Problem 2 - Algebra: Pure and Applied

Question

In Exercises 1 through 5 determine which of the indicated functions is a permutation of the indicated set. $$ f: \mathbb{R} \rightarrow \mathbb{R}, \text { where } f(x)=3 x^{2}+2 $$

Step-by-Step Solution

Answer

The function is not a permutation of \( \mathbb{R} \).

Step 1: Understanding Permutations

A function is a permutation if it's a bijection from a set to itself. This means the function must be both injective (one-to-one) and surjective (onto). Since the function is from \( \mathbb{R} \rightarrow \mathbb{R} \), we need to check both properties for all real numbers.

Step 2: Check for Injectivity

A function is injective if it assigns distinct outputs to distinct inputs. Consider two inputs \( a \) and \( b \) such that \( f(a) = f(b) \). If \( f \) is injective, then \( a = b \). For the given function \( f(x) = 3x^2 + 2 \), assume \( f(a) = f(b) \):\[ 3a^2 + 2 = 3b^2 + 2 \] Simplifying gives \( 3a^2 = 3b^2 \) and hence \( a^2 = b^2 \). This implies \( a = b \) or \( a = -b \). Therefore, the function is not injective.

Step 3: Check for Surjectivity

A function is surjective if, for every element \( y \) in the codomain, there exists at least one element \( x \) in the domain such that \( f(x) = y \). Given \( f(x) = 3x^2 + 2 \), solve for \( x \): \( y = 3x^2 + 2 \). Thus \( x^2 = \frac{y-2}{3} \). For \( x \) to be real, \( \frac{y-2}{3} \geq 0 \) which means \( y \geq 2 \). Hence, the function is not surjective over all \( \mathbb{R} \) because not every real number \( y \) can be obtained.

Step 4: Conclusion on Permutation

Since the function \( f(x) = 3x^2 + 2 \) is neither injective nor surjective over \( \mathbb{R} \), it cannot be a permutation of \( \mathbb{R} \). A permutation must be bijective, which \( f \) is not.

Key Concepts

Injective Function

In mathematics, an injective function is a type of function where distinct inputs lead to distinct outputs. This means that for a function \( f : A \to B \), if \( f(a_1) = f(a_2) \) implies that \( a_1 = a_2 \), then the function is injective. You can think of it like a one-to-one pairing where each input has a unique output.

For example, consider the function \( f(x) = 2x \). If \( f(a) = f(b) \), then \( 2a = 2b \). Simplifying gives \( a = b \). This is injective because no two different numbers will produce the same output when doubled.

However, in the case of our original exercise, with \( f(x) = 3x^2 + 2 \), the function is not injective as \( 3a^2 = 3b^2 \) leads to \( a = b \) or \( a = -b \) — meaning two different inputs can still result in the same output. Thus, it is not one-to-one.

Surjective Function

A function is surjective if it hits every element in its target set at least once. More formally, a function \( f: A \to B \) is surjective if, for every \( y \in B \), there is at least one \( x \in A \) such that \( f(x) = y \). This means the range of \( f \) covers the entire codomain \( B \).

Imagine a simple function \( f(x) = x + 1 \) where \( x \) is a real number. Given any real number \( y \), we can find \( x = y - 1 \) such that \( f(x) = y \). Thus, this function is surjective over the reals.

For the function \( f(x) = 3x^2 + 2 \) from our exercise, we determined that not every real number \( y \) can be achieved. Specifically, \( y \) needs to be greater than or equal to 2 for a corresponding \( x \) to exist. Since it doesn’t map to all real numbers, it’s not a surjective function over \( \mathbb{R} \).

Bijection

A bijection is a function that is both injective and surjective. In other words, it's a perfect pairing between the domain and the codomain where every element is covered once and only once. This means every element in the codomain is mapped to exactly once by an element in the domain.

Consider the function \( f: \mathbb{Z} \to \mathbb{Z} \), defined by \( f(x) = x + 1 \). It is both injective (since \( x \) values are distinct for distinct \( y \) values) and surjective (since every integer can be written as \( y = x + 1 \) for some integer \( x \)). Thus, this function is a bijection.

However, for the function \( f(x) = 3x^2 + 2 \), neither injective nor surjective properties are satisfied over \( \mathbb{R} \). As a result, it cannot be considered a bijection, and therefore is not a permutation of \( \mathbb{R} \).

Function

A function is a fundamental concept in mathematics that describes a relationship between sets, usually denoted as \( f: A \to B \). It assigns each element of a set \( A \) (the domain) to a single element in set \( B \) (the codomain).

Every function must satisfy two basic criteria:

• Each input from the domain is assigned to exactly one output in the codomain.
• No element in the domain is left "unmapped."

In simple terms, a function is like a machine that takes an input and gives an output based on the specific rules defined by the function.

For example, in the given exercise, the function \( f(x) = 3x^2 + 2 \) takes a real number \( x \) as an input and produces a real number as an output. But remember, not all functions are permutations, as a permutation requires the function to be a bijection. This function, with the rule \( f(x) = 3x^2 + 2 \), does not satisfy this when both injectivity and surjectivity are examined.