[HN Gopher] Why don't we define "imaginary" numbers for every "i...
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Why don't we define "imaginary" numbers for every "impossibility"?
(2012)
Author : curling_grad
Score : 85 points
Date : 2023-04-10 07:46 UTC (15 hours ago)
(HTM) web link (math.stackexchange.com)
(TXT) w3m dump (math.stackexchange.com)
| AstixAndBelix wrote:
| 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 wrote:
| So what did happen?
| shagie wrote:
| There'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_ wrote:
| 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 wrote:
| 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".
| ubj wrote:
| 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
|
| [2]: https://book.sciml.ai/notes/08-Forward-
| Mode_Automatic_Differ...
|
| [3]: https://en.m.wikipedia.org/wiki/Dual_quaternion
| tomstuart wrote:
| If anyone's interested, I wrote up an example application of
| dual numbers in Ruby: https://tomstu.art/automatic-
| differentiation-in-ruby
| contravariant wrote:
| 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 wrote:
| That is why I found algebra to be annoying, unless it was
| algebra from algebraic topology. Ring of polynomials is too
| complicated.
| alli_star wrote:
| [dead]
| ndsipa_pomu wrote:
| 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.
| iamerroragent wrote:
| Riemann Sphere:
|
| https://en.wikipedia.org/wiki/Riemann_sphere
| ndsipa_pomu wrote:
| That'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.
| hansvm wrote:
| They 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 wrote:
| "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.
| hansvm wrote:
| Take a look into division rings as a concept. The usual
| definition for division in rings and fields is via
| multiplicative inverses for some subset of the nonzero
| elements. Not all algebraic spaces have division, but
| that doesn't change what it is, especially from the
| "number theory, fields, and rings" point of view.
|
| Unless you're talking about some higher-order concept?
|
| Edit: For a bit of completeness, what's happening with
| the Riemann Sphere is that the algebraic definition is
| being extended in a way that has some useful analytic,
| topological, and quality-of-life properties, but which is
| no longer wholly compatible with the underlying algebra.
| The algebraic issues are isolated to the extra point at
| infinity, so they're not terrible to work around, but the
| operation in question is a proper extension of the
| underlying algebraic definitions -- much how the gamma
| function in no way can be defined as multiplication of
| integers but is a useful extension of the factorials
| nonetheless.
| syzarian wrote:
| Division is multiplication by the multiplicative inverse.
| Subtraction is addition by the additive inverse. Both
| division and subtraction undo their corresponding
| operation. Multiplying by a (provided it's not zero) is
| undone by dividing by a. Adding a is undone by
| subtracting a.
|
| In a ring the elements form a group under addition and
| thus every element has an additive inverse. The additive
| identity element, let's call it e, has the property that
| e _a = e and a_ e = e. For this reason we use 0 instead
| of e. In a nontrivial ring 0 can't have a multiplicative
| inverse because if it did then every element would be
| equal to the multiplicative identity (which is unique).
| iamerroragent wrote:
| Okay so if you can get to a ring without a multiplicative
| inverse and then applying that operation to the ring
| forms it into a field then wouldn't it be fair to say
| that division is not really the opposite of
| multiplication the same way that subtraction absolutely
| is for addition?
| syzarian wrote:
| The definition of division is multiplication by the
| multiplicative inverse. It may be the case that some
| elements don't have such an inverse but the definition is
| analogous to that of subtraction. The analogy is not
| perfect because every element has an additive inverse
| while not every element had a multiplicative inverse.
| Lichtso wrote:
| You 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 wrote:
| Downside is that now 0[?][?] is undefined so you've
| introduced a new 'impossibility'
| 2muchcoffeeman wrote:
| This 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.
| momentoftop wrote:
| The Isabelle/HOL theorem prover assigns 0 to x/0 for all x,
| without contradiction.
| ndsipa_pomu wrote:
| Thanks - 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 wrote:
| Ah, 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."
| brookst wrote:
| I 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 wrote:
| By specifying that x/y*y is only equal to x if y[?]0, I
| guess?
| momentoftop wrote:
| Exactly.
|
| Functions in these logics are total, so if you want
| division to be a function (and you probably do), it has
| to assign something to division by 0.
|
| It would be acceptable to assign an unspecified object
| from the domain, for which you have no non-trivial
| theorems, and so all your real theorems must have a
| precondition about the denominator being non-zero. But if
| you specify a candidate like 0, you can get some theorems
| which don't have the precondition. Consider:
|
| a/b * c/d = ac/bd.
|
| This now holds even if one of b or d is 0.
| brookst wrote:
| I appreciate the explanation and I'm in no position to
| disagree, but ugh. Seems like it would work just as well
| to define x/0 as 6, or e, or -15. I'm sure that's not the
| case. But as a long time tech person who's always
| considered underflow/overflow to be a hack to get around
| limitations of hardware, it offends be a bit to find
| conditionals in abstract math. Undefined seems cleaner,
| like null, since it implicitly says "don't treat this as
| a normal value that you can operate on".
|
| I suspect the real math people know what they're doing
| more than I do, though.
| momentoftop wrote:
| The theorem a/b * c/d = ac/bd doesn't hold if x/0 = 6,
| though.
|
| The theorem prover HOL Light is a close cousin of
| Isabelle/HOL and doesn't adopt this, and just says that
| x/0 is some unspecified number. You can't prove much
| interesting about it. You can prove, say, that x/0 * 0 =
| 0, but you can't prove whether or not x/0 is, say,
| positive or not.
|
| If you prefer null, there was a logic that allowed for
| undefined terms and partial functions that became the
| basis of the IMPS theorem prover. I found it most notable
| for the fact that it doesn't have reflexivity of
| equality: 1/0 = 1/0 is false in IMPS.
| poizan42 wrote:
| It'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))
| amitport wrote:
| "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?
| ndsipa_pomu wrote:
| The most common definition of division being the inverse of
| multiplication.
|
| if b [?] 0 then the equation a/b = c is equivalent to a = b x
| c. Assuming that a/0 is a number c, then it must be that a =
| 0 x c = 0. However, the single number c would then have to be
| determined by the equation 0 = 0 x c, but every number
| satisfies this equation, so we cannot assign a numerical
| value to 0/0
| amitport wrote:
| Thanks, 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.
| ndsipa_pomu wrote:
| There's a couple of mentions in other comments about the
| Riemann Sphere
| (https://en.wikipedia.org/wiki/Riemann_sphere) which does
| define division by zero, but sacrifices the numbers
| forming a field under addition and multiplication.
| afiori wrote:
| let [?] = 0/0 then 1*[?] = [?] = 0/0 = (0*0)/0 = 0*(0/0) =
| 0*[?] it follows 1 = 0 and thus x = x * 1 = x * 0 = 0 = y * 0
| = y * 1 = y for all x and y
| xigoi wrote:
| This is assuming that Th interacts with arithmetic
| operations the usual way (that is, R [?] {Th} is a field),
| which the person you're replying to did not say.
| afiori wrote:
| True, but the point of giving a "value" to 0/0 is to use
| it somehow.
|
| For example in the context of limits you define a whole
| lot of number like values like 0+ or 0- that are useful
| wrt operations on limits.
|
| I was trying to give an example of how R [?] {Th} has
| almost no advantages compared to just R
| tshaddox wrote:
| Sure, but the whole "problem" we were trying to solve was
| that zero doesn't interact with arithmetic operations the
| usual way.
| henry2023 wrote:
| Division 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 f(x) exists if and only if
| given any e > 0 there exists a d > 0 such that ...)[1].
|
| 2. chose a function f(x) such that on a given "a", f(a) =
| f(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%B
| 5,_%...
| [deleted]
| topaz0 wrote:
| "_____ 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.
| nextaccountic wrote:
| > 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 wrote:
| > 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.
| comte7092 wrote:
| 1 / 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.
| tshaddox wrote:
| Yeah, 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: a/b = c if and
| only if a = c*b and b!=0
|
| And with this infinity thing you just have to make
| essentially the same exception for multiplication and
| infinity: c*b = a if and only if a/b = c
| and b!=infinity and c!=infinity
| Grustaf wrote:
| You 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
| Grustaf wrote:
| On 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 wrote:
| 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.
| antognini wrote:
| An 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.
|
| [1]: https://en.wikipedia.org/wiki/Bring_radical
| rain1 wrote:
| 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 wrote:
| 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 wrote:
| 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 wrote:
| > 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 wrote:
| > 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 wrote:
| 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.
| [deleted]
| legosexmagic wrote:
| 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.
| InfiniteRand wrote:
| 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.
| heinrichhartman wrote:
| For 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: 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.
| H8crilA wrote:
| Nit: I think everywhere you write x^2-1 you actually meant
| x^2+1.
| heinrichhartman wrote:
| yes!
| civilized wrote:
| 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 wrote:
| fixed.
| red_trumpet wrote:
| > 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.
| heinrichhartman wrote:
| Yes! Fixing
| orblivion wrote:
| 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.
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