12
$\begingroup$

After recently concluding my Real Analysis course in this semester I got the following question bugging me:

Is the canonical operation of addition on real numbers unique?

Otherwise: Can we define another operation on Reals in a such way it has the same properties of usual addition and behaves exactly like that?

Or even: How I can reliable know if there is no two different ways of summing real numbers?

Naturally these dense questions led me to further investigations, like:
The following properties are sufficient to fully characterize the canonical addition on Reals?

  1. Closure
  2. Associativity
  3. Commutativity
  4. Identity being 0
  5. Unique inverse
  6. Multiplication distributes over

If so, property 6 raises the question: Is the canonical multiplication on Reals unique?
But then, if are them not unique, different additions are differently related with different multiplications?
And so on...

The motivation comes from the construction of real numbers.
From Peano's Axioms and the set-theoretic definition of Natural numbers to Dedekind and Cauchy's construction of Real numbers we haven't talked about uniqueness of operations in classes nor I could find relevant discussion about this topic on the internet and in ubiquitous Real Analysis reference books by authors as:

  • Walter Rudin
  • Robert G. Bartle
  • Stephen Abbott
  • William F. Trench

Not talking about the uniqueness of the operations, as we know them, in a first Real Analysis course seems rather common and not elementary matter.

Thus, introduced the subject and its context, would someone care to expand it eventually revealing the formal name of this field of study?

$\endgroup$
  • 2
    $\begingroup$ You ought to specify what things you're keeping the same. The operations on the rationals? The ordering? Nothing? $\endgroup$ – David Dec 16 '15 at 2:57
  • 1
    $\begingroup$ Your list of six properties for addition does not include successor, which is an (undefined) unary operation in PA. Without that, I can't even prove the set of reals is infinite-think of the three element field. It seems your question would be, given a continuum many set that has two binary operations that satisfy your list, does it have to be isomorphic to the reals with the usual addition and multiplication? Otherwise, we start with the naturals satisfying PA, construct the reals, define addition and multiplication on them, then prove these properties. $\endgroup$ – Ross Millikan Dec 16 '15 at 3:49
  • $\begingroup$ Then the question of uniqueness does not come up. This is a part of model theory. You are asking what models there are for a set of axioms and whether the model is unique up to isomorphism. $\endgroup$ – Ross Millikan Dec 16 '15 at 3:52
  • $\begingroup$ @David I am talking about the usual operations and relations up to the real numbers. $\endgroup$ – mucciolo Dec 16 '15 at 17:57
  • $\begingroup$ @RossMillikan Now I understand the content of your comment. I will be working on updating my question. $\endgroup$ – mucciolo Jun 24 '17 at 22:32
3
$\begingroup$

The proof goes in four steps:

  • Addition on the natural numbers is uniquely determined by the successor operation (proved using induction)

  • Addition on the integers is uniquely determined by addition on the natural numbers

  • Addition on the rational numbers is uniquely determined by addition of integers

  • Addition of real numbers is uniquely determined by addition of rationals, by continuity, using the fact that the reals are a complete ordered field containing the rationals as its prime subfield.

The same four steps can be used to show that multiplication on the reals is ultimately uniquely determined by the successor operation on the natural numbers.

$\endgroup$
  • 2
    $\begingroup$ The multiplication is by no means unique. Many horrible discontinuous additive automorphisms will yield a perfectly good different multiplication. $\endgroup$ – Matt Samuel Dec 16 '15 at 2:40
  • 1
    $\begingroup$ I don't see continuity mentioned anywhere in the OP. $\endgroup$ – Matt Samuel Dec 16 '15 at 2:46
  • 1
    $\begingroup$ Again, I think you might be missing the point of the construction of the reals. $\mathbb{R}$ is a complete ordered field. $\mathbb{C}$ is not orderable as a field, so cannot be isomorphic to $\mathbb{R}$ qua ordered field. The entire point of the constructions of the integers, rationals, and reals beginning with the naturals is that we can prove that each extension is unique up to the appropriate sense of isomorphism! $\endgroup$ – Carl Mummert Dec 16 '15 at 3:01
  • 1
    $\begingroup$ You have finally earned my downvote. Have you even read the question? 6 conditions are listed. They have nothing to do with order or continuity. $\endgroup$ – Matt Samuel Dec 16 '15 at 3:01
  • 2
    $\begingroup$ The canonical operation is the usual one, and the question is whether it is unique. It is not. $\endgroup$ – Matt Samuel Dec 16 '15 at 3:02
2
$\begingroup$

The short answer is no: the operation defined by $a+_3 b=(a^3+b^3)^{1/3}$ also has all the properties 1 through 6 over the reals. This distributes over cannonical multiplication: for any $a,b,z\in \mathbb{R}$,

$z(a+_3b)=z(a^3+b^3)^{1/3}=(z^3)^{1/3}(a^3+b^3)^{1/3}=((za)^3+(zb)^3)^{1/3}=(za)+_3(zb).$

It might, however, be the case (and this is entirely speculation, not necessarily true) that only operations of the form $a+_f b = f^{-1}(f(a)+f(b))$ (where $f:\mathbb{R}\to \mathbb{R}$ is bijective and fixes the origin; or stated differently, $f$ is a permutation of the real numbers and $f(0)=0$) have all the properties 1 through 6. That would mean that addition is unique up to automorphism on the real numbers.

I would agree that this is not a trivial question.

$\endgroup$
  • $\begingroup$ What multiplication are you using? You have to define it and show it distributes. $\endgroup$ – Ross Millikan Jun 19 '17 at 18:18
  • $\begingroup$ I have done so, see revisions. @RossMillikan $\endgroup$ – MathTrain Jun 19 '17 at 18:37
  • $\begingroup$ Thank you very much for such an enlightening input! In a near future I will update the question considering the answers and comments along with my progress. $\endgroup$ – mucciolo Jun 24 '17 at 22:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.