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

Re:Why don't we define “imaginary” numbers for every “imposs... by heinrichhartman on Monday April 10, 2023 @08:23AM (#35511439)

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 1 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 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.

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Re:Why don't we define “imaginary” numbers for every “imposs... by red_trumpet on Monday April 10, 2023 @08:36AM (#35511538)

> 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.

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Re:Why don't we define “imaginary” numbers for every “imposs... by heinrichhartman on Monday April 10, 2023 @09:30AM (#35512042)

Yes! Fixing

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Re:Why don't we define “imaginary” numbers for every “imposs... by hgsgm on Monday April 10, 2023 @09:17PM (#35519465)

What's wrong with negative degree monomials?

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Re:Why don't we define “imaginary” numbers for every “imposs... by red_trumpet on Wednesday April 12, 2023 @09:21AM (#35539203)

You mean like $x^{-1} = 1/x$? That's called a rational function[1], but not a polynomial, so it's not an element of the polynomial ring[2]. Of course you can also consider the algebra of rational functions, but this is a field[3] (almost by definition: you make every polynomial invertible), which means that modding out anything other than 0 yields the zero ring[4].

[1] https://en.wikipedia.org/wiki/Rational_function

[2] https://en.wikipedia.org/wiki/Polynomial_ring

[3] https://en.wikipedia.org/wiki/Field_(mathematics)

[4] https://en.wikipedia.org/wiki/Zero_ring

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Re:Why don't we define “imaginary” numbers for every “imposs... by civilized on Monday April 10, 2023 @08:27AM (#35511476)

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).

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Re:Why don't we define “imaginary” numbers for every “imposs... by heinrichhartman on Monday April 10, 2023 @09:34AM (#35512071)

fixed.

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Re:Why don't we define “imaginary” numbers for every “imposs... by H8crilA on Monday April 10, 2023 @08:49AM (#35511639)

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

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Re:Why don't we define “imaginary” numbers for every “imposs... by heinrichhartman on Monday April 10, 2023 @09:29AM (#35512039)

yes!

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