Show that $-a$ is not the inverse of $a \in N$ for the addition operation $+$ on $N$ and $\frac{1}{a}$ is not the inverse of $a \in N$ for the multiplication operation $\times$ on $N$,for $a \neq 1$.

(A) For the addition operation $+$ on $N$,the identity element is $0$. However,$0 \notin N$. Even if we consider the condition $a + (-a) = 0$,the element $-a$ must belong to the set $N$ to be an inverse. Since for any $a \in N$,$-a$ is a negative integer,$-a \notin N$. Therefore,$-a$ is not the inverse of $a$ in $N$.
For the multiplication operation $\times$ on $N$,the identity element is $1$. For an element $a \in N$ to have an inverse $b \in N$,we must have $a \times b = 1$. This implies $b = \frac{1}{a}$. For any $a \in N$ where $a \neq 1$,$\frac{1}{a}$ is not an integer,so $\frac{1}{a} \notin N$. Thus,no element $a \in N$ except $a = 1$ has a multiplicative inverse in $N$.