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Irrational numbers are very easy to find. Square roots require only a little bit more than the most basic arithmetic. So it might be that this question is impossible to answer because it presupposes a world where math looks completely different to what it really does. I am hoping this is not the case.

If we didn't have examples of irrational numbers, would $\mathbb{R}$ and $\mathbb{Q}$ be assumed to be the same set?

Can we construct a proof that irrational numbers exist without giving examples?

halrankard
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Nacht
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  • Interesting question! It's kind of related to the concept of a complete metric space. If our idea of a "number" was just "something of the form $a/b$ for an integer $a$ and a positive integer $b$", we could still do a lot of arithmetic. We could prove that there is no number $x$ such that $x^2 = 2$. However, there is a sequence $(x_n)$ such that $x_n^2$ gets as close to $2$ as we want. (eg $14/10, 141/100, \dotsc$). This begs the question "is there a number system where sequences like this converge to a value?". This is $\Bbb R$. – Izaak van Dongen Jul 09 '20 at 12:57
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    One counter-example is enough to prove that we cannot express some quantities using rational numbers. – Vasili Jul 09 '20 at 13:06
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    As a relevant point, we know there are Uncomputable Numbers though, for obvious reasons, nobody can write one down. We know this via counting arguments...there can be only countably many programs but there are uncountably many reals. – lulu Jul 09 '20 at 13:45
  • Although I gave an answer I also downvoted. If we didn't have irrational numbers, then what is $\mathbb{R}$? How are you asking the question "would $\mathbb{R}$ and $\mathbb{Q}$ be assumed to be the same? – SBK Jul 10 '20 at 08:42
  • @T_M Yes, thus my hesitation in the first part. But it turns out there are great answers, as I had hoped. – Nacht Jul 10 '20 at 23:41

3 Answers3

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From the surrounding discussion it seems that perhaps there are two questions here:

  1. What is the impetus to define the real numbers and how do we get such a definition?

  2. Given a definition of $\mathbb{R}$ can we prove that irrational numbers exist without constructing examples?

My answer addresses the latter question.

You can prove that the real numbers are uncountable (there is no bijection between the real numbers and the integers).

You can prove the rational numbers are countable (there is a bijection between the rational numbers and the integers).

So $\mathbb{R}\setminus\mathbb{Q}$ is nonempty.

Since I don't know your background I will add some more.

  1. A bijection between two sets $X$ and $Y$ is a function $f:X\to Y$ that is both one-to-one and onto, i.e, for all $x,y\in X$, $f(x)=f(y)$ implies $x=y$; and for all $y\in Y$ there is $x\in X$ such that $f(x)=y$.

  2. An infinite set is $X$ countable if there is a bijection $f:X\to \mathbb{N}$.

  3. It is a famous result of Cantor (called a diagonal argument) that $\mathbb{R}$ is uncountable.

  4. It is a standard (and good practice) exercise that $\mathbb{Q}$ is countable.

  5. It is a standard (and good practice) exercise that if $X$ and $Y$ are countable then so is $X\cup Y$. So if $\mathbb{R}\backslash \mathbb{Q}$ were countable then so would be $\mathbb{R}$.

So altogether, this actually shows that not only do irrational numbers exist, but there are more irrational numbers than rational numbers because the set $\mathbb{R}\backslash\mathbb{Q}$ must be uncountable by the above points.

By the way, the same kind of proof shows that transcendental numbers exist and there are more transcendental numbers than algebraic numbers. Indeed, the set of algebraic numbers is also countable and so its complement in $\mathbb{R}$ is uncountable.

halrankard
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    But this proof presupposes a definition of reals... Assuming (as per the OP's question) that we haven't found any irrational (that means that we are still thinking that the ration side/diagonal is expressible by some "convoluted" fraction), what are the "reals" ? – Mauro ALLEGRANZA Jul 09 '20 at 12:55
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    Dedekind cuts (ie bounded above subsets $A$ of $\mathbb{Q}$ without a greater element such if $x \in A$ and $y < x$, then $y \in A$), for instance. – Aphelli Jul 09 '20 at 12:59
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    @MauroALLEGRANZA If you don't presuppose a definition of $\mathbb{R}$ then how does one interpret the question: "Would $\mathbb{R}$ and $\mathbb{Q}$ be the same set?" – halrankard Jul 09 '20 at 13:01
  • @MauroALLEGRANZA We might notice that not every bouded set of rationals has a least upper bound. At that point we might decide that such non-existent suprema do not deserve to be called numbers and leave it at that. But we might find these non-existing numbers useful because we can do useful computations with them - similar to how "impossible" square roots of negative numbers might help find (real) roots of polynomials by "blindly" computing with them, which is how historically imaginary/complex numbers got promoted from weird auxiliary construction to actual numbers – Hagen von Eitzen Jul 09 '20 at 13:02
  • @MauroALLEGRANZA I should add that I take your point about the philosophical question of defining the reals. But there are plenty of definitions of the reals that don't require one to assume the existence of irrationals. So have that philosophical discussion and decide on a definition of $\mathbb{R}$. Then my answer (rather Cantor's) shows that irrationals exist. – halrankard Jul 09 '20 at 13:08
  • From my point of view, the "counterfactual" question in the title: "What happened if we do not discovered the proof of the irrationality of $\sqrt 2$ ?" is difficult to answer... Ancient Egyptian math "lived" about 2.000 years using fractions without "irrationals" at all. The big "divide" is between ancient "others" math and Ancient Greek math that discovered it. The interesting fact is the link with "proof": Greek math discovered the "irrationality" phenomenon and this led them to invent the "tool" named mathematical proof. – Mauro ALLEGRANZA Jul 09 '20 at 13:18
  • Thus, assuming this point of view (a sort of "alternative history" scenario), maybe we do not have the concept of mathematical proof available... – Mauro ALLEGRANZA Jul 09 '20 at 13:19
  • @MauroALLEGRANZA Yes, it is an interesting discussion and I like your points. I hope that if for some reason the human brain had a blind spot for algebra/arithmetic, we still would have developed the concept of mathematical proof through other means like geometry. – halrankard Jul 09 '20 at 13:33
  • But the proof was developed through geometry... The proof of the irrationality is a geometrical one (no algebra) :-) – Mauro ALLEGRANZA Jul 09 '20 at 13:36
  • @MauroALLEGRANZA Yes of course I know that. But the statements involving commensurability of diagonals of squares with their sides still have algebraic flavor on their face. Compare to something like Desargues' Theorem in which the algebra is much more esoteric. One could imagine a scenario in which humans could not perceive of notions like commensurability, while still making sense something like of Desargues' Theorem and proving it. – halrankard Jul 09 '20 at 13:43
  • Thanks @halrankard... I liked your original answer better, it was very simple and clear. Oh well, this is better for posterity I suppose. Like you said, my question presupposes the definition of real numbers. – Nacht Jul 10 '20 at 01:53
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halrankard's answer is good but this may help as well.

It is common to define various sets which are larger than the rational numbers $\mathbb{Q}$ yet still smaller than the reals $\mathbb{R}$. By larger and smaller here, I just mean one is a strict subset of the other.

halrankard mentions the Algebraic numbers $\mathbb{A}$. These are the roots to polynomials with rational coefficients. $\sqrt{2}$ is algebraic but $\pi$ is not.

Larger again, are the Computable numbers. Informally, these are ones that an ideal computer, e.g. a Turing machine, could calculate. $\pi$ is computable, you can calculate it to any desired precision in a finite time.

Larger again are the Definable numbers. Informally, we can specify these precisely yet we cannot even compute them. We know that there are numbers which are definable but not computable. They are necessarily rather weird. See this earlier question.

And this is still not all real numbers. All of these sets are countable. This means that although intuitively each set is bigger than the previous one, they can all be put into one to one correspondence with the smallest infinite set: the natural numbers $\mathbb{N}$. We could assign each definable number a unique natural number label without missing any.

However, we know that the real numbers are not countable so there must be undefinable numbers and that most real numbers are undefinable.

So, one of these last two sets might be like what you want: we know that they exist but we cannot point to any examples. No one will ever prove that a specific number is undefinable since, if they could specify which number they are talking about then it is definable

badjohn
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I think the answers so far miss a very basic point: Perhaps the most famous proof in mathematics shows that there is no rational number $x$ with $x^2 = 2$. This is not the same as saying "$\sqrt{2}$ is irrational". The reason they are not the same is because the former can be articulated as a statement just about rational numbers, before any development of the reals or the square root function or any concept of "irrationality". Now it is up to you whether or not you want to try to expand to a larger number system in which $x^2 = 2$ has a solution.

SBK
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