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Chapter 2: Problem 4

Is the set of all irrational real numbers countable?

Short Answer

Expert verified
Explain your answer. Answer: The set of all irrational real numbers is not countable. The proof involves assuming that the set of irrational real numbers is countable and then using the diagonalization argument to show that this assumption leads to a contradiction. The diagonalization argument constructs an irrational real number that is not in our assumed list, proving that the set of all irrational real numbers is not countable.

Step by step solution

01

Understanding irrational real numbers

Irrational real numbers are real numbers that cannot be expressed as a ratio of two integers. In other words, an irrational number cannot be written in the form p/q, where p and q are integers, and q is not equal to zero. Examples of irrational numbers include the square root of 2, 蟺 (pi), and e (Euler's number).
02

Proof by contradiction

In order to prove that the set of all irrational real numbers is not countable, we will assume that it is countable, and then attempt to show that this assumption leads to a contradiction. If we can do this, it will prove that our original assumption must be false, and therefore, that the set of all irrational real numbers is not countable.
03

Assumption of countability

Let's assume that the set of all irrational real numbers is countable. If this is true, then we should be able to create a list of all these numbers in such a way that each number on the list corresponds to exactly one natural number. For example, our list might start like this: 1. sqrt(2) 2. 蟺 3. e ... And so on.
04

Diagonalization argument

We can now use a technique called the diagonalization argument, which is originally due to Cantor, to show that our assumption of countability must be false. The diagonalization argument goes like this: Assume that we have listed all the irrational real numbers in such a way that each number corresponds to exactly one natural number. Now, consider an irrational real number R that is constructed as follows: For the n-th digit after the decimal point of R, select a digit different from the n-th digit after the decimal point of the n-th number in our list. Since we have chosen a digit different from that of each number in our assumed list at its corresponding position, the number R can never be in the list. This is a contradiction to our assumption of countability.
05

Conclusion

The diagonalization argument shows that our assumption of countability for the set of all irrational real numbers is false. Therefore, we conclude that the set of all irrational real numbers is not countable.

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Key Concepts

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

Countability
Countability refers to the ability to list elements in a set such that there is a one-to-one correspondence with natural numbers. A set is countable if you can enumerate all its elements, just like numbering them as 1, 2, 3, and so on. This means that each element from the set can be paired with a unique natural number.

When it comes to determining whether the set of irrational numbers is countable, we need to think about this concept of listing and pairing elements. Rational numbers, for example, are countable because we can systematically list them, even though it might seem complex. However, the concept of countability changes when applied to irrational numbers, as we shall see in the proofs below regarding the irrational real numbers.
Diagonalization Argument
The diagonalization argument is a clever technique used to show that certain sets cannot be counted or listed like natural numbers. This argument was famously utilized by mathematician Georg Cantor to show that the set of real numbers is not countable, meaning it is uncountably infinite.
  • Imagine trying to list all irrational numbers one after the other in a sequence.
  • For each position in the list, we'll assume we can determine some decimal value for the corresponding irrational number.
Now, by constructing a new number which differs in at least one decimal place for each number in your list (using the n-th position from the n-th number to derive this new number), we form a contradiction.

This newly formed number can't possibly be on the original list, because it deliberately differs from every listed number at a particular decimal place. Therefore, this proves our initial listing attempt wrong. The so-called diagonalization essentially builds a number not accountable within our listing, hence proving the set cannot be countable.
Proof by Contradiction
A proof by contradiction is a logical method that begins with assuming the opposite of what you want to prove. The idea is to show that this assumption inevitably leads to a contradiction, thus proving that the original statement must be true.

In the case of proving that the set of all irrational numbers is not countable, we start by assuming that it is countable. By this assumption, we should, theoretically, be able to list every irrational number with a unique natural number corresponding to each one.
  • We then use the deliberate construction of a new number through the diagonalization method.
  • This newly constructed number cannot be part of the list, as it was specifically designed to differ from each element already listed.
Once we identify this contradiction鈥攖hat there will always be at least one irrational number unlisted despite our assumption鈥攚e prove that our initial assumption was false. Thus, irrational numbers, as shown through this contradiction, form an uncountably infinite set.

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Most popular questions from this chapter

Let \(X\) be an infinite set. For \(p \in X\) and \(q \in X\), define $$ d(p, q)=\left\\{\begin{array}{ll} 1 & (\text { if } p \neq q) \\ 0 & \text { (if } p=q) . \end{array}\right. $$ Prove that this is a metric. Which subsets of the resulting metric space are open? Which are closed? Which are compact?

Define a point \(p\) in a metric space \(X\) to be a condensation point of a set \(E \subset X\) if every neighborhood of \(p\) contains uncountably many points of \(E\). Suppose \(E \subset R^{k}, E\) is uncountable, and let \(P\) be the set of all condensation points of \(E\). Prove that \(P\) is perfect and that at most countably many points of \(E\) are not in \(P\). In other words, show that \(P^{\epsilon} \cap E\) is at most countable. Hint: Let \(\left\\{V_{n}\right\\}\) be a countable base of \(R^{k}\), let \(W\) be the union of those \(V_{n}\) for which \(E \cap V_{n}\) is at most countable, and show that \(P=W^{c}\).

Let \(A_{1}, A_{2}, A_{3}, \ldots\) be subsets of a metric space. (a) If \(B_{n}=\bigcup_{n=1} A_{i}\), prove that \(B_{n}=U_{i=1} A_{i}\), for \(n=1,2,3, \ldots .\) (b) If \(B=\bigcup_{i=1} A_{i}\), prove that \(B \supset \bigcup_{i=1}^{\infty} X_{i} .\) Show, by an example, that this inclusion can be proper.

Let \(E\) be the set of all \(x \in[0.1]\) whose decimal expansion contains only the digits 4 and 7 , Is \(E\) countable? Is \(E\) dense in \([0,1] ?\) Is \(E\) compact? Is \(E\) perfect?

Let \(X\) be a metric space in which every infinite subset has a limit point. Prove that \(X\) is compact. Hint: By Exercises 23 and \(24, X\) has a countable base. It follows that every open cover of \(X\) has a countable subcover \(\left\\{G_{a}\right\\}, n=1,2,3, \ldots\) If no finite subcollection of \(\left\\{G_{n}\right\\}\) covers \(X\), then the complement \(F_{n}\) of \(G_{1} \cup \cdots \cup G_{n}\) is nonempty for each \(n\), but \(\cap F_{n}\) is empty. If \(E\) is a set which contains a point from each \(F_{n}\), consider a limit point of \(E\), and obtain a contradiction.
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