Analysis of Function Properties: One-one and Onto

Let's carefully examine each statement regarding the properties of the given functions. We will determine if each function is one-one (injective) and onto (surjective) based on its domain and codomain.

Examining Statement 1: Function f: Z → Z, f(x) = x + 1

The function is defined as \(f(x) = x + 1\) with the domain being the set of integers (\(Z\)) and the codomain also being the set of integers (\(Z\)).

Checking if the function is one-one (Injective):

A function \(f: A \rightarrow B\) is one-one if for any \(a, b \in A\), \(f(a) = f(b)\) implies \(a = b\).

• Let \(a\) and \(b\) be any two integers in the domain \(Z\).
• Assume \(f(a) = f(b)\).
• Substituting the function definition, we get \(a + 1 = b + 1\).
• Subtracting 1 from both sides of the equation, we have \(a = b\).
• Since \(f(a) = f(b)\) implies \(a = b\) for all \(a, b \in Z\), the function \(f(x) = x + 1\) is indeed one-one when mapping from \(Z\) to \(Z\).

Checking if the function is onto (Surjective):

A function \(f: A \rightarrow B\) is onto if for every element \(y\) in the codomain \(B\), there exists at least one element \(x\) in the domain \(A\) such that \(f(x) = y\).

• Let \(y\) be any integer in the codomain \(Z\).
• We need to find an integer \(x\) in the domain \(Z\) such that \(f(x) = y\).
• Setting \(f(x) = y\), we have \(x + 1 = y\).
• Solving for \(x\), we get \(x = y - 1\).
• Since \(y\) is an integer, \(y - 1\) is also always an integer.
• Thus, for every \(y \in Z\), there exists an \(x = y - 1 \in Z\) such that \(f(x) = y\).
• Therefore, the function \(f(x) = x + 1\) is onto when mapping from \(Z\) to \(Z\).

Based on this analysis, Statement 1, which says that the function \(f : Z \rightarrow Z\), defined by \(f(x) = x + 1\), is one-one as well as onto, is correct.

Examining Statement 2: Function f: N → N, f(x) = x + 1

The function is defined as \(f(x) = x + 1\) with the domain being the set of natural numbers (\(N\)) and the codomain also being the set of natural numbers (\(N\)). We typically consider the set of natural numbers \(N = \{1, 2, 3, \dots\}\).

Checking if the function is one-one (Injective):

A function \(f: A \rightarrow B\) is one-one if for any \(a, b \in A\), \(f(a) = f(b)\) implies \(a = b\).

• Let \(a\) and \(b\) be any two natural numbers in the domain \(N\).
• Assume \(f(a) = f(b)\).
• Substituting the function definition, we get \(a + 1 = b + 1\).
• Subtracting 1 from both sides of the equation, we have \(a = b\).
• Since \(f(a) = f(b)\) implies \(a = b\) for all \(a, b \in N\), the function \(f(x) = x + 1\) is indeed one-one when mapping from \(N\) to \(N\).

Checking if the function is onto (Surjective):

A function \(f: A \rightarrow B\) is onto if for every element \(y\) in the codomain \(B\), there exists at least one element \(x\) in the domain \(A\) such that \(f(x) = y\).

• Let \(y\) be any natural number in the codomain \(N\).
• We need to find a natural number \(x\) in the domain \(N\) such that \(f(x) = y\).
• Setting \(f(x) = y\), we have \(x + 1 = y\).
• Solving for \(x\), we get \(x = y - 1\).
• For \(x\) to be in the domain \(N = \{1, 2, 3, \dots\}\), \(x\) must be a natural number (greater than or equal to 1).
• Consider the element \(y = 1\) which is in the codomain \(N\).
• We need to find an \(x \in N\) such that \(f(x) = 1\), i.e., \(x + 1 = 1\).
• Solving for \(x\), we get \(x = 1 - 1 = 0\).
• However, \(0\) is not a natural number in the set \(N = \{1, 2, 3, \dots\}\).
• This means there is no element in the domain \(N\) that maps to the element \(1\) in the codomain \(N\).
• Therefore, the function \(f(x) = x + 1\) is NOT onto when mapping from \(N\) to \(N\).

Based on this analysis, Statement 2, which says that the function \(f : N \rightarrow N\), defined by \(f(x) = x + 1\), is one-one but not onto, is correct.

Conclusion on Correctness of Statements

Our analysis shows that Statement 1 is correct because \(f(x) = x + 1\) is both one-one and onto for the domain and codomain being the set of integers \(Z\). Statement 2 is also correct because \(f(x) = x + 1\) is one-one but not onto for the domain and codomain being the set of natural numbers \(N\).

Revision Table: Function Properties Summary

Additional Information: Key Function Concepts

Understanding function properties like one-one and onto is fundamental in mathematics. Here's a quick recap:

• One-one Function (Injective Function): A function \(f: A \rightarrow B\) is called one-one if distinct elements in the domain \(A\) are always mapped to distinct elements in the codomain \(B\). In other words, if \(a \neq b\) for elements \(a, b\) in the domain, then \(f(a) \neq f(b)\). Equivalently, if \(f(a) = f(b)\), then it must mean \(a = b\).
• Onto Function (Surjective Function): A function \(f: A \rightarrow B\) is called onto if every element in the codomain \(B\) is the image of at least one element in the domain \(A\). This means for any \(y \in B\), there exists an \(x \in A\) such that \(f(x) = y\). The range of an onto function is equal to its codomain.
• Bijective Function: A function is called bijective if it is both one-one (injective) and onto (surjective). Bijective functions are also called one-to-one correspondence because each element in the domain is paired with exactly one element in the codomain, and vice-versa.
• Domain and Codomain: The domain is the set of all possible input values (\(x\)) for the function. The codomain is the set of all possible output values (\(y\)) that the function could *potentially* produce. The range is the actual set of output values produced by the function for the given domain. The range is always a subset of the codomain.