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

by curling_grad on 4/10/23, 7:46 AM with 94 comments

by ndsipa_pomu on 4/10/23, 12:07 PM
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.
by ubj on 4/10/23, 11:13 AM
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
by heinrichhartman on 4/10/23, 12:23 PM
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.
by InfiniteRand on 4/10/23, 12:55 PM
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.
by legosexmagic on 4/10/23, 12:18 PM
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.
by orblivion on 4/10/23, 10:27 PM
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.
by devit on 4/11/23, 1:20 PM
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).
by AstixAndBelix on 4/10/23, 12:38 PM
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
by rain1 on 4/10/23, 9:38 AM
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.