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

Options

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

Answer

To solve the question regarding which relation is valid in Boolean algebra, we will analyze each option step by step. Let's denote the variable as \( a \) and its complement as \( \bar{a} \).

### Step 1: Define the values of \( a \) and \( \bar{a} \)
Let's assume:
- \( a = 0 \)
- Therefore, \( \bar{a} = 1 \)

### Step 2: Analyze the first option
**Option 1: \( a + a = a \)**

- **Left-Hand Side (LHS):**
\[ a + a = 0 + 0 = 0 \]
- **Right-Hand Side (RHS):**
\[ a = 0 \]

Since LHS = RHS = 0, Option 1 is valid.

### Step 3: Analyze the second option
**Option 2: \( a + 1 = 1 \)**

- **Left-Hand Side (LHS):**
\[ a + 1 = 0 + 1 = 1 \]
- **Right-Hand Side (RHS):**
\[ 1 = 1 \]

Since LHS = RHS = 1, Option 2 is valid.

### Step 4: Analyze the third option
**Option 3: \( a \cdot \bar{a} = 0 \)**

- **Left-Hand Side (LHS):**
\[ a \cdot \bar{a} = 0 \cdot 1 = 0 \]
- **Right-Hand Side (RHS):**
\[ 0 = 0 \]

Since LHS = RHS = 0, Option 3 is valid.

### Conclusion
All three options are valid in Boolean algebra. Therefore, the correct answer is that all the relations are valid.

### Final Answer
All options are valid in Boolean algebra.

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