# Complex number arithmetic

1. Dec 18, 2012

### Syrus

1. The problem statement, all variables and given/known data

A textbook of mine asserts that ℝ is a subset of ℂ. The motivation for this is drawn by defining complex addition and multiplication and then showing that these operations on complex numbers of the form (x,0), with x an element of ℝ, are isomorphic to the field ℝ witih addition and multipication as ordinarily defined there. This may be nitpicky, but in order to state that ℝ is a subset of ℂ, isn't it necessary to have that 0i = 0, so that x + 0i = x + 0 = x, which is already assumed to be in ℝ? This isn't explicitly mentioned anywhere though (since i is not a real number and hence we can't automatically deduce that 0i = 0 from the [real] field axioms).

2. Relevant equations

3. The attempt at a solution

2. Dec 18, 2012

### Kreizhn

I think that you perhaps should be wary about that addition symbol. In the nitpicky context with which you are referring, it is not designed to be an actual sum: rather, it is supposed to denote a separate coordinate over a linear space. For example, physicists like to use $\hat i,\hat j,\hat k$ as the standard basis, so that a vector $(a,b,c)$ can be written $a \hat i + b \hat j + c \hat k$. Here, the addition is quite formal. If b=0 then we keep it as $a \hat i + 0\hat j + c \hat k$ and while it is true that $(a+0) \hat i + c \hat k = a\hat i + 0\hat j + c \hat k$, this is not what we really mean when we set b=0.

In the case of the complex number, you can think of the i part as again simply denoting a placeholder. Thus when we write $x + 0i$ we simply mean that the first coordinate is x and the second is 0.

3. Dec 18, 2012

### lurflurf

That is somewhat lose language as the usual R is not a subset of C, but the subset of C with no imaginary part is isomorphic to R.

4. Dec 18, 2012

### Syrus

That's what I was thinking also lurflurf- so R isn't formally a subset of C, but since complex numbers of the form (x,0) with x in R BEHAVE (with respect to complex number addition and multiplication) as do ordinary real numbers (with respect of ordinary real number addition and multiplication), it is conventional to abuse the terminology and refer to R as a subset of C.

Surely this is what you're implying?

5. Dec 18, 2012

### Dick

Yes, it's an abuse. But it's a pretty conventional abuse. If you have a real integral and you want to compute it with the complex residue theorem you usually don't stop to say that it's ok because R is isomorphic to the line (x,0) in the complex plane. It's also a pretty harmless abuse.

6. Dec 18, 2012

### Kreizhn

As Dick says, it is a very common abuse. Anytime you have an isomorphism (choose your favourite category), you can choose to make an identification between those two objects.

Interestingly however, as everything we have mentioned in this context are fields, the identifications are actually somewhat more natural. First of all, any non-trivial field morphism is necessarily injective, meaning we can in fact always identify fields as subfields of other fields. This is why whenever discussing fields we always (abusively) talk only about field containment. Furthermore, we often formally build new fields out of old ones simply by adjoining objects simply not contained in the original field: these are called field extensions. In every such case, there is a natural identification of the original field in the new field (The complex numbers may be formed from the reals by adjoining a root of $x^2 + 1$). Hence field theory extensively abuses this concept.

7. Dec 18, 2012

### Kreizhn

I'm not sure if it will help you to understand the concept, but there is another beautiful and practical way of viewing the complex numbers. Consider the set of matrices
$$C=\left\{ \begin{pmatrix} a & b \\ -b & a \end{pmatrix}: a,b \in \mathbb R \right\} .$$
I claim that there is an isomorphism of fields (you should first check it is indeed a field, under usual addition and matrix multiplication!) $C \cong \mathbb C$ given by
$$\begin{pmatrix} a & b \\ -b & a \end{pmatrix} \mapsto a + ib.$$
Give it a try and you will see that it works. There is also an isomorphism of corresponding subfields
$$\begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix} \mapsto a$$
taking the diagonal matrices isomorphically to $\mathbb R$.

8. Dec 18, 2012

### Syrus

Ah, yes, I vaguely recall seeing something similar to this in an abstract algebra class. In this case, of course, the diagonal matrices are technically a subset (i.e. subfield) of C. I see the point you're conveying, however.

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