# binary number 2

## Subtraction

We 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 place-weight 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 negative-weight 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 negative-weight 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 place-weight.

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 twenty-five, we need at least five bits, plus a sixth bit for the negative-weight bit. Let’s start by representing positive twenty-five, 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 twenty-five by using the negative-weight (sixth) bit with a value of negative thirty-two, 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}.

## Overflow

One caveat with signed binary numbers is that of *overflow*, where the answer to an addition or subtraction problem exceeds the magnitude which can be represented with the alloted number of bits. Remember that the place of the sign bit is fixed from the beginning of the problem. With the last example problem, we used five binary bits to represent the magnitude of the number, and the left-most (sixth) bit as the negative-weight, or sign, bit. With five bits to represent magnitude, we have a representation range of 2^{5}, or thirty-two integer steps from 0 to maximum. This means that we can represent a number as high as +31_{10} (011111_{2}), or as low as -32_{10} (100000_{2}). If we set up an addition problem with two binary numbers, the sixth bit used for sign, and the result either exceeds +31_{10} or is less than -32_{10}, our answer will be incorrect. Let’s try adding 17_{10} and 19_{10} to see how this overflow condition works for excessive positive numbers:

. 17_{10} = 10001_{2} 19_{10} = 10011_{2}

.

. 1 11 < --- Carry bits
. (Showing sign bits) 010001
. + 010011
. --------
. 100100

The answer (100100_{2}), interpreted with the sixth bit as the -32_{10} place, is actually equal to -28_{10}, not +36_{10} as we should get with +17_{10} and +19_{10} added together! Obviously, this is not correct. What went wrong? The answer lies in the restrictions of the six-bit number field within which we’re working Since the magnitude of the true and proper sum (36_{10}) exceeds the allowable limit for our designated bit field, we have an *overflow error*. Simply put, six places doesn’t give enough bits to represent the correct sum, so whatever figure we obtain using the strategy of discarding the left-most “carry” bit will be incorrect.

A similar error will occur if we add two negative numbers together to produce a sum that is too low for our six-bit binary field. Let’s try adding -17_{10} and -19_{10} together to see how this works (or doesn’t work, as the case may be!):

. -17_{10} = 101111_{2} -19_{10} = 101101_{2}

.

. 1 1111 < --- Carry bits
. (Showing sign bits) 101111
. + 101101
. --------
. 1011100
. |
. Discard extra bit
.
FINAL ANSWER: 011100_{2 = +2810}

The (incorrect) answer is a *positive* twenty-eight. The fact that the real sum of negative seventeen and negative nineteen was too low to be properly represented with a five bit magnitude field and a sixth sign bit is the root cause of this difficulty.

Let’s try these two problems again, except this time using the seventh bit for a sign bit, and allowing the se of 6 bits for representing the magnitude:

. 17_{10} + 19_{10} (-17_{10}) + (-19_{10})

.

. 1 11 11 1111

. 0010001 1101111

. + 0010011 + 1101101

. ——— ———

. 0100100_{2} 11011100_{2}

. |

. Discard extra bit

.

. ANSWERS: 0100100_{2} = +36_{10}

. 1011100_{2} = -36_{10}

By using bit fields sufficiently large to handle the magnitude of the sums, we arrive at the correct answers.

In these sample problems we’ve been able to detect overflow errors by performing the addition problems in decimal form and comparing the results with the binary answers. For example, when adding +17_{10} and +19_{10} together, we knew that the answer was *supposed* to be +36_{10}, so when the binary sum checked out to be -28_{10}, we knew that something had to be wrong. Although this is a valid way of detecting overflow, it is not very efficient. After all, the whole idea of complementation is to be able to reliably add binary numbers together and not have to double-check the result by adding the same numbers together in decimal form! This is especially true for the purpose of building electronic circuits to add binary quantities together: the circuit has to be able to check itself for overflow without the supervision of a human being who already knows what the correct answer is.

What we need is a simple error-detection method that doesn’t require any additional arithmetic. Perhaps the most elegant solution is to check for the *sign* of the sum and compare it against the signs of the numbers added. Obviously, two positive numbers added together should give a positive result, and two negative numbers added together should give a negative result. Notice that whenever we had a condition of overflow in the example problems, the sign of the sum was always *opposite* of the two added numbers: +17_{10} plus +19_{10} giving -28_{10}, or -17_{10} plus -19_{10} giving +28_{10}. By checking the signs alone we are able to tell that something is wrong.

But what about cases where a positive number is added to a negative number? What sign should the sum be in order to be correct. Or, more precisely, what sign of sum would necessarily indicate an overflow error? The answer to this is equally elegant: there will *never* be an overflow error when two numbers of opposite signs are added together! The reason for this is apparent when the nature of overflow is considered. Overflow occurs when the magnitude of a number exceeds the range allowed by the size of the bit field. The sum of two identically-signed numbers may very well exceed the range of the bit field of those two numbers, and so in this case overflow is a possibility. However, if a positive number is added to a negative number, the sum will always be closer to zero than either of the two added numbers: its magnitude *must* be less than the magnitude of either original number, and so overflow is impossible.

Fortunately, this technique of overflow detection is easily implemented in electronic circuitry, and it is a standard feature in digital adder circuits: a subject for a later chapter.

## Bit groupings

The singular reason for learning and using the binary numeration system in electronics is to understand how to design, build, and troubleshoot circuits that represent and process numerical quantities in digital form. Since the bivalent (two-valued) system of binary bit numeration lends itself so easily to representation by “on” and “off” transistor states (saturation and cutoff, respectively), it makes sense to design and buil circuits leveraging this principle to perform binary calculations.

If we were to build a circuit to represent a binary number, we would have to allocate enough transistor circuits to represent as many bits as we desire. In other words, in designing a digital circuit, we must first decide how many bits (maximum) we would like to be able to represent, since each bit requires one on/off circuit to represent it. This is analogous to designing an abacus to digitally represent decimal numbers: we must decide how many digits we wish to handle in this primitive “calculator” device, for each digit requires a separate rod with its own beads.

A ten-rod abacus would be able to represent a ten-digit decimal number, or a maxmium value of 9,999,999,999. If we wished to represent a larger number on this abacus, we would be unable to, unless additional rods could be added to it.

In digital, electronic computer design, it is common to design the system for a common “bit width:” a maximum number of bits allocated to represent numerical quantities. Early digital computers handled bits in groups of four or eight. More modern systems handle numbers in clusters of 32 bits or more. To more conveniently express the “bit width” of such clusters in a digital computer, specific labels were applied to the more common groupings.

Eight bits, grouped together to form a single binary quantity, is known as a *byte*. Four bits, grouped together as one binary number, is known by the humorous title of *nibble*, often spelled as *nybble*.

A multitude of terms have followed byte and nibble for labeling specfiic groupings of binary bits. Most of the terms shown here are informal, and have not been made “authoritative” by any standards group or other sanctioning body. However, their inclusion into this chapter is warranted by their occasional appearance in technical literature, as well as the levity they add to an otherwise dry subject:

**Bit**: A single, bivalent unit of binary notation. Equivalent to a decimal “digit.”**Crumb**,**Tydbit**, or**Tayste**: Two bits.**Nibble**, or**Nybble**: Four bits.**Nickle**: Five bits.**Byte**: Eight bits.**Deckle**: Ten bits.**Playte**: Sixteen bits.**Dynner**: Thirty-two bits.**Word**: (system dependent).

The most ambiguous term by far is *word*, referring to the standard bit-grouping within a particular digital system. For a computer system using a 32 bit-wide “data path,” a “word” would mean 32 bits. If the system used 16 bits as the standard grouping for binary quantities, a “word” would mean 16 bits. The terms *playte* and *dynner*, by contrast, always refer to 16 and 32 bits, respectively, regardless of the system context in which they are used.

Context dependence is likewise true for derivative terms of *word*, such as *double word* and *longword* (both meaning twice the standard bit-width), *half-word* (half the standard bit-width), and *quad* (meaning four times the standard bit-width). One humorous addition to this somewhat boring collection of *word*-derivatives is the term *chawmp*, which means the same as *half-word*. For example, a *chawmp* would be 16 bits in the context of a 32-bit digital system, and 18 bits in the context of a 36-bit system. Also, the term *gawble* is sometimes synonymous with *word*.

Definitions for bit grouping terms were taken from Eric S. Raymond’s “Jargon Lexicon,” an indexed collection of terms — both common and obscure — germane to the world of computer programming.

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