Question

write an argument showing that the set of whole numbers can be placed in a one-to-one correspondence with the set of natural numbers. choose the correct answer below. a. for every natural number n, let it be paired with the whole number 2n. for every whole number w, let it be paired with the natural number 2w. b. for every natural number n, let it be paired with the whole number n − 1. for every whole number w, let it be paired with the natural number w + 1. c. every natural number is also a whole number. this means there must be a one-to-one correspondence between the set of whole numbers and the set of natural numbers. d. for every natural number n, let it be paired with the whole number n + 1. for every whole number w, let it be paired with the natural number w − 1.

Explanation:

To show a one - to - one correspondence between whole numbers (let's assume whole numbers start from 0: 0, 1, 2, 3, …) and natural numbers (let's assume natural numbers start from 1: 1, 2, 3, 4, …), we need a function that maps each whole number to a unique natural number and vice - versa.

• Option A: If we pair natural number \(n\) with \(2n\), when \(n = 1\), we get \(2\); \(n = 2\), we get \(4\). But we are missing natural numbers like 1, 3, 5, etc. So it's not a one - to - one correspondence that covers all natural numbers.
• Option B: If we pair natural number \(n\) with \(n - 1\), when \(n=1\), we get \(0\) (which is a whole number), but for \(n = 1\), the mapping from natural number to whole number is okay, but when we consider the reverse (mapping whole numbers to natural numbers), if we have whole number \(w\), and we try to map it to \(w + 1\), let's check. Whole numbers are \(0,1,2,\cdots\). If \(w = 0\), \(w + 1=1\) (natural number), \(w = 1\), \(w + 1 = 2\) (natural number), and so on. Also, for natural number \(n\), \(n-1\) gives a whole number (when \(n\geq1\), \(n - 1\geq0\)). Now, we need to check if it's one - to - one. Suppose \(n_1-1=n_2 - 1\), then \(n_1=n_2\), so it's one - to - one for the natural to whole mapping. For the whole to natural mapping, if \(w_1 + 1=w_2+1\), then \(w_1 = w_2\), so it's one - to - one. Also, it covers all whole numbers and natural numbers.
• Option C: Just stating that every natural number is a whole number does not show a one - to - one correspondence. It's just a statement about the set inclusion.
• Option D: If we pair natural number \(n\) with \(n + 1\), when \(n = 1\), we get \(2\); \(n=2\), we get \(3\). We are missing natural number 1. And when we map whole number \(w\) to \(w - 1\), for \(w = 0\), we get \(- 1\), which is not a natural number. So this mapping is invalid.

Answer:

B. For every natural number \(n\), let it be paired with the whole number \(n - 1\). For every whole number \(w\), let it be paired with the natural number \(w + 1\)