Problem 64 - Calculus: Graphical, Numerical, Algebraic

Question

Everywhere Discontinuous Give a convincing argument that the following function is not continuous at any real number. $$f(x)=\left\\{\begin{array}{ll}{1,} & {\text { if } x \text { is rational }} \\\ {0,} & {\text { if } x \text { is irrational }}\end{array}\right.$$

Step-by-Step Solution

Answer

The function \( f(x) \) is not continuous at any real number. The function values jump between 0 and 1 at every point depending on whether the x-value is rational or irrational, not meeting the requirement of continuity that \( \lim_{x \to a} f(x) = f(a) \) for any real number \( a \).

Step 1: Understand the Definition of Continuity

The function \( f(x) \) is said to be continuous at a point \( a \) if \( \lim_ {x \to a} f(x) = f(a) \). This means that the value of the function at \( a \) is equal to the limit of the function as \( x \) approaches \( a \). This condition must hold for every point in the domain for the function to be called continuous everywhere.

Step 2: Applying the Definition to the Function's Terms

Consider a number \( x \) such that \( x \) is irrational, and call it \( x_{0} \). Now take a sequence of rational numbers \( x_{n} \) that converge to \( x_{0} \). The limit of these numbers as \( n \) approaches to infinity \( \lim_ {n \to \infty} x _{n} \) is \( x_{0} \). Now consider the function values at these rational number sequences \( f(x _{n}) \). For each \( x_{n} \), \( f(x_{n}) \) equals 1. So, \( \lim_ {n \to \infty} f(x _{n}) = 1 \). However, \( f(x_{0}) = 0 \). Hence \( \lim_ {x \to x_{0} } f(x) \neq f(x_{0}) \), violating the definition of continuity at \( x_{0} \).

Step 3: Extending the Argument to All Real Numbers

From this argument, it can be seen that the function \( f(x) \) is not continuous at any irrational number \( x_{0} \). The same procedure can be applied with a sequence of irrational numbers converging to any rational number, to show that \( f(x) \) is discontinuous at all rational numbers as well.

Key Concepts

Continuity

Continuity is a fundamental concept in calculus and analysis. A function is continuous at a point if it has no sudden jumps. To be formal, a function \( f(x) \) is continuous at a point \( a \) if the limit of \( f(x) \) as \( x \) approaches \( a \) equals the function's value at \( a \):

\[\lim_{x \to a} f(x) = f(a)\]
This means the function must approach a specific value as \( x \) gets closer to \( a \), with no interruptions.

• No gaps or jumps in the graph
• Value at \( a \) must match the surrounding values

For the given function, defined as \( f(x) = 1 \) if \( x \) is rational and \( f(x) = 0 \) if \( x \) is irrational, there's an inherent problem with continuity. This function always oscillates between 0 and 1, because no matter how close you get to a rational point, irrational numbers are still right there, and vice versa. Thus, \( \lim_{x \to a} f(x) \) can never concretely approach any single value.

Real Numbers

Real numbers are the building blocks of calculus and analysis. They include both rational and irrational numbers:

• Rational numbers, like \( \frac{1}{2} \) or 2, can be expressed as fractions.
• Irrational numbers, such as \( \sqrt{2} \) or \( \pi \), cannot be neatly expressed as fractions.

This set of numbers is continuous, meaning that between any two real numbers, there is always another real number. This property is fundamental in understanding real functions because it impacts how limits and continuity are evaluated.

When examining a function over real numbers, like \( f(x) \) in the given exercise, both types of numbers need to be considered, highlighting the intricate dance between rationality and irrationality in defining function behavior.

Rational and Irrational Numbers

Rational and irrational numbers are distinct yet interconnected parts of the real number system.

• **Rational numbers**: Expressed as \( \frac{p}{q} \) where \( p \) and \( q \) are integers and \( q eq 0 \).
• **Irrational numbers**: Cannot be expressed as a simple fraction. Examples include \( \sqrt{2} \) and \( \pi \).

The interesting part about these numbers is how densely they populate the real number line.

• Between any two rationals, there is an irrational number.
• Between any two irrationals, there is a rational number.

This density challenges functions like \( f(x) \), which have different values for rational and irrational inputs. For instance, as you approach a number through a sequence of rationals, the function value is 1, but through irrationals, it's 0. This creates discontinuity, as observed where rational and irrational numbers converge on the same point on the real number line.