Problem 5 - A Book of Abstract Algebra

Question

Proof By exhibiting a counterexample: \(-1\) is not equal to \(f(x)\) for any \(x \in \mathbb{R}\). \(f(x)=\left\\{\begin{array}{c}x \text { if } x \text { is rational } \\ 2 x \text { if } x \text { is irrational }\end{array}\right.\)

Step-by-Step Solution

Answer

-1 is equal to \(f(x)\) for \(x = -1\), so \(-1 = f(x)\) for some \(x\).

Step 1: Define the Problem

We need to determine if there is any real number \(x\) such that the function \(f(x)\) equals \(-1\). The function is defined piecewise: it returns \(x\) if \(x\) is a rational number, and returns \(2x\) if \(x\) is an irrational number.

Step 2: Analyze the Function for Rational \(x\)

If \(x\) is a rational number, then by the function's definition, \(f(x) = x\). For the equation \(f(x) = -1\) to hold, we would need \(x = -1\). Since \(-1\) is a rational number, substituting \(x = -1\) into the function gives \(f(-1) = -1\). Thus \(-1\) is achievable for rational \(x = -1\).

Step 3: Analyze the Function for Irrational \(x\)

If \(x\) is an irrational number, then \(f(x) = 2x\). We need \(2x = -1\), which leads to \(x = -\frac{1}{2}\). However, \(-\frac{1}{2}\) is a rational number. Thus, there is no irrational \(x\) that makes \(f(x) = -1\).

Step 4: Conclusion

Since \(-1\) can be achieved by the function when \(x = -1\), which is rational, we cannot say that \(-1\) is not equal to \(f(x)\) for any \(x \in \mathbb{R}\). Thus, the premise that \(-1\) is unachievable is incorrect.

Key Concepts

Counterexample

In abstract algebra and math in general, a counterexample is a particular case that disproves a statement or proposition. When we deal with proofs or conjectures, a single counterexample is enough to demonstrate that the general statement is false.

For the particular problem at hand, the task was to prove that (-1) cannot be the value of f(x) for any x  in \(\mathbb{R}\). However, by constructing a counterexample where x = -1, the proposition is invalidated. Hence, in the case of rational x, -1 equals f(x), which serves as our counterexample. The fact that the function can produce a value of (-1) under certain conditions refutes the original statement entirely.

Piecewise Function

A piecewise function is a function that is defined by different expressions for different parts of its domain. This means that depending on the input value, the function might behave in distinct ways. It is like having multiple sub-functions within a single function definition.

In the exercise, f(x) is given as a piecewise function. This means that the output depends on whether the input x is rational or irrational. If x is rational, f(x) = x; however, if x is irrational, f(x) = 2x. This kind of function plays a pivotal role in describing relationships that can't be captured by a single formula. Piecewise functions are useful for modeling situations where abrupt changes occur, such as tax brackets or shipping charges.

Rational Numbers

Rational numbers are numbers that can be expressed as the quotient of two integers, where the denominator is not zero. In other words, any number that can be written as a fraction \(\frac{a}{b}\), where \(a\) and \(b\) are integers and \(b eq 0\). 

• Examples include numbers like -1, 0.5, 2, \(-\frac{1}{2}\), and even 0 since they can all be expressed as simple fractions.
• \(-1\) is rational because it can be represented by \(-\frac{1}{1}\).

Rational numbers are very familiar as they include all the integers, finite decimals, and repeating decimals.

In the exercise, for any rational number x, the function f(x) = x directly correlates with its rational input, proving the equation f(-1) = -1 . 

Irrational Numbers

Irrational numbers are numbers that cannot be expressed as a ratio of two integers. They have decimal expansions that neither terminate nor repeat, making them distinct from rational numbers.

• Classic examples of irrational numbers include \(\pi\), and \(\sqrt{2}\).
• They provide a deeper understanding of the number line by filling in gaps left by rational numbers.

In the exercise, if x is irrational, then the piecewise function dictates that f(x) = 2x.

The challenge comes when we attempt to find an irrational x such that 2x = -1.  As shown, this leads to x = -\frac{1}{2}, a result which is rational and not irrational. Hence, no real irrational number allows f(x) = -1  in this function.