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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 apon the axioms of the complex numbers.

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The notion of real number was "formalized" only in the third quarter of the nineteenth century. Complex numbers (of course not formalized) have been used since the late $16$th century. – André Nicolas Dec 4 '12 at 8:20
I realized the date/order in which they were formalized was irrelevent to my question, I edited it, thanks – Ethan Dec 4 '12 at 8:23
I think possibly proofs and theorems are invalidated, because your cannot define an order operation with the same properties as for reals. – Gerenuk Dec 4 '12 at 8:33
I don't understand what your saying, could you elaborate more please – Ethan Dec 4 '12 at 8:34
Also you need to extend your exponentiation rules. Otherwise, if you just use real exponential rules, you might end up with contradictions ( – Gerenuk Dec 4 '12 at 8:44
up vote 5 down vote accepted

Using the real numbers $\mathbb R$ one can construct new objects in various ways and then show that these new structures extend $\mathbb R$ by adjoining one new element $i$ satisfying $i^2=-1$, while preserving the field axioms.

This is a special case of general extension constructions and can be done in many ways. These ways depend on the existence of a model for the real numbers. You might ask how do you know the real numbers exist and the answer is that these too can be built from simpler things, all the way down to the existence of sets (which can't be proved).

In some more detail, assuming $\mathbb R$ is your favourite model of the real numbers you can construct $\mathbb C$ very directly by considering $\mathbb R \times \mathbb R$ (whose existence requires a slight set-theoretic trick) and then define addition and multiplication as follows: $(a,b)+(x,y)=(a+x,b+y)$ and $(a,b)\cdot (x,y)=(ax-by,ay+bx)$. One can then verify that the axioms of a field are satisfied, that $i=(0,1)$ satisfies $i^2=-1$ etc. This is a lengthy but completely straightforward verification.

Another possibility is to use some ring theory and consider the factor ring $\mathbb R[X]/(X^2+1)$. It follows from some basic ring theory that this quotient is a field with the desired properties.

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So your saying in order to verify that a notion such as that of complex numbers is compatible with that of the real numbers, I must check that my notion satisfies all the axioms the real numbers satisfy? – Ethan Dec 4 '12 at 8:15
What are all the axioms the real numbers satisfy? – Ethan Dec 4 '12 at 8:15
no, it's enough that you construct the new system. In my answer I gave two such constructions. The existence of a construction shows that something exists. The construction uses a model for the real and gives you a model of the complex numbers. If it is constructed then it exists. – Ittay Weiss Dec 4 '12 at 8:17
I don't understand, how do I know that I wont encounter such a contradiction as I did by defineing the opperator x. – Ethan Dec 4 '12 at 8:20
when you defined the operator x you didn't give any construction. If you had constructed something in which your operator x was shown to exist it would have been guaranteed no contradiction can exist. Notice how I did not introduce i out of thin air above but rather defined it to be (0,1). – Ittay Weiss Dec 4 '12 at 8:22

Note that, when we generalized from real numbers to complex numbers, we lost some properties. For instance, the order property. See here.

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It's true that the complex numbers are not ordered but the order on the real numbers is not lost nor contradicted by extending to the complex numbers. – Ittay Weiss Dec 4 '12 at 8:14
Complex numbers can be ordered, but not in a "natural" way. – yo' Dec 4 '12 at 8:17
However, the statement "$1$ has only one cube root" becomes false. So do many other sentences true in the reals. – André Nicolas Dec 4 '12 at 8:17
Once extended universal and existential quantifiers should become types to retain their original meaning. So 1 still has just one real cube root, and it's ok. – Ittay Weiss Dec 4 '12 at 8:19

Consider an operator $X$ adjoining to $\mathbb{R}$, where we specify that the only condition $X$ must satisfy is $X^2 = -1$. What other conditions can $X$ satisfy? Well, in arithmetic the most basic things we want to do are addition, subtraction, multiplication and division. If $X^2 = -1$, i.e. $X^2 + 1 = 0$, we expect things like $X(X^2+1)$, $\frac{X^2+1}{X+3}$ etc to be 0.

Let us try to collect all these relations $X$ must satisfy - call this set $R$ - you can see that this set is the collection of all $\frac{p(X)}{q(X)} (X^2+1)$, where $p(X),q(X)$ are polynomials in $X$ with real coefficients, and $q(X)$ is relatively prime to $X^2+1$. Note that in constructing this set of relations I am only considering the most basic operations: addition/subtraction/multiplication/division, with $X$ considered as a formal symbol. If you want to see whether contradiction arises in other occasion, you will have to define more precisely the conditions for $X$ to be compatible with $\mathbb{R}$.

Now what does it mean by contradicting arithmetic of real numbers? It means that by playing with $X$, some miraculous cancellations occur so that a non-zero real number $a$ has to be 0. In other words this number $a$ lies in the set of relations $R$. Is that possible? If it happens, then

$$\frac{p(X)}{q(X)} (X^2+1) = a$$

for some polynomial $p,q$ mentioned above (in particular $q(X)$ is relatively prime to $X^2+1$. This means that

$$p(X)(X^2+1) = a q(X)$$

This is impossible since $X^2+1$ divides left hand side but not the right hand side.

Remark: As for the other examples you mentioned, say $X^{2n} = -1$ and $X^{2n+1} = 1$ one, it is a fun exercise to try to figure out what the set of relations would be. There are more general situation where a similar process can be done, related to so-called quotient rings. You will see them when you study ring theory.

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Because we, actually, use the algebra of polynomials. Let $\mathbb R[X]$ denote all polynomials over real numbers. Let $f(X)=X^2+1$. Then this polynomial is irreducible and from the theory we know that since $\mathbb R$ is a field, the factor space $\mathbb R[X]/f$ is a field as well. And it is such field where $X^2+1=0$. This means that as well $$X^3=X^3-X\cdot 0=X^3-X(X^2+1)=X^3-X^3-X=-X.$$

All these sort of computations work because we are in $\mathbb R[X]/f$ which means that we can rewrite $0$ as $f(X)$ and vice versa. In the end, from each polynomial we can reduce to polynomial of degree one.

Notice that the identity $X^2+1=0$ is exactly the identity that "defines" complex numbers. Just you don't write the members as polynomils $Z(X)=bX+a$ but as numbers $z=a+bi$. The previous identity $X^3=-X$ then reads as $i^3=-i$ which is true.s

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I don't know what a field or factor space is, if you could explain it differently I would appreicate your help – Ethan Dec 4 '12 at 8:25
Sorry, if you don't know what field is, it is close to impossible to explain why construction of field of complex numbers works. Anyways, I'll elaborate on factor spaces. – yo' Dec 4 '12 at 8:38

A field is a generalization of the real number system. For a structure to be a field, it should fulfill the field axioms (

It is rather easy to see that the complex numbers are, indeed, a field. Proving that there isn't a paradox hiding in the complex-number theory is harder. What can be proved is this:

If number theory (natural numbers, that is) is consistent, then so is the complex number system. The main problem is that you can't prove the consistency of a theory without using a stronger theory. And then you have the problem of proving that the stronger theory is consistent, ad infinitum.

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Let's say we have defined the complex numbers as the set $\Bbb R\times\Bbb R$ together with addition $$(a,b)+(c,d)=(a+c,b+d)$$ and multiplication $$(a,b)\cdot(c,d)=(ac-bd,ad+bc).$$It is then possible to show that this structure is a field, which basically means that all the usual algebraic laws hold.

We can then define a subset $R$ which contains exactly those complex numbers with zero imaginary part. The addition and multiplication, when restricted to this set, becomes $$(a,0)+(c,0)=(a+c,0)\quad\text{and}\quad(a,0)\cdot(c,0)=(ac,0).$$ It can again be proven that $R$ is a field, it's a subfield of the complex numbers. Since every element in this set is uniquely determined by the first component, we could just write $(x,0)=\bar x$. Now it holds that $$\bar x+\bar y=(x,0)+(y,0)=(x+y,0)=\overline{x+y}\quad\text{and}\\ \bar x\cdot\bar y=(x,0)\cdot(y,0)=(xy,0)=\overline{xy}$$ where the addition in $\overline{x+y}$ is just addition of real numbers and similarly for multiplication. Thus, you can think of $\overline x$ as the real number $x$. There is no difference except for how you write it.

If we now define $i=(0,1)$ then we see that $$i^2=(0,1)\cdot(0,1)=(-1,0)=\overline{-1}$$ and so the axioms of the complex numbers are consistent (because we have a model for them). Now if you take any property which can be proven from the field axioms, for example $a^m\cdot a^n=a^{m+n}$ for integers $m$ and $n$, then this is valid for all complex numbers (because they form a field, too). Since the real numbers are just a special kind of complex numbers, we see that the complex numbers extend the reals in a way which is compatible with them in this sense.

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