Consider the identity function $I_{N}: N \rightarrow N$ defined as $I_{N}(x) = x$ for all $x \in N$. Show that although $I_{N}$ is onto,the function $I_{N} + I_{N}: N \rightarrow N$ defined as $(I_{N} + I_{N})(x) = I_{N}(x) + I_{N}(x) = x + x = 2x$ is not onto.

(N/A) The identity function $I_{N}: N \rightarrow N$ is defined by $I_{N}(x) = x$. For any $y \in N$,there exists $x = y \in N$ such that $I_{N}(x) = y$,so $I_{N}$ is onto.
Now,consider the function $f(x) = (I_{N} + I_{N})(x) = 2x$. The range of this function is the set of all even natural numbers,i.e.,$\{2, 4, 6, \dots\}$.
Since the co-domain is $N = \{1, 2, 3, \dots\}$,we can observe that an element like $3 \in N$ does not have a pre-image in the domain $N$ because $2x = 3$ implies $x = 1.5$,which is not a natural number.
Therefore,$I_{N} + I_{N}$ is not onto.