Show that the function $f: \mathbf{N} \rightarrow \mathbf{N}$, given by $f(1)=f(2)=1$ and $f(x)=x-1$, for every $x>2$, is onto but not one-one.

Step 1: Analyze the function definition

The function $f$ maps natural numbers to natural numbers. We are given its definition in three parts: $f(1)=1$, $f(2)=1$, and for any natural number $x$ greater than 2, $f(x)=x-1$. We need to determine if this function is one-one and onto.

Step 2: Check for one-one property

A function is one-one (or injective) if distinct elements in the domain map to distinct elements in the codomain. In this case, we observe that $f(1)=1$ and $f(2)=1$. Since $1 \neq 2$ but their images under $f$ are the same, $f(1)=f(2)$, the function is not one-one.

Step 3: Check for onto property

A function is onto (or surjective) if every element in the codomain has at least one pre-image in the domain. The codomain is $\mathbf{N}$, the set of natural numbers $\{1, 2, 3, \dots\}$. We need to show that for any $y \in \mathbf{N}$, there exists an $x \in \mathbf{N}$ such that $f(x)=y$.

Step 4: Case 1: $y=1$

If $y=1$, we can choose $x=1$ or $x=2$. Both $f(1)=1$ and $f(2)=1$. So, $1$ has pre-images in the domain.

Step 5: Case 2: $y>1$

If $y>1$, we need to find an $x$ such that $f(x)=y$. Let's consider $x=y+1$. Since $y$ is a natural number greater than 1, $y+1$ will be a natural number greater than 2. For such $x$, the function definition is $f(x)=x-1$. Substituting $x=y+1$, we get $f(y+1)=(y+1)-1=y$. Thus, for every $y>1$, there exists an $x=y+1$ in the domain such that $f(x)=y$.

Step 6: Conclusion for onto property

Combining both cases, we have shown that for any natural number $y$, there is at least one natural number $x$ such that $f(x)=y$. Therefore, the function $f$ is onto.

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