4 ms·

I've never liked the framing of "imaginary" numbers as "not reflecting reality" or somehow being less "real" or something. It's just an ordered pair of numbers

I've never liked the framing of "imaginary" numbers as "not reflecting reality" or somehow being less "real" or something. It's just an ordered pair of numbers with a convenient extension of multiplication that preserves nice properties of the reals. I think it's because of the name. If you called them double numbers or paired numbers nobody would say that.

amai 5y ago
The problem is not the term "imaginary", the problem is the term "number. Unlike any other numbers imaginary numbers cannot be ordered by size. So in that sense they are not numbers at all. In fact they represent rotations in two dimensions. They should be better called rotation operators instead of imaginary numbers.
mathnmusic 5y ago
Gauss complained about the name too. He suggested using "Direct, inverse and lateral numbers" instead of "positive, negative and imaginary numbers". Imaginary numbers are not pairs. Complex numbers are. Continuing Gauss, I'd rename "complex numbers" as "planar numbers".
- Natsu 5y ago
  I kinda think of them as 'orthogonal' numbers.
  - carlmr 5y ago
    That I think is the best alternative name I've heard yet. Since you describe two orthogonal dimensions. Also vector numbers could be somewhat of a useful name, since they behave like a 2D vector (or even higher dimensional vectors for e.g. quaternions)
    - ogogmad 5y ago
      "Vector numbers" for something purely 2D seems dodgy to me. There are lots of unital algebras, even in 2 and 4 dimensions, and none of them should hog the name "vector numbers" as they all have equal entitlement to the name.
  - abecedarius 5y ago
    they'll always be twirly numbers to me
- ogogmad 5y ago
  Dual numbers and double numbers are also planar. Don't know how relevant this is, but I've been thinking about a better naming scheme for hypercomplex number systems. I came to it after seeing a paper about the "dual-complex numbers", which are not a straightforward complexification of the dual numbers as one might expect. Hopefully, the scheme should be pronounceable, and without the possibility of confusing it for something else. This town needs law! I'm thinking of asking for suggestions for what it should be exactly.
  - pjbk 5y ago
    Huh? Clifford and Cayley-Dickson Algebras?
    - ogogmad 5y ago
      Those ones. What are you confused about?
- agumonkey 5y ago
  wow, `naming is hard` is way older than I thought
b0rsuk 5y ago
Imaginary numbers are just as imaginary as negative numbers. To disprove my claim, post a photo of -4 cats.
1970-01-01 5y ago
In college, our professor said "We should be saying i is for invisible numbers, because you won't see them unless you know about them. But they're as real as death and taxes."
pcrh 5y ago
Agreed. It's a similar semantic problem to the use of the term "significant" in statistics. P<0.05 as being "significant", while P>0.05 as "non-significant" has a technical meaning that doesn't equate to the common use of the term "significant".
sigstoat 5y ago
> I think it's because of the name. indeed. the name is terrible; causes a lot of folks consternation, or tricks them into thinking strangely about the complex numbers. anyone happen to know of any languages where they're not referred to as "imaginary", or anything that implies they're less "real"?
- orra 5y ago
  Interesting question. I had a scan of the articles in other languages, on Wikipedia. Almost all the Latin alphabet languages use a variant of 'imaginary'. Icelandic appears to be an exception, where it means cross-number. https://en.wiktionary.org/wiki/þvertala#Etymology https://en.wiktionary.org/wiki/þvertala#Etymology (I don't know non-Latin alphabets I can't tell you anything about them.)
dls2016 5y ago
> If you called them double numbers or paired numbers nobody would say that. I'm not sure that's true. Just think of solving x^2+k=0. It's clear that for k<0 you get two solutions and k=0 you get one solution. But when k>0 the graph doesn't touch the x-axis... so why should I expect a "double number" or "paired number" to be the solution? I'm teaching college algebra right now and introduce 'i' algebraically as a solution to x^2+1=0... but then we talk about graphing quadratics and there's no simple connection between the geometry/graph and the algebra. Even if I had the time to talk about the geometry of multiplication and such, it's still a big leap to the graph of z^2+1 and its roots in the complex plane. And it's this leap which, IMO, makes them seem "not real".
- lordnacho 5y ago
  > I'm teaching college algebra right now and introduce 'i' algebraically as a solution to x^2+1=0... but then we talk about graphing quadratics and there's no simple connection between the geometry/graph and the algebra. Take an equation like x^2 -2x + c When C is some large negative number, the roots are symmetric about x = 1. As you increase C, roots are real until C = 1, basically the two roots meet in the middle. When you increase it beyond 1, the roots become complex numbers, but they stay symmetric, they "lift off" from being real into being complex, but still symmetric (conjugates) where 1 is always the real part but the imaginary parts become sqrt(C - 1). You can conveniently imagine the complex plane on top of the real XY plane and the roots go orthonogal to the direction they were going when they were real, from the point where they met. That's kind of how I visualize it for quadratics. For higher orders you cut a complex circle into n equal pieces. Haven't quite figured it out in my head yet.
  - dls2016 5y ago
    My point is that we step off the real line to talk about the points where the function vanishes... but what about the neighborhood of the roots where we're plugging complex numbers into the function and not getting a real number as output? There's a big conceptual leap here.
    - 5y ago

amai 5y ago

The problem is not the term "imaginary", the problem is the term "number. Unlike any other numbers imaginary numbers cannot be ordered by size. So in that sense they are not numbers at all. In fact they represent rotations in two dimensions. They should be better called rotation operators instead of imaginary numbers.

mathnmusic 5y ago

Gauss complained about the name too. He suggested using "Direct, inverse and lateral numbers" instead of "positive, negative and imaginary numbers". Imaginary numbers are not pairs. Complex numbers are. Continuing Gauss, I'd rename "complex numbers" as "planar numbers".

Natsu 5y ago
I kinda think of them as 'orthogonal' numbers.
- carlmr 5y ago
  That I think is the best alternative name I've heard yet. Since you describe two orthogonal dimensions. Also vector numbers could be somewhat of a useful name, since they behave like a 2D vector (or even higher dimensional vectors for e.g. quaternions)
  - ogogmad 5y ago
    "Vector numbers" for something purely 2D seems dodgy to me. There are lots of unital algebras, even in 2 and 4 dimensions, and none of them should hog the name "vector numbers" as they all have equal entitlement to the name.
- abecedarius 5y ago
  they'll always be twirly numbers to me
ogogmad 5y ago
Dual numbers and double numbers are also planar. Don't know how relevant this is, but I've been thinking about a better naming scheme for hypercomplex number systems. I came to it after seeing a paper about the "dual-complex numbers", which are not a straightforward complexification of the dual numbers as one might expect. Hopefully, the scheme should be pronounceable, and without the possibility of confusing it for something else. This town needs law! I'm thinking of asking for suggestions for what it should be exactly.
- pjbk 5y ago
  Huh? Clifford and Cayley-Dickson Algebras?
  - ogogmad 5y ago
    Those ones. What are you confused about?
agumonkey 5y ago
wow, `naming is hard` is way older than I thought

b0rsuk 5y ago

Imaginary numbers are just as imaginary as negative numbers. To disprove my claim, post a photo of -4 cats.

1970-01-01 5y ago

In college, our professor said "We should be saying i is for invisible numbers, because you won't see them unless you know about them. But they're as real as death and taxes."

pcrh 5y ago

Agreed. It's a similar semantic problem to the use of the term "significant" in statistics. P<0.05 as being "significant", while P>0.05 as "non-significant" has a technical meaning that doesn't equate to the common use of the term "significant".

sigstoat 5y ago

> I think it's because of the name. indeed. the name is terrible; causes a lot of folks consternation, or tricks them into thinking strangely about the complex numbers. anyone happen to know of any languages where they're not referred to as "imaginary", or anything that implies they're less "real"?

orra 5y ago
Interesting question. I had a scan of the articles in other languages, on Wikipedia. Almost all the Latin alphabet languages use a variant of 'imaginary'. Icelandic appears to be an exception, where it means cross-number. https://en.wiktionary.org/wiki/þvertala#Etymology https://en.wiktionary.org/wiki/þvertala#Etymology (I don't know non-Latin alphabets I can't tell you anything about them.)

dls2016 5y ago

> If you called them double numbers or paired numbers nobody would say that. I'm not sure that's true. Just think of solving x^2+k=0. It's clear that for k<0 you get two solutions and k=0 you get one solution. But when k>0 the graph doesn't touch the x-axis... so why should I expect a "double number" or "paired number" to be the solution? I'm teaching college algebra right now and introduce 'i' algebraically as a solution to x^2+1=0... but then we talk about graphing quadratics and there's no simple connection between the geometry/graph and the algebra. Even if I had the time to talk about the geometry of multiplication and such, it's still a big leap to the graph of z^2+1 and its roots in the complex plane. And it's this leap which, IMO, makes them seem "not real".

lordnacho 5y ago
> I'm teaching college algebra right now and introduce 'i' algebraically as a solution to x^2+1=0... but then we talk about graphing quadratics and there's no simple connection between the geometry/graph and the algebra. Take an equation like x^2 -2x + c When C is some large negative number, the roots are symmetric about x = 1. As you increase C, roots are real until C = 1, basically the two roots meet in the middle. When you increase it beyond 1, the roots become complex numbers, but they stay symmetric, they "lift off" from being real into being complex, but still symmetric (conjugates) where 1 is always the real part but the imaginary parts become sqrt(C - 1). You can conveniently imagine the complex plane on top of the real XY plane and the roots go orthonogal to the direction they were going when they were real, from the point where they met. That's kind of how I visualize it for quadratics. For higher orders you cut a complex circle into n equal pieces. Haven't quite figured it out in my head yet.
- dls2016 5y ago
  My point is that we step off the real line to talk about the points where the function vanishes... but what about the neighborhood of the roots where we're plugging complex numbers into the function and not getting a real number as output? There's a big conceptual leap here.
  - 5y ago