Show that the function f:N→N, given by f(1)=f(2)=1 and f(x)=x−1, for every x>2, is onto but not one-one.
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Step-by-Step Solution
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=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 N, the set of natural numbers {1,2,3,…}. We need to show that for any y∈N, there exists an x∈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.