(a) How would you show that, for any whole number $$ a, a+0=a $$ if you were starting from our definition of addition and the audience is your university class? (b) How would you show that, for any whole number $$ a, a+0=a $$ if you were talking to your class of fourth graders? (a) Choose the correct answer below. A. Use manipulatives to show that the union of a collection with a collection with no elements has the same number of elements as the first collection. B. 0 is the cardinal number of the empty set: that is, $$ n(\varnothing)=0 $$. If the set $$ A $$ has a elements in it $$ (n(A)=a) $$, then $$ n(A)+n(\varnothing)=n(A \cup \varnothing) $$, because $$ A $$ and $$ \varnothing $$ are disjoint. But $$ A \cup \varnothing=A $$, since there are no elements in $$ \varnothing $$ to add to the set $$ A $$. Thus, $$ n(A \cup \varnothing)=n(A)=a $$. C. Start with two disjoint sets of the same cardinality $$ A $$ and $$ B $$ with $$ n(A)=n(B)=a $$. Then write the equation. Since $$ n(A)=n(B)=a, n(A \cup B)=a $$. Since they are disjoint, $$ A \cap B=\varnothing $$. Then, $$ a+0=n(A \cup(A \cap B)) $$ which is equal to $$ n(A \cup B)=a $$. D. $$ O $$ is the cardinal number of the empty set: that is, $$ n(\varnothing)=0 $$. If the set $$ A $$ has a elements in it $$ (n(A)=a) $$, then $$ n(A)+n(\varnothing)=n(A \cap \varnothing) $$, because $$ A $$ and $$ \varnothing $$ are disjoint. But $$ A \cap \varnothing=A $$, since there are no elements in $$ \varnothing $$ to add to the set $$ A $$. Thus, $$ n(A \cup \varnothing)=n(A)=a $$.

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To address the problem, we will break it down into two parts: (a) for a university class and (b) for a class of fourth graders.

### (a) University Class Explanation

To show that for any whole number $$ a $$, $$ a + 0 = a $$, we can use the concept of cardinality and set theory. The most appropriate answer from the options provided is:

**B.** 0 is the cardinal number of the empty set: that is, $$ n(\varnothing)=0 $$. If the set $$ A $$ has $$ a $$ elements in it $$ (n(A)=a) $$, then $$ n(A)+n(\varnothing)=n(A \cup \varnothing) $$, because $$ A $$ and $$ \varnothing $$ are disjoint. But $$ A \cup \varnothing=A $$, since there are no elements in $$ \varnothing $$ to add to the set $$ A $$. Thus, $$ n(A \cup \varnothing)=n(A)=a $$.

**Explanation:**
1. Define the empty set $$ \varnothing $$ and explain that it has no elements, so its cardinality is 0.
2. Let $$ A $$ be a set with $$ a $$ elements, so $$ n(A) = a $$.
3. Since $$ A $$ and $$ \varnothing $$ are disjoint, the union of these two sets is simply $$ A $$ itself: $$ A \cup \varnothing = A $$.
4. Therefore, the cardinality of the union is $$ n(A \cup \varnothing) = n(A) $$.
5. This leads to the equation $$ n(A) + n(\varnothing) = n(A) $$, which simplifies to $$ a + 0 = a $$.

### (b) Fourth Graders Explanation

To explain $$ a + 0 = a $$ to fourth graders, we can use a more visual and intuitive approach.

**Explanation:**
1. **Use Manipulatives:** Gather some objects (like blocks or counters) and show a group of $$ a $$ blocks. For example, if $$ a = 3 $$, show 3 blocks.
2. **Introduce Zero:** Now, show them an empty space or an empty box and explain that this represents 0 blocks.
3. **Combine the Groups:** Ask the students what happens if you combine the 3 blocks with the empty box. 
4. **Count Together:** Count the blocks together. They will see that there are still 3 blocks, and nothing was added from the empty box.
5. **Conclusion:** Explain that adding zero (the empty box) to any number of blocks (the 3 blocks) doesn’t change the number of blocks. So, $$ 3 + 0 = 3 $$ and in general, $$ a + 0 = a $$.

This approach uses concrete examples and counting, which is more relatable for fourth graders, helping them understand the concept of addition with zero intuitively.
(a) For a university class, the correct explanation is: **B.** 0 is the cardinal number of the empty set: that is, $$ n(\varnothing)=0 $$. If the set $$ A $$ has $$ a $$ elements in it $$ (n(A)=a) $$, then $$ n(A)+n(\varnothing)=n(A \cup \varnothing) $$, because $$ A $$ and $$ \varnothing $$ are disjoint. But $$ A \cup \varnothing=A $$, since there are no elements in $$ \varnothing $$ to add to the set $$ A $$. Thus, $$ n(A \cup \varnothing)=n(A)=a $$. (b) For a class of fourth graders, the explanation is: **Use manipulatives to show that adding zero to any number of objects doesn't change the count.** For example, if you have 3 blocks and add zero blocks, you still have 3 blocks. This demonstrates that $$ a + 0 = a $$.

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