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Chapter 1: Problem 194

Let \(f(x)=x-1 / x\). For any real number \(x_0\), consider the sequence defined by \(x_0, x_1=f\left(x_0\right), \ldots, x_{n+1}=f\left(x_n\right), \ldots\), provided \(x_n \neq 0\). Define \(x_0\) to be a \(T\)-number if the sequence terminates, that is, if \(x_n=0\) for some \(n\). (For example, \(-1\) is a T-number because \(f(-1)=0\), but \(\sqrt{2}\) is not, because the sequence $$ \sqrt{2}, \quad 1 / \sqrt{2}=f(\sqrt{2}), \quad-1 / \sqrt{2}=f(1 / \sqrt{2}), \quad 1 / \sqrt{2}=f(-1 / \sqrt{2}), \ldots $$ does not terminate.) a. Show that the set of all T-numbers is countably infinite (denumerable). b. Does every open interval contain a T-number?

Short Answer

Expert verified
T-numbers are countably infinite because they correspond to rational numbers that under function composition lead to zero. Each open interval contains a T-number because rationals are dense in real numbers.

Step by step solution

01

Understand the Function and the Sequence

The function given is \(f(x) = x - \frac{1}{x}\). The sequence \(x_0, x_1, x_2, \ldots\) is defined by \(x_{n+1} = f(x_n)\). A T-number is a number \(x_0\) such that the sequence terminates, meaning \(x_n = 0\) for some n.
02

Determine Conditions for Termination

For the sequence to terminate, there must be some \(x_n = 0\). This means for some n, \(f(x_{n-1}) = 0\). Since \(f(x) = x - \frac{1}{x}\), set this equal to 0 and solve: \[ x - \frac{1}{x} = 0 \] This gives \(x^2 = 1\), so \(x = 1\) or \(x = -1\).
03

Identify T-numbers

To see that the set of T-numbers is countably infinite, note that if we start with any rational number which can be expressed as \(\frac{p}{q}\) (where p and q are integers with q not zero), the sequence will eventually reach a value where multiplication by the inverse creates 1 or -1, given enough steps and without hitting a zero denominator prematurely. Examples include any positive or negative rational number p/q other than already the roots. Each rational number is expressed in a simplified form which obviously remains denumerable. For instance: starting from rational values like 0.5, 2, -2 etc set to their least denominator combinations.
04

Showing Open Intervals Contain T-numbers

Open intervals are open by their nature for all real numbers between the intervals. Considering the property of rationals where they are dense in real domains, construct any sample small interval, which will definitely include at least one rational number and hence according to the previous step, those rationals are in turn T numbers. Therefore, each open interval will have a T number due to rationals.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sequence Termination
To understand sequence termination, let's first think about what it means for a sequence to 'end' or 'terminate'. In our example, a sequence terminates if it eventually hits the number 0. In mathematical terms, this is when for some n, we have \(x_n = 0\). So, if we start with a number \(x_0\) and keep applying the function \(f(x) = x - \frac{1}{x}\), we need to see if we ever get to 0. If we can find such a number \(x_0\), it is called a T-number. This is because its sequence stops when hitting zero.
Rational Numbers
Rational numbers are numbers that can be expressed as the quotient or fraction \(\frac{p}{q}\) of two integers, where \(p\) and \(q\) are integers and \(q eq 0\). Examples include numbers like 1/2, -3/4, and 5.0 (since 5 can be written as \( \frac{5}{1} \)). Almost all numbers we encounter in daily life (like fractions) are rational numbers. In this problem, rational numbers play a crucial role because we are looking for T-numbers, which are forming sequences that would terminate based on the given function.
Countably Infinite Sets
A set is countably infinite if it can be put into a one-to-one correspondence with the set of natural numbers \(\{1, 2, 3, \ldots\}\). This means that even though the set is infinite, we can 'count' its elements one by one. The set of all rational numbers is countably infinite. This is important because the problem asks us to show that the set of T-numbers is countably infinite. Since rational numbers make up the T-numbers, we can list T-numbers in a way analogous to how we list natural numbers.
Open Intervals
An open interval \((a, b)\) is a range of numbers that includes all real numbers between a and b but does not include the endpoints a and b themselves. For instance, the open interval \((0, 1)\) contains numbers like 0.1 and 0.999999, but not 0 or 1. The problem states every open interval contains a T-number. Because rational numbers are densely packed in real numbers, any range of numbers we pick, no matter how small, will contain rational numbers, hence containing T-numbers. Thus, we can say that every open interval has at least one T-number.

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Most popular questions from this chapter

Note that the three positive integers \(1,24,120\) have the property that the sum of any two of them is a different perfect square. Do there exist four positive integers such that the sum of any two of them is a perfect square and such that the six squares found in this way are all different? If so, exhibit four such positive integers; if not, show why this cannot be done.

Let \(\mathbb{Z} / n \mathbb{Z}\) be the set \(\\{0,1, \ldots, n-1\\}\) with addition modulo \(n\). Consider subsets \(S_n\) of \(\mathbb{Z} / n \mathbb{Z}\) such that \(\left(S_n+k\right) \cap S_n\) is nonempty for every \(k\) in \(\mathbb{Z} / n \mathbb{Z}\). Let \(f(n)\) denote the minimal number of elements in such a subset. Find $$ \lim _{n \rightarrow \infty} \frac{\ln f(n)}{\ln n} $$ or show that this limit does not exist.

The proprietor of the Wohascum Puzzle, Game, and Computer Den has invented a new two-person game, in which players take turns coloring edges of a cube. Three colors (red, green, and yellow) are available. The cube starts off with all edges uncolored; once an edge is colored, it cannot be colored again. Two edges with a common vertex are not allowed to have the same color. The last player to be able to color an edge wins the game. a. Given best play on both sides, should the first or the second player win? What is the winning strategy? b. There are twelve edges in all, so a game can last at most twelve turns (whether or not the players use optimal strategies); it is not hard to see that twelve turns are possible. How many twelve-turn end positions are essentially different? (Two positions are considered essentially the same if one can be obtained from the other by rotating the cube.) \(\quad\)

In general, composition of functions is not commutative. For example, for the functions \(f\) and \(g\) given by \(f(x)=x+1, g(x)=2 x\), we have \(f(g(x))=2 x+1\) and \(g(f(x))=2 x+2\). Now suppose that we have three functions \(f, g, h\). Then there are six possible compositions of the three, given by \(f(g(h(x))), g(h(f(x))), \ldots\). Give an example of three continuous functions that are defined for all real \(x\) and for which exactly five of the six compositions are the same. (Reprinted with the permission of the Canadian Mathematical Society, this problem was originally published in the Mathematical Mayhem section of Crux Mathematicorum with Mathematical Mayhem, vol. 25,1999, p. 293, problem C87.)

Starting with a positive number \(x_0=a\), let \(\left(x_n\right)_{n \geq 0}\) be the sequence of numbers such that $$ x_{n+1}= \begin{cases}x_n^2+1 & \text { if } n \text { is even } \\\ \sqrt{x_n}-1 & \text { if } n \text { is odd }\end{cases} $$ For what positive numbers \(a\) will there be terms of the sequence arbitrarily close to 0 ?
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