Problem 68 - Calculus: Early Transcendentals

Question

For what values of \( x \) is \( g \) continuous? $$ g(x) = \left\\{ \begin{array}{ll} 0 & \mbox{if \( x \) is rational}\\\ x & \mbox{if \( x \) is irrational} \end{array} \right.$$

Step-by-Step Solution

Answer

The function \( g(x) \) is discontinuous for all values of \( x \).

Step 1: Understand the Function

The given function, \( g(x) \), is defined piecewise. For rational \( x \), \( g(x) = 0 \), and for irrational \( x \), \( g(x) = x \). We need to determine where this function is continuous.

Step 2: Recall the Continuity Condition

A function \( g(x) \) is continuous at a point \( x = a \) if the following three conditions are met: 1. \( g(a) \) is defined.2. The limit \( \lim_{x \to a} g(x) \) exists.3. \( \lim_{x \to a} g(x) = g(a) \).

Step 3: Check Continuity at a Rational Point

Let \( a \) be rational. Then \( g(a) = 0 \). Consider the limit of \( g(x) \) as \( x \) approaches \( a \). For irrational \( x \) near \( a \), \( g(x) = x \). Therefore, as \( x \to a \), \( \lim_{x \to a} g(x) = a \). Since the limit \( a eq g(a) = 0 \), the function is not continuous at any rational point.

Step 4: Check Continuity at an Irrational Point

Let \( a \) be irrational. Then \( g(a) = a \). Consider the limit of \( g(x) \) as \( x \to a \) from both rational and irrational sides. For rational \( x \) near \( a \), \( g(x) = 0 \). For irrational \( x \), \( g(x) = x \). Thus, \( \lim_{x \to a} g(x) \) does not exist or differs depending on the path, meaning the function is not continuous at irrational \( a \) too.

Step 5: Conclude the Values of \( x \) Where \( g \) is Continuous

Since \( g(x) \) is not continuous at any rational \( x \) due to misalignment of limits, and not continuous at any irrational \( x \) due to nonexistence of the limit, \( g(x) \) is discontinuous for all \( x \).

Key Concepts

Piecewise Functions

Piecewise functions are defined in segments, where different rules apply to different parts of the function's domain. For the function given in the exercise, which is a classic example of a piecewise function, we have two distinct conditions:

• If \( x \) is rational, then \( g(x) = 0 \).
• If \( x \) is irrational, then \( g(x) = x \).

This means that the function behaves differently based on whether \( x \) is a rational or an irrational number.
Understanding how each piece of the function works independently is crucial for determining where the function might be continuous.
Since each section of the domain might affect the result and the continuity of the function, it’s important to analyze limits separately at rational and irrational points.

Limits and Continuity

A function is continuous at a point if its behavior at that point and its behavior as it approaches that point from either side are consistent.
To determine if \( g(x) \) is continuous, we examine if it satisfies the conditions of continuity, primarily:

• The function must be defined at the point.
• The limit of the function as it approaches the point must exist.
• The limit must equal the function's value at that point.

For rational points \( x = a \), the function \( g(a) = 0 \), while the limit \( \lim_{x \to a} g(x) = a \) if approaching through irrationals, does not satisfy \( \lim_{x \to a} g(x) = g(a) \).
At irrational points, rational approaches lead to the limit \( \lim_{x \to a} g(x) = 0 \), and irrational approaches lead to \( \lim_{x \to a} g(x) = a \). These differing limits mean there is no single limit value at irrational points. Thus, \( g(x) \) fails the continuity test at all points, remaining discontinuous everywhere.

Rational and Irrational Numbers

Understanding the difference between rational and irrational numbers plays a key role in solving problems involving piecewise functions like the one given.

• Rational numbers are numbers that can be expressed as the quotient or fraction \( \frac{p}{q} \), where \( p \) and \( q \) are integers, and \( q e 0 \).
• Irrational numbers cannot be written as simple fractions. This includes numbers like \( \pi \), \( e \), and \( \sqrt{2} \).

These two types of numbers together form the real number line, with the rationals densely packed yet still unable to fill the number line without irrationals.
In the function \( g(x) \), the discontinuity arises because the function tries to represent two separate behaviors that clash at any point, whether approaching from a rational or irrational side. Hence, understanding these types of numbers helps explain why there are no points where the function is continuous.