Let $$ \mathrm{N}=\{1,2,3,4, \ldots\} $$ be the set of netural numbers and $$ S=\{1,4,9,16, \ldots\} $$ be the set of squares of the natural numbers. Then $$ \mathrm{N} \sim \mathrm{S} $$, 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 itsell Show thet each of the following pairs of sets are equivalent by carefully describing a one-fo-one correspondence befween the sets. Complete parts (a) through (c) belo (a) The whole numbers and netural numbers, $$ W=\{0,1,2,3, \ldots\} $$ and $$ \mathrm{N}=\{1,2,3,4, \ldots\} $$ 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 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.

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We need to find a one-to-one correspondence (a bijection) between the set of whole numbers  
$$
W=\{0,1,2,3,\ldots\}
$$  
and the set of natural numbers  
$$
N=\{1,2,3,4,\ldots\}.
$$

A one-to-one correspondence consists of a function $$ f: W \to N $$ that is both one-to-one (injective) and onto (surjective).

Let’s check the given options:

**Option A:**  
For each element in $$ W $$, assign the element in $$ N $$ that is double that element.  
The proposed function is  
$$
f(w)=2w.
$$  
However, $$ f(0)=2\cdot 0=0 $$, and $$ 0 \notin N $$ since $$ N $$ starts at 1. Therefore, this function does not map $$ W $$ into $$ N $$.

**Option B:**  
For each element in $$ W $$, assign the element in $$ N $$ that is 2 greater than that element.  
This gives  
$$
f(w)=w+2.
$$  
Notice that $$ f(0)=0+2=2 $$, $$ f(1)=1+2=3 $$, etc. Although this function is one-to-one, the range is $$\{2,3,4,\ldots\}$$, which does not include 1. So the function is not onto.

**Option C:**  
For each element in $$ W $$, assign the element in $$ N $$ that is 1 greater than that element.  
This function is  
$$
f(w)=w+1.
$$  
Now, $$ f(0)=0+1=1 $$, $$ f(1)=1+1=2 $$, $$ f(2)=2+1=3 $$, and so on. Every natural number is hit exactly once, so this function is both one-to-one and onto.

**Option D:**  
For each element in $$ W $$, assign the element in $$ N $$ that is 1 greater than double that element.  
That is,  
$$
f(w)=1+2w.
$$  
Here, $$ f(0)=1+2\cdot0=1 $$, $$ f(1)=1+2\cdot1=3 $$, $$ f(2)=1+2\cdot2=5 $$, etc. This function maps $$ W $$ to the set $$\{1,3,5,\ldots\}$$, which does not include any even numbers (except possibly 1 if even, but it's not even). Hence, it is not onto $$ N $$.

Thus, only **Option C** defines the desired one-to-one correspondence between $$ W $$ and $$ N $$.
The correct one-to-one correspondence between the whole numbers $$ W = \{0,1,2,3,\ldots\} $$ and the natural numbers $$ N = \{1,2,3,4,\ldots\} $$ is: For each element in $$ W $$, assign the element in $$ N $$ that is 1 greater than that element. This means: - $$ 0 \leftrightarrow 1 $$ - $$ 1 \leftrightarrow 2 $$ - $$ 2 \leftrightarrow 3 $$ - $$ 3 \leftrightarrow 4 $$ - and so on. This pairing ensures that every whole number is matched with a unique natural number without any overlaps or omissions.

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