0
$\begingroup$

I am on a journey to redefine very rigorously everything I've learnt a few years ago in Math classes, and like many others, I always had troubles with complex numbers. To be honest, I suspect that a lot of confusion arises from the symoblic notation a + ib, which has the pernicious effect of hiding issues, especially between different "=" signs. I'm currently working on category theory and abstract algebra, and they made me very skeptical about my own unwarranted use of some notations. Let me explain.

I'm currently considering complex numbers as tuples (a, b), where a and b are reals, as it is easier for me to grasp, and I believe this notation is more rigourous and less error prone. We define the addition and multiplication of such objects in the ways that you are all familiar with : you take two such tuples, and get back another tuple. Everything is stable and good. Now, in order for me to import results from the complex world to the real world, like trigonometric identities, I have to prove rigorously that if (a, b) = (a', b'), then a = a' and b = b'. To me, it's absolutely not obvious, as I believe the equal signs do not belong in the same world! This is revealed, I believe, by using tuples instead of the classic notation.

Let's take the demonstration here : Equality of complex numbers

The issue I have is with this line : (a−c)^2+(d−b)^2 = 0

Hidden behind the notation, this line assumes that the "=" sign is the same as the Real's, but it's not : it's the equal sign between two complex numbers, so we cannot just consider this to behave "just like the reals".

I believe that what we really have with this line is, using my notation : ( (a-c)^2 + (d-b)^2, 0) = (0, 0), where the symbol "=" is the complex equal, not the real's.

Therefore, I don't feel warranted to say that (a−c)^2+(d−b)^2 = 0 (with "=" being the Real's equal). Sure, the set of numbers of the form (a, 0) are isomorphic to the real numbers, but as soon as we introduce an imaginary part b not equals to 0, there is no reason for this structure not to break down with multiplication.

Now, either (a-c)^2 + (d-b)^2 = 0 (with the "=" sign of the reals), or it is not. If it is not, then I have (d, 0) = (0, 0) for every real d. It may seem absurd, but I don't see where I can get a contradiction from there. I could just say this situation wouldn't be useful nor interesting, but it still wouldn't lead to the contradiction that would be necessary to warrant my use of, say, trigonometric identities.

Please help me figure this out : I could just put it under the carpet, but then I would have the feeling that I am cheating every time I just assumes that identities we derive from complex numbers to the reals are true. How can I prove this equality using the notation above ?

$\endgroup$
1
  • 1
    $\begingroup$ if (a, b) = (a', b'), then a = a' and b = b' is a property of ordered pairs. $\endgroup$ May 11, 2020 at 16:59

5 Answers 5

1
$\begingroup$

This is my understand of what question you are asking; if I'm correct, I can answer the question (if I'm not, maybe this misinterpretation will help you clarify the question).

I believe you are starting with $\Bbb R^2$, or in other words ordered pairs of real numbers, and then defining an addition and a multiplication on that set and calling the resulting ring the complex numbers. (All perfectly reasonable). Along the way, you are asking why $(a,b)=(a',b')$ is equivalent to $a=a'$ and $b=b'$.

My answer is: that is part of the definition of ordered pairs in the first place. (In particular, it has nothing to do with the addition and multiplication rules that are subsequently introduced.) When ordered pairs are defined, part of the definition is (or should be!) the fact that $(a,b)=(a',b')$ if and only if $a=a'$ and $b=b'$.

$\endgroup$
1
  • $\begingroup$ I think you nailed it. My issue was exactly that : forgetting that my initial set is the set of the ordered pairs of real numbers, and subsequently the axioms that are derived from it. I still feel a bit shaky about this, I'm going to dive deeper into ordered sets. I'll wait for other answers before being sure and validating your answer. Thanks. $\endgroup$
    – Kerighan
    May 11, 2020 at 17:05
0
$\begingroup$

If you want an "all complex" proof,

$$(a,b)=(c,d)$$ $$(a-c,0)=(0,d-b)$$

$$((a-c)^2,0)=-(0,(d-b)^2)$$

$$((a-c)^2+(d-b)^2,0)=(0,0).$$

then you are stuck because you still need a definition of equality.

But as Matthew said, equality of ordered pairs is defined to be equality of the members.

$\endgroup$
0
$\begingroup$

What's betraying you here is not an overloaded symbol $=$, but a lack of context. You can look at an equality $(a - c)^2 + (d - b)^2 = 0$, but this statement is meaningless without some idea what $a, b, c, d$ are, and possibly what addition/multiplication is being used.

Are $a, b, c, d$ real numbers? Complex numbers? Equivalence classes modulo $5$? Matrices? Functions? What are the domain/codomains? Is the squaring composition, or pointwise multiplication?

The point is, the statement says very little without further information. If $a, b, c, d$ are reals, and $+$ and $\times$ are standard addition/multiplication on $\Bbb{R}$, then you can conclude that $a = c$ and $d = b$. If not, then anything goes.

Granted, some arguments don't specify all such context explicitly. It's not uncommon to see something like "Suppose $x + iy$ is a complex number", which is usually an indicator that the author is assuming $x$ and $y$ are real (and indeed, the addition/multiplication are the usual such operations).

A good mathematical argument is clear. It shouldn't be at all difficult for a reasonable reader to understand the context of the statements.

Ironically, out of all the symbols, $=$ is the least ambiguous. It almost always means that two objects are "the same" for all purposes. That is, $x = y$ essentially means that, $P(x)$ is true for some predicate $P$ if and only if $P(y)$ is true.

$\endgroup$
0
$\begingroup$

This famous proof is notable for not needing to impose as an axiom that our $2$-part numbers work like ordered pairs in the way others have discussed. You can't avoid such an axiom, for example, with split-complex or dual numbers.

Instead, we use the fact that complex numbers are assumed to obey the same field arithmetic as real ones, so that$$a+bi=c+di\implies a-c=(d-b)i\implies(a-c)^2=-(d-b)^2.$$At this point, we must also use the assumption $a,\,b,\,c,\,d\in\Bbb R$ (after all, if they're more general complex numbers $a+bi=c+di$ does not require $a=c,\,b=d$, because e.g. $0+(-i)i=0+0i$). We can therefore use certain properties of reals, such as $a-c\in\Bbb R,\,d-b\in\Bbb R$, so their squares are real too. The rest is trivial: $x^2+y^2=0\implies x=y=0$ in the reals, e.g. by the ordering $x^2\ge0\ge-y^2$ together with trichotomy.

As to your $=$-on-$\Bbb R$ vs. $=$-on-$\Bbb C$ point, note that in my display line above the three statements respectively equate two complex numbers, a real number and an imaginary number, and two real numbers. That imaginary numbers both square to real ones follows from the axiom $i^2=-1$. There is no distinction between two types of equality here. Each real number is equal to exactly one complex number; that complex number really is that real number, viz. $\Bbb R=\{z\in\Bbb C|\Im z=0\}$. It makes no more sense to say $\Bbb R$ isn't a subset of $\Bbb C$ than to say the imaginary numbers don't comprise a subset of it.

$\endgroup$
0
$\begingroup$

If you define a complex number as an ordered pair of real numbers, then, for you $\Bbb C=\Bbb R^2$. So, you want to prove that, in $\Bbb R^2$, $(a,b)=(c,d)$ if and only if $a=c$ and $b=d$. But then you have to define the meaning of “ordered pair”. A possible definition (it's the usual one, I think) is: $(a,b)$ means $\{\{a\},\{a,b\}\}$. So suppose that$$\{a\},\{a,b\}\}=\{\{c\},\{c,d\}\}.\tag1$$Then both sets have the same number of elements. There are two possibilities:

  • both sets are singletons: then $\{a,b\}=\{a\}$ and $\{c,d\}=\{c\}$, which means that $b=a$ and that $d=c$. But then $(1)$ means that $\{a\}=\{c\}$, which means that $a=c$. And, since $b=a$ and that $d=c$, $b=d$.
  • both sets have two elements each: then $b\ne a$ and $d\ne c$. Since $(1)$ holds, since $\{a\}$ and $\{c\}$ are singletons and since $\{a,b\}$ and $\{c,d\}$ have two elements each, $\{a\}=\{c\}$ and $\{a,b\}=\{c,d\}$. The first equality implies that $a=c$, and, since $a=c$, the second equality imples that $b=d$.
$\endgroup$
2
  • $\begingroup$ thanks for your answer, it really helped me understand my issue. To recap : my equal sign does not come after my definitions of multiplication and addition, but before, when I used the ordered pairs as the initial set of my ring. Contrary to some may be saying, this isn't that obvious, or freely given by definition of the complex numbers. I believe it mostly comes from set theory. $\endgroup$
    – Kerighan
    May 11, 2020 at 17:28
  • $\begingroup$ Yes! Equality between complex numbers comes before defining algebraic operations. $\endgroup$ May 11, 2020 at 17:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .