Let `A=ZxxZ` and `'**'` be a binary operation on A defined by :
`(a, b)**(c, d)=(ad+bc,bd)`.
Find the identity element for `'**'` in A.

Answer

To find the identity element for the binary operation defined on the set \( A = \mathbb{Z} \times \mathbb{Z} \) by the operation \( (a, b) ** (c, d) = (ad + bc, bd) \), we need to determine an element \( (e_1, e_2) \) such that for any element \( (a, b) \in A \):

\[ (a, b) ** (e_1, e_2) = (a, b) \quad \text{and} \quad (e_1, e_2) ** (a, b) = (a, b) \]

### Step 1: Set up the equation for the identity element
We start by applying the operation to \( (a, b) \) and \( (e_1, e_2) \):

\[ (a, b) ** (e_1, e_2) = (ae_2 + be_1, be_2) \]

For this to equal \( (a, b) \), we need:

1. \( ae_2 + be_1 = a \)
2. \( be_2 = b \)

### Step 2: Solve the second equation
From the second equation \( be_2 = b \), we can deduce:

- If \( b \neq 0 \), then \( e_2 = 1 \).
- If \( b = 0 \), then \( e_2 \) can be any value.

For the identity element to work for all \( b \), we can set \( e_2 = 1 \).

### Step 3: Substitute \( e_2 \) into the first equation
Now substituting \( e_2 = 1 \) into the first equation:

\[ ae_2 + be_1 = a \implies a(1) + be_1 = a \implies a + be_1 = a \]

This simplifies to:

\[ be_1 = 0 \]

### Step 4: Solve for \( e_1 \)
For the equation \( be_1 = 0 \) to hold for all \( b \):

- If \( b \neq 0 \), then \( e_1 \) must be \( 0 \).
- If \( b = 0 \), it does not impose any restriction on \( e_1 \).

Thus, we can conclude that \( e_1 = 0 \).

### Step 5: Conclusion
The identity element \( (e_1, e_2) \) is:

\[ (e_1, e_2) = (0, 1) \]

### Summary
The identity element for the binary operation \( ** \) defined on \( A = \mathbb{Z} \times \mathbb{Z} \) is \( (0, 1) \).

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