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 ?
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    $\begingroup$ How would you define multiplication on the "ordered-pair" version of the reals? $\endgroup$ – mlg4080 Feb 2 '17 at 17:17
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    $\begingroup$ 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. $\endgroup$ – Andreas Blass Feb 2 '17 at 18:08
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    $\begingroup$ 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. $\endgroup$ – Daniel Fischer Feb 2 '17 at 18:44
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    $\begingroup$ 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. $\endgroup$ – Greg Martin Feb 2 '17 at 18:58
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    $\begingroup$ 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. $\endgroup$ – Andreas Blass Feb 2 '17 at 22:31

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 $S$.

  • $\begingroup$ "Such sets exist by AoC" - are such sets constructible? $\endgroup$ – GFauxPas Feb 3 '17 at 14:39
  • $\begingroup$ @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 $V=L$.) $\endgroup$ – Andreas Blass Feb 3 '17 at 14:43
  • $\begingroup$ @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 ? $\endgroup$ – A---B Feb 3 '17 at 15:13
  • $\begingroup$ @AndreasBlass I meant in the sense of a constructive proof. What would be an example of such a set? $\endgroup$ – GFauxPas Feb 3 '17 at 18:31
  • $\begingroup$ @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. $\endgroup$ – Andreas Blass Feb 3 '17 at 18:35

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