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Why set of real numbers not a set of ordered pairs ?

We write $\mathbb{R}^2 = \mathbb{R} \times \mathbb{R}$, then we define addition and multiplication on this new set. Together with those definitions we call $\mathbb{R}^2$ as the set of complex numbers $\mathbb{C}$.

This is the gist of what I know from my book.

(I don't know much about complex numbers)

We can write any real number as $r = x + y$ ,where $x$ is a rational number and $y$ is a purely irrational number ($2\pi$ is correct value of y but not $2 + \pi$) $\qquad{(1})$.

A complex number is written as $z = x+ iy$, where $x$ and $y$ is real numbers $\qquad{(2})$.

Statement (1) and (2) are similar. We can say, by drawing a little bit of inspiration from the the definition of complex numbers, that $\mathbb{R} = \{(x,y): \ \ x\in \mathbb{Q}, y \in \mathbb{I}\} = \mathbb{Q}\times \mathbb{I}$.

Hence the set of complex numbers is a set of ordered pairs.

But this something that I have never seen anywhere.

  • I want know what is the fault in defining set of real numbers like this ?
  • Before downvoting this question as some weird math conpiracy theory please consider that I have no intention of doing that. –  Feb 02 '17 at 17:13
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    How would you define multiplication on the "ordered-pair" version of the reals? – mlg4080 Feb 02 '17 at 17:17
  • What rational and irrational you use to define the number $2$? In many contexts is true that $\Bbb R^2\equiv \Bbb C$ but in some others (by example as vector spaces themselves) are not equivalent. – Masacroso Feb 02 '17 at 17:28
  • If $r = x + y$, then $x = 0$ if $r$ is irrational, and $y = 0$ if $r$ is rational. So, it doesn't add anything new. In the complex case, you are allowing yourself to multiply $i$, which is not contained in the real numbers, by any real number. Both the rationals and irrationals are subsets of $\mathbb{R}$. – J126 Feb 02 '17 at 17:37
  • A complex number can be written as $x+iy$ for exactly one pair of real numbers $x$ and $y$ (so we can talk about its real and imaginary parts), but a real number $r$ can be written as $x+y$ with rational $x$ and irrational $y$ in infinitely many ways (unless $r$ itself is rational in which case it can't be written in that form at all), so it would make no sense to talk about its rational and irrational parts. – Andreas Blass Feb 02 '17 at 17:42
  • @JoeJohnson126 Assuming you are right. Can we go backwards, to define complex numbers not as the set of ordered pairs ? –  Feb 02 '17 at 18:02
  • @AndreasBlass An example for your claim ? –  Feb 02 '17 at 18:04
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    Example of many rational-plus-irrational expressions for the same number: $0+(\sqrt2) = 1+(\sqrt2-1) =-17+(17+\sqrt 2)$ where in each case the stuff in parentheses is an irrational number and the stuff before that is rational. – Andreas Blass Feb 02 '17 at 18:08
  • @AndreasBlass Hey, this is cheating ;). I should add that y is purely irrational like $\sqrt{2}$ or $2\sqrt{3}$ but things like $\sqrt{2} + a$ is not allowed. where a is rational. –  Feb 02 '17 at 18:25
  • @A---B What do you mean by "purely rational"? I doubt this term can be defined. – littleO Feb 02 '17 at 18:33
  • @littleO I wrote purely irrational. I would call $2\pi$ as purely irrational and $2 + \pi$ as not purely irrational because it contains a rational term. I am sure this can be defined properly. –  Feb 02 '17 at 18:38
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    I doubt that it can be defined properly. Generically, how would you select a representative of $\xi + \mathbb{Q}$ where $\xi$ is irrational? There are a few cases where we know a "special" representative, but in general, none of the elements is privileged over the others. – Daniel Fischer Feb 02 '17 at 18:44
  • In the case of the complex numbers, you have the real numbers when you define the complex numbers. In your definition, you use the irrational numbers to define the real numbers, but you need the real numbers to get the irrational numbers. – martin.koeberl Feb 02 '17 at 18:48
  • @DanielFischer My motivation in defining "purely irrational numbers" was that we can not do $1 + (1+i)i = 1i$ for complex numbers but I don't have any expertise in maths like you but if we, for a second, forget this issues is there anymore issues with this definition of real numbers. –  Feb 02 '17 at 18:51
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    @A---B Tangential comment, but you might be interested in learning about number systems such as $\mathbb Z[\sqrt{2}]$, which are studied in number theory. See this question, for example: http://math.stackexchange.com/questions/280878/the-units-of-mathbb-z-sqrt2 – littleO Feb 02 '17 at 18:51
  • @martin.koeberl That is not true I guess. A number that can't be represtented in the form $p/q$ where p, q are co-primes and $p \in \mathbb{Z}, q \in \mathbb{Z_*}$ is a irrational number. –  Feb 02 '17 at 18:53
  • @littleO Thanks for that resource. –  Feb 02 '17 at 18:54
  • Yes, but where did you get the number from? How do you guarantee it actually is a real, and not a hyperreal, surreal, imaginary number, Hamiltonian or some other mathematical object? To say that it is a real, you need to have the set of reals. – martin.koeberl Feb 02 '17 at 18:56
  • @martin.koeberl Ok but why would not this apply to rational numbers ? –  Feb 02 '17 at 18:57
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    Your basic premises about the real numbers are incorrect. "We can write any real number as $r=x+y$, where $x$ is a rational number and $y$ is a purely irrational number" is false (take $r$ to be a rational number). And as others have mentioned, there is no consistent way to define "purely irrational number" in the way you would like to. – Greg Martin Feb 02 '17 at 18:58
  • Because rational numbers can be constructed from the integers (again as pairs (in fact equivalence classes of pairs)). – martin.koeberl Feb 02 '17 at 19:02
  • @GregMartin I will take y = 0 but 0 is not an irrational number. Or I can take $y=0 * \sqrt{2}$ or $y=0 * \pi$, but this is not consistent. –  Feb 02 '17 at 22:00
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    Good luck trying to define "purely irrational" in a way that makes the rational-plus-purely-irrational representations of all (irrational) real numbers unique. The set of purely-irrational numbers in such a situation cannot be a Borel set, which makes it very unlikely to have a reasonably understandable definition. – Andreas Blass Feb 02 '17 at 22:31
  • @AndreasBlass Fair enough. Can you write an answer, I will happpily select it and upvote it. –  Feb 03 '17 at 13:19

1 Answers1

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This answer is essentially a re-organization and amplification of what I said in the comments. If one wants to represent every real number $r$ uniquely as $x+y$ with $x$ rational and $y$ taken from some specific set $S$ of irrationals, then two problems arise. The lesser problem is that a rational number $r$ cannot be the sum of a rational and an irrational number; this problem is easily circumvented by allowing $S$ to contain one rational number, say $0$. The bigger problem is to produce the rest of $S$ in a reasonable way. We need $S$ to contain exactly one member from each coset of $\mathbb Q$ in the additive group $\mathbb R$. Such sets exist as an consequence of the axiom of choice. But that's the only way to get such sets. Zermelo-Fraenkel set theory without the axiom of choice does not prove the existence of such a set, and even with the axiom of choice it's consistent with there being no definable such $S$.

Andreas Blass
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  • "Such sets exist by AoC" - are such sets constructible? – GFauxPas Feb 03 '17 at 14:39
  • @GFauxPas I assume you mean "constructible" in the sense of Gödel's consistency proof for AC+GCH. Then such sets $S$ need not be constructible without some further assumptions. In fact, if there is a constructible one then there is also a definable one, namely the first one in the standard well-ordering of the constructible sets. (Of course, it's consistent that some such $S$ is constructible, simply because it's consistent that all sets are constructible --- Gödel's $V=L$.) – Andreas Blass Feb 03 '17 at 14:43
  • @AndreasBlass I think mlg4080's comment is also a fair point- "How would you define multiplication on the "ordered-pair" version of the reals?" What do you think about it ? –  Feb 03 '17 at 15:13
  • @AndreasBlass I meant in the sense of a constructive proof. What would be an example of such a set? – GFauxPas Feb 03 '17 at 18:31
  • @GFauxPas Since the existence of such a set can't be proved in ZF (meaning classical Zermelo-Fraenkel set theory without the axiom of choice), it's hard to imagine any constructive framework that could prove the existence of such a set. – Andreas Blass Feb 03 '17 at 18:35
  • Can I ask a question as a sidestep: Is the set of ordered pairs of real numbers countable given that the set of real numbers is not? If so, how do we prove it? I have an idea that since the set of real numbers is not enumerable, and the set of ordered pairs of real numbers is not enumerable, the former being a subset of the latter makes it non-enumerable. – Cherry Blossom Bomb Apr 11 '21 at 02:49