Boolean form of a NOR gate For a 2-input NOR logic gate with inputs A and B, which expression correctly represents its output?

Introduction / Context:
NOR is a fundamental logic operator used as a universal gate. Recognizing its Boolean form enables algebraic manipulation, simplification, and design using only NOR devices.

Given Data / Assumptions:

• Two inputs: A and B.
• Symbol conventions: + means OR, · means AND, and ' means NOT (complement).

Concept / Approach:
A NOR gate outputs the complement of the OR operation. Therefore, its output is the negation of A + B. This is directly written as Y = (A + B)' using the prime to denote inversion.

Step-by-Step Solution:
Start with OR: A + B is 1 when any input is 1.Apply inversion: NOR = NOT(OR) → Y = (A + B)'.Hence the output is 1 only when both A = 0 and B = 0.

Verification / Alternative check:
Truth table check: (0,0) → OR = 0 → NOT = 1; any other case has OR = 1 → NOT = 0, matching NOR behavior.

Why Other Options Are Wrong:
Y = A' + B': that is NOT A OR NOT B, not NOR.Y = (A · B)': complement of AND is NAND, not NOR.Y = A' · B': that is the DeMorgan form of NOR only after distributing the inversion, but without the outer parentheses it implies double inversion of each input, not the NOR operator as given.

Common Pitfalls:
Mixing DeMorgan equivalents: NOR equals A' · B' only if the bar is over A + B as a whole, then distributed; writing A' · B' directly can be misread.

Final Answer:
Y = (A + B)'