Question

Do the nonzero real numbers form a group under the operation given by$\mathit{a}\mathbf{\ast }\mathit{b}\mathbf{=}\mathbf{\left|}\mathbf{a}\mathbf{\right|}\mathit{b}$, where $\mathbf{\left|}\mathbf{a}\mathbf{\right|}$is the absolute value of a?

Short answer

It is proved that $ℝ-\left\{0\right\}$does not form a group under the binary 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 element$\mathit{a}\mathbf{\in }\mathit{G}$,there exists an element${\mathit{a}}^{\mathbf{\text{'}}}$in $\mathit{G}$such that,

${\mathit{a}}^{\mathbf{\text{'}}}\mathbf{\ast }\mathit{a}\mathbf{=}\mathit{a}\mathbf{\ast }{\mathit{a}}^{\mathbf{\text{'}}}\mathbf{=}\mathit{e}$

Conclusion

$-1or1$Let,$a,b\in ℝ-\left\{0\right\}$then, we have,$|a\ast b|=\left|\left|a\right|b\right|=\left|a\right|\left|b\right|$

Clearly, in the set, $ℝ-\left\{0\right\}$, closure property, which is an associative property, holds for the binary operation$\ast$

Now, we look for the identity element. Let be the identity element of the set $ℝ-\left\{0\right\}$with respect to the absolute value function.

Then,$a\ast e=e\ast a=a$must hold

But, $\left|a\right|=|a\ast e|=\left|a\right|\left|e\right|\Rightarrow \left|a\right|=1$. So, the possible value of are $-1or1$.

Here,

$\begin{array}{l}(-1)*1=|-1|1=1;\\ 1\ast (-1)=\left|1\right|(-1)=-1\end{array}$

which shows neither of$-1or1$, the identity element.

And hence, $ℝ-\left\{0\right\}$does not form a group under the binary operationlocalid="1659165044982" $\ast$.