10 ms·

Why don't we define “imaginary” numbers for every “impossibility”? (2012)

116 points by curling_grad 3y ago

legosexmagic 3y ago
both 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.
rain1 3y ago
we 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 3y ago
  Those 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 3y ago
    The 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 3y 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 3y 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.
      - syzarian 3y ago
        How does your pedantry contribute meaningfully? If something doesn’t behave like 0 in a ring or other algebraic structure then using that label is confusing and simply not done. You are free to use any symbol you want but mathematics is a human endeavor and as such communication is important. Using the symbol 0 signifies something to those with mathematical training. Zero can’t have an multiplicative inverse because anything you call 0 that has an multiplicative inverse makes it behave like something other than zero. So no one would use 0 to describe such an element. In a ring, or abelian group, the symbol 0 is reserved for the additive identity element. Similarly, I could say snkwoo is what most people call a chair. A grammarian would say there is no word snkwoo even though I just defined it. Your original comment was wrong and bad. Instead of just admitting it or moving on you’ve decided to double down and make another bad comment.
        thaumasiotes 3y ago
        I'm having trouble following the argument from your premise "it is a stupid question to ask why I referred to a chair as a chair instead of a snkwoo" to your conclusion "it is not a stupid question to ask why, when we have no answer to a question, we don't just say that we do have one". The answer (to both of those questions!) is, of course, that we could do that, but it wouldn't accomplish anything. Asking the question just means you have no idea what you're saying. Or in other words, it's a stupid question.
ubj 3y ago
One 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 https://en.m.wikipedia.org/wiki/Dual_number [2]: https://book.sciml.ai/notes/08-Forward-Mode_Automatic_Differentiation_(AD)_via_High_Dimensional_Algebras/ https://book.sciml.ai/notes/08-Forward-Mode_Automatic_Differ... [3]: https://en.m.wikipedia.org/wiki/Dual_quaternion https://en.m.wikipedia.org/wiki/Dual_quaternion
- tomstuart 3y ago
  If anyone’s interested, I wrote up an example application of dual numbers in Ruby: https://tomstu.art/automatic-differentiation-in-ruby https://tomstu.art/automatic-differentiation-in-ruby
- contravariant 3y ago
  One 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 3y ago
    That is why I found algebra to be annoying, unless it was algebra from algebraic topology. Ring of polynomials is too complicated.
    - hgsgm 3y 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?
      - lanstin 3y ago
        matrices over reals are ok especially if you keep to SO(n) but you can get very weird maths as polynomial quotients. they do not look to me like they are very similar. complex plane and extensions of all kinds of weird. seems hacky rather than illuminating to me. but then i only really like complex numbers as a field since analytic functions are so nice
alli_star 3y ago
[dead]
ndsipa_pomu 3y ago
Well, 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.
- renewiltord 3y ago
  Yep, 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.
- amitport 3y 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 3y ago
    let 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 3y ago
      This 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.
      - 3y ago
deleted 3y ago
[deleted]
heinrichhartman 3y ago
For polynomial equations, the construction works in quite some generality, and is known as quotient ring: https://en.wikipedia.org/wiki/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: R[X]/(x^2 + 1) = C R[X]/x = R C[X]/(x^2 + 1) = C + C.x R[X]/1 = 0 # Properties - 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.
- civilized 3y ago
  Thanks 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 3y ago
    fixed.
- red_trumpet 3y 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.
AstixAndBelix 3y ago
We 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 3y ago
  So what did happen?
  - aap_ 3y ago
    Square 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 3y ago
      Specifically 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".
  - shagie 3y ago
    There's a good YouTube video on it that includes an epic math battle. Veritasium - How Imaginary Numbers Were Invented - https://youtu.be/cUzklzVXJwo https://youtu.be/cUzklzVXJwo Solving the cubic was a physical thing back then. https://www.maa.org/press/periodicals/convergence/solving-the-cubic-with-cardano-depressing-the-cubic https://www.maa.org/press/periodicals/convergence/solving-th...
InfiniteRand 3y ago
You can invent as many impossible systems as you want, but unless you can do something useful or interesting, no one will pay any attention.
orblivion 3y ago
Negative 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.
devit 3y ago
You 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).

116 points by curling_grad 3y ago

both 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.

rain1 3y ago

we 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 3y ago
Those 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 3y ago
  The 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 3y 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 3y 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.
    - syzarian 3y ago
      How does your pedantry contribute meaningfully? If something doesn’t behave like 0 in a ring or other algebraic structure then using that label is confusing and simply not done. You are free to use any symbol you want but mathematics is a human endeavor and as such communication is important. Using the symbol 0 signifies something to those with mathematical training. Zero can’t have an multiplicative inverse because anything you call 0 that has an multiplicative inverse makes it behave like something other than zero. So no one would use 0 to describe such an element. In a ring, or abelian group, the symbol 0 is reserved for the additive identity element. Similarly, I could say snkwoo is what most people call a chair. A grammarian would say there is no word snkwoo even though I just defined it. Your original comment was wrong and bad. Instead of just admitting it or moving on you’ve decided to double down and make another bad comment.
      - thaumasiotes 3y ago
        I'm having trouble following the argument from your premise "it is a stupid question to ask why I referred to a chair as a chair instead of a snkwoo" to your conclusion "it is not a stupid question to ask why, when we have no answer to a question, we don't just say that we do have one". The answer (to both of those questions!) is, of course, that we could do that, but it wouldn't accomplish anything. Asking the question just means you have no idea what you're saying. Or in other words, it's a stupid question.

ubj 3y ago

One 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 https://en.m.wikipedia.org/wiki/Dual_number [2]: https://book.sciml.ai/notes/08-Forward-Mode_Automatic_Differentiation_(AD)_via_High_Dimensional_Algebras/ https://book.sciml.ai/notes/08-Forward-Mode_Automatic_Differ... [3]: https://en.m.wikipedia.org/wiki/Dual_quaternion https://en.m.wikipedia.org/wiki/Dual_quaternion

tomstuart 3y ago
If anyone’s interested, I wrote up an example application of dual numbers in Ruby: https://tomstu.art/automatic-differentiation-in-ruby https://tomstu.art/automatic-differentiation-in-ruby
contravariant 3y ago
One 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 3y ago
  That is why I found algebra to be annoying, unless it was algebra from algebraic topology. Ring of polynomials is too complicated.
  - hgsgm 3y 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?
    - lanstin 3y ago
      matrices over reals are ok especially if you keep to SO(n) but you can get very weird maths as polynomial quotients. they do not look to me like they are very similar. complex plane and extensions of all kinds of weird. seems hacky rather than illuminating to me. but then i only really like complex numbers as a field since analytic functions are so nice

alli_star 3y ago

[dead]

ndsipa_pomu 3y ago

Well, 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.

renewiltord 3y ago
Yep, 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.
amitport 3y 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 3y ago
  let 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 3y ago
    This 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.
    - 3y ago

deleted 3y ago

[deleted]

heinrichhartman 3y ago

For polynomial equations, the construction works in quite some generality, and is known as quotient ring: https://en.wikipedia.org/wiki/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: R[X]/(x^2 + 1) = C R[X]/x = R C[X]/(x^2 + 1) = C + C.x R[X]/1 = 0 # Properties - 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.

civilized 3y ago
Thanks 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 3y ago
  fixed.
red_trumpet 3y 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.

AstixAndBelix 3y ago

We 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 3y ago
So what did happen?
- aap_ 3y ago
  Square 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 3y ago
    Specifically 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".
- shagie 3y ago
  There's a good YouTube video on it that includes an epic math battle. Veritasium - How Imaginary Numbers Were Invented - https://youtu.be/cUzklzVXJwo https://youtu.be/cUzklzVXJwo Solving the cubic was a physical thing back then. https://www.maa.org/press/periodicals/convergence/solving-the-cubic-with-cardano-depressing-the-cubic https://www.maa.org/press/periodicals/convergence/solving-th...

InfiniteRand 3y ago

You can invent as many impossible systems as you want, but unless you can do something useful or interesting, no one will pay any attention.

orblivion 3y ago

Negative 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.

devit 3y ago

You 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).