(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. 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 $$ n(A \cup B)=a $$. 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. 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 $$ n(A)+n(\varnothing)=n(A \cap \varnothing) $$, 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, $$ n(A \cup \varnothing)=n(A)=a $$. D. 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.

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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 correct 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. **Definition of Cardinality**: We start by defining the cardinality of a set, which is the number of elements in that set.
2. **Empty Set**: We establish that the empty set $$ \varnothing $$ has a cardinality of 0, i.e., $$ n(\varnothing) = 0 $$.
3. **Disjoint Sets**: We consider a set $$ A $$ with $$ a $$ elements. The union of $$ A $$ and the empty set $$ \varnothing $$ is still $$ A $$ because adding nothing to a set does not change it.
4. **Cardinality of Union**: Since $$ A $$ and $$ \varnothing $$ are disjoint, we can apply the cardinality rule: $$ n(A) + n(\varnothing) = n(A \cup \varnothing) $$.
5. **Conclusion**: Therefore, $$ a + 0 = n(A) + n(\varnothing) = n(A) = a $$.

### (b) Fourth Graders Explanation

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

**Explanation:**
1. **Using Manipulatives**: We can use physical objects like blocks or counters. 
2. **Show a Set**: Start by showing them a group of $$ a $$ blocks. For example, if $$ a = 5 $$, show them 5 blocks.
3. **Introduce Zero**: Then, show them an empty space or an empty box and explain that this represents 0 blocks.
4. **Combine the Sets**: Ask them to combine the blocks with the empty box. They will see that they still have 5 blocks.
5. **Conclusion**: Explain that adding 0 (the empty box) to the 5 blocks doesn’t change the number of blocks. So, $$ 5 + 0 = 5 $$.

This method uses concrete examples that are relatable to fourth graders, making the concept of adding zero more understandable.
(a) The correct answer is B: 0 is the cardinal number of the empty set, so adding 0 to any set $$ A $$ with $$ a $$ elements results in the same set $$ A $$, meaning $$ a + 0 = a $$. (b) To explain $$ a + 0 = a $$ to fourth graders, use physical objects like blocks. Show them a group of $$ a $$ blocks, then add an empty space (representing 0) and demonstrate that the total number of blocks remains the same, illustrating that adding 0 doesn't change the count.

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