Proof the complex numbers cannot be ordered

10 Feb 2019

In this post, we will demonstrate why the complex numbers, \(\mathbb{C}\) cannot be ordered. That is, there is no sensible way we can use the “less-than” symbol (\(\lt\)) on two members of the complex field.

Some background information on fields

You may have heard that the set of complex numbers endowed with operations for addition and multiplication \(\left( \mathbb{C}, +, \cdot \right)\) are what’s known as a field. That is, for all complex numbers \(z\), \(z_1\), \(z_2\) and \(z_3\), \(\mathbb{C}\) has:

A good exercise is to verify all of the above properties hold for \(\mathbb{C}\) with standard addition and multiplication to confirm it is indeed a field.

But why is being a field such a big deal? Knowing \(\mathbb{C}\) is a field is enough for us to accept various theorems, processes, and discussions in other areas of mathematics as applicable in \(\mathbb{C}\) as well. For example, linear algebra tells us that an element of any field can be used as a scalar in a vector space and, as a result, all of the theorems pertaining to vectors and scalars can now be accepted as true for complex numbers. Another example can be found in many helpful algebraic “tricks” or “shortcuts” we take for granted, such as using the FOIL method to multiply two binomials–this is really just making use of the distributivity of multiplication over addition. The fact is, we can do this with any field, not just \(\mathbb{R}\).

Ordered fields

That’s not to say all properties of all fields are shared, otherwise there would be no reason for others to exist. One such property we cannot take for granted from other fields such as \(\mathbb{R}\) is the notion of order.

For an ordering to exist, our field must have a notion of a positive cone, a strict subset \(P\) of our field such that the following is true:

  1. \(P\) is closed under addition and multiplication, that is, for \(z_1\) and \(z_2\) in \(P\), we have that both \(z_1 + z_2\) and \(z_1 \cdot z_2\) are also in \(P\),
  2. any element of the field multiplied by itself is in \(P\). In other words, for any element of the field \(z\), we have that \(z \cdot z\) is in \(P\),
  3. any element of the field is either in \(P\), its additive inverse is in \(P\), or it is zero. In other words, for any element of the field \(z\), either \(z\) or \(-z\) is in \(P\), or \(z = 0\).

Note that the above is indeed true in \(\mathbb{R}\) if we take \(P\) to be the natural numbers with zero and so the reals are ordered. In this article we will be examining a simple proof that the complex numbers, \(\mathbb{C}\) are not ordered.

The proof

Let’s begin by supposing \(\mathbb{C}\) is ordered. We will aim to arrive to a contradiction in the following steps to conclude that our assumption was wrong and that \(\mathbb{C}\) is not ordered.

Let’s consider the complex number \(i\), the famous “imaginary” number having the property \(i^2 = -1\).

We have our contradiction, specifically to point (3)–neither \(i\) nor \(-i\) can be in \(P\)!

Therefore, the assumption must have been wrong and \(\mathbb{C}\) must not be ordered after all.

Followup questions