Why don't we define “imaginary” numbers for every “impossibility”? (2012)
116 points by curling_grad 2 years ago | 94 comments- ndsipa_pomu 2 years agoWell, we can define mathematical objects for every gap (impossibility), but most of them will turn out to be inconsistent with our existing mathematical objects, and thus not very useful or interesting. I'd consider that mathematics is the study of consistency and what can be discovered using the simplest possible starting points (axioms).
The classic case would be if mathematicians wanted to assign a value to division by zero. It turns out that if you do allow that to take a value, then it becomes possible to "prove" that any number is equal to any other number. Quite simply, it makes maths less interesting to allow that, but instead having division by zero be undefined appears far more useful/interesting.
- nextaccountic 2 years ago> The classic case would be if mathematicians wanted to assign a value to division by zero. It turns out that if you do allow that to take a value, then it becomes possible to "prove" that any number is equal to any other number. Quite simply, it makes maths less interesting to allow that, but instead having division by zero be undefined appears far more useful/interesting.
There are multiple extensions to the real numbers that allow division by zero. One is a real projective line, which has only one infinity so that 1 / 0 = -1 / 0 = infinity
https://en.wikipedia.org/wiki/Real_projective_line
Another is the extended real number line which has positive infinity and negative infinity, so 1 / 0 = +infinity and -1 / 0 = -infinity and they are different from each other
https://en.wikipedia.org/wiki/Extended_real_number_line
Those are all perfectly fine but they still can't define 0 / 0, which is a harder problem.
- marcosdumay 2 years ago> There are multiple extensions to the real numbers that allow division by zero.
Well, the gotcha is that they redefine the operations so that none of addition, subtraction, multiplication or division are total. Those operations just break in a different number than zero.
- nextaccountic 2 years agoAddition can be total if you have a single infinity, just make infinity + n = infinity for all n
Subtraction, multiplication and division is harder
But making further operations partial isn't that of a big deal; in fields the division is already partial due to division by zero not being defined.
- nextaccountic 2 years ago
- comte7092 2 years ago1 / 0 = +infinity Implies that 0 * +infinity = 1, so it does run into make of the same issues.
There are instances that make it useful, but the extended real number line isn’t used heavily in practice.
- Grustaf 2 years agoYou might not have much use for the real projective line when tallying up prices in the grocery store, but projective geometry is definitely very useful. https://en.wikipedia.org/wiki/Projective_geometry
- tshaddox 2 years agoYeah, I don't really see what this gets you. With basic real number division you have to make the exception for zero in the definition:
And with this infinity thing you just have to make essentially the same exception for multiplication and infinity:a/b = c if and only if a = c*b and b!=0c*b = a if and only if a/b = c and b!=infinity and c!=infinity - hgsgm 2 years agoWhat is the "issue"?
"/" means "* reciprocal of".
If "infinity" is defined as "reciprocal of 0", what is the problem?
Yes it is an exception to 0*n=0.
It won't work in every setting, but it works in some settings, like inversive geometry.
- scotty79 2 years agoIn normal math 1/0 is undefined but in a math where 1/0 is defined to be inf the 0*inf is still undefined.
- Grustaf 2 years ago
- marcosdumay 2 years ago
- iamerroragent 2 years agoRiemann Sphere:
- Lichtso 2 years agoYou don't even need the complex part for this. You can do the infinity-projection trick on the real numbers alone as well: https://en.wikipedia.org/wiki/Projectively_extended_real_lin...
A similar trick (point at infinity or ideal point) is used in projective geometry to distinguish between directions (vectors) and places (points) by using coordinates only: https://en.wikipedia.org/wiki/Projective_geometry
But if you actually want to do calculations with infinities and infinitesimals the surreal numbers might be better suited for that: https://en.wikipedia.org/wiki/Surreal_number
- zeroonetwothree 2 years agoDownside is that now 0⋅∞ is undefined so you’ve introduced a new ‘impossibility’
- zeroonetwothree 2 years ago
- hansvm 2 years agoThey could have been more precise, but they probably shouldn't have to in the space of a comment. The Riemann Sphere defines a value for the expression x/0, and it's often useful, but it fails to uphold the most important property division should have -- that it undoes multiplication. Division by 0 (with some assumptions about not being in a trivially small space and how those operations behave with respect to addition) does lead to contradictions in that latter sense.
- iamerroragent 2 years ago"but it fails to uphold the most important property division should have -- that it undoes multiplication"
I'm not sure I follow that as it's most important property. I'm not sure if division could even be defined as an operation that undoes multiplication.
Number theory, fields, and rings I believe make it clear while subtraction and addition can be viewed as the same function; multiplication and division cannot.
Apologize if that's not clear as to why that is; it's been a while since I read up on those being defined.
However I recommend One, Two, Three: Absolutely Elementary Mathematics by David Berlinski that gives in my opinion pretty good layman understanding of these nuances and number theory.
- iamerroragent 2 years ago
- ndsipa_pomu 2 years agoThat's a good example of where defining division by zero leads to interesting maths, but it ends up sacrificing some of the usual rules of arithmetic, so it comes down to a choice of which is more useful in the relevant circumstance.
- 2muchcoffeeman 2 years agoThis just goes to show that you really have to be careful when slinging out math facts. I've done some under grad maths and the only line on that page that I understand is
"The extended complex numbers are useful in complex analysis because they allow for division by zero in some circumstances, in a way that makes expressions such as 1 / 0 = ∞ 1/0=\infty well-behaved."
It clearly does not satisfy a primitive understanding of 1/0.
- hgsgm 2 years ago1/0 is the limit of x/y as x approaches 1 and y approaches 0. It works fine if you choose to put a a point there called \infinity, with an appropriate notion of nearness.
- hgsgm 2 years ago
- Lichtso 2 years ago
- antognini 2 years agoAn example where this does work quite nicely has to do with Bring radicals or "ultraradicals [1]. One of the most important results from Galois theory is that the quintic equation has no solution using standard radicals. But the introduction of "Bring radicals" allows quintic equations to be formally solved. As far as I'm aware though, Bring radicals only work for quintic equations in general and don't work for 6th order or higher polynomials, so your bang for the buck is a somewhat limited.
- amitport 2 years ago"it becomes possible to "prove" that any number is equal to any other number."
There are multiple ways to define what division by zero means. Which definition leads to this outcome? How?
- afiori 2 years agolet ϴ = 0/0 then 1*ϴ = ϴ = 0/0 = (0*0)/0 = 0*(0/0) = 0*ϴ it follows 1 = 0 and thus x = x * 1 = x * 0 = 0 = y * 0 = y * 1 = y for all x and y
- xigoi 2 years agoThis is assuming that Θ interacts with arithmetic operations the usual way (that is, ℝ ∪ {Θ} is a field), which the person you're replying to did not say.
- xigoi 2 years ago
- ndsipa_pomu 2 years agoThe most common definition of division being the inverse of multiplication.
if b ≠ 0 then the equation a/b = c is equivalent to a = b × c. Assuming that a/0 is a number c, then it must be that a = 0 × c = 0. However, the single number c would then have to be determined by the equation 0 = 0 × c, but every number satisfies this equation, so we cannot assign a numerical value to 0/0
- amitport 2 years agoThanks, this definition does seem problematic. In any case, it is not the only possible definition and in a/0=c, c does not have to be defined as a real number. We can define it as similarly to complex number with new rules that do not collide with existing reals.
- amitport 2 years ago
- henry2023 2 years agoDivision by zero is not defined anywhere on math.
The closest thing you'd get to it is to
1. define a limit (lim x->a of ƒ(x) exists if and only if given any ε > 0 there exists a δ > 0 such that ...)[1].
2. chose a function ƒ(x) such that on a given "a", ƒ(a) = ƒ(a)/0.
3. prove that the limit exists and is finite.
Now if we defined division by zero it would look like this:
Axiom: For every element x of the real numbers there exists a x' in the real numbers such that x/0 = x'
I advise you to play with this new "rule" to see if it leads to something interesting. Hint: try to prove that 1/0 = 2/0
[1]: https://en.wikipedia.org/wiki/Limit_of_a_function#(%CE%B5,_%...
- topaz0 2 years ago"_____ is not defined anywhere in math"
is a kind of sentence that is almost never true, and even if it were, it would be impossible to prove that someone hadn't jotted a valid definition on a napkin somewhere. In this case it is certainly not true (as others have mentioned: https://en.wikipedia.org/wiki/Riemann_sphere ). Now, specifying a definition for division by zero does require you to be careful about how the other operations extend to this new number, but there are perfectly consistent (and useful!) ways to do so.
- 2 years ago
- topaz0 2 years ago
- afiori 2 years ago
- momentoftop 2 years agoThe Isabelle/HOL theorem prover assigns 0 to x/0 for all x, without contradiction.
- ndsipa_pomu 2 years agoThanks - I was not aware that theorem provers often allow "division" by zero.
Looking at https://xenaproject.wordpress.com/2020/07/05/division-by-zer... I see that they don't use mathematical division, but define a slightly different operator with an additional condition for handling zero. This appears to be far more convenient for theorem provers.
The trade-off would be that "division" is no longer the inverse of multiplication.
- momentoftop 2 years agoAh, thanks for the link. I suggested the reason that Isabelle/HOL does this is because it requires total functions and you don't have a convenient way to do refinement types. But that's not an adequate explanation, because Lean does allow such refinements, but it still turns out to be inconvenient for division.
I will note that setting a - b = 0 for a <= b is pretty standard, and is often called "partial subtraction."
- momentoftop 2 years ago
- brookst 2 years agoI believe you but that’s kind of mind blowing. How do they avoid the seemingly-obvious corollary that 0*0 = X, for all values of X? That is, just multiplying both sides of “x/0 = 0” by zero.
- xigoi 2 years agoBy specifying that x/y*y is only equal to x if y≠0, I guess?
- xigoi 2 years ago
- poizan42 2 years agoIt's not making a multiplicative inverse of 0 exist though, it just defines a '/' operator that is slightly different from our usual one (i.e. a/b = a*b^(-1))
- ndsipa_pomu 2 years ago
- Grustaf 2 years agoOn the contrary, the extensions can be very useful and interesting. You do typically have to sacrifice something, like commutativity in the case of quaternions, but it will often be worth it.
- renewiltord 2 years agoYep, an extension is only interesting if it is a true extension, i.e. retains the properties of the thing being extended. So complex numbers are interesting as an an extension of reals since reals are isomorphic to the subring. Likewise with quaternions and reals / complex numbers.
- JumpCrisscross 2 years ago> we can define mathematical objects for every gap (impossibility), but most of them will turn out to be inconsistent with our existing mathematical objects
Is the short answer it's not parsimonious or useful?
- paulddraper 2 years agoYou're right, though your example is weak.
Infinity and negative infinity can make a lot of sense. You can even allow imaginary infinities.
The caveats:
* You cannot multiply zero by infinity, or divide infinity by infinity
* You cannot add different infinities
- nextaccountic 2 years ago
- ubj 2 years agoOne interesting case of this is the concept of dual numbers [1], where you have the symbol \epsilon !=0 but (\epsilon)^2 = 0.
It seems contradictory, but the resulting theory is very useful for automatic differentiation [2] and for mechanics (dual quaternions) [3].
[1]: https://en.m.wikipedia.org/wiki/Dual_number
[2]: https://book.sciml.ai/notes/08-Forward-Mode_Automatic_Differ...
- contravariant 2 years agoOne thing that is interesting to note is that both dual numbers and imaginary numbers arise as quotient of the polynomial ring.
Complex numbers being equivalent to R[X]/(1+X^2) and dual numbers being equivalent to R[X]/(X^2).
- lanstin 2 years agoThat is why I found algebra to be annoying, unless it was algebra from algebraic topology. Ring of polynomials is too complicated.
- hgsgm 2 years ago"too complicated" is a weird way to say "provides a concise and consistent way to model superficially diverse phenomena and show how similar they really are" .
So you also find matrices too complicated?
- hgsgm 2 years ago
- lanstin 2 years ago
- tomstuart 2 years agoIf anyone’s interested, I wrote up an example application of dual numbers in Ruby: https://tomstu.art/automatic-differentiation-in-ruby
- contravariant 2 years ago
- heinrichhartman 2 years agoFor polynomial equations, the construction works in quite some generality, and is known as quotient ring: https://en.wikipedia.org/wiki/Quotient_ring
Given any polynomial P (e.g. x^2 + 1) over a filed F (e.g. reals) we can form: `R = F[X]/P`
This is an algebraic "set" that supports addition, substraction, multiplication and has 0,1 but not division in general. Elements are elements of F and a new symbol X that satisfies "P(X) = 0".
Examples:
# PropertiesR[X]/(x^2 + 1) = C R[X]/x = R C[X]/(x^2 + 1) = C + C.x R[X]/1 = 0- If the polynomial P is invertible, i.e. has degree 0 and is not zero, then the resulting ring is zero R[X]/P = 0. This is what happens in the example x = x-1 (which corresponds to P = x - 1 - x = -1).
- If the polynomial P has degree 1 (i.e. P=aX+b), then the equation P=0 is equivalent to x=-b/a, representing an element already present in R, hence the ring R[X]/P is equal to R.
- If the polynomial P is irreducible (i.e. not a product of two proper polynomials) then the quotient R[X]/P is a field. This happens in the case R[x]/(x^2 + 1) which results in the complex numbers.
- If the polynomial P is a product of two polynomials P1,P2 which don't have common divisors, then R[X]/P = R[X]/P1 + R[X]/P2, this happens in the case that C[X]/(x^2+1), since P = x^2 + 1 factors as (x+i)*(x-i) in C. The equivalent result for integers is known as Chinese Remainder Theorem.
- red_trumpet 2 years ago> If the polynomial P is invertible, i.e. has degree 1
Should be degree 0: only constant polynomials are invertible. E.g. x+1 is not invertible, and modding it out doesn't result in the zero ring.
The example is a bit confusing, because $x=x+1$ is equivalent to $0=1$, which has degree 0.
- hgsgm 2 years agoWhat's wrong with negative degree monomials?
- red_trumpet 2 years agoYou mean like $x^{-1} = 1/x$? That's called a rational function[1], but not a polynomial, so it's not an element of the polynomial ring[2]. Of course you can also consider the algebra of rational functions, but this is a field[3] (almost by definition: you make every polynomial invertible), which means that modding out anything other than 0 yields the zero ring[4].
[1] https://en.wikipedia.org/wiki/Rational_function
[2] https://en.wikipedia.org/wiki/Polynomial_ring
- red_trumpet 2 years ago
- heinrichhartman 2 years agoYes! Fixing
- hgsgm 2 years ago
- civilized 2 years agoThanks for this comment! Quick note - for clarity and conformity with standard notation, it would be good to have parentheses around the denominators of those ring quotients (in those cases like x^2 - 1 where they contain multiple additive terms).
- heinrichhartman 2 years agofixed.
- heinrichhartman 2 years ago
- H8crilA 2 years agoNit: I think everywhere you write x^2-1 you actually meant x^2+1.
- heinrichhartman 2 years agoyes!
- heinrichhartman 2 years ago
- red_trumpet 2 years ago
- InfiniteRand 2 years agoYou can invent as many impossible systems as you want, but unless you can do something useful or interesting, no one will pay any attention.
- legosexmagic 2 years agoboth of these are reasonable. if you have an `x` such that `x + n = x` implies that `n = 0`. (assuming x still has an additive inverse) in other words you just invented modular arithmetic which is a very reasonable thing to invent.
1/0 is maybe a bit trickier and leads you to invent projective spaces.
- orblivion 2 years agoNegative numbers are sort of imaginary to begin with come to think of it. Actually I think I'm getting flashbacks now to my childhood when my older brother blew my mind with this concept.
- 2 years ago
- devit 2 years agoYou can do that, but there's a tradeoff of losing properties that otherwise hold.
For example, by adding the imaginary numbers, there is no longer an ordering compatible with addition and multiplication (ordering compatible with multiplication means that z > 0 and x > y implies x * z > y * z: assuming that, if 0 < i, then 0 = 0 * i < i * i = -1, absurd, or if 0 > i and thus 0 < -i, then 0 = 0 * -i < -i * -i = -1, absurd).
You can certainly add a number x such that x = x + 1 (e.g. what is commonly called an infinity or NaN), but that implies no longer having additive left inverses assuming you keep associativity of addition and 0 != 1 (since otherwise 0 = -x + x = -x + (x + 1) = (-x + x) + 1 = 0 + 1 = 1).
- AstixAndBelix 2 years agoWe didn't invent 'i' to "solve sqrt(-1)". This is an extremely common misconception about maths and how it progressed that unfortunately people get led into believing by lazy teachers every day
- rtpg 2 years agoSo what did happen?
- aap_ 2 years agoSquare roots of negative numbers came up when solving cubic equations, even if the final solutions were all real. This meant the square root of a negative number was not something nonsensical the way you might claim for x^2 = -1, but actually...real in some sense.
- caf 2 years agoSpecifically I believe it involved a geometric construction for solving the cubics, which in some cases could not find a solution unless you allowed a square with "negative area".
- caf 2 years ago
- shagie 2 years agoThere's a good YouTube video on it that includes an epic math battle.
Veritasium - How Imaginary Numbers Were Invented - https://youtu.be/cUzklzVXJwo
Solving the cubic was a physical thing back then. https://www.maa.org/press/periodicals/convergence/solving-th...
- aap_ 2 years ago
- rtpg 2 years ago
- alli_star 2 years ago[dead]
- rain1 2 years agowe do, it's called the algebraic numbers!
every polynomial with algebraic coefficients has 'n' solutions (counted with multiplicity)!
so e.g. x^121 + sqrt(7)x^9 + fithroot(22)x^7 + (1+i)x^3 + 22/7 = 0 has 121 solutions. and they're all algebraic numbers: nothing weird like pi in there.
- thaumasiotes 2 years agoThose are all just normal imaginary numbers. The question is why, when we can't answer a question, we don't just invent a symbol, say it's the answer to the question, and call it a day.
It's a stupid question, but it's not related to your response.
- syzarian 2 years agoThe question has 300+ upvotes. That’s a proxy for how “good” it is. A person is curious about an aspect of mathematics and posed a well stated question. It is not a stupid question. From their perspective mathematicians appear to do something and they wonder why it can’t be done in other situations. Such a question is the basis of understanding. It is by wondering such things that enables one to gain true understanding of a topic.
Most questions asked by beginners in an area are “stupid” and few as insightful as this one. I’ve taught mathematics at a community college for 20 years and I would be delighted to have been asked this. Usually questions are mundane like, “Why did you add x to both sides?”. Here the person is trying to understand what mathematicians do, what the basis of expanding a number system really involves. This is a fantastic question.
Peoples’ curiosity ought not be labeled as stupid.
- zvmaz 2 years ago> Peoples’ curiosity ought not be labeled as stupid.
Correct. That is why I feel more comfortable asking "stupid" questions to chatGPT. I clarified a lot of concepts in economics through repeatedly asking questions about each concept that pop up in its answers and trying to push it to the limits of what can be defined, explained, etc. One cannot be sure of the truthfulness or soundness of the answers, but they may help.
- thaumasiotes 2 years ago> It is not a stupid question. From their perspective mathematicians appear to do something and they wonder why it can’t be done in other situations.
I mean, you've already gotten it wrong. This can be done in other situations. Where it isn't done, it isn't done because doing it is pointless, not because there's some bar to giving names to opaque labels.
- zvmaz 2 years ago
- syzarian 2 years ago
- thaumasiotes 2 years ago