To find all binary operations on the set $S = \{e, a\}$ for which $S$ is a semigroup and $e$ is a left identity element, we need to satisfy the following conditions:

1. Closure: The result of the binary operation on any two elements in $S$ must be an element of $S$.
2. Associativity: For all $x, y, z \in S$, $(x \ast y) \ast z = x \ast (y \ast z)$.
3. Left Identity: There exists an element $e \in S$ such that for every $x \in S$, $e \ast x = x$.

Let's define the binary operation $\ast$ on $S = \{e, a\}$ and check the conditions.

Step 1: Define the Operation Table

The binary operation can be represented as a table where the rows and columns correspond to the elements of $S$. The entries of the table represent the result of the operation $\ast$ on the corresponding elements.

Let $\ast$ be the binary operation. The operation table will look like this:

\ast & e & a \\ \hline e & e \ast e & e \ast a \\ a & a \ast e & a \ast a \\ \end{array}$$ ### Step 2: Apply the Left Identity Condition Given $$e$$ is a left identity, for all $$x \in S$$: - $$e \ast e = e$$ - $$e \ast a = a$$ So we update the table with this information: $$\begin{array}{c|cc} \ast & e & a \\ \hline e & e & a \\ a & a \ast e & a \ast a \\ \end{array}$$ ### Step 3: Determine Possible Values for $$a \ast e$$ and $$a \ast a$$ Since $$S$$ must be a semigroup, the operation must be associative: - $$a \ast e$$ can be either $$e$$ or $$a$$ (closure property). Let's analyze the possibilities: #### Case 1: $$a \ast e = e$$ 1. Update the table: $$\begin{array}{c|cc} \ast & e & a \\ \hline e & e & a \\ a & e & a \ast a \\ \end{array}$$ 2. Check associativity for $$a \ast (e \ast a)$$ and $$(a \ast e) \ast a$$: - $$a \ast (e \ast a) = a \ast a$$ - $$(a \ast e) \ast a = e \ast a = a$$ So, $$a \ast a = a$$. Final operation table for Case 1: $$\begin{array}{c|cc} \ast & e & a \\ \hline e & e & a \\ a & e & a \\ \end{array}$$ #### Case 2: $$a \ast e = a$$ 1. Update the table: $$\begin{array}{c|cc} \ast & e & a \\ \hline e & e & a \\ a & a & a \ast a \\ \end{array}$$ 2. Check associativity for $$a \ast (a \ast a)$$ and $$(a \ast a) \ast a$$: - $$a \ast (a \ast a) = a \ast a$$ - $$(a \ast a) \ast a$$ depends on the value of $$a \ast a$$. If $$a \ast a = a$$, the table satisfies associativity: Final operation table for Case 2: $$\begin{array}{c|cc} \ast & e & a \\ \hline e & e & a \\ a & a & a \\ \end{array}$$ ### Summary There are two binary operations on $$S = \{e, a\}$$ such that $$S$$ is a semigroup with $$e$$ as a left identity: 1. $$\ast$$ defined as: $$\begin{array}{c|cc} \ast & e & a \\ \hline e & e & a \\ a & e & a \\ \end{array}$$ 2. $$\ast$$ defined as: $$\begin{array}{c|cc} \ast & e & a \\ \hline e & e & a \\ a & a & a \\ \end{array}$$ Would you like more details or have any questions? Here are some questions you might consider next: 1. How do we determine the number of semigroups for larger sets? 2. What are the properties of associative operations in algebraic structures? 3. Can you explain the concept of monoids and how they relate to semigroups? 4. How can we construct Cayley tables for different algebraic structures? 5. What are some real-world applications of semigroups in mathematics? **Tip:** When working with algebraic structures like semigroups, it is helpful to draw operation tables to visualize how the binary operation behaves with each element in the set.