Romance on a Parabola: A PreTeXt Math Blog

Subsection 1.3.1 An Attempt to Define \(\mathbb{C} \times \mathbb{R}\) as a Field

In mathematics, we define a field to be a set \(S\) on which the binary operations of addition and multiplication are defined, with the additional requirement that the following list of properties hold:

1. \(a + b \in S\) and \(a \cdot b \in S\text{.}\)

2. \(a + (b + c) = (a + b) + c\) and \(a \cdot (b \cdot c) = (a \cdot b) \cdot c\text{.}\)

3. \(a + b = b + a\) and \(a \cdot b = b \cdot a\text{.}\)

4. \(a \cdot (b + c) = (a \cdot b) + (a \cdot c)\text{.}\)

5. \(\exists e_{add} \in S \: | \: a + e_{add} = a\text{.}\)

6. \(\exists e_{mul} \in S \: | \: a \cdot e_{mul} = a\text{.}\)

7. \(\forall a \in S, \exists -a \in S \: | \: a + (-a) = e_{add}\text{.}\)

8. \(\forall a \neq 0 \in S, \exists a^{-1} \in S \: | \: a \cdot a^{-1} = e_{mul}\text{,}\)

where \(a\text{,}\) \(b\text{,}\) and \(c\) are any elements of \(S\text{.}\) In this context, \(e_{add}\) is known as the additive identity and \(e_{mul}\) is known as the multiplicative identity.

We now wish to define addition and multiplication on our set \(\mathbb{C} \times \mathbb{R}\) in a way that satisfies all the properties of a field. There are a lot of potential starting options, but I would like to propose what I think is the most natural approach.

Before that, though, here's some more notation. Let

be three arbitrary elements of \(\mathbb{C} \times \mathbb{R}\text{,}\) with \(x_{abc}, \: y_{abc}, \: z_{abc} \in \mathbb{R}\)¹. We now define addition and multiplication in the following manner:

I think this is a fairly pragmatic way of going about the matter, as we are simply adding and multiplying the complex and real components of \(a\) and \(b\) separately.

The rest of this subsection is used to prove that this particular definition of addition and multiplication satisfy the aforementioned properties of a field.

----------------------------------------------------------------------------------------------------------------------

1. \(a + b \in S\) and \(a \cdot b \in S\text{.}\) The set \(S\) is closed under addition and multiplication. Per our definition of addition, \(a + b = (x_a + x_b + i(y_a + y_b), \: z_a + z_b)\text{.}\) Since \(x_a + x_b + i(y_a + y_b) \in \mathbb{C}\) and \(z_a + z_b \in \mathbb{R}\text{,}\) it follows that \(a + b \in S\text{.}\) A similar argument shows that \(a \cdot b \in S\text{.}\)

----------------------------------------------------------------------------------------------------------------------

2. \(a + (b + c) = (a + b) + c\) and \(a \cdot (b \cdot c) = (a \cdot b) \cdot c\text{.}\) Addition and multiplication are associative. Observe that

and

Therefore addition and multiplication are associative².

----------------------------------------------------------------------------------------------------------------------

3. \(a + b = b + a\) and \(a \cdot b = b \cdot a\text{.}\) Addition and multiplication are commutative. Observe that

and

Therefore addition and multiplication are commutative.

----------------------------------------------------------------------------------------------------------------------

4. \(a \cdot (b + c) = (a \cdot b) + (a \cdot c)\text{.}\) Multiplication is distributive over addition. We have

----------------------------------------------------------------------------------------------------------------------

5. \(\exists e_{add} \in S \: | \: a + e_{add} = a\text{.}\) There exists an additive identity \(e_{add}\) such that for every element \(a\) in \(S\) , \(a + e_{add} = a\text{.}\) Let \(e_{add} = (0 + 0i, \: 0)\text{.}\) Then

----------------------------------------------------------------------------------------------------------------------

6. \(\exists e_{mul} \in S \: | \: a \cdot e_{mul} = a\text{.}\) There exists a multiplicative identity \(e_{mul}\) such that for every element \(a\) in \(S\) , \(a \cdot e_{mul} = a\text{.}\) Let \(e_{mul} = (1 + 0i, \: 1)\text{.}\) Then

----------------------------------------------------------------------------------------------------------------------

7. \(\forall a \in S, \exists -a \in S \: | \: a + (-a) = e_{add}\text{.}\) Every element \(a\) in \(S\) has an additive inverse \(-a\) such that \(a + (-a) = e_{add}\text{.}\) Let \(-a = (-x_a - iy_a, \: -z_a)\text{.}\) Then

----------------------------------------------------------------------------------------------------------------------

8. \(\forall a \neq 0 \in S, \exists a^{-1} \in S \: | \: a \cdot a^{-1} = e_{mul}\text{.}\) Every non-zero element \(a\) in \(S\) has a multiplicative inverse \(a^{-1}\) such that \(a \cdot a^{-1} = e_{mul}\text{.}\) Let \(d = a^{-1} = (x_d + iy_d, \: z_d)\text{.}\) We now solve for \(x_d\text{,}\) \(y_d\text{,}\) and \(z_d\) in terms of \(x_a\text{,}\) \(y_a\text{,}\) and \(z_a\text{.}\) Observe that

We then are presented with the following three equations:

The first two equations form a nice system of equations. After some algebra, we see that

Hence, we see that

Unfortunately, this is where tragedy strikes. You may have noticed that all elements of the form \((x_a + iy_a, \: 0)\) lack a multiplicative inverse, as \(\frac{1}{0}\) is not defined.

----------------------------------------------------------------------------------------------------------------------

So, this definition of addition and multiplication was very close to satisfying all the properties of a field, but could not fulfill this last requirement. If it's any consolation, we have at least shown that \(\mathbb{C} \times \mathbb{R}\) is a commutative ring under these addition and multiplication operations.

There may exist a definition of addition and multiplication which makes the set \(\mathbb{C} \times \mathbb{R}\) into a field, but I am not sure. This is somewhat similar to how \(\mathbb{R}^3\) is not a field when multiplication is defined component-wise. Indeed, there seem to be many analogues between the set \(\mathbb{C} \times \mathbb{R}\) and \(\mathbb{R}^3\text{.}\)

Speaking of which...

Subsection 1.3.2 The Cardinality of \(\mathbb{C} \times \mathbb{R}\)

There are obviously an uncountably infinite amount of elements in \(\mathbb{C} \times \mathbb{R}\text{,}\) but we can still get a feel for the cardinality of this set, \(|\mathbb{C} \times \mathbb{R}|\text{.}\) In particular, I claim that \(|\mathbb{C} \times \mathbb{R}| = |\mathbb{R}|\text{.}\)

There is more than one way to show that this is true, and we will explore one such method involving bijective maps. By definition, two sets have the same cardinality if there exists a bijection from one set to the other.

Therefore, finding a bijection from \(\mathbb{C} \times \mathbb{R}\) to \(\mathbb{R}\) will prove that these two sets share the same cardinality. However, such a task is a little difficult. Instead, we are going to be a little cheeky and make use of a proven fact that \(|\mathbb{R}^3| = |\mathbb{R}|\)³.

So instead, we will find a bijective map \(f\text{,}\) where \(f: \mathbb{C} \times \mathbb{R} \rightarrow \mathbb{R}^3\text{.}\) The most sensible thing to first try is, in my opinion, the map \((x + iy, \: z) \mapsto (x, \: y, \: z)\text{.}\) We now show that \(f\) is injective and surjective.

Let \(a = (x_a + iy_a, \: z_a)\) and \(b = (x_b + iy_b, \: z_b)\) represent two arbitrary elements in \(\mathbb{C} \times \mathbb{R}\text{,}\) and suppose \(f(a) = f(b)\text{.}\) It then follows that \((x_a, y_a, z_a) = (x_b, y_b, z_b)\) and so we have

in which case \(a = b\text{.}\)

Next, suppose \(C = (x_c, y_c, z_c)\) is an arbitrary element of \(\mathbb{R}^3\text{.}\) Observe that the element \(c = (x_c + iy_c, \: z_c)\) in \(\mathbb{C} \times \mathbb{R}\) satisfies \(f(c) = C\text{.}\)

Since \(f\) is both injective and surjective, it follows that \(f\) is bijective and hence \(|\mathbb{C} \times \mathbb{R}| = |\mathbb{R}^3| = |\mathbb{R}|\text{.}\) If you are unfamiliar with infinite sets, such a result probably seems bizarre, but this is just how infinite sets are.

Does the set \(\mathbb{C} \times \mathbb{R}\) have any practical applications? Maybe. But for now, I think I'll stick to the usual \(\mathbb{C}\) and \(\mathbb{R}\) we all know and love (?).