# Proofs with steps of division and setting things not equal to zero

I am self studying from A Book of Abstract Algebra by Charles C. Pinter. In chapter 3 problem set B problem 4 it asks to show whether $(a,b)\star(c,d)=(ac-bd,ad+bc)$ on the set $\{(x,y) \in \mathbb R^2 | (x,y) \not= (0,0)\}$ is a group. Currently I am in the middle of showing that there is an inverse. This is the equation I am trying to solve to show that there is an inverse.

$(a,b) \star (a',b')=(1,0)=(aa' - bb',ab' + ba')$

I am pretty sure I have the answer for the first entry of the inverse element but I pass through steps with division. In particular I am wondering whether $[1]$ to $[2]$ is a valid step and if it is why. In other words if I exclude $0$ at one step do I have to keep $0$ excluded for the rest of the proof? If its not legal would I just have to split the problem into a couple of cases?

$[1]$ $\frac{aa'-1}{b} =\frac{-ba'}{a}$ where $a\not=0$ and $b\not=0\\$

$[2]$ $a^2 a' - a = -b^2 a'$ (note I did not exclude $0$)

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Are you allowed to notice that this is just the multiplicative group of nonzero complex numbers? In the right solution the only divisor is $a^2+b^2$ which is known to be nonzero because $a$ and $b$ are not both $0$. –  Henning Makholm Jun 28 '12 at 20:15
I would say no because complex numbers have not been introduced in the text and its an exercise to learn how to prove something is a group. –  Disciple of Barney Jun 28 '12 at 20:45
Note: $(a,b)\neq (0,0)$ does not mean $a\neq 0$ and $b\neq 0$. It means either $a\neq 0$ or $b\neq 0$. So I don't think that [1] is valid. –  Arturo Magidin Jun 28 '12 at 20:49
@Paul: If what you're supposed to learn is just how to prove it is a group, then it is allowed simply to pull the correct expression for the inverse out of a hat and then prove that it happens to satisfy the required condition. You don't have to derive it from scratch. The hat might involve lucky guesses and/or previous knowledge of complex numbers; that doesn't affect the validity of the resulting proof. –  Henning Makholm Jun 28 '12 at 20:53
@HenningMakholm While it may not make a proof invalid the question is still an exercise and looking up the answer and showing that the answer works is not very helpful. If I had proved that property of complex numbers (which I guess I currently am) before I would consider using that knowledge. –  Disciple of Barney Jun 28 '12 at 21:04

You want $a'$ and $b'$ such that $aa'-bb'=1$ and $ab'+ba'=0$. In effect you want to show that the system

\left\{\begin{align*} &ax-by=1\\ &bx+ay=0 \end{align*}\right.\tag{1}

always has a solution with $\langle x,y\rangle\ne\langle 0,0\rangle$ provided that $\langle a,b\rangle\ne\langle 0,0\rangle$. You can do this in several ways. If you already know the relevant linear algebra, you can note that $$\det\pmatrix{a&-b\\b&a}=a^2+b^2\ge 0\;,$$

with equality iff $a=b=0$, so $(1)$ always has a unique solution, which clearly (by virtue of the first equation of $(1)$) cannot be $x=y=0$.

Alternatively, you can work out the solution by whatever technique appeals to you and then show directly that it exists and is a non-$\langle 0,0\rangle$ solution whenever $a\ne 0\ne b$.

Your approach, however, does require you to consider separate cases, since it’s entirely possible that one side of $[1]$ is undefined.

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Hint: Let $a=r\cos\theta$, $b=r\sin\theta$, and let $a'=r'\cos\theta'$, $b'=r'\sin\theta'$. (As usual choose $r$, $r'$ positive.) It is easy to see that $r'=1/r$. Now you can identify $\theta'$ by recalling some trigonometric identities.

Or else if you are familiar with complex numbers, there is a much more elegant version of the same thing. We have $(a+bi)(c+di)=ac-bd++(ad+bc)i$. So $$\frac{1}{a+bi}=\frac{a-bi}{(a+bi)(a-bi)}=\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i.$$

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