Problem 76 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(f(x)=0\) if \(x\) is irrational and let \(f(x)=1 / q\) if \(x\) is the rational number \(p / q\) in reduced form \((q>0)\). (a) Sketch (as best you can) the graph of \(f\) on \((0,1)\). (b) Show that \(f\) is continuous at each irrational number in \((0,1)\), but is discontinuous at each rational number in \((0,1)\).

Step-by-Step Solution

Answer

\( f(x) \) is continuous at all irrationals and discontinuous at rationals.

Step 1: Understanding the Function

The function \( f(x) \) is defined such that it evaluates to \( 0 \) for all irrational numbers in the interval \((0,1)\), and to \( \frac{1}{q} \) for rational numbers expressed in lowest terms as \( \frac{p}{q} \). This means that \( f(x) \) takes small positive values for rational numbers and zero for irrationals.

Step 2: Graphing the Function: Visualizing Discontinuity

When sketching the graph of \( f(x) \) on \((0,1)\), note that for any irrational value \(x\), \( f(x) = 0 \). For rational values, \( f(x) \) is a positive fraction determined by the denominator \(q\) in \(x = \frac{p}{q}\). Points corresponding to rational numbers appear as isolated dots above the x-axis, whereas irrational points form a continuous line along the x-axis at height 0.

Step 3: Showing Continuity at Irrational Numbers

For continuity at an irrational number \(c\), consider any \( \epsilon > 0 \). We need to show there is a \( \delta > 0 \) such that whenever \( |x - c| < \delta \), then \( |f(x) - f(c)| < \epsilon \). Since \( f(c) = 0 \), this simplifies to showing \( |f(x)| < \epsilon \). Given rational numbers \( x = \frac{p}{q} \) in \((c-\delta, c+\delta)\), we can choose \( \delta \) small enough so \( q > \frac{1}{\epsilon} \). Thus \( |f(x)| = \frac{1}{q} < \epsilon \), proving continuity at irrationals.

Step 4: Proving Discontinuity at Rational Numbers

For a rational \(c = \frac{p}{q}\), we need to show there exists an \( \epsilon > 0 \) for which no \( \delta > 0 \) makes \( |f(x) - f(c)| < \epsilon \) whenever \( |x - c| < \delta \). Choose \( \epsilon < \frac{1}{2q} \). For any \( \delta > 0 \), there exist irrational numbers within \((c-\delta, c+\delta)\), having \(f(x) = 0\). The difference \( |f(x) - f(c)| = \frac{1}{q} \), greater than \( \epsilon \), proving discontinuity at rationals.

Key Concepts

Rational and Irrational Numbers

Let's dive into the concepts of rational and irrational numbers. Rational numbers are any numbers that can be expressed as the quotient of two integers, where the denominator is not zero. In other words, they can be written in the form \(\frac{p}{q}\), with integers \(p\) and \(q\), and \(q > 0\). This includes fractions like \(\frac{1}{2}\) or whole numbers like 3 (which is \(\frac{3}{1}\)).

Irrational numbers, on the other hand, cannot be expressed as a simple fraction. These numbers have non-repeating, non-terminating decimal expansions. Examples include \(\pi\) and \(\sqrt{2}\). In the context of the function \(f(x)\) discussed, all irrational numbers on the interval \((0,1)\) lead the function to evaluate to zero. For rational numbers, however, the function relies on the denominator of their fractional form to determine its value.

Graph Analysis

Analyzing the graph for the given function \(f(x)\) offers insight into how it behaves across the interval \((0,1)\). For every irrational number in this interval, \(f(x)\) is zero, meaning the graph sits entirely on the x-axis at these points. However, at rational numbers, the function value is expressed as a fraction \(\frac{1}{q}\) where \(x = \frac{p}{q}\).

When graphing this function, you will observe that:

• Irrational number evaluations create a continuous line along the x-axis, indicating zero.
• Rational numbers manifest as discrete points above the x-axis, often appearing as dots or tiny line segments, since \(f(x)\) returns small fractions.

These discrete points highlight discontinuities, making the function's graph appear "dotted" with discrete jumps at every rational point, while remaining flat and unbroken wherever irrational numbers occur.

Epsilon-Delta Definition

The epsilon-delta definition of continuity is a fundamental concept in calculus to determine whether a function is continuous at a given point. For this function, we particularly emphasize proving continuity and discontinuity.

For any irrational number \(c\), the goal is to show that for every \(\epsilon > 0\), there is a corresponding \(\delta > 0\) such that if \(|x - c| < \delta\), then \(|f(x) - f(c)| < \epsilon\). Given \(f(c) = 0\), we simplify to ensuring \(|f(x)| < \epsilon\). Thus, by selecting \(\delta\) so small that any rational number \(x = \frac{p}{q}\) within \((c-\delta, c+\delta)\) has \(q > \frac{1}{\epsilon}\), we ensure \(|f(x)| = \frac{1}{q} < \epsilon\), proving continuity at irrational points.

Conversely, discontinuity at a rational number \(c = \frac{p}{q}\) requires showing that for some \(\epsilon > 0\), no \(\delta > 0\) exists such that \(|f(x) - f(c)| < \epsilon\) if \(|x - c| < \delta\). Selecting \(\epsilon < \frac{1}{2q}\), and knowing irrationals exist within any neighborhood \((c-\delta, c+\delta)\), we ascertain \(|f(x) - f(c)| = \frac{1}{q} > \epsilon\), thereby proving its discontinuity at rational points.