Binary place value — In a pure binary number, what is the weight (place value) of the least significant bit (LSB)?

Introduction:
Understanding positional weights is foundational in digital logic, number systems, and data conversion. The least significant bit (LSB) determines the smallest increment that can be represented in an unsigned integer and directly relates to resolution in digital-to-analog and analog-to-digital conversions when the binary number is mapped to a physical quantity.

Given Data / Assumptions:

• Unsigned binary integers, no fractional bits.
• Positional weighting follows powers of two.
• LSB is the rightmost bit of the integer portion.

Concept / Approach:

Binary is a base-2 system. From right to left, the weights are 2^0, 2^1, 2^2, and so on. Therefore, the LSB has weight 2^0 = 1. Each step left doubles the weight; each step right (into fractional bits) halves it. This concept generalizes: in fixed-point schemes, the first fractional bit has weight 2^(−1) = 0.5, but that is not the LSB of an integer.

Step-by-Step Solution:

Verification / Alternative check:

Consider the binary number 0001_2 = 1_10; toggling only the LSB changes the value by exactly 1. Similarly, adding 1 to any integer changes only the LSB (with carries) confirming the unit weight at that position.

Why Other Options Are Wrong:

• 2 / 3 / 4: These correspond to higher-order bit weights (2^1, 2^? not valid for 3, 2^2).
• 0.5: This is the weight of the first fractional bit 2^(−1), not an integer LSB.

Common Pitfalls:

• Mixing integer and fixed-point fractional interpretations.
• Assuming LSB means “smallest decimal step”; the base dictates weights.

Final Answer:

1