Before, the concept of imaginary numbers, the number $i = \sqrt{-1}$ was shown to have no solution among the numbers that we had. So we declared $i$ to be a new type of number. How come we don't do the same for other "impossible" equations, such as $x = x + 1$, or $x = 1/0$?

Edit: OK, a lot of people have said that a number $x$ such that $x = x + 1$ would break the rule that $0 \neq 1$. However, let's look at the extension from whole numbers to include negative numbers (yes, I said that I wasn't going to include this) by defining $-1$ to be the number such that $-1 + 1 = 0$. Note that this breaks the "rule" that "if $x \leq y$, then $ax \leq ay$", which was true for all $a, x, y$ before the introduction of negative numbers. So I'm not convinced that "That would break some obvious truth about all numbers" is necessarily an argument against this sort of thing.

Simon Fraser
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Lily Chung
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    Well, if we had a number such that $x = x + 1$, then I could subtract $x$ from both sides so that $0 = 1$, and then I could prove that $y = z$ for any $y$ and $z$ whatsoever. So this doesn't seem very useful, unless we give up, say, subtraction. – Zhen Lin Dec 15 '12 at 23:44
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    See also the answers to [this related question](http://math.stackexchange.com/questions/85111/) proposing a different hypothetical "imaginary". – hmakholm left over Monica Dec 15 '12 at 23:47
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    Basically, we don't create other sorts of solutions to "impossible problems" because they don't have much use. It should be noted, negative numbers are also an attempt to solve an "impossible problem," when all you know are positive numbers... – Thomas Andrews Dec 15 '12 at 23:58
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    People “invented” complex numbers not because they wanted to solve an “impossible” equation $x^2=1$ but rather because they realized that in order to find real roots of polynomials they need to perform intermediate computations in $\mathbb C$. Later, complex numbers have turned out to be very useful in many areas of math. If there was a nice non-trivial way to extend real numbers so that the equation $x=x+1$ had a solution, people would do that. – Yury Dec 16 '12 at 00:00
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    @ThomasAndrews I was going to add that to the original question, but I didn't feel like it helped it. – Lily Chung Dec 16 '12 at 00:01
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    @Arafinwe So, are you saying $0 = 1$ is an acceptable state of affairs? – Zhen Lin Dec 16 '12 at 00:24
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    You can look up "Real Numbers Modulo 1". This is basically taking any real number and looking only at the fractional part of it. Since $0$ and $1$ have *only* integer parts, they will be equal in $\mathbb{R}/<1>$. – Andrew Maurer Dec 16 '12 at 00:52
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    I wrote [a fairly extensive explanation of what happens if you try to define division by zero](http://math.stackexchange.com/questions/125186/why-not-to-extend-the-set-of-natural-numbers-to-make-it-closed-under-division-by/125208#125208) in response to a similar question. You may find it interesting. The summary is that you can make it work, but what you get is generally uninteresting—for interesting reasons. [IEEE floating-point arithmetic](http://math.stackexchange.com/a/125212/25554) *does* define division by zero in this way, but loses a number of important mathematical properties. – MJD Dec 16 '12 at 01:51
  • (This question might be a duplicate of [Why not to extend the set of natural numbers to make it closed under division by zero?](http://math.stackexchange.com/questions/125186/why-not-to-extend-the-set-of-natural-numbers-to-make-it-closed-under-division-by) ) – MJD Dec 16 '12 at 01:54
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    @Zhen Lin: Actually, Georg Cantor's Theory of Transfinite numbers, describes numbers with just such properties as `x = x+1`. – RBarryYoung Dec 16 '12 at 03:40
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    $\infty=1+\infty$, so yes, you can even define certain elements with this property ($x=1+x$) and indeed you may find such elements in ordinal numbers' theory. – Pantelis Sopasakis Dec 16 '12 at 04:46
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    Also $i^2 = -1$, **not** $i = \sqrt{-1}$ – Ruud Dec 16 '12 at 11:31
  • We have, for example, x=1/0 if we extend the real numbers by adding infinity, and we specify 0 as -0 or +0 depending from which direction we take the limit. – vsz Dec 16 '12 at 12:08
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    @RuudvA: that depends on what you mean by that. $i$ is a square root of $-1$. Just as $-i$ is. And just as $2$ and $-2$ are square roots of $4$... – tomasz Dec 16 '12 at 22:14
  • Perhaps I'll answer this question another time, but I want to point out that there are other "impossible" equations over $\mathbb{R}$ aside from $x = x + 1$ and $x^2 + 1 = 0$. For instance, there's $x^4 + 1 = 0$ and $e^x = 0$ and $\sin x = 2$. As I read it, the intent of the question is: are these last three examples actually subsumed by both $x = x+1$ and $x^2 + 1 = 0$? If so, why? Answers should address this. – Jesse Madnick Dec 16 '12 at 22:54
  • There is a number system where $k = k+1$ has a solution, that is the trivial ring $\{0\}$ where $0 = 1$. – tghyde Jun 03 '13 at 23:49
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    What you are saying makes sense. I once tried to define $1/0$ as a number called **j** it did not work out well, it was confusing trying to define addition, subtraction, multiplication, and division. Worse, powers, square roots. It must have been hard enough to define all properties of the imaginary constant **i**, and because of that, I don't think mathematicians will be making any undefined things constants anytime soon. – Anonymous Pi Jun 03 '13 at 23:38

23 Answers23


Here's one key difference between the cases.

Suppose we add to the reals an element $i$ such that $i^2 = -1$, and then include everything else you can get from $i$ by applying addition and multiplication, while still preserving the usual rules of addition and multiplication. Expanding the reals to the complex numbers in this way does not enable us to prove new equations among the original reals that are inconsistent with previously established equations.

Suppose by contrast we add to the reals a new element $k$ postulated to be such that $k + 1 = k$ and then also add every further element you can get by applying addition and multiplication to the reals and this new element $k$. Then we have, for example, $k + 1 + 1 = k + 1$. Hence -- assuming that old and new elements together still obey the usual rules of arithmetic -- we can cheerfully subtract $k$ from each side to "prove" $2 = 1$. Ooops! Adding the postulated element $k$ enables us to prove new equations flatly inconsistent what we already know. Very bad news!

Now, we can in fact add an element like $k$ consistently if we are prepared to alter the usual rules of addition. That is to say, if we not only add new elements but also change the rules of arithmetic at the same time, then we can stay safe. This is, for example, exactly what happens when we augment the finite ordinals with infinite ordinals. We get a consistent theory at the cost e.g. of having cases such as $\omega + 1 \neq 1 + \omega$ and $1 + 1 + \omega = 1 + \omega$.

Peter Smith
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    Of all the posts, only this one hits the key point, in my opinion: Suppose you extend the real numbers to add a solution to $x = x+1$. How does this change what happens when you restrict back to being regular real numbers? The answer is - it changes everything - you've made all numbers equal! On the other hand, suppose you extend the real numbers by adding a solution to $x^2+1 = 0$. Now, restrict this back to the real numbers and say "ok, what's changed"? The answer: nothing. – Jason DeVito Dec 16 '12 at 00:52
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    @JasonDeVito, "you've made all numbers equal" - that's not what you should have done! You should have extended the set instead of collapsing it, so you must reject the injectivity of addition instead. Otherwise that's not an extension, as you've shown. – Rotsor Dec 17 '12 at 03:09
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    @Rostor: I'm not exactly sure I understand. Of course, one way around this is to declare that subtraction is a "partial" operator. One can add a solution $x$ to $x = x+1$ but then say "but you can never subtract by $x$" and still save the real numbers (I think), but this seems unnatural. – Jason DeVito Dec 17 '12 at 04:07
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    Subtraction can still be total if it's defined that $\bot - x = \bot; x - \bot = \bot$ where $\bot$ is the solution for $x = x + 1$. It reminds me of data type lifting used in Haskell to give total semantics to computations with failures. – Rotsor Dec 17 '12 at 13:03
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    I do not agree with this answer. The proposal is not to add to the reals an element $k$ such that $k+1=k$. Instead the proposal is to imagine a set of which the reals are a proper subset and an element k such that $k+1=k$ is also a member. In the same way, the set of complex numbers includes the set of real numbers as a proper subset. $i$ is also a member of the complex numbers, but not the reals. – emory Dec 18 '12 at 16:08
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    Several years later, this answer makes a little more sense to me in terms of adjoining elements to rings :) – Lily Chung Apr 21 '16 at 07:36

In ordinal arithmetic we have $1+\omega=\omega$. There is an algebraic downside: it turns out that $\omega+1\ne \omega$.

André Nicolas
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    Of course, in cardinal arithmetic, $1 + \aleph_0 = \aleph_0 + 1 = \aleph_0$. –  Dec 16 '12 at 06:24
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    @Hurkyl Not to mention that you can't figure out what $2^{\aleph_0}$ is at all. – Ryan Reich Jun 03 '13 at 23:50
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    @RyanReich The problem is not figuring out what $2^{\aleph_0}$ is. You can even give explicitly a representative of it, the reals, or the sequences of natural numbers (cardinals are equivalence classes by bijections). The problem is matching it to the ordinal sequence. – OR. Jul 06 '13 at 04:48

The short answer is that you can add any made up solution to any equation you want and extend whatever number system (or any system) you have to a larger one.

The slightly longer answer is that in mathematics it is usually with some aim in mind that an extension is made. Particularly for the imaginary numbers you mentioned, the square root of $-1$ was contemplated because it simplified manipulations on polynomials when looking for their roots.

The irrationals are added to the rational numbers since the rationals do not suffice for measuring distances (i.e., the hypotenuse of a triangle with sides equal to $1$ is $\sqrt2$).

Infinitesimals are added to the real numbers in order to make rigorous heuristic arguments using such entities.

Infinitely large natural numbers are added to the ordinary natural numbers in order to construct certain models showing the independence of certain axioms from others.

Infinite sets are added to the more tame finite sets since it is convenient to be able to talk about infinite collections of, say, numbers.

100-150 years ago 'function' assumed a very narrow meaning (not well defined) basically what we today would call: a function that is analytic everywhere except possibly at isolated points. There were even attempts to prove that every continuous function must be differentiable at almost all points. Gradually, the more exotic beasts - functions that are continuous but nowhere differentiable - entered the scene. Thus extending the study of functions from the narrow class of almost everywhere differentiable ones to the class of continuous ones. This was necessitated again by applications since such functions occur as uniform limits of analytic functions.

There are many more such examples where some extension is made fueled by some applications or a need to better understand the axiomatics of some system.

Ittay Weiss
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    Excellent answer. We can say that the numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 are the only numbers that exist because they correspond to the fingers on our hands. The number 11 is an extension based on the imaginary notion of a 3rd hand. You can count to 11 if you use your imagination and imagine your third hand. – emory Dec 18 '12 at 16:55
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    You wouldn't necessarily count on your fingers, you could just count 11 apples. – Ovi Mar 31 '13 at 22:14

Adding an "imaginary" solution to a previous "impossible" equation always breaks existing rules (by definition, because one of the existing rules was that the impossible equation was impossible). The question is whether the gain of the new solution is worth the loss. In the case of extending reals to complex numbers, you lose the usual ordering property (an ordering $\le$ that is compatible with $+$ and $\cdot$ must have all squares nonnegative), but the resulting gain is huge because you can solve so many equations you couldn't before.

In your example of going from nonnegative numbers to all numbers, you give up the property $x \le y$ implies $ax \le ay$, but it's easy enough to fix up slightly, namely, to add the condition that $a \ge 0$ (and perhaps to say that the inequality is reversed if $a < 0$). This is also a small change.

If you add a solution to $x=x+1$, then as others have mentioned, you either have to give up $0 \ne 1$ or else give up subtraction. The first one would pretty much makes the new system useless. The second can be useful under certain circumstances. For example (as is done in measure theory, among other places), you can introduce a symbol $\infty$ that satisfies $\infty=\infty+1$. You can also define addition involving $\infty$, and most multiplications, and even most subtractions. A problem arises when you try to define the difference $\infty - \infty$, or the product $\infty \cdot 0$, so you leave those undefined. You have given up the ability to always subtract or multiply, but in some contexts that is okay. You just have to remember those restrictions when you're working in those contexts.

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    I know this is a very late comment, but since you bring up measure theory, it's worth mentioning $\infty \cdot 0 = 0$ is common. – Brevan Ellefsen Jun 12 '18 at 23:25
  • Besides losing ordering, suddenly logarithms, roots and many other functions become multi-valued. – Anixx Mar 03 '21 at 23:48
  • @BrevanEllefsen: IMHO, many fields could benefit from distinguishing between the concepts of "multiply a member of a group/ring/field/etc. by another of that same structure" versus "iterate addition or subtraction N times", and likewise for exponentiation versus iterated multiplication. Using the iterative style of multiplication with a repetition count of zero yields the Additive Identity in any structure, even in cases where the multiplication by a structure's additive identity would not. – supercat Aug 23 '21 at 09:44
  • @Anixx: True, but in a way e.g. roots "should always have been" multivalued. In $\mathbb C$, every nonzero number has $n$ different $n$-th roots; I find that far more sensible than the situation over the reals, where for odd $n$, *every* number has *one* $n$-th root, whereas for even $n$, *only positive* numbers have $n$-th roots, but then *two* of them (just by convention we choose the positive one, forcing it to be single-valued). In other words, just looking at the graphs of $x^n$ for even vs. odd $n$ strongly suggests something more is going on, and complex numbers explain it beautifully. – Torsten Schoeneberg Aug 23 '21 at 14:28

If you can use sets to define a structure that has some properties (like $\exists x[x=x+1]$, of course one has to know what $1$ is.), then we are done. Formal Constructions using sets is what is used to make the natural numbers, integers, rationals, real numbers,....

This part uses abstract algebra:

The ring $R[x]/\langle x^2+1\rangle$ has solutions to the equation $x^2+1=0$. (Where $1$ is the multiplicative identity of $R[x]/\langle x^2+1\rangle$)

In a similar way, the ring $R[x]/\langle1\rangle$ has solutions to the equation $x+1=x$.

This is the trivial ring. However, the trivial ring is not really interesting.

Michael Albanese
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To your edit:

So I'm not convinced that "That would break some obvious truth about all numbers" is necessarily an argument against this sort of thing.

You're right, of course. Some obvious truths simply need to be bent or broken in order for mathematics to be useful - if we insisted that every rule we learnt in primary school held for every concept we ever came across in more advanced mathematics, we'd never encounter new breakthroughs. And in the same way, $i^2 = -1$ breaks the 'obvious truth' that all numbers have non-negative square. So we're not scared of doing it - it would have held back both mathematics and physics hugely to be scared of $i$ just because it was a little unfamiliar.

However, the people who are telling you that a new number $x$ with the property that $x=x+1$ would break the rules have a point too. The introduction of this new number - with some exceptions - is a gratuitous breaking of the rules. That is, there is little sense behind it - it's not motivated by a mathematical (in the sense of 'mathematics research') question or observation, it screws arithmetic right up, and it doesn't seem to have much benefit. But don't take my word for it - spend a few minutes (a few hours, a few years) playing with the concept. The thing is, if you do so, you'll soon realise one of two things:

  1. You come up with a system like ordinal arithmetic (look it up!), a very interesting system of numbers with well-defined notions of $+$ and $\times$, in which there is a number $\omega$ with the property that $\omega = 1+\omega$. Unfortunately, you're really really unlikely to come up with ordinals by approaching them from this angle: after all, ordinals are weird things that are only really mentioned in very specific (and rather esoteric, if I may say so!) areas of study. Anyway, here are some reasons why ordinals are probably too weird to come up with from this angle: in ordinal arithmetic, $+$ and $\times$ are non-commutative (look it up), for a start, and $-$ and $\div$ don't exist. And you'd have to get rid of negative numbers and stick a whole shedload of infinities in there too. More than just introducing a number, you're having to break almost all the rules of arithmetic to even begin to talk about these things! Or even...

  2. You come up with a weird and useful system that nobody has ever discovered before. Except nobody's ever discovered it before, so it's probably even harder to find and weirder than ordinals. Okay, much more likely...

  3. The thing you come up with is ugly and useless. It's no coincidence that ordinals look so different to ordinary arithmetic: if you try to retain too many of the properties of ordinary arithmetic but add in your new number, it will simply all collapse in on itself. Let's take an example. Suppose $x = x+1$, and your system of arithmetic allows me to subtract $x$. Then suddenly $0 = 1$, and we can multiply both sides of that equation by $a$ to get that $0 = a$, for any number $a$. And even $0 = x$. So everything is equal to zero, and the equation "$x = x+1$" simply says "$0 = 0+0$". Oops!

The problem with questions like this is that they are invariably not fruitful directions to follow. That's not to say that they won't eventually have interesting and useful answers, but rather the process of answering them probably will come indirectly - usually via trying to answer a different, much more specific mathematically-motivated question.

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It's because you'd be forced to break other rules. Say $\frac 10=j$. Then

$$\frac jjj=j$$ $$\frac {\frac 10}{\frac 10}\frac 10=\frac 10$$

Actually do out the fractional division first and you get

$$\frac 00=\frac 10$$ $$0=1$$

You can prove, once you've introduced this number, that all numbers are equal. So you've inadvertently collapsed the whole number line, in a sense. The difference between $0$ and $1$ in this system is the same as the difference between $4$ and the roman IV: one of notation, not of value. All "numbers" are really the same quantity (i.e. equal), and this quantity adds, subtracts, and multiplies to be itself (this is the trivial ring). It's all consistent when looked at in this way, but it's also very boring. If you want to add $\frac 10$ and not run into this boring situation, then you have to drop/change enough algebraic rules to remove your ability to prove $0=1$. This happens in say, wheel theory.

On the other hand, introducing $i^2=-1$ poses no such problem. In fact, we can keep essentially everything about the real numbers that we knew previously. So this addition doesn't destroy structure, it creates it. The idea is this: say you want to extend the real numbers. Then defining the new system:

$$\text{"The real numbers, but with $i$"}$$

is a perfectly consistent statement to make.

$$\text{"The real numbers, but with $j$"}$$

with $j$ as defined above, does not. They're contradictory ideas. You can declare whatever kinds of numbers you want, so long as you have consistent rules. And if you want those numbers to extend the real numbers without seriously modifying them, that requirement of "having consistent behaviour with what we know about $\mathbb{R}$" puts constraints on the kinds of extensions you can define.

Robert Mastragostino
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    Actually, rather than proving all numbers are equal, you've proved not all 0 are equal. This is reasonable. First postulate there is a number j such that 1/0 can equal it. Prove that 0 cannot always equal 0 (ie, there is more than one number that we've been calling zero). In consequence, there is more than one solution to j=1/0. Call the various j "infinite numbers", and call the various 0 "infinitessimal numbers." Wheel theory is not altogether incompatible with this insight. – Aleksandr Dubinsky Dec 16 '12 at 11:36
  • @AleksandrDubinsky is there a better way to phrase this then? My understanding is that $\mathbb{Q}$ and its extensions, taken as *fields*, are incompatible with this sort of extension. I know that you can extend any commutative ring into a wheel, but I was filing that under "changing enough algebraic rules". – Robert Mastragostino Dec 16 '12 at 17:22
  • @RobertMastragostino Infinite numbers always require an expansion of the rules, but what could be more egregious than calling x "a number" when x+2=x, x*2=x, etc (which is what most formulations of infinity require). My proposal keeps numbers as numbers (and algebra as algebra), and explains things by proposing we've been fudging quite a bit by calling a whole set of numbers simply 0 or simply infinity. I'm not a mathematician, but I'm absolutely sure this is the correct way to understand infinity. I've wanted for a long time to put the idea on a formal footing, but don't know where to start. – Aleksandr Dubinsky Dec 16 '12 at 21:55
  • @AleksandrDubinsky I'm not a mathematician either, so I can't quite say. But what about your proposal stops me from proving $0=1$? If there are multiple $0$s, say $0_1$ and $0_2$, what is $1+0_1$? $1+0_2$? They can't both be $1$, otherwise by the group axioms I could prove that $0_1=0_2$ and we haven't extended anything. I'm not sure how "multiple zeroes" works when any group has a unique additive identity (how are you defining $0$ if not that?). Accepting all the field axioms means that you cannot divide by the additive identity, as far as I'm aware. You'd have to drop a rule somewhere. – Robert Mastragostino Dec 16 '12 at 22:13
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    @Aleksandr: I am confident that whatever idea you have isn't the idea that mathematicians have in mind when they talk about 0. I find it plausible that whatever idea you have in mind could be made rigorous. I am nearly certain that you will never manage to do it if you keep thinking your idea is about 0. –  Dec 16 '12 at 23:36
  • @RobertMastragostino You are right. It must mean there is more than one 1 as well (as well as every number). 1 plus one kind of infinitesimal. 1 plus another. Etc. But that is how it must be. How it is. However, you cannot prove some 1 = some 0 because adding any finite amount of infinitesimals will never add up to something finite (a distance greater than between any real). Adding an infinity of them (some 0 * some infinity) might, but it depends very precisely on what kind of 0 and what kind of infinity were used. – Aleksandr Dubinsky Dec 17 '12 at 07:35
  • @RobertMastragostino Absolute additive identity does have to be discarded as a number. Absolute zero is not a number but a concept, the inverse of "the largest possible infinity" which isn't a number either (hence LPI+1=LPI+2 etc, exactly like a NaN in programming). But this need not be a practical problem. If you restrict your operations to a certain plane of infinitessimals/infinites, you can use use smaller infinitessimals as identity (but should you try to divide by it, you bump up to a higher plain and have to reconsider every place you used it). Algebra becomes relative. – Aleksandr Dubinsky Dec 17 '12 at 07:47
  • @Hurkyl I will. Just tell me where to start. – Aleksandr Dubinsky Dec 17 '12 at 07:49
  • @AleksandrDubinsky I just did prove that "some $1$ equals some $0$" as above. If you want this system to work, you need to know what about your system prevents the above proof (or similar proofs) from working. I think you might want to check out non-standard analysis, since it seems more similar to what you're looking for. Infinitesimals are still distinct from 0 though. – Robert Mastragostino Dec 17 '12 at 15:32
  • @RobertMastragostino Can you expand your proof? I'm not sure what you mean by "do out the fractional division." – Aleksandr Dubinsky Dec 17 '12 at 15:40
  • @AleksandrDubinsky just standard division of fractions: $\frac{\frac 10}{\frac 10}=\frac10\times\frac01=\frac 00$. – Robert Mastragostino Dec 17 '12 at 16:24
  • @RobertMastragostino But there was another $\frac{1}{0}$ in your equation. Nor do I see why if $\frac{0}{0} = \frac{1}{0}$, do you prove $0 = 1$. In fact $\frac{0}{0}$ can equal [anything](http://in.answers.yahoo.com/question/index?qid=20061127061647AAKmCg7), which agrees with my saying that there are different 0s. – Aleksandr Dubinsky Dec 17 '12 at 18:49
  • @RobertMastragostino $\frac{0_j}{0_j} \frac{1_a}{0_j} = \frac{1_a 0_j}{{0_j}^2} \neq 0_j$ (or any 0) Since $\frac{0_j}{0_j} = 1$ (we no longer have ambiguity in this case of $\frac{0}{0}$), it equals precisely $\frac{1_a}{0_j}$, which we defined as j. No second way to solve this. Sorry, you can't disprove the truth. ;) – Aleksandr Dubinsky Dec 17 '12 at 19:08
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    @AleksandrDubinsky A while ago you weren't a mathematician, and now you're talking about "disproving the truth"? You've already acknowledged that you don't know abstract algebra: that's fine, but in that case approaching the subject with some humility (or with the length of this conversation, opening up a separate question) is probably a better approach. What is your definition of zero? You've written down symbols, but I can't judge the correctness of what you've just written if you don't have a definition for what "different $0_j$" are. – Robert Mastragostino Dec 17 '12 at 19:12
  • @AleksandrDubinsky Take a number $m$ such that $a+m=a$ for any $a$, then $m+m=m$ and $a-a=m$, so $ma=(m+m)a=ma+ma$. Subtracting $ma$ gives $m=ma$. $a/m=x$ means $a=mx$, which is never well defined for $x$. Normally $m=0$. If you have *any* element that does this (i.e. functions likes $0$ normally does), you will run into these problems with it. If you don't, then you lose group structure, which is very dramatic. I really think you want infinitesimals here, rather than multiple zeroes. You can't divide by zero, but you can divide by infinitely small numbers to get infinitely large numbers. – Robert Mastragostino Dec 17 '12 at 19:20
  • @RobertMastragostino You're right, what I am really doing is referring to the infinitesimals. I have an idea that you can draw a line between numbers and number-like concepts, and that zero would land on the non-number side of that line (and that in every instance where we've been referring to zero we would have to refer to some infinitesimal instead). But I suppose this isn't the place to discuss that. The biggest favor you could do is suggest to me some reading material about infinitesimals, infinites, and the methods of formally proving things about them. I'd like to become a mathematician. – Aleksandr Dubinsky Dec 17 '12 at 20:01
  • @RobertMastragostino I think my ideas about infinites/infinitesimals are distinct, because the formulations of infinites I've read about all share the property that they don't behave like normal numbers (ie, $\infty_a+1=\infty_a$). I don't believe that has to be the case. Infinites can be numbers. And I'd like to build a theory around that. – Aleksandr Dubinsky Dec 17 '12 at 20:05
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    @AleksandrDubinsky I would start out with some abstract algebra, since you can't rigorously formulate something new unless you know what you're rejecting. "Abstract Algebra" by Dummit and Foote seems to be the standard here. I'm only partway through that book, but it's very good. Chapman's "Real Mathematical Analysis" seems good for real analysis, which you'd need before non-standard analysis. Infinite set theory would be crucial. I would open a question looking for resources on these areas, since I don't know of very many good resources offhand. – Robert Mastragostino Dec 18 '12 at 00:04
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    @AlexsandrDubinsky Check out the [surreal numbers](https://en.wikipedia.org/wiki/Surreal_number). – Jay Jun 19 '13 at 00:12
  • Step from $\frac 00=\frac 10$ to $0=1$ is simply wrong. – Anixx Mar 03 '21 at 23:44
  • @Anixx of course. I'm proving by contradiction that we can't define 1/0 in a way that behaves according to usual arithmetic, and that includes the rule $a/c=b/c \implies a=b$. If division by zero isn't cancellable then it can't be an inverse to multiplication by zero. – Robert Mastragostino Mar 05 '21 at 02:28

The reason why the complex numbers are so special is that they are the end of a chain of questions of the form "How can we solve this equation?" or "What are the roots of this equation?".

We start with the positive integers.

We get the positive rationals by asking "How can we solve $a*x = b$ for $x$ ($a \ne 0$)?"

From the positive rationals, we get all the rationals by asking "How can we solve $a+x=b$ for $x$?"

From the rationals we get the algebraic numbers by asking "How can we solve $\sum a_i x^i = 0$ for x?"

We get the reals from the rationals (one of a number of ways) by asking "What is $\lim_{n \to \infty} a_n$?"

We get the complex numbers from the reals by asking "What are the roots of $\sum a_i x^i = 0$?"

But here it stops. All the roots of $\sum a_i x^i = 0$, where the $a_i$ are complex are complex - no new types of numbers need to be introduced.

For the details of this, I recommend Landau's "Foundations of Analysis".

marty cohen
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    +1 That's the most concise way I've ever seen of summarizing the least upper bound axiom. – Mario Carneiro Dec 16 '12 at 07:25
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    This is very misleading, because you only reach the complex numbers as an endpoint if you restrict yourself to certain questions, impose certain relations on the answers, and make a prior assumption of the arithmetic of those answers. For example, your limit question could have easily lead to the extended real numbers, or to number systems with infinitesimals if asked differently. –  Dec 16 '12 at 08:02
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    Hurkyl's comment is quite correct, in that branching differently could lead to different numbers. My goal was to show why the complex numbers were the end of a reasonable number-generating path. – marty cohen Dec 17 '12 at 00:02
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    In addition, what about things like quaternions? – Joe Z. Feb 15 '13 at 13:41

The complex numbers were an excellent and highly useful abstraction. It is because exploration of them yielded significant and productive mathematical results.

It's the usefulness of an abstraction that matters.

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  • Yes, complex numbers are excellent and useful _in hindsight_. The question is: why won't other extensions be valuable? – Jesse Madnick Dec 16 '12 at 22:51
  • Some have. The extended real numbers greatly simplify real analysis. Projective numbers are invaluable in complex analysis and algebraic geometry. Cardinal and ordinal numbers are of great importance in set theory. –  Dec 16 '12 at 22:55
  • You only find out by exploring. The abstraction should yield results that have some mathematical richness. It is possible they might not be immediately useful, but we all know "usefulness" is a shaky criterion. – ncmathsadist Dec 16 '12 at 23:12
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    @Hurkyl, measure theory aside, do you really think the extended numbers *greatly* simplify analysis? I'm not sure they're such a big deal. – goblin GONE Feb 17 '14 at 11:40
  • @user18921: Well, yes. It's pretty much the same thing as projective number example. –  Feb 17 '14 at 16:39

$x=x+1$ will define the trivial ring as other people have discussed.

However, it also defines an abelian (commutative) group over the set $[0, 1)$ with the operator $+$. The elements of this group are the equivalence classes of real numbers with the same fractional part.

For example, $3/4+1/3=13/12=1+1/12=1/12$        ($=2+1/12 = 3+1/12...$)

The inverse in this group is $x^{-1}=1-x$ The identity element is $0$

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There is a problem with that: not all impossibilities behave as nicely as imaginary numbers. $i \cdot i $ will always result in a nice and solid -1 no matter how you got the two $i$. Meanwhile, ${\infty \over \infty} $ may turn out pretty much damned everything depending on how you obtained these two infinities.

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Another factor not yet mentioned is that in the physical sciences, many real-world systems behave in ways which are beautifully described by complex numbers. For example, Ohm's law dictates that a certain resistance R through which current I is flowing will drop voltage E=IR. Although the law was written to work only with DC resistances, it can also describe the behavior of any fixed network of resistors, capacitors, and inductors, at any fixed frequency; all one needs to do is define the impedance of capacitors and inductors as being imaginary numbers (of one sign for capacitors, and the other sign for inductors), and define real voltages and currents as being in phase with a reference frequency, and imaginary ones as being 90 degrees out of phase. When things are defined in that way, using complex arithmetic to perform the same calculations one would perform using real numbers if one was limited to DC signals and resistors, the rules of complex arithmetic will properly capture the interactions between resistances, capacitances, and inductances, and the ways in which they will affect the phase of the voltage and current waveforms.

I doubt that the first people who invented and worked with complex numbers knew that they would be so useful in the physical sciences, but it turns out that the analysis of many real-world things is greatly facilitated by the use of complex numbers, notwithstanding the "imaginary" name.

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So much discussion...

In fact it's very easy to construct a setting where x=x+1 by making them angles, postulating your "1" to be 360deg and only considering the modulus 360 of each number :)

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    Yes this should be obvious. Or the case when you do +12 and each number is an hour on a clock you will get the same number. – mathreadler Dec 18 '15 at 07:30
  • Indeed. You don't even need to think of the values as angles. If you had $\mathbb{R}$ and stated that for some number $x$, $x=x+1$, then you now have $\mathbb{R}/\mathbb{Z}$. – Rosie F Jul 15 '18 at 11:02

I like marty cohen's answer above but I am going to extend it a bit. In the beginning there were only the natural counting numbers


But then we couldn't solve equations like $x+4=4$ so the concept of zero was grasped and slowly adopted. Quick side note, the concept of zero was not trivial at all and in Europe wasn't even fully accepted until after the dark ages. Anyway, after adding zero, our number system is


But then we couldn't solve equations like $x+4=2$ so negative integers are added and now our number system is


But then we couldn't solve equations like $3x=1$ so then rational number are added so now we have the all numbers which can be written as a ratio of two integers without the denominator being zero. But then we couldn't solve equations like $x^2=2$ and $\cos(x)=0$ so then we added all of the irrationals. This step can be broken into algebraic and transcendental numbers but I am just including both in a single step. Now we have all of the real numbers but now we can't solve equations like $x^2=-1$ so then the imaginary unit $i$ is added to the real numbers preserving all of the old operations like addition, subtraction, multiplication, division, and so on. And just by adding a single number $i$ to the real line gives us the entire complex plane.

Here I do disagree that this is the "end". This is not the end and there are still many "impossible" equations and depending on what you want to solve, how do you "want" the solution to "look", and if the extension will be useful and consistent with the "number system" we have in the past, you can throw in more solutions and keep expanding. An example I can give you is, even with complex numbers, we still cannot solve an equation like $xy-yx=1$ so now we have quaternions (matrices are another number system where "impossible" equations like $AB\neq BA$ hold but quaternions are a direct extension of the complex numbers). The article on wikipedia on quaternions is very nicely written and I would urge you to read at least the history part of it which explains how Hamilton pondered the problem of expanding the complex plane and defining multiplication and division so that it would stay consistent with what we have in the complex plane.

And by the way in case you are interested, after quaternions we do have octonions too.

So to answer your question, yes we can define imaginary numbers for all "impossible" equations but the trick is to try to expand the "old" number system, then to expand it in such a way as to stay consistent with what we have in the "old" number system, and then have the expansion be useful in some way.

Fixed Point
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    Rational numbers were invented much before zero and negatives. – Anixx Mar 03 '21 at 23:26
  • Actually, the logical extension of numbers is adding divergent integrals and series. Besides this, of course, functions, then operators... There is also interval arithmetic. It would be interesting to find an equation or property to which we could not find a solution even by extending the number set. – Anixx Mar 03 '21 at 23:34
  • There is even arithmetic with $0/0$ as an additional element. – Anixx Mar 03 '21 at 23:41

There are two ways in which modifying a system can "break a rule about all numbers": you can break them in the original system, or only in an extension.

  1. Suppose we have the natural numbers with addition, multiplication and ordering. If we add a solution to the equation $x=x+1$ then, if we still want subtraction to work, $0=1$. Since $0$ and $1$ are natural numbers, this breaks the rule about natural numbers which says that $0\ne 1$. If we want multiplication to also behave as before, we get that $x=1\cdot x=0\cdot x=0$ for all (natural) numbers $x$ which means that instead of extending the system, we have collapsed it into a zero. So we have broken pretty much every rule of the original system.

  2. But what if we instead add negative numbers? You said that the rule "$x\le y\implies ax\le ay$ for all $a,x,y$" breaks. And formulated like this, it does, but if you think about it closer, the original statement wasn't about all numbers but all natural numbers. And in this form it still holds. In fact, if you only speak about natural numbers in the new system, no rules have changed. A similar thing happens when you add a solution to $x^2=-1$ in the reals. In these cases we only "break rules" in the extended part of the system.

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We already have such a symbol for x = x + 1 ~ the infinity symbol

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$x = x + 1$ means that $1=0$, as others have pointed out. You can do this, but you don't get anything interesting. What you get is called the trivial ring, and it only contains one element, $0$ (or $1$, since they are the same thing). That's because $x = 1x = 0x = 0$ for any $x$.

As others have pointed out also, you can define $\frac{1}{0} = \infty$. This is actually quite useful as a formality if you are dealing with infinites, but you have to be careful, because there is no consistant way to define things like $\frac{0}{0}$, $\frac{\infty}{\infty}$ or $\infty - \infty$. This is called either the extended real numbers or the extended complex numbers, depending on what other kinds of numbers you allow.

The common theme here is that we define things because they are useful. We "invent" $i = \sqrt{-1}$ because this formality gives us some useful things. In particular, any nonzero complex number $a + bi$ has a multiplicative inverse when defined this way, so that it gives a field. And in this field, very nice properties happen, for example, any polynomial has a root, so that we no longer have equations like $x^2 + 1 = 0$ that don't have solutions.

So what we do is first assert that some object exists by satisfying some equation (like $x^2 + 1 = 0$ or $x = 1/0$ or $x = x + 1$), and then we see what that implies given the other operations that we still want to hold, and what kinds of things can be well-defined and what can't. For $x^2 + 1 = 0$ (i.e., $i$), we get something very nice, a field. For $x = 1/0$, we get something useful, but not as nice (not all operations can be consistently defined on $\infty$, so that it is not a field). For $x = x + 1$, we do get a field, but it is so simple that there is absolutely no use for it, other than as a field that exists for the sake of completeness.

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The problem here is you ignoring the context of x in two states and the fact that it can't be generalized in the triplets of numbers or quadruples and etc that has a nice story to it.

$x\in R$

When one says $x^2 = -1$ is not solvable, He truncates the rest of the statement of the problem and assumes we know what x is, and what does operation $x^2$ do.

The short answer is when $x$ is a one-dimensional real-valued vector isn't possible, it wasn't, and it won't be. But if we change the definition of x to be a two-dimensional vector of real values, with another kind of product it is possible.

$x\in R^2$

We expanded the definition of $x$ to a bigger set and more general product to reach $x*x = -1$.
But with the previous definition, it still isn't possible. some detail follows.

$x \in R^3$

It can't be done, Hamilton famously worked on the problem extending the definition of common summation and product to the triplets of Real numbers for years, his son would ask him every day:"Did you find a solution for dividing the triplets?" and he would answer "No". A found "near" answer to that question for quadruples of Real numbers, that to day are call "Quaternions".

Extension of Common Sum and Product

What we mean by that is, it should form an algebriac construct called a Field1, that is the magic word that would "make it work", if a set with operations doesn't form a Field.

The Result

$R$ and $C$ are both examples of Fields.

There can't be any other $n$ such that $R^n$ creates a Field when $n\ne 1$ and $n\ne 2$.

This result where found after Hamilton found a way to divide quadruples that resembles a field, basically there was a frenzy between the mathematican to find other dimensions that "work", but the quadruple itself doesn't form a Field since by Hamilton's definition two Quaternions $AB=-BA$ that differs from the Real or Complex case where $AB=BA$ and is required to form a field.

The result was found out after a lot of mathematcians tried to find Fields with different dimensions of $R$, then people tried to prove general results about different dimensions of $R$, which in turn gave way to the stated result that there isn't any.

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    Great answer, relates the idea of complex numbers as 'directed magnitudes'. I believe that Hamilton also looked upon Complex numbers as simply pairs of numbers. I was wondering if you could provide any insight on how this 2-dimensional understanding of complex numbers relates to finding zeros of a cubic polynomial? – keithphw Sep 16 '20 at 05:58
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    I should have added, I meant the real zeroes of a cubic polynomial where all 3 exist. Seems strange that we need to journey into the complex numbered second dimension to get 3 real roots existing in the one real number dimension. – keithphw Sep 17 '20 at 01:15
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    @keithphw I added a little bit of story about Hamilton. Is it related to the delta method for cubic equations? – FazeL Sep 18 '20 at 02:47
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    Very interesting additions, thank you! I don't know anything about the delta method I'm sorry but I will look it up. – keithphw Sep 19 '20 at 03:07

If you make $x + 1 = x$ and still want the ring axioms to hold, you end up with the (very uninteresting) ring $\mathbb{Z}/1 \mathbb{Z}$ (just like postulating $x + 3 = x$ gives $\mathbb{Z}/3 \mathbb{Z}$).

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What your asking really depends on the scope of possibility depending on certain assumptions. Taking the squareroot of -1 is, in fact, not an impossibility if the polynomial ring with coefficients the ground field in consideration contains a polynomial with roots ±√(-1), for instance, the polynomial x^2+1 is in said ring. If this is not the case, then we can always extend our field to contain this new element. In particular, if we extend the field of reals by appending ±√(-1), then we get the complex numbers. The fundamental theorem of algebra states that the complex numbers are algebraically closed, meaning all algebraic equations have solutions in the extension field (which is now the complex numbers). The point is that from the point of view of the reals, √(-1) is a new element and not the solution of some "impossible equation". Similarly, there are such things as transcendental extensions. It should be noted that the first of your examples is not such an equation, since unless arithmetic is performed modulo 1, or we are using ordinal arithmetic or some other similar nonstandard arithmetic scheme, it is an absurdity. The second is of course an infinite element, and there are different magnitudes of infinity.

There are a few problems with answering your question in complete generality. First of all, we may require that we know of an equation with no solution in the ambient structure, but if we are just choosing random equations, the extended structure may not have nice behavior, and it also may not be descriptive enough to give real intuition about any new properties. The better approach is of course to first determine the desired properties that we would like the extended structure to have that the ground structure does not, and in this case, we then have to find the new elements giving rise to this desired structure. In general they may not be constructible or may not even exist. This second approach is called a forcing notion. The idea is that we want to minimally expand the current structure to allow for a desired property, while retaining a relative consistency with the ground structure. THIS was the rationality behind the definition of the imaginary number i=√(-1). The desired property was algebraic closure.


Instead of being content with having only 1 number having square equal to -1, we have 3 different such numbers. Then we can build the quaternions where $k^2 = j^2 = i^2 = -1$. And why not have 7 different such numbers? Then we could build the octinions and even more complicated can of course be built in a similar way.

Also we can define many different operations on our numbers. We are used to having addition and multiplication. This is usually the case for an algebraic field. But some types of numbers have only one operation and some can have many many operations defined on them.

Also instead of defining operations with numbers to the left or right on a line, we could define operations on a grid. Each operation having not only two elements in a particular order but a whole set of neighbours as an input.

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i should not be defined as a solution to x^2=-1 (i.e. as an artificial way to find a "solution" to a previously considered "impossible" equation. Complex numbers are defined as orderer pairs (a,b) of real numbers. You define arithmetic rules (+,x,-,/) for such ordered pairs which can be done in a way that makes (a,0) equivalent with the ordinary real number a. Complex numbers (0,b) will be found to satisfy (0,b)^2=(-b^2,0). Voila, we have found a certain kind of numbers (0,b) also called "imaginary" numbers which when squared will give a complex number (-b^2,0) equivalent with the negative real number -b^2. A similar construction of a new type of number x satisfying x=x+1 cannot be made with preservation of established arithmetic rules.


I was browsing the posts flaired Mathematics at r/askscience, and lighted upon the same question.

I replicate GOD_Over_Djinn's comment, as it helps me more than the top-voted one by wazoheat.

Let's go from your $1/m = 0$ identity. Adding 1/m to both sides, we have 1/m + 1/m = 1/m. Now, multiplying both sides by $m$, we have 1 + 1 = 1. So you can see how such a system yields inconsistent results pretty quickly.

To dig a bit further more into where the problem lies, recall the definition of division: $k = a/b$ if k is the unique number such that $kb = a$. So, setting b to 0, we get that whatever a/0 is, it is the unique solution to $0 \times k = a$. Now there are two possibilities: either a=0 or a≠0. If a≠0 then there can be no k that satisfies this equation—it is easy to show that in any sensible number system, 0*anything=0. On the other hand, if a=0, then every k solves this equation. Either way, by defining a number a/0, you're asserting that there is a unique solution to $0 \times k = a$, which is impossible. Assigning a number to be equal to a/0 is inherently contradictory by the definition of division.

As a comment on the difference your idea and defining $i = \sqrt{-1}$, it's a little different. The fundamental properties of number systems1 imply that the equation $0 \times k = 0$ holds for any k in any number system. But it turns out (and it took us awhile to be sure about this) they do not preclude a solution to $x^2 = -1$. In the real numbers there is no solution to that equation, but it turns out that introducing a solution is completely consistent with all the rules about what we call numbers systems. There's nothing in the definition of squaring which forces a number squared to be positive, whereas it is the very definition of division itself which prevents division by 0.

  1. When I talk about "number systems", I mean rings.
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