Question

Prove that the set of nonzero real numbers is a group under the operation$\mathbf{\ast }$ defined by$\mathit{a}\mathbf{\ast }\mathit{b}\mathbf{=}\mathbf{\{}\begin{array}{c}\mathbf{a}\mathbf{b}\mathbf{,}\mathbf{a}\mathbf{>}\mathbf{0}\\ \mathbf{a}\mathbf{/}\mathbf{b}\mathbf{,}\mathbf{a}\mathbf{<}\mathbf{0}\end{array}$.

Short answer

It is proved that$ℝ-\left\{0\right\}$ forms a group under the operation $\ast$.

Step-by-step solution

Group

A non-empty set$\mathit{G}$ is said to form a group under binary composition$\mathbf{\ast }$ if,

• $\mathit{G}$is closed under$\mathbf{\ast }$ .
• $\mathbf{\ast }$is associative.
• There exists an element$\mathit{e}$ in$\mathit{G}$ such that$\mathit{e}\mathbf{\ast }\mathit{a}\mathbf{=}\mathit{a}\mathbf{\ast }\mathit{e}\mathbf{=}\mathit{a}$ for all$\mathit{a}\mathbf{\in }\mathit{G}$ .
• For each$\mathit{a}\mathbf{\in }\mathit{G}$ element ,there exists an element ${\mathit{a}}^{\mathbf{\text{'}}}$in$\mathit{G}$ such that,${\mathit{a}}^{\mathbf{\text{'}}}\mathbf{\ast }\mathit{a}\mathbf{=}\mathit{a}\mathbf{\ast }{\mathit{a}}^{\mathbf{\text{'}}}\mathbf{=}\mathit{e}$

.

Proof

1. Let $a,b\in ℝ-\left\{0\right\}$be arbitrary, then, $a\ast b=abora\ast b=\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$b$}\right.\Rightarrow a\ast b\in ℝ-\left\{0\right\}$, whether$a>0ora<0$(since$a\ne 0,b\ne 0$ ),which shows that $ℝ-\left\{0\right\}$is closed under
2. Let $a,b,c\in ℝ-\left\{0\right\}$then,

$(a\ast b)\ast c=\{\begin{array}{c}a\ast \left(bc\right),b>0\\ a\ast \left(\frac{b}{c}\right),b<0\end{array}=\{\begin{array}{c}abc,a>0andb>0\\ \frac{a}{bc},a<0andb>0\\ \frac{ab}{c},a>0andb<0\\ \frac{ac}{b},a<0andb<0\end{array}$

$(a\ast b)\ast c=\{\begin{array}{c}\left(ab\right)\ast c,a>0\\ \left(\frac{a}{b}\right)\ast c,a<0\end{array}=\{\begin{array}{c}abc,a>0andb>0\\ \frac{ab}{c},a>0andb<0\\ \frac{ac}{b},a<0andb<0\\ \frac{a}{bc},a<0andb>0\end{array}$

Which shows that,$(a\ast b)\ast c=a\ast (b\ast c)$ , and hence, is associative.

3. Now, since, $1\in ℝ-\left\{0\right\}$,

$1\ast a=1\cdot a=a(\because 1>0)$and,

$a\ast 1=\{\begin{array}{c}a\cdot 1,a>0\\ a/1,a<0\end{array}=a$

which proves that $1$is the identity element of,$ℝ-\left\{0\right\}$ .

4. Let $a\in ℝ-\left\{0\right\}$be arbitrary. Then,

Let if $a>0$, the inverse of$a$ is$\frac{1}{a}$ .

And, if $a<0$, then inverse of$a$ is$a$ .

Then,

.$a>0,a\ast \left(\frac{1}{a}\right)=a\cdot \frac{1}{a}=1$ Also, $\frac{1}{a}>0$, which gives$\frac{1}{a}\ast a=\frac{1}{a}\cdot a=1$,

This shows that$1$ is the inverse of$a$ . So, every element in $ℝ-\left\{0\right\}$has an inverse.

This proves that$(ℝ-\left\{0\right\},\ast )$ forms a group under the binary operation$\ast$ .