Consider the identity function $\mathrm{I}_{\mathbf{N}}: \mathbf{N} \rightarrow \mathbf{N}$ defined as $\mathrm{I}_{\mathbf{N}}(x)=x \forall x \in \mathbf{N}$. Show that although $\mathrm{I}_{\mathrm{N}}$ is onto but $\mathrm{I}_{\mathrm{N}}+\mathrm{I}_{\mathrm{N}}: \mathbf{N} \rightarrow \mathbf{N}$ defined as $$\left(\mathrm{I}_{\mathbf{N}}+\mathrm{I}_{\mathrm{N}}\right)(x)=\mathrm{I}_{\mathbf{N}}(x)+\mathrm{I}_{\mathbf{N}}(x)=x+x=2 x \text { is not onto. }$$

Step 1: Understand the definition of an onto function

A function is considered 'onto' or surjective if every element in the codomain (the set where the function's outputs lie) is mapped to by at least one element from the domain (the set of inputs). In simpler terms, the range of the function must be equal to its codomain.

Step 2: Analyze the identity function $\mathrm{I}_{\mathbf{N}}(x) = x$

The identity function $\mathrm{I}_{\mathbf{N}}(x) = x$ maps each natural number to itself. For any natural number $y$ in the codomain, we can always find an $x$ in the domain (which is $y$ itself) such that $\mathrm{I}_{\mathbf{N}}(x) = y$. Therefore, the range of $\mathrm{I}_{\mathbf{N}}$ is $\mathbf{N}$, which is equal to its codomain, proving that $\mathrm{I}_{\mathbf{N}}$ is onto.

Step 3: Analyze the function $(\mathrm{I}_{\mathbf{N}}+\mathrm{I}_{\mathbf{N}})(x) = 2x$

Now, let's consider the function $f(x) = (\mathrm{I}_{\mathbf{N}}+\mathrm{I}_{\mathbf{N}})(x) = 2x$. The domain and codomain for this function are both the set of natural numbers, $\mathbf{N} = \{1, 2, 3, \dots\}$.

Step 4: Check if $f(x) = 2x$ is onto

To check if $f(x) = 2x$ is onto, we need to see if every element in the codomain $\mathbf{N}$ can be reached by the function. Let's pick an odd natural number from the codomain, for example, $y=1$. If $f(x) = 1$, then $2x = 1$, which means $x = \frac{1}{2}$.

Step 5: Conclusion for $f(x) = 2x$

Since $x = \frac{1}{2}$ is not a natural number, there is no element in the domain $\mathbf{N}$ that maps to $1$ in the codomain. This means that odd natural numbers are not in the range of $f(x) = 2x$. Since the range (even natural numbers) is not equal to the codomain (all natural numbers), the function $f(x) = 2x$ is not onto.

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