Live data from Hacker News

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

math.stackexchange.com

11–20 of 98 posts

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

#11
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.

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

#12
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.

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

#13

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 satisf…

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).

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

#14

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 divisi…

"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?

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

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

#15

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 satisf…

> 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.

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

#16

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 divisi…

"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?

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 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,_%...

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

#18

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 satisf…

Nit: I think everywhere you write x^2-1 you actually meant x^2+1.

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

#19

Earlier quoted context omitted.

"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?

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 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 numbe…

[deleted]
Post reply on HN