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> 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
by 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 agoMy 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.
- lordnacho 5y agoOh yeah, Veritasium has some great visualizations of that too. I cant remember what the video was called but it was about fractals and the Newton Raphson method, and he had these input/output complex planes left and right to illustrate it.
- dls2016 5y agoSure we can locally visualize a conformal mapping. But in the historical development of complex numbers, or pedagogically in a college algebra class… this would be putting the cart before the horse. Again why would anyone posit that a pair of numbers suddenly appears when trying to solve x^2+k=0 as k goes from negative to positive?
- lordnacho 5y agoI guess you mean a pair of pairs, as we expected to have two solutions that were normally real numbers. Yeah it's one of those times when the veil lifts and you find out you'd not seen the imaginary part until now. I guess enough has been said about completeness by other commentators, but there's no obvious answer as to why adding just one number solves your problem. Why won't it simply create new problems once you use those new complex numbers as coefficients in an expression? Surprisingly it's all we need.
- dls2016 5y agoI hate to “pull rank” but I have a PhD in mathematics so I’m personally well aware of all the nuances. Again my point is historical/pedagogical. Why should “pair numbers” or “double numbers” be the answer as the OP suggested? It’s not straightforward without getting into conformal mappings. And why is two dimensions enough for third and higher degree equations? Were there a simpler geometric connection, you’d probably have a nicer proof of the Jordan curve theorem… but you don’t. Not sure what you mean by pair of pairs… the OP said that calling imaginary numbers “double numbers” would clear up a lot of issues but that is not clear at all to me.