Binary-to-hexadecimal conversion technique: When converting a binary value to hexadecimal, is it valid to pad the left side (most significant side) with leading zeros to form complete groups of 4 bits before translating each group to a hex digit?

Difficulty: Easy

Correct Answer: Correct

Explanation:


Introduction / Context:
Hexadecimal is a compact way to express binary numbers by grouping bits in sets of four, since 16 = 2^4. This question tests whether you may add leading zeros on the most significant side of a binary number to complete 4-bit groups prior to conversion to hexadecimal digits.


Given Data / Assumptions:

  • A binary number may not have a length that is a multiple of 4.
  • Leading zeros do not change the value of an unsigned number.
  • Hex conversion maps each 4-bit nibble to one hex digit (0–F).


Concept / Approach:
Adding zeros to the left (toward the MSB) simply increases the bit width without altering numeric value. Grouping bits into 4-bit nibbles aligns perfectly with hexadecimal, enabling a one-to-one mapping from each nibble to a hex character. This is standard practice in debugging, memory dumps, and digital design documentation.


Step-by-Step Solution:

Take the binary sequence and partition it into 4-bit chunks from the LSB side.If the leftmost chunk has fewer than 4 bits, prepend zeros until it has 4.Translate each 4-bit group to its hex equivalent (0000→0, 1111→F).Join the hex digits to form the final hexadecimal value.


Verification / Alternative check:
Convert back from the hex digits to binary and verify that dropping any leading zeros restores the original bit pattern; the numeric value remains unchanged.


Why Other Options Are Wrong:
Signedness does not affect leading zeros for magnitude; requiring MSB = 1 is unnecessary; octal uses 3-bit groups, not 4-bit groups.


Common Pitfalls:
Mistaking leading zeros for sign extension (that applies to two’s complement when extending sign with 1s for negative numbers); confusing octal grouping (3 bits) with hex grouping (4 bits).


Final Answer:
Correct

More Questions from Number Systems and Codes

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion