Question

(21. In the group {1,3,7,9} under 10%
(3 0,67-1)-1 =
4) 7
TB) 3
(C) 9
(D) 1​

Answer 1

Answer:

option (C) is correct

Explanation:

In the group {1,3,7,9} under *10, (3*10 7^-1)-1 =

In the group {1,3,7,9}

$3\,*_{10}\,7=1$

$\textbf{Inverse of 3 is 7}$

$\implies\,3^{-1}=7$

$\text{Now,}$

$(3\,*_{10}\,7^{-1})^{-1}$

$\text{Using,the property}$

$\boxed{\bf\,(a*b)^{-1}=b^{-1}*a^{-1}}$

$=(7^{-1})^{-1}\,*_{10}\,3^{-1}$

$\text{Using,the property}$

$\boxed{\bf\,(a^{-1})^{-1}=a}$

$=7\,*_{10}\,3^{-1}$

$=7\,*_{10}\,7$

$=9$

$\implies\,\boxed{(3\,*_{10}\,7^{-1})^{-1}=9}$

Answer 2

Answer:

Answer:

option (C) is correct

Explanation:

In the group {1,3,7,9} under *10, (3*10 7^-1)-1 =

In the group {1,3,7,9}

3\,*_{10}\,7=13∗107=1

\textbf{Inverse of 3 is 7}Inverse of 3 is 7

\implies\,3^{-1}=7⟹3−1=7

\text{Now,}Now,

(3\,*_{10}\,7^{-1})^{-1}(3∗107−1)−1

\text{Using,the property}Using,the property

\boxed{\bf\,(a*b)^{-1}=b^{-1}*a^{-1}}(a∗b)−1=b−1∗a−1

=(7^{-1})^{-1}\,*_{10}\,3^{-1}=(7−1)−1∗103−1

\text{Using,the property}Using,the property

\boxed{\bf\,(a^{-1})^{-1}=a}(a−1)−1=a

=7\,*_{10}\,3^{-1}=7∗103−1

=7\,*_{10}\,7=7∗107

=9=9

\implies\,\boxed{(3\,*_{10}\,7^{-1})^{-1}=9}⟹(3∗107−1)−1=9