To solve the problem step by step, we will first prove that the operation \( ** \) is closed on \( A = \mathbb{Q} \times \mathbb{Q} \), then find the identity element, and finally determine the invertible elements.
### Step 1: Proving Closure
**Definition of Closure**: A binary operation \( ** \) is said to be closed on a set \( A \) if for every \( (a, b), (c, d) \in A \), the result of \( (a, b) ** (c, d) \) is also in \( A \).
**Operation Definition**: The operation is defined as:
\[
(a, b) ** (c, d) = (ad + b, ac)
\]
**Proof of Closure**:
1. Let \( (a, b), (c, d) \in A \), where \( a, b, c, d \in \mathbb{Q} \).
2. Compute \( (a, b) ** (c, d) \):
\[
(a, b) ** (c, d) = (ad + b, ac)
\]
3. Since \( a, b, c, d \in \mathbb{Q} \), both \( ad + b \) and \( ac \) are also in \( \mathbb{Q} \) (because the set of rational numbers is closed under addition and multiplication).
4. Therefore, \( (ad + b, ac) \in A \).
**Conclusion**: The operation \( ** \) is closed on \( A = \mathbb{Q} \times \mathbb{Q} \).
### Step 2: Finding the Identity Element
**Definition of Identity Element**: An element \( e \in A \) is an identity element for the operation \( ** \) if for every \( (a, b) \in A \):
\[
(a, b) ** e = (a, b) \quad \text{and} \quad e ** (a, b) = (a, b)
\]
**Let \( e = (e_1, e_2) \)**. We need to find \( e_1 \) and \( e_2 \) such that:
\[
(a, b) ** (e_1, e_2) = (a, b)
\]
This gives:
\[
(ad + b, ac) = (a, b)
\]
From this, we have:
1. \( ad + b = a \)
2. \( ac = b \)
**Solving the equations**:
- From \( ac = b \), if we let \( a = 1 \), then \( c = b \).
- From \( ad + b = a \), substituting \( a = 1 \) gives \( d + b = 1 \), leading to \( d = 1 - b \).
Thus, we can choose \( e = (1, 0) \):
- \( (a, b) ** (1, 0) = (0 + b, a) = (a, b) \)
- \( (1, 0) ** (a, b) = (0 + 0, 1 \cdot b) = (a, b) \)
**Conclusion**: The identity element is \( e = (1, 0) \).
### Step 3: Finding the Invertible Elements
**Definition of Invertible Element**: An element \( (a, b) \in A \) is invertible if there exists \( (c, d) \in A \) such that:
\[
(a, b) ** (c, d) = (1, 0)
\]
This leads to:
\[
(ad + b, ac) = (1, 0)
\]
From this, we have:
1. \( ad + b = 1 \)
2. \( ac = 0 \)
**Solving the equations**:
- From \( ac = 0 \), either \( a = 0 \) or \( c = 0 \).
- If \( a \neq 0 \), then \( c = 0 \). Substituting \( c = 0 \) into the first equation gives \( b = 1 \).
- If \( a = 0 \), then \( (0, b) ** (c, d) = (0, 0) \) which cannot equal \( (1, 0) \).
Thus, the only invertible element occurs when \( a \neq 0 \) and \( c = 0 \):
- We can express \( d \) as \( d = \frac{1 - b}{a} \).
**Conclusion**: The invertible elements are of the form \( (a, b) \) where \( a \neq 0 \) and \( (c, d) = \left(0, \frac{1 - b}{a}\right) \).
### Summary of Results
1. **Closure**: The operation \( ** \) is closed on \( A \).
2. **Identity Element**: The identity element is \( (1, 0) \).
3. **Invertible Elements**: The invertible elements are of the form \( (a, b) \) where \( a \neq 0 \).
