Question

In the group $G=\{0,1,2,3,4,5\}$ under addition modulo $6$, ${(2+3^{-1}+4)}^{-1}=$

• A. Zero
• B. $2$
• C. $3$
• D. $5$

Answer 1

The correct option is C $3$

$G$ is a group under addition modulo $6$

$0$ is the identity

For $a,b\in G$

If $a+b=0,$ then $a^{-1}=b$

Now, $3+3=0⟹3^{-1}=3$ ..... $\left(i\right)$

Also, $9+3=12=0⟹9^{-1}=3$ ....... $\left(ii\right)$

Consider, ${(2+3^{-1}+4)}^{-1}$

$=(2+3+4{)}^{-1}$

$=9^{-1}$

$=3$

Hence, ${(2+3^{-1}+4)}^{-1}=3$