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.
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.
There are instances that make it useful, but the extended real number line isn’t used heavily in practice.
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
"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.
There are multiple ways to define what division by zero means. Which definition leads to this outcome? How?
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
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,_%...
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.
Is the short answer it's not parsimonious or useful?
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
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...
Complex numbers being equivalent to R[X]/(1+X^2) and dual numbers being equivalent to R[X]/(X^2).
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
- 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.
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.
1/0 is maybe a bit trickier and leads you to invent projective spaces.
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).
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...
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.
It's a stupid question, but it's not related to your response.
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.