Exercise 1: Adding 10 + 11 With a Ripple-Carry Chain — Possible Solution
====================================================================

GIVEN
------------------------------
10 (binary, decimal 2) + 11 (binary, decimal 3), using a 2-bit
ripple-carry adder built from two full adders.

STAGE 0: THE LOW BITS (a0=0, b0=1), cin=0 (nothing carries into
the very first column)
------------------------------
full_adder(0, 1, 0):
  SUM = 0 xor 1 xor 0 = 1
  COUT = (0.1) + (0.(0 xor 1)) = 0 + 0 = 0

Stage 0 result: SUM=1, COUT=0

STAGE 1: THE HIGH BITS (a1=1, b1=1), cin = stage 0's own COUT = 0
------------------------------
full_adder(1, 1, 0):
  SUM = 1 xor 1 xor 0 = 0
  COUT = (1.1) + (0.(1 xor 1)) = 1 + 0 = 1

Stage 1 result: SUM=0, COUT=1

ASSEMBLE THE FINAL RESULT
------------------------------
Final carry-out (from stage 1): 1
Stage 1's SUM (the high result bit): 0
Stage 0's SUM (the low result bit): 1

Result: 101 (binary)

CONFIRMING
------------------------------
101 in binary = 4+0+1 = 5. 2+3=5. Confirmed correct.

WHY THIS WORKS AS AN ANSWER
------------------------------
Each stage's full adder is computed explicitly using this chapter's
own SUM/COUT formulas, the carry is shown ripping from stage 0's
own COUT into stage 1's own cin exactly as this chapter's chaining
rule specifies, and the final 3-bit result is assembled and checked
against the actual decimal sum rather than just asserted.