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

The function is continuous at irrationals and discontinuous at rationals.

Step 1: Understanding the Function

The function \(f(x)\) is defined such that \(f(x)=0\) when \(x\) is irrational, and \(f(x)=\frac{1}{q}\) when \(x\) is a rational number expressed in the form \(\frac{p}{q}\) with \(p\) and \(q\) being coprime and \(q>0\). Our goal is to graphically represent \(f\) on the interval \((0,1)\) and determine its continuity at both irrational and rational numbers.

Step 2: Graphing the Function

For any irrational \(x\), \(f(x) = 0\), so the graph contains the line \(y = 0\) along all irrational points in \((0,1)\). For rational \(x = \frac{p}{q}\), the value \(f(x) = \frac{1}{q}\), indicating vertical spikes or dots at rational points with heights depending inversely on \(q\). These spikes become denser as \(x\) approaches any point in \((0,1)\).

Step 3: Continuity at Irrational Points

To check continuity at an irrational \(c\), note by definition \(f(c) = 0\). For any \(\epsilon > 0\), choose \(\delta > 0\) such that the interval \((c - \delta, c + \delta)\) contains only points \(x\) with \(f(x)\) approaching \(0\), as \(\frac{1}{q} < \epsilon\) retains closeness regardless of rational approximations with large \(q\). Thus, \(f(x)\) is continuous at \(c\).

Step 4: Discontinuity at Rational Points

Consider a rational \(c = \frac{p}{q}\), then \(f(c) = \frac{1}{q}\). For any chosen \(\epsilon > 0\), near extreme \(\delta > 0\) neighborhoods include irrational numbers with \(f(x) = 0\), deviating from \(\frac{1}{q}\), making continuity false, as \(|f(x) - \frac{1}{q}| \geq \frac{1}{q}\) for any chosen small \(\epsilon\). Thus, \(f(x)\) is discontinuous at \(c\).

Key Concepts

Discontinuous Functions

Discontinuous functions often have small breaks or jumps in their graph. This can make them tricky to visualize and understand in a calculus context. In the case of the given function \( f(x) \), it is defined specifically to be discontinuous at each rational number within the interval \((0,1)\).

A function is considered continuous at a point if you can draw it without lifting your pencil from the paper. This means there's no jump or break at that point. Here, for each rational number \( \frac{p}{q} \), the function value \( f(x) = \frac{1}{q} \) creates a spike at that point, indicating a discontinuity. This is because for any small neighborhood around a rational point, there are irrational numbers where \( f(x) = 0 \), creating a gap from the spike back to zero. Thus, the function cannot be "smoothly connected" at these rational points. It "jumps" between zero and the value \( \frac{1}{q} \).

• Notice how as \( q \) becomes larger, these spikes become closer to zero, but there is always a jump wherever a rational number is present in reduced form.
• This characterizes the visual "spiky" nature of the graph of \( f(x) \) on \((0,1)\).

Rational and Irrational Numbers

Understanding the distinction between rational and irrational numbers is key to grasping the continuity and discontinuity of functions like \( f(x) \).

**Rational Numbers:** These are numbers that can be expressed as the fraction \( \frac{p}{q} \), where \( p \) and \( q \) are integers, and \( q eq 0 \). For our function, this means applying a specific rule \( f(x) = \frac{1}{q} \) when \( x \) is a rational number.

• Examples: \( \frac{1}{2}, \frac{3}{4}, 1 \)

**Irrational Numbers:** These numbers cannot be expressed as a simple fraction, as their decimal forms go on indefinitely without repeating. Such numbers are pervasive on the number line, even between any two rational numbers.

• Examples: \( \sqrt{2}, \pi, e \)

In the context of \( f(x) \), irrational numbers always take on the value \( 0 \). This constant value at irrational points creates a "rest" or baseline level of the function amidst the spikes of rational values.

• Recognizing whether a number is rational or irrational is crucial. It dictates whether \( f \) "spikes" or "rests" at zero, directly affecting the function's continuity.

Graphing Functions

Graphing functions like \( f(x) \) provides visual insight into their behavior over an interval.

**Step 1: Identify Behaviors at Irrational and Rational Points**
For irrational \( x \) in \((0,1)\), plot the line \( y = 0 \). These are easy to graph as a continuous flat line, representing the constant zero output for every irrational input.
For rational \( x = \frac{p}{q} \), plot a point at \( \left( x, \frac{1}{q} \right) \). These points create spikes since \( f(x) \) suddenly jumps from 0 to \( \frac{1}{q} \) only at rational inputs.

**Step 2: Connect the Points**
These spikes do not connect horizontally due to the discontinuity at rational numbers, but they show the frequent "jumps" or "gaps" that occur all over the interval.

This graph is full of dots and lines but no connecting curves between the spikes at rational numbers—illustrating the discontinuous nature. Trying to graphically represent such a function in its entirety is challenging due to the infinite amount of points in \((0,1)\) but visualizing at a small segment can often convey the concept effectively.

• Use software or graph paper for precision in plotting these dense/unusal graphs.
• Focus on visualizing the density of the spikes and the flatness of the line at zero to understand the function's nature.