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: Recall the standard laws of Boolean Algebra relevant to the options.
Boolean Algebra has a commutative law, \(a + b = b + a\) for OR and \(a \cdot b = b \cdot a\) for AND, and a complement law, which states that a variable combined with its own complement (NOT) using OR always gives 1, \(a + a' = 1\), while combining a variable with its own complement using AND always gives 0, \(a \cdot a' = 0\).

Step 2: Check option (A), \(a + b = b + a\).
This is the commutative law for OR, which is a genuine, well-known property of Boolean Algebra, true for all values of \(a\) and \(b\).

Step 3: Check option (B), \(a \cdot a' = 1\).
By the complement law, ANDing a variable with its own complement always gives 0, not 1, since at any point exactly one of \(a\) and \(a'\) is 0, so their AND is always 0. The statement \(a \cdot a' = 1\) contradicts this and is false, it is not a valid property of Boolean Algebra.

Step 4: Check option (C), \(a + a' = 1\).
By the complement law, ORing a variable with its own complement always gives 1, since at any point at least one of \(a\) and \(a'\) is 1. This statement is correct and is indeed a standard property of Boolean Algebra.

Step 5: Check option (D), \(a \cdot b = b \cdot a\).
This is the commutative law for AND, mirroring option (A), also a genuine, well-known property of Boolean Algebra, true for all values of \(a\) and \(b\).

Step 6: Conclusion.
Options (A), (C), and (D) are all correct, standard Boolean Algebra laws. Option (B) states \(a \cdot a' = 1\), which mixes up the AND-complement identity with the OR-complement identity, the true AND-complement identity is \(a \cdot a' = 0\), so (B) is the one statement that is not an actual property of Boolean Algebra.
\[ \boxed{a \cdot a' = 1 \text{ (option B) is not a valid property}} \]