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Chapter 9: Problem 7

Prove Theorem 9.18 . The set \(\mathbb{Q}\) of all rational numbers is countable.

Short Answer

Expert verified
To prove that the set \(\mathbb{Q}\) of all rational numbers is countable, we need to establish a surjective function from the set of natural numbers to the set of rational numbers. We sort all pairs of integers \((a,b)\) as a two-dimensional grid and follow a diagonal path, mapping each pair to a unique natural number. After skipping over redundancies to ensure each rational number corresponds to only one natural number, we establish a surjective function, verifying Theorem 9.18 and proving that \(\mathbb{Q}\) is countable.

Step by step solution

01

Sort All Rational Numbers As A Sequence

A simple method to ensure that we cover all rational numbers is to sort all pairs of integers \((a,b)\), where \(b \neq 0\), as a two-dimensional grid. The first row and column are zero, and the rest of the numbers follow in order, both in the positive and negative directions. To code this, we can generate integer pairs (a,b) such that they generate all possible rational numbers.
02

Map Every Pair to a Unique Natural Number

To establish a bijection between the pairs of integers and natural numbers, we have to map every pair of integers \((a,b)\) to a unique natural number. This can be done by following a diagonal path on our two-dimensional grid of integers. Begin from the top left corner, follow the diagonal from top-right to bottom-left, then move one step to the right, and repeat.
03

Skipping Over Redundancies

The constructed sequence from step 2 contains redundancies since different integer pairs can represent the same rational number. For instance, \((1,2)\), \((2,4)\), \((3,6)\) all represents the rational number 0.5. These same rational numbers appear multiple times in our sequence. To ensure each rational number corresponds to a unique natural number, these redundancies need to be skipped.
04

Establish the Surjective Function

The unique identification of each rational number to a natural number finalizes our surjective function from natural numbers to rational numbers. By definition, if every element from the set of rational numbers corresponds to at least one natural number, then the set of rational numbers is countable. Therefore, we have now proved that the set of all rational numbers \(\mathbb{Q}\) is countable, hence verifying Theorem 9.18.

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Key Concepts

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

Bijection Between Rational Numbers and Natural Numbers
To grasp the concept of a bijection between the rational numbers and natural numbers, first consider what each type of number represents. Rational numbers, denoted as \(\mathbb{Q}\), include all fractions where both the numerator and the denominator are integers, and the denominator is not zero. Natural numbers, on the other hand, are the simple counting numbers that begin with 1 and go on indefinitely (\(1, 2, 3, ...\)).

A bijection, also known as a bijective function, is a connection between two sets where every element in the first set is paired with exactly one element in the second set, and vice versa. This means there is a perfect one-to-one match-up. In the context of rational numbers and natural numbers, establishing a bijection involves creating a system where every rational number corresponds to a unique natural number. This is done by arranging the rational numbers in a specific sequence, often in the form of an infinite grid. By following the diagonal paths and skipping redundancies, we can systematically assign a unique natural number to each rational number, proving that they can be paired off in a one-to-one fashion. Thus, remarkably, even though there are infinitely many of both, their quantities can be matched up perfectly.
Countable Sets
The notion of countability in mathematics has a precise meaning. A set is called countable if its elements can be counted one by one, that is, if there exists a bijection between the set and the natural numbers. Informally, a countable set can either be finite or infinite, but crucially, if it is infinite, you must be able to list its elements in a sequence like \(1, 2, 3, ...\), even though the list never ends.

There are different kinds of infinity, and not all infinite sets are countable. A key distinction is between countable infinity, like the rational numbers, and uncountable infinity, like the real numbers. This comes as a surprise to many, as it seems counter-intuitive that some infinities are 'larger' than others. When proving that the rational numbers are countable, the method of creating a bijection to the natural numbers must ensure that every rational number gets 'counted', so to speak. This concept challenges our notion of size and quantity, especially when dealing with infinite collections.
Surjective Function
In identifying the countability of rational numbers, the concept of a surjective function plays an integral role. A function is said to be surjective, or onto, if every element of the function's codomain (the set that the function maps to) is an image of at least one element from the domain (the set from which the function maps). In simpler terms, in a surjective function, every possible 'output' is covered by the function mapping.

When considering the surjective function between natural numbers and rational numbers, the focus is on ensuring that every rational number is represented as an 'output' of some natural number 'input'. If we can demonstrate this, then we have effectively paired each rational number with a natural number and vice versa, in a surjective manner. Thus, the rational numbers' countability hinges on the existence of a surjective function to the natural numbers and is a fundamental reason why we can claim the set of rational numbers is truly countable.

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Most popular questions from this chapter

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

Let \(B\) be a finite, nonempty set and assume that \(f: B \rightarrow A\) is a surjection. Prove that there exists a function \(h: A \rightarrow B\) such that \(f \circ h=I_{A}\) and \(h\) is an injection. Hint: Since \(B\) is finite, there exists a natural number \(m\) such that \(\mathbb{N}_{m} \approx B\). This means there exists a bijection \(k: \mathbb{N}_{m} \rightarrow B .\) Now let \(h=k \circ g,\) where \(g\) is the function constructed in Exercise (9).

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

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.

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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