Problem 47 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(S=\left\\{x: x\right.\) is rational and \(\left.x^{2}<2\right\\}\). Convince yourself that \(S\) does not have a least upper bound in the rational numbers, but does have such a bound in the real numbers. In other words, the sequence of rational numbers \(1,1.4,1.41,1.414, \ldots\), has no limit within the rational numbers.

Step-by-Step Solution

Answer

The set \( S \) has no least upper bound in the rationals but does in the reals (\( \sqrt{2} \)).

Step 1: Understanding the Problem

The problem is asking us to find out if the set \( S = \{ x : x \text{ is rational and } x^2 < 2 \} \) has a least upper bound (lub) in the rational numbers and in the real numbers.

Step 2: Analyze Bounds in Rational Numbers

Let's identify a rational upper bound for \( S \). Notice that every rational number \( x \) such that \( x > \sqrt{2} \) is an upper bound of \( S \). However, \( \sqrt{2} \) is not rational, so there is no smallest rational upper bound.

Step 3: Sequence Evaluation

Consider the sequence of rational numbers \( 1, 1.4, 1.41, 1.414, \ldots \). Each term of this sequence squares to approach closer or less than 2, but never reaches 2 because \( \sqrt{2} \) is irrational.

Step 4: Conclusion for Rational Numbers

Since there is no smallest rational number greater than all elements of \( S \), we conclude that \( S \) does not have a least upper bound in the rationals.

Step 5: Analyze Bounds in Real Numbers

In the real numbers, \( \sqrt{2} \) is a real number and is the least upper bound of \( S \). This is because for any \( \epsilon > 0 \), there exists a rational number \( r \) such that \( \sqrt{2} - \epsilon < r < \sqrt{2} \).

Key Concepts

Rational Numbers

Rational numbers are quite intriguing due to their fundamental properties in mathematics. They are the numbers that can be expressed as the quotient or fraction of two integers, say \( \frac{p}{q} \), where \( p \) and \( q \) are integers and \( q eq 0 \).
This means that any number you can write as a simple fraction is rational. For example, \( \frac{1}{2} \), \( -3 \), and \( 4.5 \) (which is \( \frac{9}{2} \)) are all rational numbers. It's important to know that the decimal representation of a rational number either stops eventually or repeats.
For instance, \( \frac{1}{4} \) equals 0.25 (terminating), and \( \frac{1}{3} \) equals 0.333... (repeating).In the context of our problem, rational numbers play a crucial role in exploring bounds. We are tasked with determining if there is a smallest rational upper limit for a set composed of rational numbers less than \( \sqrt{2} \).
Understanding the limitations of rational numbers helps in finding such bounds.

Real Numbers

Real numbers encompass both the set of rational numbers and irrational numbers, forming a continuous set of values.
Think of the real numbers as points lying on an infinitely long line. This line includes all the fractions we're familiar with (rationals) and numbers that cannot be expressed as fractions (irrationals). When talking about least upper bounds, real numbers are particularly significant because they fill in the gaps between rational numbers.
In our exercise, while \( \sqrt{2} \) is not a rational number, it is definitely a real number. This enables it to act as the least upper bound of the set \( S \), whereas no such rational number exists. In essence, real numbers allow us to precisely capture limits and bounds in calculations, which rational numbers alone cannot achieve. Hence, \( \sqrt{2} \) being real is what allows \( S \) to have a least upper bound in the reals.

Irrational Numbers

Irrational numbers are those numbers that cannot be expressed as a simple fraction of two integers. Their decimal expansions are neither terminating nor repeating.
Examples include numbers like \( \pi \), \( e \), and the square root of non-perfect squares, such as \( \sqrt{2} \).They are critical in mathematics because they fill up the 'gaps' that rational numbers leave behind.
In our problem's context, \( \sqrt{2} \) cannot be precisely equated to any fraction, yet it is crucial because it serves as the least upper bound for set \( S \) in the real numbers.Understanding irrational numbers is essential to fully grasp differences between rational and real number-related limits, especially when we're dealing with concepts like upper bounds or convergence in sequences.

Sequence Convergence

The idea of sequence convergence is central in analyzing bounds and limits. A sequence is simply a list of numbers in a specific order, and it is said to converge if it approaches a specific value as it progresses.
In our scenario, the sequence \( 1, 1.4, 1.41, 1.414, \ldots \) represents rational numbers getting closer to \( \sqrt{2} \).However, in the realm of rational numbers, this sequence will never "reach" \( \sqrt{2} \) since \( \sqrt{2} \) is irrational. The sequence converges towards \( \sqrt{2} \), but its limit is not a rational number.
Thus, there is no least upper rational bound for this sequence within the rational numbers.The concept of convergence illuminates why rationals alone can't provide a least upper bound, highlighting the necessity of real numbers in more advanced mathematical analyses.