|
|
Re: Non-Euclidean Arithmetic
Posted:
Sep 10, 2012 9:15 PM
|
|
Joe:
Glad to learn that you were amused by the interpretation that
"multiplication" of a+bi by 7i means multiplying the -b+ai "complex i(mage)
of a+bi ,by 7.
Accordingly, you might also get a chuckle out of this one. The parabolic
function -5(x-10)(x+4) has roots at 3 [+/-] 7 ... and has polynomial
formula -5x^2+30x+200 ... and parabolic formula -5(x-3)^2+155 ...
down-cupped from its vertex at (3,155).
>From the same-vertex, the up-cupped parabola, 5(x-3)^2+155 , has no roots.
So what does it mean when popular textbooks declare that 5(x-3)^2+155 has
"complex roots" 3+7i and 3-7i?
What we now call "complex numbers" were earlier called "imaginary" numbers.
For sure, it is quite complex to try to imagine any "fictitious" roots of
parabolic functions. Indeed, it is mathematically absurd ... because the
5(x-3)^2+155 parabola is not a complex function ...
and the 5(z-3)^2+155 complex function is far too complex to be a parabola.
So is there any way to imag-ine a real-istic meaning behind textbook claims
that 5x^2+30x+200 ...
... which is 5(x-3)^2+155 ... has "complex roots", 3 [+/-] 7i? Sure.
Just recognize that
5(x-3)^2+155 and -5(x-3)^2+155 are (vertical) "images" of each other. If
either has two roots, the other has none.
Thereby, to say that the roots of 5(x-3)^2+155 are 3 [+/-] 7i ... or
better, that they are (3 [+/-] 7)i ... just to say that 3[+/-] 7 are the
roots of the image of 5(x-3)^2+155. The drawing of the double-parabola
eliminates all confusion about what it means to "find imaginary roots" of
quadratics. In this context, "i" means a kind of "inversion."
Just imag-ine how that reduces the complex-ity of solving parabolic
quadratics. ((:-)>
[Yes, dozens of times.]
Cordially,
Clyde
- --------------------------------------------------
From: "Joe Niederberger" <niederberger@comcast.net>
Sent: Monday, September 10, 2012 10:36 AM
To: <math-teach@mathforum.org>
Subject: Re: Non-Euclidean Arithmetic
> Clyde says:
>>How about (3+7i) of those there a+bi complex numbers?
> If you take 3 of those there a+bi complexes, you can have the 7i(mage) of
> them, for free ... today only!
>
> Thanks for joining in the fun! But, seriously, I get your point. Its
> interesting how that little two letter word does so much work. I'm not
> quite sure what to make *of* it though ;-)
|
|
