0

$\frac{(a,b)}{(c,d)} = \left ( \frac{ac + bd}{c^2 + d^2}, \frac{bc - ad}{c^2 + d^2} \right )$ if $(c,d) \ne (0,0)$

I'm dealing with quaternions in my book..complex numbers. how do I derive..or proof (a,b)/(c,d)

Shraddheya Shendre
  • 2,431
terry
  • 71
  • Your question appears to be about complex numbers. Why do you mention quaternions (which are completely different)? – Will Orrick Jul 06 '16 at 01:14
  • quaternions can be viewed as a generalization of the complex numbers; – terry Jul 06 '16 at 01:17
  • 1
    except that quaternions do not satisfy commutativity, so $\frac{x}{y}$ it is ambiguous whether you mean $\frac{1}{y}\cdot x$ or if you mean $x\cdot \frac{1}{y}$. Those could possibly be different things. For now, it is implied that you are working with complex numbers, not quaternions, since you have it written as a fraction. Although you might be studying quaternions later on, you are not right this second and so mentioning them is only going to confuse people. – JMoravitz Jul 06 '16 at 01:20
  • @user3513743 quaternions are an extension used to deal with an undefined operation. That is not an issue here. – user64742 Jul 06 '16 at 01:26
  • 1
    @TheGreatDuck I understand saying that about the complex numbers, but which undefined operation do the quaternions resolve? – Matt Samuel Jul 06 '16 at 01:55
  • @MattSamuel it is a series of statements that are very difficult for me to recall. The answer is somewhere in the comments here: http://math.stackexchange.com/questions/1832047/are-there-important-situations-where-we-study-false-statements-as-if-they-were-t – user64742 Jul 06 '16 at 02:34

1 Answers1

2

$(a,b)/(c,d)$ is defined as $(a,b)$ times the inverse of $(c,d)$.

Let the inverse be $(p,q)$. Then you need $$(c,d)(p,q)=1$$ $$(cp-dq,cq+dp)=(1,0)$$

First if $d=0$, then $c\ne 0$ and $p=1/c$, $q=0$.

Otherwise, $$p=-\frac{cq}{d}$$ $$c\left(-\frac{cq}{d}\right)-dq=1$$ $$-\frac{c^2+d^2}{d}q=1$$ $$q=-\frac{d}{c^2+d^2}$$ $$p=\frac{c}{c^2+d^2}$$

It turns out the above expression also hold for the first case of $d=0$, as a special case.

Hence $$\frac{(a,b)}{(c,d)}=(a,b)\left(\frac{c}{c^2+d^2},-\frac{d}{c^2+d^2}\right)$$ $$=\left(\frac{ac+bd}{c^2+d^2},\frac{bc-ad}{c^2+d^2}\right)$$

velut luna
  • 9,961