I am trying to derive a proof of the associative property of addition of complex numbers using only the properties of real numbers.

I found the following answer but was hoping someone can explain why it is correct, since I am not satisfied with it (From Using the properties of real numbers, verify that complex numbers are associative and there exists an additive inverse):

\begin{equation} \begin{aligned} z_1 + (z_2 + z_3) &= (a + bi) + [(c+di) + (e+fi)] \\ &= (a+bi)+[(c+e)+(di+fi)] \\ &= [a+(c+e)]+[bi+(di+fi)] \\ &= [(a+c)+e]+[(bi+di)+fi] \\ &= [(a+c)+(bi+di)]+(e+fi) \\ &= [(a+bi)+(c+di)]+(e+fi) \\ &= (z_1+z_2)+z_3 \end{aligned} \end{equation}

I justify step 1 by the definition of complex numbers, step 2 and 3 by commutativity in R, step 4 by associativity in R, step 5 and 6 by commutativity in R.

I don't see how step 2 and 3 are commutativity in R. Wouldn't these steps require commutativity and associativity in C as well? For example, in step 2, I am presuming the author of the proof did the following rearrangement:

\begin{equation} [(c + di) + (e + fi)] \\ [c + di + e + fi] \\ [c + e + di + fi] \\ [(c + e) + (di + fi)] \\ \end{equation}

But doesn't the following rearrangement

\begin{equation} [(c + di) + (e + fi)] \\ [c + di + e + fi] \\ \end{equation}

require associativity in C? And doesn't the following

\begin{equation} [c + di + e + fi] \\ [c + e + di + fi] \\ \end{equation}

require commutativity in C?

There are some other steps I am confused by as well, like step 4 only requiring associativity in R when clearly there is an associativity in C occurring on the right end of the equation.

Can anyone provide insight into why this proof is correct?

  • $\begingroup$ As fleablood points out, the OP in the other question should've written $(x+y)i$ rather than $xi+yi$ wherever that appears in the proof which is where some of the confusion stems $\endgroup$ – Shuri2060 Jul 23 '17 at 8:40

The person who did that solution just made a mess of it. S/he seems to be believing we can treat the imaginary $bi$ and real components $a$ and the imaginary unit $i$ as real summands and rearrange them and distribute them. We can. But that has to be proven.

A better proof:

$z_1 + (z_2 + z_3) = \{a + bi)\} + (\{c+di\} +\{e+fi\})$ Notation of complex numbers. (Am adding unconventional $\{\}$ brackets to clarify what exactly are the inseparable "units" of complex numbers. At this time we cannot break the two compenents of $\{a + bi\}$ into $a$ and $bi$ and shuffle them about as though they were real summands.)

$= \{a + bi\} + \{(c+e) + (d+f)i\}$ Definition of complex addition. (Note we can't, at this time treat $di$ and $fi$ as components and so $(c + di) + (e+fi) = (c+e) + (di+fi)$ via commutivity of addition on reals. Yet. We don't have that the "+" in $c + di$ actually means $di$ added to $c$ in the same sense that $c + e$ has $e$ added to $c$ would mean.

$= \{(a + (c+e)) + (b + (d+f))i\}$ Definition of complex addition.

$=\{((a+c) + e) + ((b + d) +f\}i\}$ Associativity of addition in the reals.

$=\{(a+c) + (b+d)i\} + \{e + fi\}$ Definition of addition of complex numbers. (but going in the opposite direction. $\{a + bi\} + \{c+di\} = \{(a+b) + (b+d)i\}$ is summands to sum. This is the other direction: $\{(a+b) + (b+d)i\}= \{a + bi\} + \{c+di\} $... sum to summands.)

$= (\{a + bi\} + \{c + di\}) + \{e + fi\}$ Definition of complex addition.

$= (z_1 + z_2) + z_3$. Notation.

A better way, would be to simply not use $a + bi$ notation at all but define a complex number as an ordered pair $a + bi = (a,b)$ and $\{a + bi\} + \{c + di\} = (a,b) + (c,d) := (a+c, b+d)$.

Then the argument is:

$z_1 + [z_2 + z_3]= (a,b) + [(c,d) + (e,f)]$ Notation

$(a,b) + (c+e,d+f)$ Def. Addition.

$(a + [c+e], b + [d+f])$ Def. Addition.

$([a+c] + e, [b+d]+f)$ Assoc. of Addition on reals.

$(a+c, b+d) + (e,f)$ Def. Addition.

$[(a,b) + (c,d)] + (e,f)$ Def. Addition.

$[z_1+ z_2] + z_3$ Notation.

Associativity of multiplication is similar but harder:

$(a,b)[(c,d)*(e,f)] = $

$(a,b)(ce - df, de + cf) =$

$(a(ce - df) - b(de + cf), b(ce - df) + a(de+cf))=$

$(ace - adf -bde - bcf, ade + acf + bce - bdf)$.

And $[(a,b)*(c,d)]*(e,f)=$

$(ac - bd, bc + ad)(e,f) = $

$((ac - bd)e - (bc + ad)f, (bc+ad)e + (ac-bd)f) = $

$(ace - bde - bcf + adf, ade + afc +bce = bdf)=(a,b)[(c,d)*(e,f)]$

  • $\begingroup$ Great answer and explanation! I knew something was up with the proof I linked... this one makes so much more sense! And cheers for the additional proof of multiplication $\endgroup$ – wizardjoe Jul 23 '17 at 1:40

I've annotated the proof below

$$\begin{aligned} &\quad\ \ z_1 + (z_2 + z_3)\\ &= (a + bi) + [(c+di) + (e +fi)] \\ &= (a+bi)+[(c+e)+(d+f)i)]\quad \text{by definition of addition in}\ \Bbb C\\ &= [a+(c+e)]+[(b+(d+f))i)] \!\quad \text{by definition of addition in}\ \Bbb C\\ &= [(a+c)+e]+[((b+d)+f)i] \,\quad \text{by associativity of addition in}\ \Bbb R\\ &= [(a+c)+(b+d)i)]+(e+fi) \quad \text{by definition of addition in}\ \Bbb C\\ &= [(a+bi)+(c+di)]+(e+fi) \quad \text{by definition of addition in}\ \Bbb C\\ &= (z_1+z_2)+z_3 \end{aligned}$$

  • $\begingroup$ Makes so much more sense than the proof I linked. Thanks! $\endgroup$ – wizardjoe Jul 23 '17 at 1:41

I think the key here is to differentiate between the different '$+$' we're seeing here - there are actually $3$ different kinds, $+:\Bbb R\times\Bbb R\rightarrow\Bbb R, +:\Bbb C\times\Bbb C\rightarrow\Bbb C$ and finally the '$+$' in $a+bi$.

The '$+$' in $a+bi$ is just used in representing a complex number (but there are good reasons for why '$+$' is used, as @Bill Dubuque comments and also see this question). You could perhaps consider this '$+$' as $+:\Bbb R\times\Bbb I\rightarrow\Bbb C$ to represent the complex number. For clarity, maybe it'd be more helpful if you replace it with something like $*$ just for visualisation, or even denote a complex number $a+bi$ as $(a,b)$.

The most important thing to remember here is that $+:\Bbb C\times\Bbb C\rightarrow\Bbb C$ is defined to be $(a, b)+(c, d)=(a+c, b+d)$.

The proof is rewritten as (using $(a,b)$ to represent a complex number $a+bi$ and $+_A$ for $+:A\times A\rightarrow A$):

\begin{equation} \begin{aligned} z_1 +_{\Bbb C} (z_2 +_{\Bbb C} z_3) &= (a, b) +_{\Bbb C} [(c, d) +_{\Bbb C} (e,f)] \quad\quad\quad\,\,\text{representation of complex numbers}\\ &= (a, c)+_{\Bbb C}(c+_{\Bbb R}e,d+_{\Bbb R}f) \,\quad\quad\quad \text{complex addition}\\ &= (a+_{\Bbb R}(c+_{\Bbb R}e),b+_{\Bbb R}(d+_{\Bbb R}f)) \quad \text{complex addition}\\ &= ((a+_{\Bbb R}c)+_{\Bbb R}e, (b+_{\Bbb R}d)+_{\Bbb R}f) \quad\text{associativity of $+$ in $\mathbb R$}\\ &= [(a+_{\Bbb R}c,b+_{\Bbb R}d)]+_{\Bbb C}(e,f) \,\,\,\quad\quad\text{complex addition}\\ &= [(a,b)+_{\Bbb C}(c,d)]+_{\Bbb C}(e,f) \quad\quad\quad\,\text{complex addition}\\ &= (z_1+_{\Bbb C}z_2)+_{\Bbb C}z_3 \,\,\quad\quad\quad\quad\quad\quad\,\,\,\text{representation of complex numbers} \end{aligned} \end{equation}

  • $\begingroup$ Actually the $+$ in $a+bi$ can be ring addition depending on how one constructs $\,\Bbb C,$ e.g. see here. @OP See also this answer for more about the Hamilton pair constructtion of $\,\Bbb C\,$ (and Hankel's scathing critique of Cauchy's informal construction of $\Bbb C,\,$ which sheds much light on the meaning of $+$ in $\,a+bi).\,$ $\qquad$ $\endgroup$ – Bill Dubuque Jul 22 '17 at 22:34
  • $\begingroup$ This was very insightful. I know + can mean different operations depending on the operands, but I never thought about its meaning in the context of a complex number a + bi. $\endgroup$ – wizardjoe Jul 23 '17 at 1:43

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