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Chapter 9: Problem 2

Is the set of irrational numbers countable or uncountable? Prove that your answer is correct.

Short Answer

Expert verified
The set of irrational numbers is uncountable. To prove this, we can use Cantor's diagonal argument. We first assume that the set of irrational numbers is countable and organize them in a list. Then, we create a new number by taking each diagonal element and adding 1 to it. This new number differs from every number in our list by at least one digit, meaning it's not included in the list. This contradiction leads us to conclude that the set of irrational numbers is uncountable.

Step by step solution

01

Understanding Countable and Uncountable Sets

A set is considered countable if its elements can be put into one-to-one correspondence with the set of all natural numbers or a subset of natural numbers. An uncountable set, on the other hand, does not have such a one-to-one correspondence with the set of natural numbers or any subset of natural numbers. In other words, countable sets can be listed, whereas uncountable sets cannot be listed.
02

Recall the Properties of Irrational Numbers

An irrational number is a real number that cannot be expressed as a fraction of two integers. Examples of irrational numbers include square roots of non-perfect square numbers, \(蟺\) (pi), and \(e\) (Euler's number). Irrational numbers are often formed as the result of adding, subtracting, multiplying, or dividing rational and irrational numbers.
03

Using Cantor's Diagonal Argument

To prove that a set is uncountable, we can use Cantor's diagonal argument. This argument says that if there is a way to create a new element that is not in the given list of elements, then the set must be uncountable. In this case, we'll assume that the set of irrational numbers is countable and create a contradiction using Cantor's diagonal argument. Suppose we have a list of all irrational numbers and organize them as follows: \(x_1\)=0. x鈧佲倎 x鈧佲倐 x鈧佲們 x鈧佲倓 ... \(x_2\)=0. x鈧傗倎 x鈧傗倐 x鈧傗們 x鈧傗倓 ... \(x_3\)=0. x鈧冣倎 x鈧冣倐 x鈧冣們 x鈧冣倓 ... \(x_4\)=0. x鈧勨倎 x鈧勨倐 x鈧勨們 x鈧勨倓 ... . . . Now, we can create a new number by taking each diagonal element (x鈧佲倎, x鈧傗倐, x鈧冣們, ...) and adding 1 to it (if the digit is 1 or 9, add -1 to avoid going out of bounds), forming a new number: \(y\)=0. (x鈧佲倎+1) (x鈧傗倐+1) (x鈧冣們+1) (x鈧勨倓+1) ... This new number y is different from every number in our list by at least one digit, which means it is not included in the list. This contradicts our initial assumption that the list included all irrational numbers, and leads us to the conclusion that the set of irrational numbers must be uncountable.

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Key Concepts

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

Cantor's diagonal argument
One of the most fascinating topics in set theory is Cantor's diagonal argument, a simple yet powerful tool used to show that certain sets are uncountable. To understand it, let's explore a thought experiment: Imagine trying to list all the irrational numbers in a giant table where each number stretches out to an infinite decimal. If you could list them all, they would be countable, each corresponding to a natural number.

However, Cantor's surprising twist was to create a new irrational number that is guaranteed not to be on the list. By changing the nth digit of the nth number on the list, you create a number that differs from every single number listed in at least one decimal place. The creation of this new number鈥攐ur proof by contradiction鈥攕hows that no matter how comprehensive your list may seem, there will always be an irrational number that's not on it, proving that the set of irrational numbers is uncountable. This elegant argument shows there are levels of infinity, and some infinities are larger than others.
Uncountable sets
Understanding uncountable sets opens up the mind-boggling reality of different infinities. An uncountable set, as mentioned in the earlier exercise, is so vast that it cannot be paired up with the natural numbers. In other words, you cannot list all the elements because for every attempt to count them, there are always elements left out.

This realization not only applies to irrational numbers but also extends to the real numbers, of which irrationals are a subset. The concept of uncountability challenges our intuitive understanding of size and quantity. It throws open the doors to a world where some infinities are truly untamable by our usual tools of counting and listing, showcasing the richness and complexity of the mathematical universe.
Properties of irrational numbers
Irrational numbers are far more enigmatic than our everyday fractions and whole numbers, because they cannot be represented by a simple ratio of two integers. They go on forever without repeating, which is a beautiful yet perplexing property.

Common examples like the square root of 2, 蟺, or Euler's number 'e', show us that irrationals pop up quite naturally in geometry and calculus. They contain endless unpatterned numbers that never repeat, which not only defies our expectation of finding a pattern in everything, but also defines a remarkable characteristic that differentiates them from rational numbers. The 'non-repeating' aspect is crucial in the argument about uncountability because it's what prevents us from finding a neat correlation between the irrational numbers and natural numbers.

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Most popular questions from this chapter

For this activity, we will consider subsets of \(\mathbb{N}_{30}\) that contain eight elements. (a) One such set is \(A=\\{3,5,11,17,21,24,26,29\\} .\) Notice that $$ \\{3,21,24,26\\} \subseteq A \quad \text { and } \quad 3+21+24+26=74 $$ \\{3,5,11,26,29\\}\(\subseteq A\) and \(\quad 3+5+11+26+29=74\) Use this information to find two disjoint subsets of \(A\) whose elements have the same sum. (b) Let \(B=\\{3,6,9,12,15,18,21,24\\}\). Find two disjoint subsets of \(B\) whose elements have the same sum. Note: By convention, if \(T=\\{a\\},\) where \(a \in \mathbb{N},\) then the sum of the elements in \(T\) is equal to \(a\). (c) Now let \(C\) be any subset of \(\mathbb{N}_{30}\) that contains eight elements. i. How many subsets does \(C\) have? ii. The sum of the elements of the empty set is \(0 .\) What is the maximum sum for any subset of \(\mathbb{N}_{30}\) that contains eight elements? Let \(M\) be this maximum sum. iii. Now define a function \(f: \mathcal{P}(C) \rightarrow \mathbb{N}_{M}\) so that for each \(X \in\) \(\mathcal{P}(C), f(X)\) is equal to the sum of the elements in \(X\). Use the Pigeonhole Principle to prove that there exist two subsets of \(C\) whose elements have the same sum. (d) If the two subsets in part (11(c)iii) are not disjoint, use the idea presented in part (11a) to prove that there exist two disjoint subsets of \(C\) whose elements have the same sum. (e) Let \(S\) be a subset of \(\mathbb{N}_{99}\) that contains 10 elements. Use the Pigeonhole Principle to prove that there exist two disjoint subsets of \(S\) whose elements have the same sum.

Prove that if \(A\) is countably infinite and \(B\) is finite, then \(A-B\) is countably infinite.

Another Proof that \(\mathrm{Q}^{+}\) Is Countable. For this activity, it may be helpful to use the Fundamental Theorem of Arithmetic (see Theorem 8.15 on page 432 ). Let \(Q^{+}\) be the set of positive rational numbers. Every positive rational number has a unique representation as a fraction \(\frac{m}{n},\) where \(m\) and \(n\) are relatively prime natural numbers. We will now define a function \(f: \mathbb{Q}^{+} \rightarrow \mathbb{N}\) as follows: If \(x \in Q^{+}\) and \(x=\frac{m}{n},\) where \(m, n \in \mathbb{N}, n \neq 1\) and \(\operatorname{gcd}(m, n)=1,\) we write $$ \begin{array}{l} m=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{r}^{\alpha_{r}} \\ n=q_{1}^{\beta_{1}} q_{2}^{\beta_{2}} \cdots q_{s}^{\beta_{x}} \end{array} $$ where \(p_{1}, p_{2}, \ldots, p_{r}\) are distinct prime numbers, \(q_{1}, q_{2}, \ldots, q_{s}\) are distinct prime numbers, and \(\alpha_{1}, \alpha_{2}, \ldots, \alpha_{r}\) and \(\beta_{1}, \beta_{2}, \ldots, \beta_{s}\) are natural numbers. We also write \(1=2^{0}\) when \(m=1\). We then define $$ f(x)=p_{1}^{2 \alpha_{1}} p_{2}^{2 \alpha_{2}} \cdots p_{r}^{2 \alpha_{r}} q_{1}^{2 \beta_{1}-1} q_{2}^{2 \beta_{2}-1} \cdots q_{3}^{2 \beta_{s}-1} $$ If \(x=\frac{m}{1},\) then we define \(f(x)=p_{1}^{2 \alpha_{1}} p_{2}^{2 \alpha_{2}} \cdots p_{r}^{2 \alpha_{r}}=m^{2}\). (a) Determine \(f\left(\frac{2}{3}\right), f\left(\frac{5}{6}\right), f(6), f\left(\frac{12}{25}\right), f\left(\frac{375}{392}\right),\) and \(f\left(\frac{2^{3} \cdot 11^{3}}{3 \cdot 5^{4}}\right)\). (b) If possible, find \(x \in \mathbb{Q}^{+}\) such that \(f(x)=100\). (c) If possible, find \(x \in \mathrm{Q}^{+}\) such that \(f(x)=12\). (d) If possible, find \(x \in Q^{+}\) such that \(f(x)=2^{8} \cdot 3^{5} \cdot 13 \cdot 17^{2}\). (e) Prove that the function \(f\) is an injection. (f) Prove that the function \(f\) is a surjection. (g) What has been proved?

Do two uncountable sets always have the same cardinality? Justify your conclusion.

Complete the proof of Theorem 9.17 by proving the following: Let \(A\) and \(B\) be disjoint countably infinite sets and let \(f: \mathbb{N} \rightarrow A\) and \(g: \mathbb{N} \rightarrow\) \(B\) be bijections. Define \(h: \mathbb{N} \rightarrow A \cup B\) by $$ h(n)=\left\\{\begin{array}{ll} f\left(\frac{n+1}{2}\right) & \text { if } n \text { is odd } \\ g\left(\frac{n}{2}\right) & \text { if } n \text { is even. } \end{array}\right. $$ Then the function \(h\) is a bijection.
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