How do you prove algebraiclly that the NAND operation is notassociative?
The correct answer and explanation is :
To prove algebraically that the NAND operation is not associative, we need to show that for three Boolean variables (A), (B), and (C), the following does not hold:
[
(A \text{ NAND } B) \text{ NAND } C \neq A \text{ NAND } (B \text{ NAND } C)
]
Step 1: Define NAND operation
The NAND operation is the negation of the AND operation. For two Boolean variables (A) and (B), the NAND operation is defined as:
[
A \text{ NAND } B = \neg(A \land B)
]
This means the output is false only when both (A) and (B) are true, and true in all other cases.
Step 2: Evaluate both sides for specific values of (A), (B), and (C)
Let’s test with specific values for (A), (B), and (C). We will choose (A = 1), (B = 1), and (C = 0).
Left-hand side: ((A \text{ NAND } B) \text{ NAND } C)
- First, calculate (A \text{ NAND } B):
[
A \text{ NAND } B = 1 \text{ NAND } 1 = \neg(1 \land 1) = \neg 1 = 0
] - Now, calculate ((A \text{ NAND } B) \text{ NAND } C):
[
0 \text{ NAND } 0 = \neg(0 \land 0) = \neg 0 = 1
]
Thus, the left-hand side equals 1.
Right-hand side: (A \text{ NAND } (B \text{ NAND } C))
- First, calculate (B \text{ NAND } C):
[
B \text{ NAND } C = 1 \text{ NAND } 0 = \neg(1 \land 0) = \neg 0 = 1
] - Now, calculate (A \text{ NAND } (B \text{ NAND } C)):
[
1 \text{ NAND } 1 = \neg(1 \land 1) = \neg 1 = 0
]
Thus, the right-hand side equals 0.
Step 3: Conclusion
We have shown that for (A = 1), (B = 1), and (C = 0), the left-hand side of the equation evaluates to 1, while the right-hand side evaluates to 0. Since the two sides are not equal, this demonstrates that the NAND operation is not associative.
Explanation
The NAND operation is not associative because the order in which operations are performed affects the result. This is due to the fact that the negation (NOT) operation applied to the AND of two operands changes based on how the operands are grouped. In contrast, associative operations (like AND or OR) yield the same result regardless of how operands are grouped. The failure of associativity in NAND is a key characteristic that distinguishes it from such operations.