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Problem 47 (Can We Use Division To Create A Group)

Let $G=\mathbb{R}$ and $H=\mathbb{R}\setminus \{0\}$.

  1. Show that division $a\div b$ is not a
    binary operation
    Definition (Binary Operation)

    Let $G$ be a set. A binary operation on $G$ is a way of combining two elements of $G$ to obtain a new element in $G$. Formally, we just say that a binary operation $*$ is function $*:G\times G\to G$, and we use the notation $a*b$ to represent the function notation $*(a,b)$.

    on $G$.
  2. Show that division $a\div b$ is a binary operation on $H$.
  3. Since division is a binary operation on $H$, determine if $(H,\div)$ is a
    group
    Definition (Group)

    Let $G$ be a nonempty set, and let $*$ be a binary operation on $G$, which means for every $x,y\in G$ we have $x*y\in G$ $\textbf{[Closure]}$. The structure $\mathbb{G} = (G,*)$ is called a $\textdef{group}$ if the following hold.

    1. $\textbf{[Associativity]}$ For all $x,y,z\in G$ we have $(x* y)* z = x* (y* z)$.
    2. $\textbf{[Identity]}$ There is an $e\in G$ such that for all $x\in G$ we have $x * e = e* x = x$.
    3. $\textbf{[Inverses]}$ For all $x\in G$ there is a $y\in G$ such that $x* y = y* x = e$.

    We usually simply write $G$ when referring to the entire structure $\mathbb{G}=(G,*)$. The element $e$ from the second point is called the $\textdef{identity}$. The element $y$ from the third point is called the $\textdef{inverse}$ of $x$ and is usually denoted $x^{-1}$. One often simply writes $xy$ in place of $x*y$, and for every positive integer $n$, we'll write $x^n$ as shorthand for $x* x* \cdots * x$ ($n$ times).

    .
    • Does $e=1$ satisfy the property of being an identity?
    • If $x\in H$, find an inverse $x^{-1}\in H$ or explain why none exists.
    • Is the operation $\div$ associative?

Solution

Let $G=\mathbb{R}$ and $H=\mathbb{R}\backslash \{0\}$.

1. We must show that division $a\div b$ is not a binary operation on $G$. Let $a=5$ and $b=0$. It is shown that $a\div b$ is undefined. This means $a\div b\not\in G$. Thus, $a\div b$ is not a binary operation on $G$.

2. We must now show that $a\div b$ is a binary operation on $H$. Let $a,b\in H$. Since $a,b\in\mathbb{R}$ we know that $a\div b\in\mathbb{R}$. This is because the division $a\div b$ of any two real numbers $a$ and $b$ results in a real number $c$ as long as $b\neq0$. Since $0\not\in H$, then $b\neq 0$. Thus, it is shown that $a\div b\in H$ proving that $a\div b$ is a binary operation on $H$.

3. It will now be shown that $(H, \div )$ is not a group. Let $a=8$, $b=4$, and $c=2$. Since $a,b,c\in H$ it will be shown that $(a\div b)\div c\neq a\div (b\div c)$. We compute,

$(8\div 4)\div 2=(2)\div 2=1$.

Similarly,

$8\div (4\div 2)=8\div (2)=4$.

Thus,

$(8\div 4)\div 2\neq 8\div (4\div 2)$.

This means that associativity does not hold and therefore proves that $(G,\div )$ is not a group.

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