Let $$ N=\{1,2,3,4, \ldots\} $$ be the set of natural numbers and $$ S=\{1,4,9,16, \ldots\} $$ be the set of squares of the natural numbers. Then $$ N-S $$, since we have the one-to-one correspondence $$ 1 \leftrightarrow 1,2 \leftrightarrow 4,3 \leftrightarrow 9,4 \leftrightarrow 16, \ldots, n \leftrightarrow n^{2} $$. (This example is interesting, since it shows that an infinite set can be equivalent to a proper subset of itself.) Show that each of the following pairs of sets are equivalent by carefully describing a one-fo-one correspondence between the sets. Complete parts (a) through (c) below. (a) The whole numbers and natural numbers, $$ W=\{0,1,2,3, \ldots\} $$ and $$ N=\{1,2,3,4, \ldots\} $$ Which of the following describes a one-to-one correspondence between the two sets? A. For each element in $$ W $$, there is an element in $$ N $$ that is double that element. B. For each element in W , there is an element in N that is 2 greater than that element. C. For each element in $$ \mathbf{W} $$, there is an element in N that is 1 greater than that element. D. For each element in W , there is an element in N that is 1 greater than double that element. Write a general rule for the correspondence described in the previous step using w as an element in W. $$ \square $$

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UpStudy AI · Accepted Answer

**Choice:** C

**Step-by-step explanation:**

1. The set of whole numbers is given by  
   $$
   W = \{0, 1, 2, 3, \ldots\}
   $$
   and the set of natural numbers is  
   $$
   N = \{1, 2, 3, 4, \ldots\}.
   $$

2. To establish a one-to-one correspondence between $$ W $$ and $$ N $$, we need to find a function $$ f: W \to N $$ that is both one-to-one and onto.

3. Option C specifies that for each element in $$ W $$, there is an element in $$ N $$ that is 1 greater than that element. That suggests the function
   $$
   f(w) = w + 1
   $$
   where $$ w \in W $$.

4. **Verification of Properties:**
   - *Injectivity (One-to-One):*  
     Suppose $$ f(w_1) = f(w_2) $$. Then  
     $$
     w_1 + 1 = w_2 + 1.
     $$
     Subtracting 1 from both sides gives  
     $$
     w_1 = w_2,
     $$
     showing that $$ f $$ is one-to-one.
     
   - *Surjectivity (Onto):*  
     For every $$ n \in N $$, there exists a $$ w \in W $$ such that  
     $$
     n = w + 1.
     $$
     Solving for $$ w $$ gives  
     $$
     w = n - 1.
     $$
     Since $$ n \ge 1 $$, we have $$ w \ge 0 $$, so $$ w \in W $$. Thus $$ f $$ is onto.

5. Since $$ f(w) = w + 1 $$ is both one-to-one and onto, it is a bijection (one-to-one correspondence) between $$ W $$ and $$ N $$.

**Answer in box:**  
$$
\boxed{f(w)=w+1}
$$
**Answer:** - **Choice:** C - **Correspondence Rule:** $$ f(w) = w + 1 $$ **Explanation:** To pair each whole number with a natural number one-to-one, we add 1 to each whole number. This ensures every natural number is matched uniquely without missing any. **Simplified Answer:** Each whole number $$ w $$ is paired with $$ w + 1 $$ in the natural numbers.

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