Binary subtraction rule — evaluate the statement: “If you borrow from a position that contains a 0, you must borrow from the next more significant 1; all 0s up to that point become 1s, and the digit last borrowed from becomes 0.”

Introduction / Context:
Manual binary subtraction mirrors decimal subtraction with borrowing, but carries and borrows are base dependent. In binary, a 1 can be borrowed from a more significant position to add 2 to a lower position. This question confirms whether the described cascading-borrow rule is accurate when the immediate higher-order bit is 0 and further searching is needed for a 1.

Given Data / Assumptions:

• Base-2 subtraction of unsigned numbers.
• Borrow operation when the minuend bit is smaller than the subtrahend bit.
• Possible runs of zeros among higher-order bits before encountering a 1.

Concept / Approach:
When borrowing in binary, if the next higher bit is 0, you must continue to propagate the borrow until a 1 is found. The found 1 becomes 0, the intermediate zeros turn into 1s (because each 0 receives a borrow making it 2 in base 2, then one is used to settle the lower position, leaving 1 to pass down), and the original position gains 2 to complete the subtraction. This is the classic cascaded-borrow mechanism.

Step-by-Step Solution:

Verification / Alternative check:
Work an example: subtract 0001 from 1000. You must borrow across the run of zeros: 1000 − 0001 = 0111. The higher 1 becomes 0; the intervening zeros become ones, consistent with the rule.

Why Other Options Are Wrong:

• Incorrect: Contradicts the standard borrow algorithm.
• “Only for hexadecimal” / “Only for signed magnitude”: Borrowing logic applies to binary subtraction irrespective of signed representation or other bases.

Common Pitfalls:
Losing track of bit positions during multi-step borrows; forgetting that intermediate zeros become ones as the borrow propagates.

Final Answer:
Correct — the described cascaded-borrow rule accurately reflects binary subtraction mechanics.