The operation $ (a,b)(c,d)=(ac-bd,ad+bc) $ on $\Bbb R\times\Bbb R\smallsetminus\{\, (0,0)\,\}$ yields a group

Question

Here is the binary operation $ *: \mathbb{R}\times \mathbb{R} \smallsetminus \{(0,0)\} $ defined by $ (a,b)*(c,d)=(ac-bd,ad+bc) $. My idea is that to show this is a group ($\mathbb{R}\times \mathbb{R} \smallsetminus\{\, (0,0)\,\}, * $), I need to show that $ * $ is well-defined and associative and then show it has an identity and inverse. I am struggling to do the first part. How do I show $ * $ is well-defined (and is the first part required)? Is showing that $ ac-bd=0,ad+bc=0 $ will only be true if $a=b=c=d=0$ sufficient?

Answer 1

How do I show $∗$ is well-defined (and is the first part required)?

This is indeed required, but it is almost always true because it is very obvious very fast if an operation is well-defined.

In general, for a function $f : A \to B$ to be well-defined, these must hold:

1. To each value $a \in A$, the value $f(a)$ must be defined
2. Moreover, that value must lie in $B$ (i.e. $f(a) \in B$ for all $a \in A$)
3. Moreover, that value must be unique, i.e. you cannot send $a \in A$ to two distinct values. Hence if $b,b' \in B$ are such that $b=f(a)$ and $b'=f(a)$, then $b=b'$

So, for instance, some examples:

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• Consider the mapping $$\begin{align*} f : \mathbb{R} &\to \mathbb{R} \\ x &\mapsto f(x) := \frac 1 x \end{align*}$$ This violates criterion $1$ above: $f(0)=1/0$ is not defined.

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• Consider the mapping $$\begin{align*} f : \mathbb{R} &\to \mathbb{R} \\ x &\mapsto f(x) := \sqrt{x} \end{align*}$$ This violates criterion $1$ above: $\sqrt{-1}$ is not defined. Up to certain preference, you could argue $\sqrt{-1}$ is a non-real complex number $i$, and hence criterion $2$ is violated.

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• Consider the mapping $$\begin{align*} f : \mathbb{R} &\to \mathbb{Z} \\ x &\mapsto f(x) := x \end{align*}$$ This identity map on $\mathbb{R}$ should be sending its elements back into $\mathbb{R}$, not $\mathbb{Z}$. For instance, $f(1.5) \not \in \mathbb{Z}$. Criterion $2$ is hence violated.

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• Consider the mapping $$\begin{align*} f : \mathbb{Z} &\to \mathbb{Z} \\ x &\mapsto f(x) := \frac x 2 \end{align*}$$ Not every output is an integer. For instance, $\frac 1 2 = f(1) \not \in \mathbb{Z}$.

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• Consider the mapping $$\begin{align*} f : \mathbb{R} &\to \mathbb{R} \cup \{\infty\} \\ x &\mapsto f(x) := \text{the sum of the digits of $x$} \end{align*}$$ This violates criterion $3$. Since $1=0.999\cdots$, we would get different answers for $f(1)$ based on its representation.

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• Consider the mapping $$\begin{align*} f : \mathbb{Q} &\to \mathbb{Z} \\ \frac p q &\mapsto p+q \end{align*}$$ This likewise violates criterion $3$. Notice that $\frac 1 2 = \frac 2 4$, but these representations map to different values. (Really, a lot of examples of such "multivalued mappings" can be made by making the map dependent on presentation.)

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So, in short: to verify a map is well-defined:

• Show it defines an output for every input
• Show that every output lies in the claimed codomain
• Show that the output is unique for a given input (no input has two or more possible outputs)

Answer 2

Hint $$(a^2 + b^2)(c^2 + d^2) = (a c - b d)^2 + (a d + b c)^2 .$$

Answer 3

It may be obvious that * is well-defined, but it still needs to be shown. Since $\mathbb{R}$ is a field and $a,b,c,d$ are in $\mathbb{R}$, their products $ab$ and $cd$ are well defined elements of $\mathbb{R}$, and so are the sum and difference of these products. Hence $ac-bd$ and $ad+bd$ are elements of $\mathbb{R}$, and so $(ac-bd,ad+bc)$ is in $\mathbb{R}\times\mathbb{R}$. Now you need to show that * is closed, ie that if $(a,b)\ne(0,0)$ and $(c,d)\ne(0,0)$ then $(a,b)*(c,d)\ne(0,0)$

Answer 4

You have to prove that: $$((a\ne0)\vee (b\ne0))\wedge ((c\ne0)\vee (d\ne0)) \Longrightarrow (ac-bd\ne0)\vee (ad+bc\ne0)$$ which is true iff (contrapositive): $$(ac-bd=0)\wedge (ad+bc=0)\Longrightarrow\neg((a\ne0)\vee(b\ne0))\vee \neg((c\ne0)\vee (d\ne0))$$ namely: $$(ac-bd=0)\wedge (ad+bc=0)\Longrightarrow((a=0)\wedge(b=0))\vee ((c=0)\wedge (d=0))$$ which in turn is true by @TravisWillse's hint, because for general propositions $P$ and $Q$, $P\wedge Q\Longrightarrow P\vee Q$.

Answer 5

Well-definedness means, as you said, that $(ad-bc,ac+bd)\ne(0,0)$ if $(a,b)\ne(0,0)$ and $(c,d)\ne(0,0)$.

Suppose $(c,d)\ne(0,0)$. Then we need to prove that $$ ac-bd=0 \qquad ad+bc=0 $$ has only the trivial solution $a=b=0$. This is a linear system with determinant $c^2+d^2$, which is nonzero.

Note that, essentially the same linear system, in the form $$ ac-bd=c \qquad ad+bc=d $$ provides the identity, the only solution being $a=1$ and $b=0$. Also the inverse comes from the linear system $$ ac-bd=1 \qquad ad+bc=0 $$ will give you the inverse of $(c,d)$ and Cramer's rule provides $$ a=\dfrac{c}{c^2+d^2} \qquad b=\dfrac{-d}{c^2+d^2} $$

Answer 6

Q: "Here is the binary operation $ *: \mathbb{R}\times \mathbb{R}$ $\backslash$ $\{(0,0)\} $ defined by $ (a,b)(c,d)=(ac-bd,ad+bc) $. My idea is that to show this is a group ($\mathbb{R}\times \mathbb{R} \backslash (0,0), * $), I need to show that $ * $ is well-defined and associative and then show it has an identity and inverse. I am struggling to do the first part. How do I show $ * $ is well-defined (and is the first part required)? Is showing that $ ac-bd=0,ad+bc=0 $ will only be true if $a=b=c=d=0$ sufficient?"

A: You may embed your group into $GL(2,k)$ ($k$ the real numbers) as follows: Let

$$\rho: (k^2)^* \rightarrow GL(2,k)$$

be defined as follows:

\begin{align*} \rho(a,b):= \begin{pmatrix} a & b \\ -b & a \end{pmatrix}. \end{align*}

This realize your group as a subgroup of $GL(2,k)$. If $k[GL(2,k)]:=k[x_{ij}, 1/D]$ is the coordinate ring of $GL(2,k)$, it follows you group is the "algebraic subgroup" $U$ defined by the equations $x_{11}-x_{22}=0, x_{12}+x_{21}=0$. If you define $G:=Spec(k[GL(2,k)]$, it follows your group is the group of $k$-points $U(k)$ of $U$.

Answer 7

As explained here, these pairs comprise an algebra isomorphic to the complex numbers via $\,(a,b)\sim a+b\,i,\,$ so we can prove your claims more conceptually by simply mimicking the obvious norm-based complex number proof that a product is zero iff some factor is zero, i.e.

$\alpha = (a,b),\ \bar\alpha := (a,-b)\,\Rightarrow\, \alpha\bar\alpha = (|\alpha|,0),\,$ with $|\alpha| := a^2\!+\!b^2\!=\! 0\!\color{#c00}\iff\! \alpha\!=\!(0,0)\!=:\!\bf 0$

$\alpha\beta\!=\! {\bf 0} \overset{\!\times\,\bar\alpha\bar\beta }\Longrightarrow {\bf 0}\!=\! \alpha\bar\alpha(\beta\bar\beta)\!=\! (|\alpha|,0)(|\beta|,0)\!=\! (|\alpha||\beta|,0)\!$ $\!\iff\!\! |\alpha|\,{\rm or}\,|\beta|\!=\!0\!\color{#c00}\iff\! \alpha\,{\rm or}\,\beta\!=\! {\bf 0}$

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Similarly inverting complex numbers can be reduced to inverting reals by the following general

Key Idea multiplying by a conjugate allows us to rationalize (here $\color{#c00}{\textit{real-ize}}$) denominators, lifting "existence of inverses of $\,r\ne 0$" from $\mathbb R$ to $\mathbb C,\,$ i.e. since $\mathbb R$ is a field, $\rm\ 0\ne r\in \mathbb R\, \Rightarrow\, r^{-1}\in \mathbb R,\,$ so $$\rm 0\ne\alpha\in\mathbb C\ \ \Rightarrow\ \ 0\ne{\alpha\bar\alpha} = r\in \mathbb R\ \ \Rightarrow \underbrace{\frac{1}\alpha\, =\, \frac{1\ \bar \alpha}{\alpha\,\bar\alpha}\, =\, \frac{\bar\alpha}r\in\mathbb C}_{\textstyle{\color{#c00}{\textit{real-ize}}\ the\ denominator}}\qquad\qquad$$

Note how the above reduces division by a complex number $\alpha$ to the simpler problem of division by a real number $\,\alpha\bar\alpha$. The prior linked post shows how this idea works more generally, and how it can be viewed as a special case of the method of simpler multiples.

Alternatively we can use inverses do the above: $\,\alpha\beta = 0\Rightarrow 0 = (\bar\alpha\alpha)\beta = r\beta\underset{\times\ r^{-1}}\Longrightarrow \beta = 0$