Problem 1 - A Book of Abstract Algebra

Question

For each integer \(n\), define \(f_{n}\) by: \(f_{n}(x)=x+n .\) Prove the following: For each integer \(n, f_{n}\) is a permutation of \(\mathbb{R}\), that is, \(f_{n} \in S_{\mathrm{H}}\)

Step-by-Step Solution

Answer

Function \( f_{n}(x) = x + n \) is a bijection on \( \mathbb{R} \), hence a permutation.

Step 1: Understand Definition of a Permutation

A permutation of a set is a rearrangement of its elements. If function \( f \) is a permutation of the set \( \mathbb{R} \) (the real numbers), then \( f \) must be bijective—that is, it must be both injective (one-to-one) and surjective (onto).

Step 2: Establish Injectiveness of \( f_{n} \)

To show that \( f_{n} \) is injective, assume \( f_{n}(x_1) = f_{n}(x_2) \), which implies \( x_1 + n = x_2 + n \). Subtracting \( n \) from both sides gives \( x_1 = x_2 \). Hence, \( f_{n} \) is injective because no two different inputs can be mapped to the same output.

Step 3: Establish Surjectiveness of \( f_{n} \)

To show that \( f_{n} \) is surjective, for any real number \( y \), we can find a real number \( x \) such that \( f_{n}(x) = y \). Specifically, if we let \( x = y - n \), then \( f_{n}(x) = (y - n) + n = y \). Hence, every real number \( y \) has a pre-image, proving surjectiveness.

Step 4: Conclude \( f_{n} \) is a Permutation

Since \( f_{n} \) is both injective and surjective on \( \mathbb{R} \), it is bijective. Therefore, \( f_{n} \) is a permutation of \( \mathbb{R} \), meaning it is an element of \( S_{\mathrm{H}} \).

Key Concepts

Bijective Functions

Understanding bijective functions is critical as these are both injective and surjective. This combination means every element in the domain is paired with a unique element in the codomain, and vice versa.
In simple terms:

• A function is bijective if each input corresponds to exactly one output, and every possible output is matched to an input.
• For any set, a bijective function perfectly pairs elements from two sets without leaving any element unmated and without repetition in the matches.

The exercise shows that the function \(f_n(x) = x + n\) is bijective because it is both injective and surjective. This means it rearranges the whole set \(\mathbb{R}\) of real numbers perfectly, just like reshuffling cards without duplicating or omitting any card.

Injective Functions

Injective functions, or one-to-one functions, have a unique output for each input. No two inputs will ever map to the same output, ensuring distinctness across mappings.
To check if a function \(f\) is injective:

• Assume two outputs of the function are equal, \(f(x_1) = f(x_2)\).
• If it leads to \(x_1 = x_2\), then the function is injective.

For example, in the exercise, we proved \(f_n(x_1) = f_n(x_2)\) implies \(x_1 = x_2\). Thus, \(f_n\) is injective. Injectivity provides assurance that every input maps to a unique output, vital when considering permutations.

Surjective Functions

Surjective functions or onto functions cover their entire target set. For every element in the codomain, there is an input from the domain that maps to it.
In proving surjectiveness of \(f_n(x) = x + n\), it was shown that:

• Start by selecting any element \(y\) from the codomain, \(\mathbb{R}\).
• Find an input \(x\) in the domain such that \(f_n(x) = y\); here, \(x = y - n\).

Each element \(y\) has a corresponding pre-image \(x\), showing the function is surjective. This ensures complete coverage, essential for a function to be a permutation over \(\mathbb{R}\).