Given a binary operation $: A \times A \rightarrow A$ with the identity element $e$ in $A$, an element $a \in A$ is said to be invertible with respect to the operation * if there exists an element $b$ in A such that $a^* b=e=b^* a$ and $b$ is called the inverse of $a$ and is denoted by ${a}^{-1}$.

Question

Given a binary operation $: A \times A \rightarrow A$ with the identity element $e$ in $A$, an element $a \in A$ is said to be invertible with respect to the operation * if there exists an element $b$ in A such that $a^* b=e=b^* a$ and $b$ is called the inverse of $a$ and is denoted by $\mathrm{a}^{-1}$.

Answer 1

SOLUTION: Zero is identity for the addition operation on $\mathrm{R}$ but it is not identity for the addition operation on $\mathrm{N}$ as $0 \notin \mathrm{N}$. In fact the addition operation on $\mathrm{N}$ does not have any identity.

One further notices that for the addition operation $+: R \times R \rightarrow R$, given any $a \in R$, there exists $-a$ in $R$ such that $a+(-a)=0$ (identity for ' $+')=(-a)+a$.

Similarly for the multiplication operation on $R$ given any $a \neq 0$ in $R$, we can choose $1 / a$ in $R$ such that $a \times 1 / a=1($ identity for ' $x$ ' $)=1 / a \times a$. This leads to the existence of inverse.