When adding two positive binary numbers using the 2's complement method, if…

Concept: Two's complement addition is carried out as ordinary unsigned binary addition, column by column, across all n bits including the sign bit. Because the representation is a fixed-width, modulo-2ⁿ number system, any carry produced beyond the most-significant bit falls outside the n available bit positions and is simply discarded — the result is read from the remaining n bits alone. This is the rule that distinguishes two's complement addition from one's complement addition, which instead uses an "end-around carry": it re-adds a bit carried past the MSB back into the LSB to correct for one's complement having two representations of zero.

Application: Walking a concrete 4-bit example through the rule, adding -3 and 5:

1. Represent -3 in 4-bit two's complement as 1101, and 5 as 0101.

2. Add column by column from the LSB: 1101 + 0101 = 10010, a 5-bit result with a carry generated out of the sign-bit column.

3. Apply the rule: discard that carry bit rather than feeding it back anywhere in the number, keeping only the remaining 4 bits, 0010.

4. Read 0010 as +2, which matches the true sum (-3 + 5 = 2) — confirming that discarding the carry, not re-adding it, gives the correct result.

Cross-check: For two positive operands specifically, a genuine carry-out of the sign-bit column cannot occur while the sum stays in range (0+0 in that column produces at most a 1, with no further carry); an out-of-range positive sum instead shows up as overflow — a carry into the sign column without a matching carry out of it, which flips the sign bit incorrectly. Either way, the procedural rule governing two's complement addition never re-adds a bit carried past the MSB anywhere else in the number — it is dropped. Among the offered choices, only discarding the carry reflects this rule: re-adding it to the LSB is the one's-complement procedure, and directing a further addition or subtraction step at the sign bit or MSB afterward is not part of any standard binary addition procedure.

Result: Ignoring the carry is the correct procedure.