Question

Prove that ${ℝ}^{\ast }\times ℝ$is a group under the operation$\mathbf{\ast }$ defined by $\mathbf{(}\mathbf{a}\mathbf{,}\mathbf{b}\mathbf{)}\mathbf{\ast }\mathbf{(}\mathbf{c}\mathbf{,}\mathbf{d}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{a}\mathbf{c}\mathbf{,}\mathbf{b}\mathbf{c}\mathbf{+}\mathbf{d}\mathbf{)}$.

Short answer

It is proved that $({ℝ}^{\ast }\times ℝ,\ast )$forms a group

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 thatrole="math" localid="1659173223286" $\mathit{e}\mathbf{\ast }\mathit{a}\mathbf{=}\mathit{a}\mathbf{\ast }\mathit{e}\mathbf{=}\mathit{a}$ for all$\mathit{a}\mathbf{\in }\mathit{G}$ .
• For each element ,$\mathit{a}\mathbf{\in }\mathit{G}$there exists an element ${\mathit{a}}^{\mathbf{\text{'}}}$in$\mathit{G}$ such that,$a^{\text{'}}\ast a=a\ast a^{\text{'}}=e$

.

Proof

1. As we know, $ℝ$is closed under addition and multiplication. It also forms an integral domain so${ℝ}^{\ast }\times ℝ$is closed under$\ast$

2. Let$(a,b),(c,d),(e,f)\in {ℝ}^{\ast }\times ℝ$then,

$\begin{array}{l}(a,b)\ast ((c,d)\ast (e,f))=(a,b)\ast (ce,de+f)\\ =(ace,bce+de+f)\\ =(ac,bc+d)\ast (e,f)\\ =((a,b)\ast (c,d))\ast (e,f)\end{array}$

which shows that$\ast$is associative.

3. Now,

.$\begin{array}{l}(a,b)\ast (1,0)=(a1,b1+0)=(a,b);\\ (1,0)\ast (a,b)=(1a,0a+b)=(a,b).\end{array}$

which proves that,$(1,0)$is the identity element of.${ℝ}^{\ast }\times ℝ$.

4. Since,.localid="1659418823881" $\left(\frac{1}{a},-\frac{b}{a}\right)\in {ℝ}^{\ast }\times ℝ$(Since$a\ne 0$),then,

localid="1659418869421"

This shows thatlocalid="1659418883933" $\left(\frac{1}{a},-\frac{b}{a}\right)\in {ℝ}^{\ast }\times ℝ$is the inverse oflocalid="1659418889377" $(a,b)$. So, every element in${ℝ}^{\ast }\times ℝ$has an inverse.

This proves that${ℝ}^{\ast }\times ℝ$forms a group under the binary operation$\ast$.