Question

Which one of the following options is not a property of Boolean Algebra?
Note: + is the OR operation, \(\cdot\) is the AND operation, and \('\) is the NOT operation.

• A. \(a + b = b + a\)
• B. \(a \cdot a' = 1\)
• C. \(a + a' = 1\)
• D. \(a \cdot b = b \cdot a\)

Hint:

The complement law gives a . a' = 0 and a + a' = 1; option B swaps these two identities.

Answer 1

The Correct Option is B

Step 1: Boolean variables only ever take the value $0$ or $1$, so any claimed "law" can be checked by simply plugging in both possible values of $a$ (and $b$ where needed) and seeing if the equation holds in every case.

Step 2: Test option (B), $a \cdot a' = 1$. If $a = 0$, then $a' = 1$ (NOT of 0 is 1), so $a \cdot a' = 0 \cdot 1 = 0$, not $1$. The claimed equation already fails for this single value of $a$, so option (B) cannot be a valid property, since a true law of Boolean Algebra must hold for every possible value of its variables, not just some.

Step 3: For comparison, test option (C), $a + a' = 1$, the same way. If $a = 0$, $a' = 1$, so $a + a' = 0 + 1 = 1$, correct. If $a = 1$, $a' = 0$, so $a + a' = 1 + 0 = 1$, correct again. Both cases give $1$, so $a + a' = 1$ holds for every value of $a$, confirming it is a genuine law, and by contrast highlighting that (B) uses the wrong operation for this identity.

Step 4: Test option (A), $a + b = b + a$, and option (D), $a \cdot b = b \cdot a$, across all four combinations of $a, b \in \{0,1\}$. OR and AND are each symmetric operations, swapping the order of the two inputs never changes the output for either operation in any of the four cases, so both (A) and (D) hold in every case and are valid commutative laws.

Step 5: Since (A), (C), and (D) hold for every possible substitution of their variables, and (B) fails for $a=0$, option (B), $a \cdot a' = 1$, is the one option that is not an actual property of Boolean Algebra, the correct identity being $a \cdot a' = 0$.

\[ \boxed{a \cdot a' = 1 \text{ (option B) is not a valid property}} \]