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In $\quad$ Ine $\quad(Z, *), \quad$ if $a^{*} b=a+b-n, \forall a, b \in Z, \quad$ where $n$ is a fixed integer, then the inverse of $(-n)$ is
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Verified Answer
The correct answer is:
$3 n$
Given,
In a group $\left(Z,{ }^{*}\right)$
$a^{*} b=a+b-n, \forall a_{1}, b \in z...(i)$
where, $n$ is a fixed number
$\begin{array}{llr}\therefore & a^{*} e=a & \\ \Rightarrow & a+e-n=a \quad \text { [from Eq. (i)] } \\ \Rightarrow & e=n & \text { (identity) }\end{array}$
For finding inverse,
$\alpha^{*}(-n)=n$
$\Rightarrow \quad \alpha-n-(-n)=n \quad$ [from Eq. (i)]
$\Rightarrow \quad \alpha-n-n=n$
$\Rightarrow \quad \alpha=3 n$
So, inverse of $(-n)$ is $3 n$.
In a group $\left(Z,{ }^{*}\right)$
$a^{*} b=a+b-n, \forall a_{1}, b \in z...(i)$
where, $n$ is a fixed number
$\begin{array}{llr}\therefore & a^{*} e=a & \\ \Rightarrow & a+e-n=a \quad \text { [from Eq. (i)] } \\ \Rightarrow & e=n & \text { (identity) }\end{array}$
For finding inverse,
$\alpha^{*}(-n)=n$
$\Rightarrow \quad \alpha-n-(-n)=n \quad$ [from Eq. (i)]
$\Rightarrow \quad \alpha-n-n=n$
$\Rightarrow \quad \alpha=3 n$
So, inverse of $(-n)$ is $3 n$.
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