I posted this question nearly 10 days ago, but am still really not satisfied with the answers I got, I have no prior education in abstract algebra, group theory, or other abstractions, and most of the answers I got involved defining sets, groups and other things I am unfamiliar with, my question was,

How does one know the notion of real numbers is compatible with the axioms defined for complex numbers, ie how does one know that by defining an operator '$i$' with the property that $i^2=-1$, we will not in some way contradict some statement that is an outcome of the real numbers.

For example if I defined an operator x with the property that $x^{2n}=-1$, and $x^{2n+1}=1$, for all integers n, this operator is not consistent when used compatibly with properties of the real numbers, since I would have $x^2=-1$, $x^3=1$,thus $x^5=-1$, but I defined $x^5$ to be equal to 1.

How do I know I wont encounter such a contradiction based upon the axioms of the complex numbers.

I reiterate here again, Please don't use terminology, or abstract math, that hasn't been taught in say a thorough collage level calculus 1 course.

Here's a link to the same question I asked 8 or so days ago: The notion of complex numbers

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    $\begingroup$ Please link to the other question-I don't see it. $\endgroup$ – Ross Millikan Dec 12 '12 at 16:35
  • $\begingroup$ If you know about $2\times2$ matrices, then you can see the complex numbers concretely as a set of $2\times2$ matrices with real coefficients. See en.wikipedia.org/wiki/…. $\endgroup$ – lhf Dec 12 '12 at 16:38
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    $\begingroup$ From the late sixteenth century, when they were introduced, to the beginnings to the twentieth, complex numbers were treated informally. Relatively few mistakes were made, though there were some. With the twentieth century came formal treatment. There seems to be no good way to answer your question without introducing some formalization that goes beyond the bounds you have set. $\endgroup$ – André Nicolas Dec 12 '12 at 16:40
  • $\begingroup$ What education would I have to acquire, to understand these formilizations. $\endgroup$ – Ethan Dec 12 '12 at 16:44
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    $\begingroup$ It depends how formal you want to be. At a reasonably satisfactory level, one can define complex numbers as ordered pairs of reals, with special addition or multiplication. Or else as special $2\times 2$ matrices. Then prove that the ordered pairs $(a,b)$ with $b=0$ under the special addition, multiplication are a structure isomorphic to the reals. At the technical level this is easy to show. So since you can find a "copy" of the reals in the complexes, there can be no incompatibility. $\endgroup$ – André Nicolas Dec 12 '12 at 16:59

Since we start with an axiom C: $(\exists i)i^2=-1$ which does not explicitly contradict any other axiom in $\mathbb R$ (since $i$ does not appear in other axioms), there is no way to directly disprove it. But in a slight modification to the aforementioned, there are some axioms which are actually disproven by (C), like trichotomy:

$$(\forall x)[x<0\vee x=0\vee x>0]$$

(read "every number is either negative, positive, or zero".) Because $i^2=-1$ and $0^2=0$ implies $i\neq0$, $i>0$ implies $i^2=-1>0$ implies $1<0$ implies $1^2>0$ which is a contradiction, and $i<0$ implies $i^2=-1>0$ similarly. What we do, then, is we just drop the offending axioms. Importantly, $\mathbb C$ does not contradict the "important" axioms, like commutativity and associativity of addition and multiplication, and the existence of inverses to non-zero elements. The important part is that although we can't keep everything, we can keep some things, and we can work usefully with what remains.

The problem with your example is that you have not just defined a new element $x$, but you have also defined how it multiplies, and your definition contradicts another one which has already been defined. In $\mathbb C$, we just define the part that can not otherwise be derived, and let the other axioms "figure out the rest", so that we don't have any danger of redefining things incorrectly.

By the way, you asked on the previous question what the axioms of the reals are, so I thought I'd list them here, courtesy of Spivak's Calculus.

  1. (addition is associative) $(\forall a,b,c)\ a+(b+c)=(a+b)+c$
  2. (additive identity) $(\exists0)(\forall a)\ a+0=a$
  3. (additive inverse) $(\forall a)(\exists b)\ a+b=0$
  4. (addition is commutative) $(\forall a,b)\ a+b=b+a$
  5. (multiplication is associative) $(\forall a,b,c)\ a(bc)=(ab)c$
  6. (multiplicative identity) $(\exists1)(\forall a)\ a\cdot1=a$
  7. (multiplicative inverse) $(\forall a\ne0)(\exists b)\ ab=1$
  8. (multiplication is commutative) $(\forall a,b)\ ab=ba$
  9. (distributive law) $(\forall a,b,c)\ a(b+c)=ab+ac$
  10. (trichotomy) $(\forall x)[x<0\vee x=0\vee x>0]$
  11. (positives closed under addition) $(\forall a>0,b>0)[a+b>0]$
  12. (positives closed under multiplication) $(\forall a>0,b>0)[ab>0]$
  13. (least upper bound) $(\forall S\subseteq\mathbb R)[\mbox{if }S\mbox{ has an upper bound, then }S\mbox{ has a least upper bound}]$

The first four just define all the "nice" properties of addition, the next four do the same for multiplication, number 9 covers the distributive property, 10-12 cover the semantics of an order relation, and number 13 "fills in the gaps" in the rational numbers. That last one is a bit complicated to write in symbols, but it allows you to say that all sorts of numbers like $\sqrt2$ and $\pi$ are actually numbers. The relationship with $\mathbb C$ is that numbers 10-12 are thrown away, but the first 9, which define a field, are still okay. Number 13 makes explicit reference to $\mathbb R$, and it needs an ordering to work properly, so we just leave that one as-is. That way you can still have numbers like $i\pi$ and identify them as elements of $\mathbb C$.

I know you didn't want so much notation all at once, but nothing in here is too out there to get from Intro Calc. (By the way, $\forall$ means "for all", $\exists$ means "there exists", and $\vee$ means "or", in case you haven't seen those before.)

  • $\begingroup$ Great answer, I appreicate the time you spent writing this. $\endgroup$ – Ethan Dec 13 '12 at 17:02
  • $\begingroup$ @Ethan: one way of showing appreciation is to "upvote" an answer by clicking the arrow to the left of the question (it's right by the check mark that you clicked to accept this answer). $\endgroup$ – The Chaz 2.0 Jan 7 '13 at 16:49
  • $\begingroup$ Very instructive answer. Directly pointing to the core concept! +1 $\endgroup$ – Eric Jul 25 '18 at 7:19

Basically, you are asking for consistency of theories* in the logical sense. There is in general no easy way to spot if a theory is consistent, and in fact, as Gödel showed, any interesting theory has questions that are left unanswered.

If I read your question right, then your question is if addition of some axiom to an existing theory can make it inconsistent. That is a very valid question. To prove that it doesn't, you need to get your hands dirty - there is no easy way to see that some specific axiom can freely be added.

*A theory is, simply said, just a list of axioms.


First, $i$ and $x$ are not operators, they are new numbers to be added to the set of reals. You need to specify what properties of the reals are supposed to be maintained when you add these numbers. For example, in adding $i$ to get $\mathbb C$ you give up the ordering on the reals. I think you want to maintain the field structure. To maintain the field structure, you need to add other numbers to maintain the closure of the operations. In the complex example, this leads you to all the numbers of the form $a+bi$ and definitions of all the field operations. Now you can check the field axioms are all satisfied using the usual field properties of the reals. This is a relative consistency proof-you have shown that if the reals are consistent, then the complex numbers are, too.

For your example with $x$, since $x^2=-1$ it has to have all the same properties as $i$ because those were forced in the construction, but already $i^3 \neq 1$. Your definition of $x$ has more restriction than the definition of $i$, so it is not surprising that it fails when $i$ works.


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