2.4. SubtractionWe can subtract one binary number from another by using the standard techniques adapted for decimal numbers (subtraction of each bit pair, right to left, "borrowing" as needed from bits to the left). However, if we can leverage the already familiar (and easier) technique of binary addition to subtract, that would be better. As we just learned, we can represent negative binary numbers by using the "two's complement" method and a negative placeweight bit. Here, we'll use those negative binary numbers to subtract through addition. Here's a sample problem: Subtraction: 7_{10}  5_{10} Addition equivalent: 7_{10} + (5_{10}) If all we need to do is represent seven and negative five in binary (two's complemented) form, all we need is three bits plus the negativeweight bit: positive seven = 0111_{2} negative five = 1011_{2} Now, let's add them together: . 1111 < Carry bits . 0111 . + 1011 .  . 10010 .  . Discard extra bit . . Answer = 0010_{2} Since we've already defined our number bit field as three bits plus the negativeweight bit, the fifth bit in the answer (1) will be discarded to give us a result of 0010_{2}, or positive two, which is the correct answer. Another way to understand why we discard that extra bit is to remember that the leftmost bit of the lower number possesses a negative weight, in this case equal to negative eight. When we add these two binary numbers together, what we're actually doing with the MSBs is subtracting the lower number's MSB from the upper number's MSB. In subtraction, one never "carries" a digit or bit on to the next left placeweight. Let's try another example, this time with larger numbers. If we want to add 25_{10} to 18_{10}, we must first decide how large our binary bit field must be. To represent the largest (absolute value) number in our problem, which is twentyfive, we need at least five bits, plus a sixth bit for the negativeweight bit. Let's start by representing positive twentyfive, then finding the two's complement and putting it all together into one numeration: +25_{10} = 011001_{2} (showing all six bits) One's complement of 11001_{2} = 100110_{2} One's complement + 1 = two's complement = 100111_{2} 25_{10} = 100111_{2} Essentially, we're representing negative twentyfive by using the negativeweight (sixth) bit with a value of negative thirtytwo, plus positive seven (binary 111_{2}). Now, let's represent positive eighteen in binary form, showing all six bits: . 18_{10} = 010010_{2} . . Now, let's add them together and see what we get: . . 11 < Carry bits . 100111 . + 010010 .  . 111001 Since there were no "extra" bits on the left, there are no bits to discard. The leftmost bit on the answer is a 1, which means that the answer is negative, in two's complement form, as it should be. Converting the answer to decimal form by summing all the bits times their respective weight values, we get: (1 x 32_{10}) + (1 x 16_{10}) + (1 x 8_{10}) + (1 x 1_{10}) = 7_{10} Indeed 7_{10} is the proper sum of 25_{10} and 18_{10}.
