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It seems to me that the set of all numbers really used by mathematics and physics is countable, because they are defined by means of a finite set of symbols and, eventually, by computable functions.

Since almost all real numbers are not computable, it seems that real numbers are a set too big and most of them are unnecessary.

So why do mathematicians love this plethoric set of numbers so much?

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    $\begingroup$ Mathematicians use only positive integers below 10 most of the time, so why would anyone even need rationals? $\endgroup$ – flawr Dec 23 '14 at 10:45
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    $\begingroup$ Numbers are not defined by the symbols used to represent them. $\endgroup$ – chepner Dec 23 '14 at 14:27
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    $\begingroup$ Has anyone compiled a list of all the previous times this has been asked here? $\endgroup$ – GEdgar Dec 23 '14 at 15:03
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    $\begingroup$ @GEdgar I suspect it's countable and finite but unbounded. $\endgroup$ – fluffy Dec 23 '14 at 16:57
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    $\begingroup$ Pure mathematicians only use three numbers anyway, zero, one and eight, the last of which is written on its side, pronounced "infinity" and means "a lot." Can't remember where I first heard that, but always loved it! $\endgroup$ – Avrohom Yisroel Dec 23 '14 at 21:37

13 Answers 13

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It is much harder to avoid "unnecessary" real numbers than to accept them.

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    $\begingroup$ My point in a catchy elevator pitch! +1 $\endgroup$ – Asaf Karagila Dec 23 '14 at 20:48
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    $\begingroup$ @Hurkyl I agree (+1). But there is some attempt, apart constructivism, to do this hard work? $\endgroup$ – Emilio Novati Dec 25 '14 at 20:59
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    $\begingroup$ @Emilio: While there are always people pursuing their pet philosophies, I do not believe there is a mainstream project to do so. There does not appear to be any physical content in the hypothesis, so all that matters is what is simpler to work with. And since theoretical physics is already an enormously complicated affair pushing the boundaries of available mathematics, adding in another thing to make it even harder sounds like a bad idea. I opine that if physics does ever go that way, it won't be because people were trying to go that route, it will just be a consequence of the math used. $\endgroup$ – Hurkyl Dec 25 '14 at 22:00
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    $\begingroup$ But a disclaimer: I am not a physicist, and do not fully keep up on physics research. $\endgroup$ – Hurkyl Dec 25 '14 at 22:01
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    $\begingroup$ @Emilio: Also, there are mathematicians that do study computable analysis, if that's what you were asking. I'm just not aware of it done in the context of theoretical physics. $\endgroup$ – Hurkyl Dec 27 '14 at 2:11
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From a physics point of view, all you ever produce as the outcome of a measurement is a rational number. More accurately, any well-defined physics experiment is reproducible, and thus one can always produce a finite sequence of rationals out of a potentially infinite sequence rationals of outcome measurements per experiment. Each measurement is an approximation to the 'actual outcome', whatever it may be. Now, if all goes well, the entire countable sequence of rationals is a Cauchy sequence which is the actual outcome.

Now, mathematically it is convenient to be able to speak of this 'actual outcome' as a single object in a nice system of measurements. One way to do so is define the real numbers. Then the outcome of an experiment really is a real number. Now, when you do that you find that are actually created lots and lots of new real numbers, most of which can't ever be obtained as the outcome of anything. That's not a big deal, and its a rather small price to pay for having a really convenient system of numbers of great relevance to physics to work with. The fact that you only use a fraction of those numbers is not particularly relevant.

Another reason for introducing the real numbers is that they exhibit pleasant computational properties. Perhaps most importantly is that they form a complete metric space. Now, without that property most of analysis fails, so we really need the reals to be complete, and that necessarily means lots of numbers out of our reach. Again, this is a small price to pay for having a fantastic accompanying mathematical theory.

Finally, don't forget that it is not always a single number that is sought as the result of a computation. Sometimes what one needs to know is whether or not an integral converges, but the actual value where it converges to is immaterial, and whether or not it is a computable number in any sense is irrelevant. Then again it is crucial to have a system where you can use lots of tools and the reals provide that.

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    $\begingroup$ " The fact that you only use a fraction of those numbers is not particularly relevant. " -- strictly speaking, you don't even use a fraction of those numbers. ;) Still, that's a good answer, +1. :) $\endgroup$ – tomasz Dec 23 '14 at 15:44
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    $\begingroup$ I would say that in physics measurements produce distributions, which are then sampled over rationals. That is to say if you measure something with a ruler your result will be something that looks like a Gaussian around whatever you read off the scale, and then you'll say something like "the value is between 0.55m and 0.56m with 95% certainty". $\endgroup$ – Sean D Dec 23 '14 at 17:07
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    $\begingroup$ Most physicists would actually say exactly the opposite of your first sentence: that all measurements are irrational, albeit that no measuring device is actually capable of exactly representing a measurement. It goes to show the difference between a physics point of view and a mathematician's view of physics. (Of course @SeanD's comment is probably an even better description.) $\endgroup$ – David Z Dec 23 '14 at 19:58
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    $\begingroup$ I think we are saying the same thing. Which physicist will look at the reading off a device, something like 0.5434466, and say "that's irrational"? What you read off a device is a rational approximation to the actual outcome (which is often irrational). This is exactly what @SeanD is saying and what I tried to say in my answer. $\endgroup$ – Ittay Weiss Dec 23 '14 at 20:54
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    $\begingroup$ If I make a measurement with my protractor, the result is going to be a rational multiple of $\pi$, rather than a rational number. If I determine the height of a tree by making a few easy measurements and then invoking trigonometry, my answer is surely going to be a transcendental number unless I decide to round it. It's pretty easy to make a ruler whose readings are complex numbers if I was so inclined. That we might opt to give results as rational numbers is merely a convention. $\endgroup$ – Hurkyl Dec 25 '14 at 3:23
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Real numbers include all the numbers you might want to use - including limits of convergent sequences of rational numbers. It is true that only countably many of these can ever be defined, but we don't know which we might want in advance.

These numbers are packaged in a convenient form which enables results to be proved for all the numbers we might encounter on our mathematical travels. It is particularly useful to know that the real numbers are essentially a unique model for the key defining properties, so when we prove theorems we know we are all talking about the same thing.

Admitting countable limits of rationals, rather than simply finite limits, turns out to have far-reaching consequences - for example, we can prove the intermediate value theorem. Without this, demonstrating the existence of a point relevant to the problem in hand may require a more complex case-by-case analysis.

A similar enriching of possibilities occurs in measure theory, where countably additive measures make a huge difference.

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Almost all physical models use PDEs - partial differential equations (and ODEs). Without real numbers, how do you do PDEs? It is one thing to measure quantities in the laboratory, but without PDEs, our mathematical descriptions of nature would be very poor indeed.

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    $\begingroup$ As you say, PDEs are a model. One could imagine using descriptions based on discrete-time difference equations instead. If the time-scale is fine enough, they will be undistinguishable. $\endgroup$ – Federico Poloni Dec 23 '14 at 16:07
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    $\begingroup$ Yes, but the computational time needed to solve the discrete time difference equations would be prohibitive. Besides, PDEs and ODEs are solved using discrete time. That's the idea behind a "step size" in your solver. You decrease the step size in order improve the accuracy of the simulation. $\endgroup$ – FundThmCalculus Dec 23 '14 at 19:42
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    $\begingroup$ @FedericoPoloni You could, for argument's sake, do that. But it would be a nightmare (and not just computationally) because of the lack of existence theorems. If the physically correct answer to a problem is $\pi$, would it or would it not exist as a solution to your rational - difference equation system? $\endgroup$ – user_of_math Dec 24 '14 at 5:06
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    $\begingroup$ @user_of_math I like this argument ("we need reals because we want to ensure existence of solutions") much more, but it is a completely different one than what you wrote in your answer ("we need reals because we want to formulate models as PDEs"). $\endgroup$ – Federico Poloni Dec 24 '14 at 8:46
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    $\begingroup$ Differential operators can be defined over any [commutative] ring, which doesn't have to include reals. A related MathOverflow question: mathoverflow.net/questions/36976/… $\endgroup$ – Akater Dec 25 '14 at 12:24
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Mathematicians don't compute things. Computers compute things.

Mathematicians are not limited by computation, much like Turing machines are not bounded by $2^{1024}$ gigabits of tape.

The real numbers are a tool, and it turns out to be useful. Since mathematicians are utilitarians, they like useful things. So they use the real numbers. If you want to ask why do physicists need the real numbers, you should ask that on Physics.SE or some other community centered around physics.

Why do mathematicians like the real numbers? Because they are closed under taking limits, which makes them ideal for talking about approximations by rational numbers, something we can easily comprehend (this is true at least in principle the notion of a rational number).

Since these are closed we don't have to worry and keep track whether or not our sequence is convergent, is the sequence itself is computable, do these properties hold or not. We have an arbitrary sequence, if we know it's a Cauchy sequence, then we know that it has a limit.

This is why mathematicians tend to embrace the axiom of choice when they work outside "the countable domain". When you want to make a statement about "all commutative rings with a unit", you don't want to start adding conditions on and on and on, that will help your proof go through. You want to say "Take any such and such object, then we can do this and that". Period. Utilitarianism galore.

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Do not be compelled into thinking uncountable sets are necessary for doing mathematics. Even though the traditional set-theoretic foundation is as valid as anything from a technical standpoint, I think people have come up with more onthologically satisfactory answers by working in computable universes.

You might like to look into:

Ron Maimon's answer to "Was mathematics invented or discovered?" (for a classic but computable approach)

Type Theory this is the foundational theory I find most satisfactory

Even the main theory used by mathematicians (first-order logic + ZFC) has countable models, which means that there is a set which is countable (in meta-theory) but which satisfies the axioms of real numbers. The diagonal argument then becomes the statement that it is not possible to construct a bijection (inside ZFC) between $\omega$ and $\mathbb{R}$. To me there couldn't be a stronger sign that uncountable sets are inadequate when our system of reasoning simply can't filter them unambiguously.

Regarding the impossibility of working with differential equations, this is just not the case. To solve a differential equation is to find an algorithm which approximates the solution arbitrarily. This could be a closed form, an integral, or some numerical method. The $\epsilon-\delta$ definiton of limit is algorithmic: if you give me an $\epsilon$, I must have a procedure for finding a $\delta$ that approximates the limit well enough. Every single function and procedure treated in classical calculus is computable.

Topology is precisely the work of taming uncountable spaces into chunks with invariants in such a way we can use finitistic arguments to prove theorems. The book Classical Topology and Combinatorial Group Theory champions this point of view, giving a proof of the Jordan Curve Theorem which is much easier than the standard one.

As you can see, most interesting mathematics takes place in computable sets, and the topics that don't are also dealt with with finistic reasoning (the proofs in functional analysis and set theory fit inside books after all). Their 'uncountability' can be seen as just a useful abbreviation.

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    $\begingroup$ Not to take anything from Ron Maimon as a physicist, but whenever I read his posts about mathematics, especially foundations of mathematics I feel like smashing the keyboard against the monitor, so I won't have to continue reading it (at that point I usually just close the window, since that's the sane option). $\endgroup$ – Asaf Karagila Dec 23 '14 at 20:35
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    $\begingroup$ @AsafKaragila I was actually very interested to know your thoughts on his view of Hilbert's program and ordinal analysis, as discussed by someone else here: mathoverflow.net/questions/164148/… I'm aware you disagree with him on inaccessible cardinals and probably other things. $\endgroup$ – Felipe Jacob Dec 23 '14 at 21:38
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    $\begingroup$ Any attempt to explain to Ron how set theory works is futile and can only end in his suspension (because he will resort to calling you all sort of names quite quickly). Ron, for example, would insist that the necessity of an inaccessible cardinal for Solovay's result is unneeded. Surely Ron knows set theory much better than me, but I highly doubt that he knows set theory much better than Shelah. His repeated claims on this matter cause me to take pretty much everything he says about mathematics with a boulder of salt (because grains are too small for this). Including these things. $\endgroup$ – Asaf Karagila Dec 23 '14 at 21:43
  • $\begingroup$ Doesn't it mean for a set S to be countable that there is a bijection between S and N? Then, if the diagonal argument means that "it is not possible to construct a bijection (inside ZFC) between ω and R", how can they both be countable? Either you can count their elements, from 1 onwards, or you can't. $\endgroup$ – Tobia Dec 26 '14 at 1:22
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    $\begingroup$ @Tobia: The point is that internally the model doesn't know about such bijection, but externally the universe knows the set of reals in the model is countable, so there is a bijection. Although I have to admit that seeing people that are truly affected by this "paradox" is the same as seeing people who are unable to discern "true" and "provable" in the context of the incompleteness theorems. $\endgroup$ – Asaf Karagila Dec 26 '14 at 6:58
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We don't need real numbers, in the sense that we could do without them if we really had to.

That doesn't mean that everything we ever want to speak of is rational or even algebraic, of course -- for example, if we're talking about harmonic oscillations, then the ratio between the maximal instantaneous speed and the amplitude is the phase velocity, and the ratio between the phase velocity and the frequency of the oscillation is $2\pi$, which is transcendental. Or, in exponential decay, the ratio between the instantaneous relative rate of decay and the half-life is $\log 2$, another transcendental.

Still one can have a purely rational mathematics, where instead of talking about transcendental ratios as numbers in themselves, we speak about processes that can produce ever finer rational bounds for the ratios, to any desired precision. So instead of having a single number $\pi$ we would have a succession of rational intervals, plus a theorem you can get the circumference of a circle with diameter 1 to fit into any of these intervals if you compute it precisely enough. We would then go on to prove a lot of theorems about how such sequences-of-ever-tighter-rational-bounds relate to each other and behave in different situations.

However, during this proof program we would find that there is a lot of standard boilerplate arguments about such sequences that we find ourselves repeating again and again. Being mathematicians, our instinct is to generalize, so we'd want to combine those boilerplate arguments into a set of clean reusable lemmas with assumptions that are easy to work with.

Once we completed this work, and after polishing by a few generations of students and textbook authors, we would find that we had reinvented the reals, in the guise of a general theory of sequences-of-ever-tighter-rational-bounds. Even though our eventual interest would still be the rational bounds, the general theory is easier to apply in arguments.

We find that our general theory applies to more sequences-of-ever-tigher-rational-bounds than the computable ones. That doesn't really bother us, because it still produces true results when applied to sequences that we can compute, and explicitly excluding the noncomputable ones would be more work for no clear benefit. In the few cases where it is of particular interest whether the resulting sequence is computable or not, we can go beyond the theory and do more careful work closer to the purely-rational substratum. But in most cases that is not important enough to us to make the increased complexity of doing so seem worthwhile.

It's not as if the multitude of reals that we don't need cost anything as long as we're not using them -- the theorems about reals that we need wouldn't be any easier to prove if we excluded the unnecessary ones.

In a few cases it would be technically convenient to have an enumeration of the reals to work with -- but it is usually routine to replace that with an enumeration of a dense subset of reals plus an assumption of continuity. Certainly that bit of footwork is much less work than what it would take to add in explicit restrictions to "interesting reals" in every theorem about reals we prove.

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    $\begingroup$ Sounds like one of those "I don't need to drink coffee, just in mornings when I don't have any I'm all groggy and sluggish..." argument. :-) $\endgroup$ – Asaf Karagila Dec 26 '14 at 18:47
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    $\begingroup$ @Asaf: Well, that always seems the case when the question is "do we really need such-and-such abstraction?" We can almost always do without, treating each concrete instance of it separately. It would just be insufferably groggy and sluggish. $\endgroup$ – Henning Makholm Dec 26 '14 at 19:10
  • $\begingroup$ I'm not disagreeing. ;-) $\endgroup$ – Asaf Karagila Dec 26 '14 at 19:11
  • $\begingroup$ Another instance of “don’t like ℝ? eat ℚ!” propaganda fluff. But, this time, with 6 upvotes ☹ $\endgroup$ – Incnis Mrsi Apr 21 '15 at 7:38
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Your argument, which can be rephrased as:

There are only countably many real numbers which could ever be defined, so all the rest are useless

is fallacious. There are plenty of times in mathematics when we want to deal with large sets, without caring whether we can explicitly define all of them.

As an example: compact Hausdorff spaces have properties which make them interesting to mathematicians. But any countable example must have an isolated point, and we'd like to know if there are examples without isolated points.

An obvious example of a compact Hausdorff space with no isolated points is given by closed interval of real numbers, such as $[0,1]$. We can't define every element of $[0,1]$, but we can still say interesting things about it - for example, it is compact.

On the other hand, let $A$ be the set of all definable elements of $[0,1]$. Then $A$ and $[0,1]$ have precisely the same sets of definable elements, but $A$ is not compact (by the Baire category theorem). So even though we can't define any elements of $[0,1]\setminus A$ explicitly, these undefinable numbers do have an effect which we can define.

Why is this? One answer comes from talking about induction principles. You're probably familiar with the principle of mathematical induction - if $T$ is some set of natural numbers such that $1\in T$ and $n\in T\Rightarrow n+1\in T$, then $T=\mathbb N$ - but there are versions which apply to much larger sets than the natural numbers, and even to objects like the class of all sets which are too large to be sets. The $\in$-induction scheme is the following:

Let $\phi$ be a property of sets such that whenever $\phi$ holds for all elements of a set $x$, it must hold for $x$ as well. Then $\phi$ holds for all sets.

If you're used to thinking about induction in the following way:

We have $1\in T$, so we must have $2\in T$, and then we must have $3\in T$, and so on...

then it might seem confusing to induct over all sets. There are certainly examples of sets that can't be defined; indeed, any real number has a unique representation as a set. Yet we can still induct over all sets; it seems as if mathematics is somehow allowing us to make infinitely many calculations at once.

The reason we can do this is that induction principles are proved using arguments by contradiction, which turns out to be an extremely powerful tool. I believe that in constructivist logic (where statements can be neither true nor false) then there really is no extra benefit that you can get by including numbers that can't be defined, so your question does get to the heart of some pretty deep mathematical logic.

For further reading, have a look at this fictional dialogue written by Timothy Gowers.

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  • $\begingroup$ Your claim about countable compact metric spaces is very wrong. All the countable successors ordinals are compact metric spaces. I do agree that the general case is far more important, though. $\endgroup$ – Asaf Karagila Dec 26 '14 at 18:40
  • $\begingroup$ @AsafKaragila Thanks - yes, of course that's right. Ill edit this to make it right. $\endgroup$ – John Gowers Dec 26 '14 at 18:50
  • $\begingroup$ @Donkey thanks for the link, it's very interesting. Can you give me some link to better understand the $\in-$ induction scheme ? $\endgroup$ – Emilio Novati Dec 27 '14 at 14:16
  • $\begingroup$ @EmilioNovati See, for example, page 39 of tartarus.org/gareth/maths/notes/ii/Logic_and_Set_Theory.pdf $\endgroup$ – John Gowers Dec 27 '14 at 18:31
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Because the completeness of real numbers, which make sure limit has closure, then one could use calculus.

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    $\begingroup$ However, you can argue that computable numbers have sufficient completeness properties: Each computable sequence of computable numbers has a computable limit. $\endgroup$ – yo' Dec 23 '14 at 15:32
  • $\begingroup$ Sorry, could you explain the computable numbers? I have no idea about what you are saying. And if you try to define something, you have to proof it is well defined, not just say we can define. $\endgroup$ – Matheart Dec 24 '14 at 12:03
  • $\begingroup$ @Matheart computable numbers are well defined: en.wikipedia.org/wiki/Computable_number $\endgroup$ – Emilio Novati Dec 24 '14 at 18:09
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    $\begingroup$ well, I am not ask for the computable numbers, I am asking for the way you define limit over it. $\endgroup$ – Matheart Dec 24 '14 at 18:23
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    $\begingroup$ @Matheart $\epsilon-\delta$ machinery is computable. (see Felice Jacob answer) $\endgroup$ – Emilio Novati Dec 25 '14 at 20:49
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One example: the Hasse principle. Without $\mathbb{R}$ one can't express this theorem.

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    $\begingroup$ The Hasse Principle is not a theorem. If you mean the Hasse-Minkowksi Theorem for quadratic forms: it can indeed be expressed without leaving $\mathbb{Q}$. The Archimedean condition is simply that the quadratic form take both positive and negative values. $\endgroup$ – Pete L. Clark Dec 23 '14 at 18:12
  • $\begingroup$ All right! So is there a theorem in arithmetic that really need $\mathbb{R}$? $\endgroup$ – Macadam Dec 24 '14 at 13:59
  • $\begingroup$ @PeteL.Clark "satisfies the Hasse principle" is a well defined property of algebraic varieties over Q. Which (of course) requires only finite computations with the valuations, or sequences of such, and not the completions of Q with respect to those valuations. So this is another boring repetition of the refrain, that nothing concrete really requires "actual infinite" objects. $\endgroup$ – zyx Dec 24 '14 at 14:56
  • $\begingroup$ @zyx: I agree. In fact, model theoretic results imply that one could work with real algebraic numbers instead equivalently for all Diophantine applications: these are constructible. So this is not a good argument for the necessity of $\mathbb{R}$. (Of course many other good answers to the question -- and to various equivalent questions -- have already been given here.) "[N]othing concrete really requires 'actual infinite' objects" Well, I didn't say that and don't really agree with it. I think of the real numbers as being one of the most concrete mathematical objects! $\endgroup$ – Pete L. Clark Dec 24 '14 at 19:53
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Suppose you have a square of area 1 cm.You want to calculate length of the diagonal.If you don't have notion of real number then you can not find length of the diagonal.Another way to say this that in a countable system you can always find some equation that has no solution in that system.For example in rational number system there exist no solution of the equation $x^2-2=0$. Note:

Mathematician always concern with mathematical structure in which every equation has solution.In natural number system there exist no solution of the equation $X+2=1$ .So mathematician extended system of natural number to set of integers.Again there exist no solution of the equation $x^2-2=0$ in rational number system.So mathematician extended the rational number system to real number system.Again the equation$X^2+1=0$ has no solution in real number.So mathematician extended real number system to complex number system.Indeed in complex number system every equation has solution.So they stopped...:)

Hope this helps.

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    $\begingroup$ Actually, when the area is 2 square cm, the sides can't be found without irrational numbers, while the diagonal actually can: it's 2 cm. $\endgroup$ – Ian Dec 24 '14 at 4:55
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    $\begingroup$ Algebraic numbers are countable. Even the set of numbers definable through equations in elementary functions should be countable. $\endgroup$ – Incnis Mrsi Dec 24 '14 at 5:20
  • $\begingroup$ @Ian you are right...sorry that was my mistakes..Actually I wanted to write area is 1 c.m...Now I corrected that mistakes..thanks.. $\endgroup$ – Ripan Saha Dec 24 '14 at 6:38
  • $\begingroup$ @IncnisMrsi you are right set of algebraic number is countable....but we have many numbers which can not defined through by algebraic equation....for example take $\pi$....It is proved that it is not an algebraic number... $\endgroup$ – Ripan Saha Dec 24 '14 at 6:43
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    $\begingroup$ @Ripan Saha the point is that all numbers we use, algebraic or transcendental, form a countable set. $\endgroup$ – Emilio Novati Dec 24 '14 at 8:23
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If you want to take a serious look at 'Math without Reals', you could do worse than to read Errett Bishop's book, 'Foundations of Constructive Analysis'. Over the history of modern math, there have been several people who have set out to extirpate the reals, as it were: Kronecker, Brouwer, etc. Bishop is known for delivering the maximum of useful results with the minimum of philosophical decoration.

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Simply put, because they are the most powerful natural descriptor of quantity that we have.

To elaborate only briefly, we can easily describe certain real numbers that we encounter regularly ($\sqrt{2}$ or $\pi$). Throw in the fact that most of our theorems are more powerful and general when proven for the Reals (and often simpler) and you have reasons to include them and none to remove them.

Indeed, Reals are a very natural thing to conceive of - that alone would suffice. Go outside and tell me that the world is merely made up of Rationals.

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  • $\begingroup$ This demonstrates a propaganda trick known as a false dilemma, technically a logical fallacy. ― Poor original poster… if you have doubts about real numbers, try to eat that crappy rationals! $\endgroup$ – Incnis Mrsi Apr 20 '15 at 11:11
  • $\begingroup$ I disagree with the classification of false dilemma. I said that it is clear that the Rationals are insufficient. I do claim that the Reals are better but not the only alternative. $\endgroup$ – Logan Apr 21 '15 at 3:43
  • $\begingroup$ Does a person knowing the “computable function” term need opinions about sufficiency of ℚ? Read original poster’s comments in other “answers” like this one. $\endgroup$ – Incnis Mrsi Apr 21 '15 at 7:44

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