Question

If R is contained in a field K and $\mathit{a}\mathbf{/}\mathit{b}\mathbf{=}\mathit{c}\mathbf{/}\mathit{d}$ in F, show that $\mathit{a}{\mathit{b}}^{\mathbf{-}\mathbf{1}}\mathbf{=}\mathit{c}{\mathit{d}}^{\mathbf{-}\mathbf{1}}$ in K. [Hint: $\mathit{a}\mathbf{/}\mathit{b}\mathbf{=}\mathit{c}\mathbf{/}\mathit{d}$ implies $\mathit{a}\mathit{d}\mathbf{=}\mathit{b}\mathit{c}$in K.]

Short answer

It is proved that, $a{b}^{-1}=c{d}^{-1}$.

Step-by-step solution

By using hint

$$\begin{gathered} a/b=c/d \\ \Rightarrow ad=bc \end{gathered}$$

Given that R is contained in a field K and$a/b=c/d$ in F.

Prove that ab-1=cd-1

Here, R is contained in a field K .

Therefore, $b,d\in R\Rightarrow b^{-1}{d}^{-1}\in R$.

Now, we have, $ad=bc$.

Simplify$ad=bc$ as:

$\begin{array}{rcl}ad& =& bc\\ ad{d}^{-1}& =& bc{d}^{-1[Multiplying{d}^{-1}frombothside]}\\ b^{-1}a& =& b^{-1}bc{d}^{-1}[Multiplying{b}^{-1}frombothside]\\ a{b}^{-1}& =& c{d}^{-1}\end{array}$

Hence, it is proved that$a{b}^{-1}=c{d}^{-1}$ .