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Chapter 7: Problem 5

Let \(25 \mathbf{Z}\) be the set of all integers that are multiples of 25 . Prove that \(25 \mathbf{Z}\) has the same cardinality as \(2 \mathbf{Z}\), the set of all even integers.

Short Answer

Expert verified
To prove that the sets \(25\mathbf{Z}\) and \(2\mathbf{Z}\) have the same cardinality, define a function \(f: 25\mathbf{Z} \to 2\mathbf{Z}\) such that \(f(25n) = 2n\) for every \(n \in \mathbf{Z}\). Prove that \(f\) is injective by showing that if \(f(25n_1) = f(25n_2)\), then \(n_1 = n_2\). Prove that \(f\) is surjective by showing that for each element \(2n \in 2\mathbf{Z}\), there exists an element \(25n' \in 25\mathbf{Z}\) such that \(f(25n') = 2n\). Since \(f\) is both injective and surjective, it is bijective, proving that the sets \(25\mathbf{Z}\) and \(2\mathbf{Z}\) have the same cardinality.

Step by step solution

01

Define the sets

Write down the definition of the sets: \(25\mathbf{Z}\) = \(\{25n : n \in\mathbf{Z}\}\) - the set of all multiples of 25. \(2\mathbf{Z}\) = \(\{2n : n \in\mathbf{Z}\}\) - the set of all even integers.
02

Define a function

Define a function \(f: 25\mathbf{Z} \to 2\mathbf{Z}\) such that \(f(25n) = 2n\) for every \(n \in \mathbf{Z}\). This function maps each element in \(25\mathbf{Z}\) to a corresponding element in \(2\mathbf{Z}\).
03

Prove that the function is injective

To prove that the function is injective (one-to-one), we need to show that if \(f(25n_1) = f(25n_2)\), then \(n_1 = n_2\). So let's assume that \(f(25n_1) = f(25n_2)\): \(2n_1 = 2n_2 \implies n_1 = n_2\) Since \(n_1 = n_2\), we have shown that the function is injective.
04

Prove that the function is surjective

To prove that \(f\) is surjective (onto), we need to show that for each element \(2n \in 2\mathbf{Z}\), there exists an element \(25n' \in 25\mathbf{Z}\) such that \(f(25n') = 2n\). Let's take an arbitrary even integer \(2n\). We can find an element \(25n'\) in \(25\mathbf{Z}\) that maps to \(2n\) as follows: \(n' = \frac{n}{25}\) Therefore, \(f(25n') = f(25 \cdot \frac{n}{25}) = 2n\). Since we can find such an element \(25n'\) for any \(2n \in 2\mathbf{Z}\), we have shown that the function is surjective.
05

Conclude that the function is bijective

Since we have proven that function \(f\) is both injective and surjective, we can conclude that \(f\) is bijective. This demonstrates that the sets \(25\mathbf{Z}\) and \(2\mathbf{Z}\) have the same cardinality.

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Key Concepts

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

Cardinality
Cardinality refers to the size of a set, or how many elements it contains. In set theory, two sets are said to have the same cardinality if there is a one-to-one correspondence between the elements of the two sets. This means you can pair up every element from one set with exactly one element from the other set, leaving no elements unpaired.

Understanding cardinality helps us compare different sets and explore their properties. For example, even though the set of all integers and the set of all even integers both appear infinite, they actually have the same cardinality. This is because we can create a function that pairs each integer with an even integer, showing a one-to-one correspondence.

In the exercise above, we establish that the set of multiples of 25 and the set of even integers have the same cardinality, demonstrating our ability to establish such a mapping.
Injective Function
An injective function, also known as a one-to-one function, ensures that different inputs map to different outputs. In other words, if you apply the function to two distinct elements from the domain, you'll get two distinct elements in the range. To prove a function is injective, we need to show that similar outputs came from the same inputs.

For instance, considering the function defined in the exercise: \( f(25n) = 2n \) maps from the set of multiples of 25 to the set of even integers. To confirm this function is injective, we prove that if \( f(25n_1) = f(25n_2) \), then \( n_1 = n_2 \). As we see, this holds true because \( 2n_1 = 2n_2 \) implies \( n_1 = n_2 \). Thus, the function is injective, ensuring that no two different multiples of 25 map to the same even number.
Surjective Function
A surjective function, or onto function, is one where every element in the codomain (the set we're mapping to) is the image of at least one element from the domain (the set we're mapping from). In simpler terms, a function is considered surjective if for every possible output, there is at least one input that results in that output.

In the given exercise, we show that the function \( f(25n) = 2n \) is surjective by demonstrating that for any even integer \( 2n \) in the set of even integers, we can find a corresponding multiple of 25 that satisfies the function. By choosing \( n' = \frac{n}{25} \), we are assured that for any \( 2n \), there is a \( 25n' \) such that \( f(25n') = 2n \). This verification confirms that the function translates every element of the codomain.
Bijective Function
A bijective function is one that is both injective and surjective. This means every element from the domain is paired with a unique element from the codomain and every element from the codomain is paired with an element from the domain. A bijective function establishes a perfect "pair up" between two sets.

In the problem, after proving the function \( f(25n) = 2n \) is both injective (one-to-one) and surjective (onto), we deduce it is bijective. This confirms that each multiple of 25 gets a unique match in the even integers, and vice versa. Therefore, the two sets, \( 25\mathbf{Z} \) and \( 2\mathbf{Z} \), can perfectly be paired, demonstrating they have the same cardinality.

Bijectivity shows a fascinating aspect of infinite sets: two infinite collections can have identical sizes when we find such a comprehensive pairing between them.

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Most popular questions from this chapter

Let \(S\) be the set of all strings in \(a\) 's and \(b\) 's, and define \(C: S \rightarrow S\) by \(C(s)=a s, \quad\) for all \(s \in S .\) ( \(C\) is called concatenation by \(a\) on the left.) a. Is \(C\) one-to-one? Prove or give a counterexample. b. Is \(C\) onto? Prove or give a counterexample.

Explain how it follows from the definition of logarithm that a. \(\log _{b}\left(b^{x}\right)=x\), for all real numbers \(x\). b. \(b^{\log _{b} x}=x\), for all positive real numbers \(x\).

How many integers must you pick in order to be sure that at least two of them have the same remainder when divided by \(7 ?\)

A permutation on a set can be regarded as a function from the set to itself. For instance, one permutation of \(\\{1,2,3,4\\}\) is 2341 . It can be identified with the function that sends each position number to the number occupying that position. Since position 1 is occupied by 2,1 is sent to 2 or \(1 \rightarrow 2\); since position 2 is occupied by 3,2 is sent to 3 or \(2 \rightarrow 3\); and so forth. The entire permutation can be written using arrows as follows: $$ \begin{array}{llll} 1 & 2 & 3 & 4 \\ \downarrow & \downarrow & \downarrow & \downarrow \\ 2 & 3 & 4 & 1 \end{array} $$ a. Use arrows to write each of the six permutations of \(\\{1,2,3\\}\). b. Use arrows to write each of the permutations of \(\\{1,2,3,4\\}\) that keep 2 and 4 fixed. c. Which permutations of \(\\{1,2,3\\}\) keep no elements fixed? d. Use arrows to write all permutations of \(\\{1,2,3,4\\}\) that keep no elements fixed.

In Example \(7.2 .8\) a one-to-one correspondence was defined from the power set of \(\\{a, b\\}\) to the set of all strings of 0 's and 1's that have length 2 . Thus the elements of these two sets can be matched up exactly, and so the two sets have the same number of elements. a. Let \(X=\left\\{x_{1}, x_{2}, \ldots, x_{n}\right\\}\) be a set with \(n\) elements. Use Example \(7.2 .8\) as a model to define a one-to-one correspondence from \(\mathscr{P}(X)\), the set of all subsets of \(X\), to the set of all strings of 0 's and l's that have length \(n\). b. Use the one-to-one correspondence of part (a) to deduce that a set with \(n\) elements has \(2^{n}\) subsets. (This provides an alternative proof of Theorem 5.3.5.)
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