Question

One-to-One Functions If \(f\) is a one-to-one function and \(f(x)\) is never zero, prove that \(g(x)=1 / f(x)\) is also one-to-one.

Short answer

The function \(g(x) = 1 / f(x)\) is also one-to-one function, based on the proof shown in the solution steps.

Step-by-step solution

Confirming the definition of one-to-one function

First, recall the definition of a one-to-one function. A function \(f\) is one-to-one, or injective, if and only if for every \(x_1\) and \(x_2\) such that \(x_1 \neq x_2\), we have \(f(x_1) \neq f(x_2)\). That is, distinct inputs lead to distinct outputs.

Understanding function \(g(x)\)

Now, consider the function \(g(x) = 1 / f(x)\), given that \(f(x)\) is never zero. It is essential to ensure that \(f(x)\) never equals to zero to circumvent any undefined issues in \(g(x)\) when taking the reciprocal.

Showing if \(g(x_1) = g(x_2)\)

Assume we have \(g(x_1) = g(x_2)\). That will imply that \(1 / f(x_1) = 1 / f(x_2)\), if we are to consider the definition of \(g(x)\). By cross multiplying, we will then have \(f(x_2) = f(x_1)\).

Applying properties of \(f(x)\)

As we know that \(f\) is a one-to-one function, if \(f(x_2) = f(x_1)\) it means that \(x_1 = x_2\), due to the injective property.

Conclusion

Thus, we have shown that if \(g(x_1) = g(x_2)\), then \(x_1 = x_2\), meaning that \(g(x)\) is also a one-to-one function as required by the definition.

Key concepts

Injective Function

Understanding what it means for a function to be injective is crucial when exploring advanced mathematical concepts. An injective function, also known as a one-to-one function, has a very specific characteristic: each element of the function's domain (the set of all possible inputs) is mapped to a unique element in the function's codomain (the set of all possible outputs). In simpler terms, no two different inputs will ever lead to the same output.

In the context of the exercise, the function denoted as \(f\) is injective. Based on this information, it can be proven that another function \(g(x) = 1 / f(x)\) is also injective. This conclusion ensures that for every unique input into \(f\), there is a unique reciprocal output in \(g\), highlighting the direct relationship between the two functions.

Function Reciprocals

In mathematics, the reciprocal of a number is simply defined as \(1/x\), where \(x\) is that number. When this concept is applied to functions, you end up with a new function that is made up of the reciprocals of the output values of the original function. Suppose we have a function \(f\) that never outputs zero. Its reciprocal function, which we can denote as \(g(x) = 1 / f(x)\), will consist of the multiplicative inverses of the original function's outputs.

The exercise showcases a scenario where \(f\) is an injective, non-zero function, meaning that \(g\), comprised entirely of reciprocals, also inherits injectivity under proper conditions. This is possible because injectivity is preserved through the operation of taking reciprocals, as long as the function \(f\) does not allow zero in its range.

Function Properties

Functions have various properties that can be used to understand their behavior and the relationship between their inputs and outputs. The most relevant properties when discussing injective functions are as follows:

• Injectivity: As we've discussed, an injective function means that different inputs will always produce different outputs. No two distinct elements of the domain share an output value.
• Domain and Codomain: These are core concepts in understanding functions. The domain is the set of all possible inputs, while the codomain includes all potential outputs. For injectivity to make sense, it is essential to clarify that each element of the domain corresponds distinctly to an element in the codomain.
• Non-zero Outputs: For the purpose of creating function reciprocals, a function must not have zero in its output values. This ensures that taking reciprocals is a valid operation.

Having a firm grasp of these properties assists in comprehending the nature of a one-to-one function and in proving the injectivity of its reciprocal.

Proof by Contradiction

Proof by contradiction is a powerful method in mathematics for establishing the validity of a statement. Essentially, you begin by assuming that the statement you want to prove is false. From that incorrect assumption, you logically derive consequences until you reach a statement that is clearly contradictory or nonsensical. This contradiction implies that the initial assumption (i.e., the statement being false) must be wrong, and hence, the statement is true after all.

Applying proof by contradiction to our exercise, if we assumed that the function \(g(x) = 1 / f(x)\) is not injective, we would eventually land at a contradiction with the given fact that \(f\) is injective. As per the solution steps provided, showing that \(g(x_1) = g(x_2)\) leads to \(x_1 = x_2\) counters the hypothetical situation where \(g\) could be non-injective. Therefore, the contradiction solidifies our proof that \(g\) must be injective, precisely because \(f\) is injective and never outputs zero. It is through this indirect line of reasoning that proof by contradiction works to validate the injectivity of \(g\).