 '12 at 04:33
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    Does it matter that there is zero probability of dart hitting the line? – default locale Jun 08 '12 at 07:59
  • @defaultlocale Fair question, but no, it doesn't. If the dart hits above or below the line we would just consider the intersection of the number line and the line perpendicular to it that contains the point that the dart hit. – regularmike Jun 08 '12 at 13:17
  • @ThomasE. "if you consider no other options than tail or heads, then C is empty." And this implies that it cannot happen? This sounds to me as if you can make an event possible by merely considering it. – Stefan Jun 08 '12 at 20:42
  • @StefanWalter In the mathematical model you are creating, that is true. In a fair coin toss, we are modeling the physical coin toss using a simplified mathematical model in which there exist - by our own choice - two possibilities: heads or tails. Each of these have - again by our choice - an exact 50% chance of occurring. In reality is this the case? No, of course there is the chance of the coin blowing away or landing on it's side, and of course a real coin cannot be perfectly fair, but our mathematical model is not made to include those events. – mboratko Jun 08 '12 at 23:00
  • @MichaelBoratko: My model excludes those events too, by assigning them the probability zero. If I wanted to account for the chance of the coin blowing away, I would have to write something like $p_C=\epsilon>0$. – Stefan Jun 08 '12 at 23:41
  • @Glougloubarbaki: A random number picked by a human being is far from uniformly distributed on $[0,1]$. – Stefan Jun 09 '12 at 00:05
  • @StefanWalter: I only told how it works in the $\sigma$-algebra generated by this random variable: I didn't assign anything to them myself. No matter how much we discuss about it, an event having probability zero will not imply it is impossible. To conclude to the last example, lets say we have the option of wind blowing the coin away with probability zero. If you're inside the house, then it's impossible (because there is no wind!) but outside, eventhough it's zero probability, the wind exists and it not impossible. It is only almost sure (a.s.) that the wind does not blow the coin away. – T. Eskin Jun 09 '12 at 22:14
  • @ThomasE.: If being outside makes it possible that the wind blows the coin away, the corresponding event should have positive probability and a model assigning it probability zero is not accurate. This is so obvious to me that it is hard to give a reason for it. – Stefan Jun 10 '12 at 08:30
  • @StefanWalter: To get back to the topic: each individual number on the real line has probability zero when throwing a dart, yet you always hit one of them. Once you did hit one, an event with probability zero occured. Being likely to happen and being possible to happen are two different things. – T. Eskin Jun 10 '12 at 09:55
  • @ThomasE.: Before we get back to the other example: Do you agree with me on the coin toss and the wind or not? – Stefan Jun 10 '12 at 10:26
  • @StefanWalter: I don't. – T. Eskin Jun 10 '12 at 10:55
  • @ThomasE.: Well, then it's pointless to discuss a more complicated example. – Stefan Jun 10 '12 at 11:24

6 Answers6

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Mathematicians are strange in that we distinguish between "impossible" and "happens with probability zero." If you throw a magical super sharp dart at the number line, you'll hit a rational number with probability zero, but it isn't impossible in the sense that there do exist rational numbers. What is impossible is, for example, throwing a dart at the real number line and hitting $i$ (which isn't even on the line!).

This is formalized in measure theory. The standard measure on the real line is Lebesgue measure, and the formal statement Seife is trying to state informally is that the rationals have measure zero with respect to this measure. This may seem strange, but lots of things in mathematics seem strange at first glance.

A simpler version of this distinction might be more palatable: flip a coin infinitely many times. The probability that you flip heads every time is zero, but it isn't impossible (at least, it isn't more impossible than flipping a coin infinitely many times to begin with!).

Qiaochu Yuan
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  • So do we agree that it's not accurate to say "never"? – regularmike Jun 07 '12 at 14:46
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    @regularmike: well, one can interpret this as a little artistic license on the author's part, and in any case the English language is not precise enough to make the kind of distinctions that mathematicians make (it is hardly precise enough to make the kind of distinctions that non-mathematicians make!). It depends on whether you interpret "never" as "impossible" or "probability zero." – Qiaochu Yuan Jun 07 '12 at 14:48
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    Fair enough, but I still think it's misleading to say "never" unless we think that even if everyone in the world were throwing darts at a number line for a few hours every day that we still wouldn't hit one in anyone's lifetime. Do you think the rationals are safe enough from darts to warrant such confidence? – regularmike Jun 07 '12 at 15:02
  • @regularmike YES, probability zero means exactly that the rationals are safe enough in that sense. – Phira Jun 07 '12 at 15:05
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    @regularmike: yes. For any finite number of throws, the probability that any of those throws hits a rational number is still zero. We could fill the universe with dart-throwing nanites and force them to throw darts at a number line a Planck length away at the speed of light until the heat death of the universe and the probability would still be zero. (This is all, again, under the "magical super sharp dart" hypothesis.) – Qiaochu Yuan Jun 07 '12 at 15:07
  • @QiaochuYuan I like this idea a lot. But it doesn't necessarily require super sharp darts. Couldn't you just use the center point of the dart's tip? So I make a big fat hole on the number line with my dart that covers a range of values, but maybe it's still exactly centered at 0. – regularmike Jun 07 '12 at 15:12
  • Yes. The point is that you're hitting precisely one number at a time, and not any larger interval of them (which would clearly always contain rational numbers). Though by the above point it will never (probability 0% never) be "exactly centred at 0" like you say because 0 is rational. – Robert Mastragostino Jun 07 '12 at 15:28
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    @regularmike: it doesn't matter. Looking at the exact center of a fat dart is the same as looking at the point of a magical super sharp dart. – Qiaochu Yuan Jun 07 '12 at 15:31
  • @QiaochuYuan Right, I'm just saying it doesn't have to be completely hypothetical. We could actually perform the experiment right now, but just probably not be able to measure any results. – regularmike Jun 07 '12 at 15:43
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    @regularmike: there's no way to actually find the "exact center" of a physical dart. Once you zoom in too far Heisenberg's uncertainty principle kicks in and you can't have complete knowledge of the position of all of the atoms in the dart without having zero knowledge of their momenta... more generally, exact positions don't really have physical meaning. – Qiaochu Yuan Jun 07 '12 at 15:46
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    I’ve gotten the feeling that physicists too don’t understand the difference between “probability zero” and “impossible”. – Lubin Jun 07 '12 at 21:06
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    @Lubin: well, for physics the distinction is pretty much irrelevant. – Qiaochu Yuan Jun 07 '12 at 21:15
  • @QiaochuYuan what languages are better at making these distinctions? – Caleb Jares Jun 08 '12 at 01:02
  • @cable: I have no idea! I don't know much about languages other than English. – Qiaochu Yuan Jun 08 '12 at 01:36
  • @QiaochuYuan, without being at all expert in these matters, I think the physicists’ misunderstanding is the source of some extremely serious philosophical and even scientific errors. I’d love to discuss this with you elsewhere: could you e-mail me? – Lubin Jun 08 '12 at 04:38
  • @Lubin as a physicist I would say that the problem is more with confusing "low probability" with "impossible", rather than "measure zero" with "impossible". But keep in mind that physicist usually handle low probabilities well enough (think of computation involving quantum mechanical crossing of barriers). And physicist often also handle the subtleties of hypothesis testing reasonably well. Now, surely "often" doesn't mean that the set of cases where some error may happen has measure zero :-) – Francesco Jun 08 '12 at 10:08
  • @cable729: No languages are better at this. The purpose of mathematical technical vocabulary is precisely that it is invented language for making precise distinctions that the 'host' language does not. – Tynam Jun 08 '12 at 12:39
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    @Xnyyrznaa: you are welcome to write your own answer if you feel there is something lacking in the existing answers. I'm not sure how to respond to your accusation that I am trying to accrue upvotes. I certainly don't think this answer is worth 48 upvotes, but in answering questions on SE sites I think a philosophy of "rapid prototyping" makes more sense than anything else. Can you be more specific about what you find confusing about this answer? – Qiaochu Yuan Jun 09 '12 at 18:44
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Note that if you randomly (i.e. uniformly) choose a real number in the interval $[0,1]$ then for every number there is a zero probability that you will pick this number. This does not mean that you did not pick any number at all.

Similarly with the rationals, while infinite, and dense and all that, they are very very sparse in the aspect of measure and probability. It is perfectly possible that if you throw countably many darts at the real line you will hit exactly all the rationals and every rational exactly once. This scenario is highly unlikely, because the rational numbers is a measure zero set.

Probability deals with "what are the odds of that happening?" a priori, not a posteriori. So we are interested in measuring a certain structure a set has, in modern aspects of probability and measure, the rationals have size zero and this means zero probability.

I will leave you with some food for thought: if you ask an arbitrary mathematician to choose any real number from the interval $[0,10]$ there is a good chance they will choose an integer, a slightly worse chance it will be a rational, an even slimmer chance this is going to be an algebraic number, and even less likely an transcendental number. In some aspect this is strongly against measure-theoretic models of a uniform probability on $[0,10]$, but that's just how life is.

Asaf Karagila
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    True. Though I daresay if you ask an arbitrary mathematician, the chance is actually not so slight that they'll choose a nonrational number, most likely one of the obvious $\sqrt{2},e,\pi,2\pi,\ldots$. – leftaroundabout Jun 07 '12 at 17:55
  • We can run an experiment, it's not too hard. – Asaf Karagila Jun 07 '12 at 18:04
  • What do you mean by 'highly unlikely'? – TROLLHUNTER Jun 07 '12 at 20:47
  • @AsafKaragila: And what exactly do you mean by "uniformly choosing a real number"? – krlmlr Jun 07 '12 at 21:02
  • @Xnyyrznaa: That there is a probability of zero of such thing happening. – Asaf Karagila Jun 07 '12 at 21:06
  • @user946850: There is a well-defined notion of a continuous uniform probability distribution. I mean that. – Asaf Karagila Jun 07 '12 at 21:07
  • @AsafKaragila: Over the whole real line? (Please forgive my pickiness...) – krlmlr Jun 07 '12 at 21:30
  • @user946850: No. Uniformly means that an interval $[a,b]$ has the measure of $\tau\cdot(b-a)$. We can show by standard arguments that there cannot be such $\tau$. I will stress this in my post. – Asaf Karagila Jun 07 '12 at 21:38
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    The last paragraph is, of course, a comment on mathematicians, not mathematics, but it's highly pertinent. Short form: the human brain is a terrible uniform random number generator. It doesn't visualise distribution well. Which is also why it's so hard to understand that the probability of hitting a rational being 0 means that you'll never hit one with a finite number of darts - your brain can find rationals easily, so it has no intuitive grasp of what it means that they are sparse. – Tynam Jun 08 '12 at 12:49
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    A mathematician naming an undefinable number, on the other hand, is both impossible and has probability zero. – MartianInvader Jun 08 '12 at 19:19
  • @MartianInvader: This is just because the use of "definable number" is a bit murky. If you fix a language which may very well include elementary functions, integrals, all the rationals, and even Turing machines, you can still only define (internally) a countable number of reals inside a given model of ZFC. I can take an enumeration of the real numbers and pick the first one not definable by the above relations. This is not contradictory because the well-order itself was not definable, nor it was a part of the language allowed to define real numbers. – Asaf Karagila Jul 03 '12 at 21:39
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    (cont.) You could argue that I did not specifically name any number at all, but then again... what is a name of a number? I could augment the language by a constant $m$ and add the countable set of axioms which say that $m$ is not any of the numbers definable by the aforementioned functions, now $m$ itself is definable but in a richer language, and not in the original. So we're still fine! :-) – Asaf Karagila Jul 03 '12 at 21:41
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I think the author is exaggerating a bit in order to convey the idea. This is more clearly noticed with the phrase "the infinity of the rationals is nothing more than a zero", which is certainly not true when taken literally. What does happen, as Qiaochu says, is that the Lebesgue measure of the set of rational numbers is zero, because it's a countable set, and the probability of getting a rational number when picking a random number on the real line is indeed zero. However, that doesn't mean it's not possible to get a rational number; you can get "really lucky" and pick any of the infinite rational numbers. However, it's very unlikely, in a specific sense that you will learn from measure theory and probability theory.

talmid
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    I don't think they are exagerating--the accepted answer seems correct. Another way to think about it, how do you write the number "0" exactly? "0" is just an abrevation/truncation. It's 0.00000... but you have to keep on writing. Any time you write anything but a zero, you can stop (because n != 0) but you'd have to write infinitly long for it to actually be zero, and nobody can write infinitly long, so you can't actually even truly write "0", how could you hit it with a dart? – Bill K Jun 08 '12 at 00:30
  • @BillK If your argument held, one could say the same for any real number and then it would be impossible to get any real number by picking a random real number, but my dart must hit somewhere, right? It is certainly possible to pick any one of the elements in $\mathbb{R}$ (and not $i$ or $x^2 + 3x$ for example), but each one has probability zero. This is a counterintuitive idea, but if each real number had probability $\epsilon \gt 0$ (all numbers should have the same probability of being chosen), then the probability of picking any number, summing all probabilities, would not be $1$. – talmid Jun 08 '12 at 02:10
  • @BillK And yes, the author is exaggerating. "the infinity of the rationals is nothing more than a zero" is a nice example because it specifically says $\aleph_0 = 0$, if taken literally. So I understand the author is speaking figuratively. One could say, without exaggerating: "the infinity of the rationals is nothing more than that of the natural numbers". – talmid Jun 08 '12 at 02:12
  • @BillK "0" is not an abbreviation/truncation for 0.00000..., the two are equivalent but this is for an entirely different reason and has nothing at all to do with the reason you can't hit 0 with a dart. – mboratko Jun 08 '12 at 03:38
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    @MichaelBoratko It has everything to do with the reason you can't hit 0 with a dart. Obviously, your dart could hit 0 to within a millimeter, or a nanometer, or a picometer. In fact, the dart could hit within any arbitrary precision you specify, and if that's the best you can measure, you'd say that the dart hit 0. But if you could measure more precisely still—adding more and more digits—eventually you'd find that you had not hit 0 after all. (But this cannot happen in the real world, because of Heisenberg uncertainty. So in fact, in the real world, you will ALWAYS hit rationals.) – Nick Matteo Apr 14 '13 at 00:22
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No.
The probability of hitting a specific number is 0, whether it's rational or not. However, when we throw the dart, we'll inevitably hit a specific number. Thus hitting this specific number was not impossible.

user606723
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One very useful way to think about probability is in terms of betting. Suppose someone offers you a payoff of 1 dollar if event X happens, and 0 dollars if event X does not happen. What's the largest amount of money that you're willing to pay to play this game? That amount is the probability of X happening. (Probably I need to be a bit more careful, but this is roughly the idea.)

So what does it mean to say that an event has probability zero? It doesn't mean that it can't happen, it just means that you wouldn't be willing to play that game for 1 cent, or a tenth of a cent, or any actual non-zero amount of money.

If you want to read more about this way of thinking about probability, you can search for "Dutch book."

Noah Snyder
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  • Dear Noah, "smallest amount" should probably read "largest amount", and I guess this is correct when the probability of $X$ is small. Regards, – Matt E Jun 08 '12 at 02:39
  • Thanks for the correction. Not sure where small would come in... I think the technical bit that I don't have totally right is that it's best phrased as "there's no way for you to beat the house betting on either side" rather than what you would be willing to play. – Noah Snyder Jun 08 '12 at 03:13
  • Dear Noah, I think I'm just wrong regarding "small" (I got confused), so ignore that! Cheers, – Matt E Jun 08 '12 at 03:20
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    If only we where that rational, then the stock markets would not have crashed. But the probability of throwing a dart in to a trading pit at the stock market (or your pension providers office), and hitting a rational is also zero. – ctrl-alt-delor Jun 08 '12 at 10:11
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It's possible, but it would take you an infinite amount of time to verify that you actually hit a rational number because you would have to keep "zooming in" forever.

Thomas
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    If we're already hypothesizing that we can somehow work with a physical number line that actually contains all real numbers and also hypothesizing that we have darts sharp enough to hit a single number on this line, why not hypothesize that we can verify that we did or did not hit a rational number instantaneously? – Qiaochu Yuan Jun 07 '12 at 20:57
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    If you're using a physical number line then you would always hit a rational, because physical space is quantized - there would always be an integral number of atoms (or planck lengths perhaps) on both sides of the point you hit. – evil otto Jun 07 '12 at 22:30
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    @QiaochuYuan I was assuming that the only hypotheses being made were that the dart was sharp enough to hit a single point, that the number line contained all real numbers, and that neither of those hypotheses implied that it would be possible to verify if the dart hit a rational in a finite amount of time. I was trying to offer insight as to why one might think a dart could never hit a rational, because the intuitive thing to do after the dart hits is to check to see where it hit. – Thomas Jun 08 '12 at 00:43
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    @Thomas: okay, but it would also take an infinite amount of time to verify that you hit an irrational number in this framework, and the probability of that happening is still $1$. This issue is unrelated to the issue the OP is asking about. – Qiaochu Yuan Jun 08 '12 at 01:37