Question

Prove that the function \(f: \mathbb{N} \rightarrow \mathbb{Z}\) defined as \(f(n)=\frac{(-1)^{n}(2 n-1)+1}{4}\) is bijective.

Short answer

The function \(f: \mathbb{N} \rightarrow \mathbb{Z}\) defined as \(f(n)=\frac{(-1)^{n}(2 n-1)+1}{4}\) is bijective because it is shown to be both injective and surjective.

Step-by-step solution

Proving Injectivity

To prove injectivity, assume \(f(x) = f(y)\) for some \(x, y \in \mathbb{N}\). So we have: \(\frac{(-1)^{x}(2 x-1)+1}{4} = \frac{(-1)^{y}(2 y-1)+1}{4}\). This simplifies to: \((-1)^{x}(2 x-1) = (-1)^{y}(2 y-1)\). When \(x\) and \(y\) are equal, this is obviously true. When \(x\) and \(y\) are not equal, since one of them is even and other is odd, \((-1)^{x}\) and \((-1)^{y}\) have different signs and so the equation can't hold. So, the function is injective.

Proving Surjectivity

To prove that the function is surjective, we need to show that for every \(y \in \mathbb{Z}\), there is an \(x \in \mathbb{N}\) such that \(f(x) = y\). Or in other words, for all \(y \in \mathbb{Z}\), when \(y < 0\), choose \(x = -2y\), then \(f(x) = -y\). And for \(y >= 0\), choose \(x = 2y + 1\), then \(f(x) = y\) So the function is surjective.

Key concepts

Injectivity

When we talk about a function being injective, we're essentially saying that no two different elements from the domain (in this case, the natural numbers \( \mathbb{N} \)) are mapped to the same element in the codomain (here, the integers \( \mathbb{Z} \)).

Let's consider two natural numbers, \( x \) and \( y \). If \( f(x) = f(y) \), and this implies that \( x = y \), then our function is injective. With the function \( f(n)=\frac{(-1)^{n}(2 n-1)+1}{4} \), we use this concept. If we substitute two different natural numbers for \( x \) and \( y \) and the result is the same, due to the nature of the function (alternating signs for even and odd inputs), it means that \( x \) must equal \( y \) for the equation to hold true. This conclusion validates that the function is injective, as each input number has its unique output, never duplicating across the natural number set.

Surjectivity

A function is surjective when every element from the codomain (our integer numbers \( \mathbb{Z} \)) has at least one corresponding element in the domain (the natural numbers \( \mathbb{N} \)) that is mapped to it. In simpler terms, this means that the function covers the entire codomain; every integer has a natural number that links to it.

Using our function \( f(n) \), to demonstrate surjectivity, we show that for every integer \( y \) we can find a natural number \( x \) that makes \( f(x) = y \). For negative integers (when \( y < 0 \) ), we find that setting \( x = -2y \) works perfectly. Similarly, for non-negative integers (when \( y \geq 0 \) ), choosing \( x = 2y + 1 \) gives us the desired result. In both scenarios, the function associates each integer with a specific natural number, thus confirming that it is surjective.

Functions from Natural to Integer Numbers

Our discussion revolves around functions which take the set of natural numbers \( \mathbb{N} \) as their domain and map them to the set of integer numbers \( \mathbb{Z} \), known as the codomain. These functions from natural to integer numbers can exhibit complex patterns and it's fascinating to construct them to comply with certain properties like injectivity or surjectivity.

The function in our exercise, \( f(n) \), with its alternating sign and specific structure, successfully maps each natural number to a unique integer. This mapping serves as an example of how mathematical creativity can be applied to bridge distinct sets of numbers, demonstrating the versatility and the power of functions in expressing mathematical ideas.

Mathematical Induction

Mathematical induction is a powerful proof technique used in mathematics to establish that a given statement holds for all natural numbers. It has two basic steps: the base case, where the statement is verified for the initial value (usually \( n = 1 \)), and the induction step, where we assume the statement holds for \( n = k \) and then prove it for \( n = k + 1 \).

Although our function's bijectivity doesn't explicitly require an induction proof, the principle of mathematical induction often aids in demonstrating the properties for functions mapping natural numbers. By validating the base case and showing the inductive step, we solidify the belief that a statement or property is true indefinitely as we move through the natural number set. This approach exemplifies the stepwise process of climbing an infinite ladder, where the validity of one rung ensures the stability of the next.