Question

\(\operatorname{Let} B=\left\\{2^{n}: n \in \mathbb{Z}\right\\}=\left\\{\ldots, \frac{1}{4}, \frac{1}{2}, 1,2,4,8, \ldots\right\\} .\) Show that the function \(f: \mathbb{Z} \rightarrow B\) defined as \(f(n)=2^{n}\) is bijective. Then find \(f^{-1}\).

Short answer

Given function \(f: \mathbb{Z} \rightarrow B\) defined as \(f(n)=2^{n}\) is bijective. The inverse function \(f^{-1}: B \rightarrow \mathbb{Z}\) is \(f^{-1}(x) = \log_2(x)\).

Step-by-step solution

Show \(f\) is Injective

To show the function is injective, let's assume \(n_{1}\) and \(n_{2}\) are integers such that \(f(n_{1}) = f(n_{2})\). If \(f(n_{1}) = 2^{n_{1}} = 2^{n_{2}} = f(n_{2})\), taking the logarithm base 2 to both sides gives \(n_{1} = n_{2}\). So, for all \(n_{1}, n_{2} \in \mathbb{Z}\), if \(f(n_{1}) = f(n_{2})\), then \(n_{1} = n_{2}\), meaning the function \(f\) is injective.

Show \(f\) is Surjective

To prove the function is surjective, we must show that for every \(b \in B\), there exists an integer \(n\) such that \(2^n = b\). Since B by definition is the set \(B=\{2^{n}: n \in \mathbb{Z}\}\) and since \(b\) is a member of this set, by definition there is an integer \(n\) such that \(2^n = b\). Hence, the function is surjective.

Find the Inverse Function

Now that we've shown the function \(f: \mathbb{Z} \rightarrow B\) is bijective, an inverse function must exist. To find the inverse function \(f^{-1}: B \rightarrow \mathbb{Z}\), we should replace \(2^n = b\) by \(2^y = x\), then solve for \(y\), finding \(y = \log_2(x)\). As such the inverse function is \(f^{-1}(x) = \log_2(x)\).

Key concepts

Injective Function

An injective function, also known as a one-to-one function, guarantees that different inputs map to different outputs. In simpler terms, if you have two different numbers and you apply an injective function to both, you'll end up with two different numbers.

Let's consider the given problem. We have the function f(n) = 2^n where each n is an integer. To test for injectivity, you compare two outputs and see if their corresponding inputs were different. Interestingly, in our case, since powers of two are distinct for different exponents, if f(n1) = f(n2), it's clear that the only way this equation holds true is if the exponents n1 and n2 are equal. This uniqueness of the powers of two solidifies that our function f is indeed injective.

Surjective Function

A surjective function, which is sometimes called an onto function, ensures that every element in the target set has a pre-image in the domain. To visualize this, think of a function as a matchmaking service where every individual in one town (the range) is paired up with someone from another town (the domain).

Back to our specific case, the set B contains all possible outputs of our function f(n) since it consists of all numbers of the form 2^n, where n is any integer. This means, for any number you pick from set B, you can find an integer n to plug into our function to get that number. This is exactly what being surjective is about, and our function f satisfies this condition perfectly.

Inverse Function

The inverse function basically reverses a function's action. Imagine a path where instead of going forward, you backtrack exactly to where you started—that's what an inverse function does in math. An inverse function only exists if the original function is bijective, meaning both injective and surjective.

In the exercise, once we established that f was bijective, we could then find its inverse f^{-1}. The process involves switching the roles of the inputs and outputs. For our f(n) = 2^n, to find the inverse, we swap n and 2^n. The operation that 'reverses' exponentiation with base 2 is taking the logarithm with base 2. So, f^{-1}(b) = log_2(b) is our inverse function. Thus, for every number in set B, applying f^{-1} gets you back to the original integer n.

Exponential and Logarithmic Functions

Let's dive deeper into exponential and logarithmic functions, which are closely related mathematical concepts. An exponential function involves raising a number, typically called the base, to the power of the variable. For instance, 2^n is an exponential function where 2 is the base and n is the exponent.

The logarithm is the inverse operation to exponentiation. If you're given 2^n = x, and you want to uncover what n is, you'd use a logarithm. The equation shifts to n = log_2(x). This interplay between the two functions is beautifully showcased in the exercise as we use logarithms to find the inverse of the exponential function f(n) = 2^n. These functions are fundamental due to their properties and applications in various fields such as compound interest in finance, population growth in biology, and the behavior of transistors in electronics.