A mapping from mathbb{N} to mathbb{N} is defi | vedclass

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(D) Step $1$: Check for one-to-one (injective) property. Let $f(n_1) = f(n_2)$.$(n_1+5)^2 = (n_2+5)^2$.Since $n_1, n_2 \in \mathbb{N}$, $n_1+5 > 0$ an

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$f$ is one-to-one but not onto

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(D) Step $1$: Check for one-to-one (injective) property. Let $f(n_1) = f(n_2)$.$(n_1+5)^2 = (n_2+5)^2$.Since $n_1, n_2 \in \mathbb{N}$, $n_1+5 > 0$ and $n_2+5 > 0$. Taking the positive square root on both sides, we get $n_1+5 = n_2+5$, which implies $n_1 = n_2$.Therefore, $f$ is one-to-one.Step $2$: Check for onto (surjective) property. For $f$ to be onto, for every $y \in \mathbb{N}$, there must exist $n \in \mathbb{N}$ such that $f(n) = y$.Let $f(n) = (n+5)^2 = y$. Since $n \ge 1$, the minimum value of $f(n)$ is $(1+5)^2 = 36$.Thus, values like $1, 2, 3, \dots, 35$ in the codomain $\mathbb{N}$ do not have any pre-image in the domain $\mathbb{N}$.For example, there is no $n \in \mathbb{N}$ such that $(n+5)^2 = 1$.Therefore, $f$ is not onto.Conclusion: $f$ is one-to-one but not onto.

$A$ mapping from $\mathbb{N}$ to $\mathbb{N}$ is defined as follows: $f: \mathbb{N} \rightarrow \mathbb{N}$ where $f(n) = (n+5)^2$ for all $n \in \mathbb{N}$ (where $\mathbb{N}$ is the set of natural numbers). Then:

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