What allows us to use imaginary numbers?

Question

What axiom or definition says that mathematical operations like +, -, /, and * operate on imaginary numbers?

In the beginning, when there were just reals, these operations were defined for them. Then, i was created, literally a number whose value is undefined, like e.g. one divided by zero is undefined.

Does anyone know about how mathematical operations' ranges and domains were expanded to include imaginaries?

EDIT: An interesting comment notes a first use of complex numbers where,

those values would cancel in the end.

But can I refute that with, "from an inconsistency, anything is provable"?

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A corollary question: Could I define a new number z which is 1/0 and simply begin using it? Seems ludicrous.

Answer 1

We can do anything we want!

Specifically, we can define anything we want (as long as our definitions don't contradict each other). So if we want to allow ourselves to use imaginary numbers, all we have to do is write something like the following:

Define a complex number as an ordered pair of the form $(a, b)$, where $a$ and $b$ are real numbers.

Define $i$ as the complex number $(0, 1)$.

If $(a, b)$ and $(c, d)$ are complex numbers, define $(a, b) + (c, d)$ as $(a + c, b + d)$.

If $(a, b)$ and $(c, d)$ are complex numbers, define $(a, b) \cdot (c, d)$ as $(ac - bd, ad + bc)$.

And define subtraction and division in similar ways.

Is that it? Are we done? No, there's still more that we want to do. There are a lot of useful theorems about real numbers that also apply to the complex numbers, but we don't know that they apply to the complex numbers until we prove them. For example, one very useful theorem about real numbers is:

Theorem: If $a$ and $b$ are real numbers, then $a + b = b + a$.

The analogous theorem about the complex numbers is:

Theorem (not yet proven): If $a$ and $b$ are complex numbers, then $a + b = b + a$.

This theorem is, in fact, true, but we didn't know that it was true until somebody proved it.

Once we've proved all the theorems that we want to prove, then we can say that we're "done."

(Do we have to prove these theorems? No, we don't have to if we don't want to. But without these theorems, complex numbers aren't very useful.)

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As for your corollary question:

Could I define a new number $z$ which is $1/0$ and simply begin using it? Seems ludicrous.

Yes, you absolutely can! All you have to do is write:

Assume that there is a value $z$. Define $1/0$ as $z$.

And that's perfectly valid; this definition doesn't contradict any other definitions. This is completely legal, acceptable and proper.

Is that it? Are we done? Probably not; there's more we'd like to do. For example, what do you suppose $z \cdot 0$ is? There are a couple of theorems here we might like to use, but we can't. Let's take a look at them:

Theorem: If $x$ is a real number, then $x \cdot 0 = 0$.

Theorem: If $x$ and $y$ are real numbers, and $y \ne 0$, then $(x / y) \cdot y = x$.

Do you see why we can't use these theorems?

Does the first theorem tell us that $z \cdot 0 = 0$? No, because we don't know that $z$ is a real number. So the first theorem doesn't apply.

How about the second theorem? We know that $z = 1/0$. Does the second theorem tell us that $(1 / 0) \cdot 0 = 1$ (and therefore $z \cdot 0 = 1$)? No, because the second theorem is only applicable when the denominator is not $0$, and here, the denominator is $0$. So the second theorem doesn't apply, either.

If we want, we can add more definitions and maybe make some of these theorems "work" for $z = 1/0$, just like we have a lot of theorems that "work" for the complex numbers. But when we do this, we encounter a lot of problems. Rather than dealing with these problems, most mathematical writers simply refuse to define $1/0$. (That's what the sentence "$1/0$ is undefined" means: the expression $1/0$ is an undefined expression, because we have refused to define it.)

Answer 2

At the risk of sounding like a postmodernist: all numbers are imaginary.

Long ago, someone abstracted: what is the thing that this collection of sheep has in common with the number of fingers on my left hand, and called that thing "five." No inconsistencies were introduced and there were great simplifications to be made.

Someone asked how to divide two pies among three people and the abstraction of fractions was born. Someone thought about debt and the abstraction of negative numbers was born. Someone realized that positive and negative fractions didn't describe the intuitive nature of a continuum and the abstraction of the reals was born.

And eventually someone abstracted solutions to $x^2 + 1 = 0$; they're not any more imaginary than any of the other abstractions, they're all products of human imagination. The name "imaginary numbers" is unfortunate.

You say: why can't I abstract a solution to $0*z = 1$, i.e. $1/0$? The problem is that your abstraction will be incompatible with your other abstractions, i.e. you will break arithmetic. But there are areas of geometry (e.g. Mobius transformations of the plane) where there is a consistent way to do a little arithmetic with an idea of $1/0 = \infty$ (although one has to be careful to remain consistent).

Answer 3

What axiom or definition says that mathematical operations like +, -, /, and * operate on imaginary numbers?

It is set theory that allows us to give a rigorous foundation for complex numbers. In particular, as explained here the axiom of pairing plays a crucial role, It allows us to construct the product set $\,\Bbb R^2\,$ and then reduce complex arithmetic to arithmetic on pairs of reals - as Hamilton did when he gave the first rigorous construction of $\,\Bbb C,\,$ representing $\,a + b\,i $ by the pair $\,(a,b)\,$ with operations

$$\begin{align} (a\!+\!bi) + (c\!+\!di) &=\ \, a\!+\!c\!+\! (b\!+\!d)i\\[.2em] \rightsquigarrow\, (a,\ \ b)\ + (c,\ \ d)\ &= (a\!+\!c,\ \ \ b+d)\\[.4em] (a\!+\!bi)\times (c\!+\!di) &= \ ac\!-\!bd\!+\!(ad\!+\!bc)i\\[.2em] \rightsquigarrow\, (a,\ \ b) \ \times\ (c,\ d)\, \ &= (ac\!-\!bd,\ \ ad\!+\!bc) \end{align}\qquad\qquad$$

This reduces the consistency of $\,\Bbb C\,$ to the consistency of $\,\Bbb R\,$ i.e. any contradiction derived in $\,\Bbb C\,$ would yield a contradiction on such pairs of reals, so a contradiction in $\,\Bbb R.$

Further, a major accomplishment of the set-theoretical construction of $\,\Bbb C\,$ (and algebraic structures) is that it eliminates imprecise syntax and semantics in informal approaches. The imprecise term $\, a + b\, i\, $ is replaced it by its rigorous set-theoretic representation $\,(a,b)\,$ - which eliminates many ambiguities, e.g. doubts about the meaning of symbols $\,i\,$ and $\,+\,$ and $\,=\,$ in complex arithmetic. Such questions were rampant in the early development of complex numbers, and without set theory or any other rigorous foundation it was difficult to provide convincing precise answers. For example below is how Cauchy tried to explain them.

In analysis, we call a symbolic expression any combination of symbols or algebraic signs which means nothing by itself but which one attributes a value different from the one it should naturally be [...] Similarly, we call symbolic equations those that, taken literally and interpreted according to conventions generally established, are inaccurate or have no meaning, but from which can be deduced accurate results, by changing and altering, according to fixed rules, the equations or symbols within [...] Among the symbolic expressions and equations whose theory is of considerable importance in analysis, one distinguishes especially those that have been called imaginary. -- Cauchy, Cours d'analyse,1821, S.7.1

It's no surprise that Cauchy's peers were not persuaded by such handwaving, e.g. Hankel replied

If one were to give a critique of this reasoning, we can not actually see where to start. There must be something "which means nothing," or "which is assigned a different value than it should naturally be" something that has "no sense" or is "incorrect", coupled with another similar kind, producing something real. There must be "algebraic signs" - are these signs for quantities or what? as a sign must designate something - combined with each other in a way that has "a meaning." I do not think I'm exaggerating in calling this an unintelligible play on words, ill-becoming of mathematics, which is proud and rightly proud of the clarity and evidence of its concepts. $\quad$-- Hankel

Hamilton's elimination of such "meaningless" symbols - in favor of pairs of reals - served as a major step forward in placing complex numbers on a foundation more amenable to his contemporaries. Although there was not yet any theory of sets in which to rigorously axiomatize the notion of pairs, they were far easier to accept naively - esp. given the already known closely associated geometric interpretation of complex numbers.

See said answer for further discussion of this and related topics (above is excerpted from there).

Answer 4

The "necessary and sufficient" axioms to define the complex numbers are

$$(a,b)+(a',b')=(a+a',b+b')$$

$$(a,b)\cdot(a',b')=(aa'-bb',ab'+a'b).$$

(Subtraction and division can be defined as the inverses of addition and multiplication, as usual.)

In particular,

$$(a,b)+(0,0)=(a,b)$$ so that $(0,0)$ is the zero and

$$(a,b)\cdot(1,0)=(a,b)$$ so that $(1,0)$ is the unity.

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As you can check, $(a,b)$ can also be represented as the expression $a+ib$, where $i$ is a reserved symbol, with the usual computation rules on polynomials (with $i$ seen as the variable). Using this notation, $$(0,1)\cdot(0,1)=(-1,0)$$

translates to the famous

$$i^2=-1.$$

As you can check, the "pair" representation and the "$i$" representation are completely interchangeable. $i$ has a simple geometric interpretation: in a 2D plane, multiplication by $i$ corresponds to a rotation around the origin by a quarter turn.

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Note that there are absolutely no undefined operations here.

Answer 5

Why "adding a new element" $i$ with $i^2 = -1$ to $\Bbb R$ "works", the technical cause.

What means "works": we want a field extension of $\Bbb R$. Intuitively, a set greater than $\Bbb R$ (this is the easy part) that is also a field: we have two operations $+$ and $\cdot$ extending the operations sum and product of $\Bbb R$ and verifying the same properties (see https://en.wikipedia.org/wiki/Field_(mathematics)). This is the hard part.

The "addition of a new element" $i$ with $i^2 = -1$ makes perfect sense in algebra: we are taking the quotient $$\Bbb R[x]/(x^2 + 1)$$ where $\Bbb R[x]$ is the polynomial ring in one indeterminate and $(x^2 + 1)$ is the ideal generated by $x^2 + 1$.

The essential fact: a quotient like this will be a field iff the ideal is maximal. You can check easily that $(x^2 + 1)$ is maximal because $x^2 + 1$ is a polynomial of degree 2 without roots in $\Bbb R$. Also important: the quotient contains (an isomorphic copy of) $\Bbb R$ strictly.

Two examples where "adding a new element" does not "work":

• $\Bbb R[x]/(x)$ is a field because $(x)$ is a maximal ideal, but the quotient is isomorphic to $\Bbb R$.

• $\Bbb R[x]/(x^2)$ isn't a field because $(x^2)$ isn't a maximal ideal.

Answer 6

In the beginning, when there were just reals, these operations were defined for them. Then, i was created, literally a number whose value is undefined, like e.g. one divided by zero is undefined.

Not quite: in the beginning were the natural numbers, and addition over the natural numbers, and everything was good.

But then people wanted to "undo" addition, and subtraction turned out to be really useful, and related to many different real-world examples (taking things away from a collection). And suddenly a problem arose: you could add any two natural numbers and get a third natural number, but when you subtracted, the result might be undefined. What natural number is five minus seven?

Some people were okay with this being undefined, but others weren't, and the ones who weren't started wondering: what if we define a new quantity that's equal to five minus seven? And so the negative numbers were born! (And for quite a long time, mainstream mathematicians derided them and didn't consider them real numbers, because can you show me negative two apples, or people, or coins? No you can't.)

Similarly, division led to the introduction of rational numbers ("one divided by two is defined to be a new type of number, which I'll just write as 1/2"), and square roots led to the introduction of irrational numbers, and eventually we ended up with the real numbers that we all know and love today. "Imaginary" numbers are just another extension of the concept of "number", giving it new capabilities.

So then, would you like to define 1/0? Go for it! Look at the projectively extended real numbers, which do exactly that: augmenting the real numbers with a single "point at infinity". In the projectively-extended reals, 1/0=∞, and there's no problem there. As Tanner Swett put it, you can do anything you want! The real question in mathematics is: does this lead to anything interesting? And the only way to answer that is, try it and see!

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P.S. The summary in this answer isn't meant to be chronologically accurate—the Greeks were dealing with "real numbers" in the form of lengths long before anyone came up with the Peano axioms. Instead, take this as a summary of how axiomatic definitions of "numbers" have evolved over time.

P.P.S. There's a legend that one ancient mathematician was either murdered or struck down by the gods for the hubris of constructing an irrational number. I haven't heard of this happening with the projectively-extended reals, but watch your back, just in case.

P.P.P.S. If you're curious about the complex numbers in particular, this website has a better explanation than I could ever give. Geometrically, $i$ can be considered as a 90-degree rotation in the plane; in geometric algebra, it would be called a unit bivector. But that's a can of worms too big for an answer here to encompass.

Answer 7

We can always use the usual rules for doing arithmetic with real numbers on complex numbers too, provided we always substitute $-1$ for $i^2$ whenever we encounter it. It follows that once we allow the imaginary unit, we have a consistent algebra which obeys the usual laws.

Thus, we discover that if we extend all the usual laws of addition and multiplication on binomials of the form $a+ib,$ with $i^2=-1,$ everything goes on smoothly. One way to formally do this is by using ordered pairs, as William Hamilton first did, but the idea I think you need is that if we allow ourselves to calculate with complex numbers as we did with real numbers, only remembering to replace $i^2$ with $-1,$ then we have a consistent algebra. Mathematicians usually call the system so defined a field. There are many other fields apart from the ones formed by the real or complex numbers with addition and multiplication as usually defined, but that's another story. The gist is just that we can define two operations over the complex numbers similar in behavior to the usual addition and multiplication over the real numbers -- and with a certain lightness we may think of these as extensions of the usual addition and multiplication. Thus, the $+$ and $×$ of the complex system is technically different from that of the reals, but of course for symbolic economy (and also because they behave very similarly), we retain the same symbols. In general we use these symbols for the operations in any field, too, whether the members of the field are numbers or people.

Answer 8

What axiom or definition says that mathematical operations like +, -, /, and * operate on imaginary numbers?

The definition of the complex numbers? There are really no new axioms involved since complex numbers are defined in terms of real ones and their behavior is completely derivable from definitions.

In the beginning, when there were just reals, these operations were defined for them.

See, there is your mistake. "there were just reals". Reals are imaginary constructs with no inherent existence. They are the consequence of choosing convenient definitions (and some axioms).

Then, i was created, literally a number whose value is undefined, like e.g. one divided by zero is undefined.

That is a common misconception and actually untenable. You cannot "define" i as "the square root of -1" since -1 has two square roots in the complex numbers. Instead, complex numbers are defined via their arithmetic properties and then the real numbers are embedded into the complex numbers. Since that embedding retains basically all operations on the real numbers (logarithms and exponentiation get a lot more iffy than they are in the reals, though, even though exp(x) remains perfectly well defined), one can continue using the same symbols and operators without introducing trouble, just like embedding whole numbers into rational numbers and rational numbers into reals worked without necessitating different operators for writing down relations.

Answer 9

Several answers have already explained about the definitions involved in setting up the complex number system and its arithmetical operations. I'd like to add a little about the word "use" in the title of your question, because using numbers can involve lots of different things. For example, when dealing with real numbers, we can invoke various algebraic laws (associative, commutative, distributive, etc.) to prove things like $a^2-b^2=(a+b)(a-b)$. And we can invoke laws about the ordering relation and multiplication to prove things like $a^2\geq0$. Some of these things continue to work with complex numbers (for example $a^2-b^2=(a+b)(a-b)$) but others don't ($i$ is a counterexample to $a^2\geq0$).

So the title of your question should really be "which of the familiar laws for the real number system continue to work for complex numbers, and why?" The answer to "which ... continue to work" is essentially that laws that are exclusively about addition, subtraction, multiplication, and division continue to work whereas laws about the ordering relation ($<$ or $\leq$) do not. (One doesn't normally even try to define, say, whether $i<1$ or $i\geq1$.)

And why do the laws about $+,-,\times,/$ continue to work? That needs to be proved --- and it has been proved (a few centuries ago).

If you think about all the formulas about $+,-,\times,/$ that you had to learn in high-school algebra, proving them all again for complex numbers might look like an awful lot of work. But fortunately, they can all be deduced from a rather small subcollection of laws (as I mentioned above for $a^2-b^2=(a+b)(a-b)$). So only that subcollection needs to be checked specifically for the case of complex numbers; the rest of the high-school laws then follow for complex numbers just as they did for real numbers with no additional work. The relevant subcollection, from which the other laws about $+,-,\times,/$ follow, is called the field axioms, and that's why several of the previous answers talk about the complex numbers being a field, i.e., satisfying the field axioms.

Answer 10

• What allow us to use anything in mathematics are the axioms. Real numbers form a complete ordered field, if you take the set of real numbers and "add" $i$ to it (considering its usual behavior), you lose this order (that is: They won't be a complete ordered field anymore). You can still define some order but it will no longer be compatible with the ordering of the reals. (It is a good exercise to verify what axioms this "addition" of $i$ violates. See this or this).

• In the previous case, the natural ordering was broken but mathematicians found a way to cope with it by using a new system in which the broken axioms were removed. These axioms give us a plethora of stuff we can do in the real numbers, it was natural to think: "In this new system, how can we do something that is at least remotely similar to what we could do with the ordering in the real numbers?"

• There is no "moral ground" to define and use things in mathematics, you can define anything you want. There is - for example - a number system in which division by zero is meaningful. In the same way, you may run into trouble (such as the aforementioned loss of ordering) when you compare some two systems.

Answer 11

I think you already have a good interpretation of complex numbers algebraic view. So I propose here the geometric approach of this numbers, following Hamilton, Clifford and Grassmann geometric view of algebra.

First, we need to separate the meaning about the number (1) and the number (-1). You can do this for trying to count negative numbers as one does with positive numbers, e.g. one cup, two notes, three boxes, minus four houses???

The later seems meaningless because we have not added a good definition about (-). For this, we need to think in the number not only as a quantification, but as a quantification AND a direction. So (-1) is not just a number as (4), but a number that one can interpret as a quantity in a certain direction.

So the geometric view of (+) and (-) is that of symbols whose when fixed with number defines a direction in a certain quantity, e.g. If (+3) means 3 steps of unity in the certain direction, say right, then (-3) means 3 steps in the opposite direction. This could be demonstrated with the axiomatic approaches that my colleagues presented above, but I want to propose here just a geometric view of all this.

In the Euclidean Geometry, we can assert more than the opposite notion. If we suppose a segment $AB$, with a certain magnitude, when applying (-1) in the segment, we rotate it $\pi$ rad. So $AB=-BA$.

This notion can be extended to "when applying positive number to a segment, it contracts (when between 0 and 1), maintain or stretch this segment. But when negative number are applied, besides the homothety, it rotates through $\pi$ rad."

It's just a geometric view, I'm not proposing here any rigorous justifications.

But this is just in the $\mathbb{R}$ line. Let's extend this thought to $\mathbb{R}^2$ plane.

We already know that $(-1)$ means rotates in $\pi$ rad, but as we now have this two lines orthogonally positioned, we also need to define an application (a number) which could rotate the segment $\pi/2$ rad. Let's suppose a new type of number, whose faces i don't see, but algebraically i can give him a name (a letter), say $i$. So as we can see (and you could prove it using the axioms of above answers), when applying twice $i$ to this segment, it rotates to $\pi$ rad and we can assert that $i^2=-1$. In this view, the meaning $i=\sqrt{-1}$ is just a consequence of $i^2=-1$.

Now, as Kline say here, Hamilton pointed out that

A complex number $a+bi$ is not a genuine sum in the sense that $2 + 3$ is. The use of the plus sign is a historical accident and $bi$ cannot be added to $a$. The complex number $a + bi$ is no more than an ordered couple $(a, b)$ of real numbers.

i suppose this answer your saying "In the beginning, when there were just reals, these operations were defined for them. Then, i was created, literally a number whose value is undefined"

This allows us to use the knowledge of vectors. Indeed, Hamilton knew that a complex number is nothing more that a ratio between two segments. If $AB$ and $AC$ are two segments conveniently positioned at the origin $A$, a complex number $z=a+bi=(a,b)=e^{i\theta}$, with $\theta$ the angle between the complex number $z$ and the origin, is the operation necessary to rotate and stretch the segment $AB$ to give $AC$, or $AB\cdot z=AC\to z=\frac{AC}{AB}$, which allows us to use complex numbers as a type of rotational numbers, as the negative numbers are a directional system of numbers.

And Hamilton gave latter a full usefulness of the imaginary numbers to also rotate in $\mathbb{R}^3$, where the Quaternions, a four dimensional algebra, was born. And you could ask "why a four dimensional system of numbers is required to rotate three dimensional directed segments (vectors)?"

I'm not allowed to answer this question here, but i can tell you to read about the Clifford Algebras, the Geometric Algebra, which are an excellent extension of Hamilton's Quaternions with the concept of Grassmann's exterior algebra. In the geometric algebra, real numbers, complex numbers, quaternions, biquaternions, and other n-dimensional systems of numbers are synthesized in a rigorous way giving to all this abstract algebraic views a good geometrical interpretation.

Answer 12

Well, coming from the algebraic side, consider field extensions of the form $${\Bbb Q}(\sqrt n)=\{a+b\sqrt n\mid a,b\in{\Bbb Q}\},$$ where $n\ne0,1$ is a square-free integer. The addition is ''componentwise'', $$(a+b\sqrt n) + (c+d\sqrt n) = (a+c) + (b+d)\sqrt n$$ and the multiplication is given by noting that $\sqrt n^2 = n$, $$(a+b\sqrt n) \cdot (c+d\sqrt n) = (ac + bdn) + (ad + bc)\sqrt n.$$ Here the case $n=-1$ comes up naturally and writing $i=\sqrt{-1}$, we obtain ${\Bbb Q}(i)=\{a+bi\mid a,b\in{\Bbb Q}\}$ and also ${\Bbb C} = {\Bbb R}(i)=\{a+bi\mid a,b\in{\Bbb R}\}$, the field of complex numbers.