Which of the following relation is valid in Boolean algebra :-

Options

A. `A + bar(A) = 0`
B. `A+A = 2A`
C. `A + bar(A) = 1`
D. `A + bar(A)=A`

Answer

To determine which relation is valid in Boolean algebra, we will evaluate each option step by step.

### Step 1: Evaluate the first option: \( a + a' \)
- **Assume \( a = 0 \)**:
- \( a' = 1 \) (since the complement of 0 is 1)
- \( LHS = a + a' = 0 + 1 = 1 \)

- **Assume \( a = 1 \)**:
- \( a' = 0 \) (since the complement of 1 is 0)
- \( LHS = a + a' = 1 + 0 = 1 \)

**Conclusion**: The first option \( a + a' = 1 \) is valid.

### Step 2: Evaluate the second option: \( a + a = 2a \)
- **Assume \( a = 0 \)**:
- \( LHS = a + a = 0 + 0 = 0 \)
- \( RHS = 2a = 2 \times 0 = 0 \)

- **Assume \( a = 1 \)**:
- \( LHS = a + a = 1 + 1 = 1 \) (in Boolean algebra, this is simplified to 1)
- \( RHS = 2a = 2 \times 1 = 2 \)

**Conclusion**: The second option \( a + a = 2a \) is not valid because for \( a = 1 \), LHS ≠ RHS.

### Step 3: Evaluate the third option: \( a + a' = 1 \)
- **Assume \( a = 0 \)**:
- \( a' = 1 \)
- \( LHS = a + a' = 0 + 1 = 1 \)

- **Assume \( a = 1 \)**:
- \( a' = 0 \)
- \( LHS = a + a' = 1 + 0 = 1 \)

**Conclusion**: The third option \( a + a' = 1 \) is valid.

### Step 4: Evaluate the fourth option: \( a + a' = a \)
- **Assume \( a = 0 \)**:
- \( LHS = a + a' = 0 + 1 = 1 \)
- \( RHS = a = 0 \)

**Conclusion**: The fourth option \( a + a' = a \) is not valid because LHS ≠ RHS.

### Final Conclusion
The only valid relation in Boolean algebra from the given options is:
- **Option 3**: \( a + a' = 1 \)