> [!thm] Show that $\alpha + \beta = \beta + \alpha \;for\;all \;\alpha, \;\beta \in \mathbb{C}.$
>[!SOLUTION]
>Suppose
>$
>\alpha =a+ bi \; and\; \beta = c + di,
>$
where a, b, c, d $\in$ $\mathbb{R}$. Then
>$
>\begin{align}
>\alpha + \beta &= (a + bi) + (c + di)\\&= (a + c) + (b + d)i\\&= (c+ a) + (d + >b)i\\&= (c + di) + (a + bi)\\&= B + a,
>\end{align}
>$
where the second and fourth equalities hold because of the definition of addition in $\mathbb{C}$
>and the third equality holds because addition is commutative on $\mathbb{R}$.
---
Commentary:
Here is the LaTeX conversion of the PDF, with detailed Commentary and Example sections added:
$\textbf{Exercise 1.}$ Show that $z + w = w + z$ for all $z, w\in \mathbb{C}$.
$\textbf{Solution 1.}$ Suppose $z = a + bi \text{ and } w = c + di,$ where $a, b, c, d\in \mathbb{R}$. Then $\begin{align*} z + w &= (a + bi) + (c + di) \\ &= (a + c) + (b + d)i \\ &= (c + a) + (d + b)i \\ &= (c + di) + (a + bi) \\ &= w + z, \end{align*}$ where the second and fourth equalities hold because of the definition of addition in $\mathbb{C}$ and the third equality holds because addition is commutative on $\mathbb{R}$.
$\textit{Commentary:}$ This exercise demonstrates the commutative property of complex addition.
It shows that for any two complex numbers $z$ and $w$, the sum $z+w$ is equal to $w+z$. The proof relies on expressing the complex numbers in terms of their real and imaginary parts and using the commutative property of real addition.
This property is fundamental to the algebraic structure of the complex numbers. It ensures that the order in which complex numbers are added does not affect the result. Commutativity is one of the key axioms that define a field, and the complex numbers form a field under addition and multiplication.
$\textit{Example:}$ Let $z = 3 + 2i$ and $w = -1 + 5i$. Then: $\begin{align*} z + w &= (3 + 2i) + (-1 + 5i) = 2 + 7i \\ w + z &= (-1 + 5i) + (3 + 2i) = 2 + 7i \end{align*}$ As expected, $z+w = w+z$.
In $\mathbb{R}^2$, if we interpret complex numbers as vectors with the real part as the $x$-coordinate and the imaginary part as the $y$-coordinate, then complex addition corresponds to vector addition. The commutative property of complex addition is then analogous to the commutative property of vector addition in $\mathbb{R}^2$.
In $\mathbb{F}_p$ where $p$ is prime, addition is commutative because $\mathbb{F}_p$ forms a field under addition and multiplication modulo $p$. For example, in $\mathbb{F}_5$, we have $2 + 3 \equiv 0 \pmod{5}$ and $3 + 2 \equiv 0 \pmod{5}$, demonstrating commutativity.
This commutativity property extends to addition in any field, including fields of functions, matrices, polynomials, and so on. It is a key ingredient that allows fields to have a well-behaved algebraic structure.
---