Bit shifting equivalence: In binary arithmetic, shifting a number left by one bit position is equivalent to which arithmetic operation on an unsigned value (ignoring overflow)?

Introduction / Context:
Shifts are among the most efficient arithmetic operations in digital systems. Understanding their equivalence to multiply/divide by powers of two helps optimize algorithms, reduce resource usage in hardware, and reason about fixed-point scaling in DSP pipelines.

Given Data / Assumptions:

• Binary, unsigned interpretation (or signed two’s complement without overflow).
• One-bit left shift (logical or arithmetic behaves identically on positive magnitudes).
• Overflow and bit growth are ignored for the equivalence statement.

Concept / Approach:
In base-2, each left shift moves every bit to a position with double the weight. Therefore, a left shift by k positions multiplies the numeric value by 2^k. For k = 1, the factor is 2. The lowest bit becomes 0, and the most significant bit may be lost if the width is fixed, which is the typical source of overflow concerns.

Step-by-Step Solution:

Verification / Alternative check:
Example: 0011₂ (3) << 1 = 0110₂ (6) = 3 * 2. For k shifts, V << k = V * 2^k. The rule mirrors decimal left-shifts multiplying by 10 when in base-10 systems.

Why Other Options Are Wrong:

• Multiplying by four: That corresponds to left shift by 2.
• Dividing by two/four: These correspond to right shifts by 1 or 2, not left shifts.

Common Pitfalls:
Forgetting fixed-width overflow; assuming the rule applies to signed negative values with arithmetic edge cases; confusing logical vs arithmetic shift (relevant for right shifts, not left).

Final Answer:
multiplying by two