Let A= {1,2,3,4,5}. Define an operative V by a Vb= max. {a,b}. Prepare its composition table.

Answer

To solve the problem, we need to prepare a composition table for the set A = {1, 2, 3, 4, 5} using the operator V defined as \( a \, V \, b = \max(a, b) \).

### Step-by-step Solution

1. **Define the Set and Operator**:
- Let \( A = \{1, 2, 3, 4, 5\} \).
- The operation \( a \, V \, b \) is defined as \( \max(a, b) \).

2. **Create a Composition Table**:
- We will create a table where the rows and columns represent the elements of set A, and each cell will contain the result of the operation \( a \, V \, b \).

3. **Fill the Table**:
- The table will look like this:

\[ \begin{array}{c|ccccc} V & 1 & 2 & 3 & 4 & 5 \\ \hline 1 & 1 & 2 & 3 & 4 & 5 \\ 2 & 2 & 2 & 3 & 4 & 5 \\ 3 & 3 & 3 & 3 & 4 & 5 \\ 4 & 4 & 4 & 4 & 4 & 5 \\ 5 & 5 & 5 & 5 & 5 & 5 \\ \end{array} \]

- **Explanation of Each Cell**:
- For example, in the cell corresponding to \( (1, 2) \), we calculate \( 1 \, V \, 2 = \max(1, 2) = 2 \).
- Similarly, for \( (2, 3) \), we have \( 2 \, V \, 3 = \max(2, 3) = 3 \).
- This process continues for all pairs of elements in the set.

4. **Check Closure Property**:
- The closure property states that for all \( a, b \in A \), \( a \, V \, b \) must also be in A.
- Since all results in the table are from the set A, the closure property is satisfied.

5. **Check Symmetry**:
- The operation is symmetric if \( a \, V \, b = b \, V \, a \) for all \( a, b \in A \).
- We can see from the table that \( \max(a, b) = \max(b, a) \), hence the operation is symmetric.

6. **Identify the Identity Element**:
- An identity element \( e \) must satisfy \( a \, V \, e = a \) and \( e \, V \, a = a \) for all \( a \in A \).
- From the table, we can see that \( 1 \) acts as the identity element since \( \max(a, 1) = a \) for all \( a \in A \).

7. **Check for Inverses**:
- An inverse for an element \( a \) must satisfy \( a \, V \, a^{-1} = e \).
- In this case, the only element that satisfies this condition is \( 1 \) itself, as \( 1 \, V \, 1 = 1 \).
- There are no other inverses since no other element combined with itself gives the identity.

### Final Composition Table

\[ \begin{array}{c|ccccc} V & 1 & 2 & 3 & 4 & 5 \\ \hline 1 & 1 & 2 & 3 & 4 & 5 \\ 2 & 2 & 2 & 3 & 4 & 5 \\ 3 & 3 & 3 & 3 & 4 & 5 \\ 4 & 4 & 4 & 4 & 4 & 5 \\ 5 & 5 & 5 & 5 & 5 & 5 \\ \end{array} \]