Number Systems

Abstract

Logic circuits are used to generate and transmit 1’s and 0’s to compute and convey information. This two-valued number system is called binary. As presented earlier, there are many advantages of using a binary system; however, the human brain has been taught to count, label, and measure using the decimal number system. The decimal number system contains 10 unique symbols (0 → 9) commonly referred to as the Arabic numerals. Each of these symbols is assigned a relative magnitude to the other symbols. For example, 0 is less than 1, 1 is less than 2, etc. It is often conjectured that the 10-symbol number system that we humans use is due to the availability of our 10 fingers (or digits) to visualize counting up to 10. Regardless, our brains are trained to think of the real world in terms of a decimal system. In order to bridge the gap between the way our brains think (decimal) and how we build our computers (binary), we need to understand the basics of number systems. This includes the formal definition of a positional number system and how it can be extended to accommodate any arbitrarily large (or small) value. This also includes how to convert between different number systems that contain different numbers of symbols. In this chapter, we cover 4 different number systems: decimal (10 symbols), binary (2 symbols), octal (8 symbols), and hexadecimal (16 symbols). The study of decimal and binary is obvious as they represent how our brains interpret the physical world (decimal) and how our computers work (binary). Hexadecimal is studied because it is a useful means to represent large sets of binary values using a manageable number of symbols. Octal is rarely used but is studied as an example of how the formalization of the number systems can be applied to all systems regardless of the number of symbols they contain. This chapter will also discuss how to perform basic arithmetic in the binary number system and represent negative numbers. The goal of this chapter is to provide an understanding of the basic principles of binary number systems.

Logic circuits are used to generate and transmit 1’s and 0’s to compute and convey information. This two-valued number system is called binary. As presented earlier, there are many advantages of using a binary system; however, the human brain has been taught to count, label, and measure using the decimal number system. The decimal number system contains 10 unique symbols (0 → 9) commonly referred to as the Arabic numerals. Each of these symbols is assigned a relative magnitude to the other symbols. For example, 0 is less than 1, 1 is less than 2, etc. It is often conjectured that the 10-symbol number system that we humans use is due to the availability of our 10 fingers (or digits) to visualize counting up to 10. Regardless, our brains are trained to think of the real world in terms of a decimal system. In order to bridge the gap between the way our brains think (decimal) and how we build our computers (binary), we need to understand the basics of number systems. This includes the formal definition of a positional number system and how it can be extended to accommodate any arbitrarily large (or small) value. This also includes how to convert between different number systems that contain different numbers of symbols. In this chapter, we cover 4 different number systems: decimal (10 symbols), binary (2 symbols), octal (8 symbols), and hexadecimal (16 symbols). The study of decimal and binary is obvious as they represent how our brains interpret the physical world (decimal) and how our computers work (binary). Hexadecimal is studied because it is a useful means to represent large sets of binary values using a manageable number of symbols. Octal is rarely used but is studied as an example of how the formalization of the number systems can be applied to all systems regardless of the number of symbols they contain. This chapter will also discuss how to perform basic arithmetic in the binary number system and represent negative numbers. The goal of this chapter is to provide an understanding of the basic principles of binary number systems.

FormalPara Learning Outcomes—After completing this chapter, you will be able to:

• 2.1 Describe the formation and use of positional number systems.

• 2.2 Convert numbers between different bases.

• 2.3 Perform binary addition and subtraction by hand.

• 2.4 Use two’s complement numbers to represent negative numbers.

2.1 Positional Number Systems

A positional number system allows the expansion of the original set of symbols so that they can be used to represent any arbitrarily large (or small) value. For example, if we use the 10 symbols in our decimal system, we can count from 0 to 9. Using just the individual symbols, we do not have enough symbols to count beyond 9. To overcome this, we use the same set of symbols but assign a different value to the symbol based on its position within the number. The position of the symbol with respect to other symbols in the number allows an individual symbol to represent greater (or lesser) values. We can use this approach to represent numbers larger than the original set of symbols. For example, let’s say we want to count from 0 upward by 1. We begin counting from 0, 1, 2, 3, 4, 5, 6, 7, 8, to 9. When we are out of symbols and wish to go higher, we bring on a symbol in a different position with that position being valued higher and then start counting over with our original symbols (e.g., …, 9, 10, 11,... 19, 20, 21,...). This is repeated each time a position runs out of symbols (e.g., …, 99, 100, 101… 999, 1000, 1001,…).

First, let’s look at the formation of a number system. The first thing that is needed is a set of symbols. The formal term for one of the symbols in a number system is a numeral. One or more numerals are used to form a number. We define the number of numerals in the system using the terms radix or base. For example, our decimal number system is said to be base 10 or have a radix of 10 because it consists of 10 unique numerals or symbols.

$$ \mathbf{Radix}=\mathbf{Base}\equiv \mathbf{the}\ \mathbf{number}\ \mathbf{of}\ \mathbf{numerals}\ \mathbf{in}\ \mathbf{the}\ \mathbf{number}\ \mathbf{system} $$

The next thing that is needed is the relative value of each numeral with respect to the other numerals in the set. We can say 0 < 1 < 2 < 3 etc. to define the relative magnitudes of the numerals in this set. The numerals are defined to be greater or less than their neighbors by a magnitude of 1. For example, in the decimal number system, each of the subsequent numerals is greater than its predecessor by exactly 1. When we define this relative magnitude, we are defining that the numeral 1 is greater than the numeral 0 by a magnitude of 1; the numeral 2 is greater than the numeral 1 by a magnitude of 1, etc. At this point we have the ability to count from 0 to 9 by 1’s. We also have the basic structure for mathematical operations that have results that fall within the numeral set from 0 to 9 (e.g., 1 + 2 = 3). In order to expand the values that these numerals can represent, we need define the rules of a positional number system.

2.1.1 Generic Structure

In order to represent larger or smaller numbers than the lone numerals in a number system can represent, we adopt a positional system. In a positional number system, the relative position of the numeral within the overall number dictates its value. When we begin talking about the position of a numeral, we need to define a location to which all of the numerals are positioned with respect to. We define the radix point as the point within a number to which numerals to the left represent whole numbers and numerals to the right represent fractional numbers. The radix point is denoted with a period (i.e., “.”). A particular number system often renames this radix point to reflect its base. For example, in the base 10 number system (i.e., decimal), the radix point is commonly called the decimal point; however, the term radix point can be used across all number systems as a generic term. If the radix point is not present in a number, it is assumed to be to the right of number. Fig. 2.1 shows an example number highlighting the radix point and the relative positions of the whole and fractional numerals.

Fig. 2.1

Definition of radix point

Next, we need to define the position of each numeral with respect to the radix point. The position of the numeral is assigned a whole number with the number to the left of the radix point having a position value of 0. The position number increases by 1 as numerals are added to the left (2, 3, 4…) and decreased by 1 as numerals are added to the right (−1, −2, −3). We will use the variable p to represent position. The position number will be used to calculate the value of each numeral in the number based on its relative position to the radix point. Figure 2.2 shows the example number with the position value of each numeral highlighted.

Fig. 2.2

Definition of position number (p) within the number

In order to create a generalized format of a number, we assign the term digit (d) to each of the numerals in the number. The term digit signifies that the numeral has a position. The position of the digit within the number is denoted as a subscript. The term digit can be used as a generic term to describe a numeral across all systems, although some number systems will use a unique term instead of digit which indicates its base. For example, the binary system uses the term bit instead of digit; however, using the term digit to describe a generic numeral in any system is still acceptable. Figure 2.3 shows the generic subscript notation used to describe the position of each digit in the number.

Fig. 2.3

Digit notation

We write a number from left to right starting with the highest position digit that is greater than 0 and end with the lowest position digit that is greater than 0. This reduces the number of numerals that are written; however, a number can be represented with an arbitrary number of 0’s to the left of the highest position digit greater than 0 and an arbitrary number of 0’s to the right of the lowest position digit greater than 0 without affecting the value of the number. For example, the number 132.654 could be written as 0132.6540 without affecting the value of the number. The 0’s to the left of the number are called leading 0’s and the 0’s to the right of the number are called trailing 0’s. The reason this is being stated is because when a number is implemented in circuitry, the number of numerals is fixed, and each numeral must have a value. The variable n is used to represent the number of numerals in a number. If a number is defined with n = 4, that means 4 numerals are always used. The number 0 would be represented as 0000 with both representations having an equal value.

2.1.2 Decimal Number System (Base 10)

As mentioned earlier, the decimal number system contains ten unique numerals (0, 1, 2, 3, 4, 5, 6, 7, 8, and 9). This system is thus a base 10 or a radix 10 system. The relative magnitudes of the symbols are 0 < 1 < 2 < 3 < 4 < 5 < 6 < 7 < 8 < 9.

2.1.3 Binary Number System (Base 2)

The binary number system contains two unique numerals (0 and 1). This system is thus a base 2 or a radix 2 system. The relative magnitudes of the symbols are 0 < 1. At first glance, this system looks very limited in its ability to represent large numbers due to the small number of numerals. When counting up, as soon as you count from 0 to 1, you are out of symbols and must increment the p + 1 position in order to represent the next number (e.g., 0, 1, 10, 11, 100, 101, …); however, magnitudes of each position scale quickly so that circuits with a reasonable amount of digits can represent very large numbers. The term bit is used instead of digit in this system to describe the individual numerals and at the same time indicate the base of the number.

Due to the need for multiple bits to represent meaningful information, there are terms dedicated to describing the number of bits in a group. When 4 bits are grouped together, they are called a nibble. When 8 bits are grouped together, they are called a byte. Larger groupings of bits are called words. The size of the word can be stated as either an n-bit word or omitted if the size of the word is inherently implied. For example, if you were using a 32-bit microprocessor, using the term word would be interpreted as a 32-bit word. For example, if there was a 32-bit grouping, it would be referred to as a 32-bit word. The leftmost bit in a binary number is called the most significant bit (MSB). The rightmost bit in a binary number is called the least significant bit (LSB).

2.1.4 Octal Number System (Base 8)

The octal number system contains eight unique numerals (0, 1, 2, 3, 4, 5, 6, 7). This system is thus a base 8 or a radix 8 system. The relative magnitudes of the symbols are 0 < 1 < 2 < 3 < 4 < 5 < 6 < 7. We use the generic term digit to describe the numerals within an octal number.

2.1.5 Hexadecimal Number System (Base 16)

The hexadecimal number system contains 16 unique numerals. This system is most often referred to in spoken word as “hex” for short. Since we only have ten Arabic numerals in our familiar decimal system, we need to use other symbols to represent the remaining six numerals. We use the alphabetic characters A–F in order to expand the system to 16 numerals. The 16 numerals in the hexadecimal system are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, and F. The relative magnitudes of the symbols are 0 < 1 < 2 < 3 < 4 < 5 < 6 < 7 < 8 < 9 < A < B < C < D < E < F. We use the generic term digit to describe the numerals within a hexadecimal number.

At this point, it becomes necessary to indicate the base of a written number. The number 10 has an entirely different value if it is a decimal number or binary number. In order to handle this, a subscript is typically included at the end of the number to denote its base. For example, 10₁₀ indicates that this number is decimal “ten.” If the number was written as 10₂, this number would represent binary “one zero.” Table 2.1 lists the equivalent values in each of the four number systems just described for counts from 0₁₀ to 15₁₀. The left side of the table does not include leading 0 s. The right side of the table contains the same information but includes the leading zeros. The equivalencies of decimal, binary, and hexadecimal in this table are typically committed to memory.

Table 2.1 Number system equivalency

Concept Check

CC2.1 The base of a number system is arbitrary and is commonly selected to match a particular aspect of the physical system in which it is used (e.g., base 10 corresponds to our 10 fingers, base 2 corresponds to the 2 states of a switch). If a physical system contained 3 unique modes and a base of 3 was chosen for the number system, what is the base 3 equivalent of the decimal number 3?

• A) 3₁₀ = 11₃   B) 3₁₀ = 3₃   C) 3₁₀ = 10₃   D) 3₁₀ = 21₃

2.2 Base Conversion

Now we look at converting between bases. There are distinct techniques for converting to and from decimal. There are also techniques for converting between bases that are powers of 2 (e.g., base 2, 4, 8, 16, etc.).

2.2.1 Converting to Decimal

The value of each digit within a number is based on the individual digit value and the digit’s position. Each position in the number contains a different weight based on its relative location to the radix point. The weight of each position is based on the radix of the number system that is being used. The weight of each position in decimal is defined as:

$$ \mathbf{Weight}={\left(\mathbf{Radix}\right)}^{\mathbf{p}} $$

This expression gives the number system the ability to represent fractional numbers since an expression with a negative exponent (e.g., x^−y) is evaluated as one over the expression with the exponent change to positive (e.g., 1/x^y). Figure 2.4 shows the generic structure of a number with its positional weight highlighted.

Fig. 2.4

Weight definition

In order to find the decimal value of each of the numerals in the number, its individual numeral value is multiplied by its positional weight. In order to find the value of the entire number, each value of the individual numeral-weight products is summed. The generalized format of this conversion is written as:

$$ \boldsymbol{Total}\ \boldsymbol{Decimal}\ \boldsymbol{Value}=\sum \limits_{\boldsymbol{i}={\boldsymbol{p}}_{\boldsymbol{min}}}^{{\boldsymbol{p}}_{\boldsymbol{max}}}{\boldsymbol{d}}_{\boldsymbol{i}}\bullet {\left(\boldsymbol{radix}\right)}^{\boldsymbol{i}} $$

In this expression, p_max represents the highest position number that contains a numeral greater than 0. The variable p_min represents the lowest position number that contains a numeral greater than 0. These limits are used to simplify the hand calculations; however, these terms theoretically could be +∞ to −∞ with no effect on the result since the summation of every leading 0 and every trailing 0 contributes nothing to the result.

As an example, let’s evaluate this expression for a decimal number. The result will yield the original number but will illustrate how positional weight is used. Let’s take the number 132.654₁₀. To find the decimal value of this number, each numeral is multiplied by its positional weight, and then all of the products are summed. The positional weight for the digit 1 is (radix)^p or (10)². In decimal this is called the hundred’s position. The positional weight for the digit 3 is (10)¹, referred to as the ten’s position. The positional weight for digit 2 is (10)⁰, referred to as the one’s position. The positional weight for digit 6 is (10)⁻¹, referred to as the tenth’s position. The positional weight for digit 5 is (10)⁻², referred to as the hundredth’s position. The positional weight for digit 4 is (10)⁻³, referred to as the thousandth’s position.

When these weights are multiplied by their respective digits and summed, the result is the original decimal number 132.654₁₀. Example 2.1 shows this process step-by-step.

Example 2.1

Converting decimal to decimal

This process is used to convert between any other base to decimal.

2.2.1.1 Binary to Decimal

Let’s convert 101.11₂ to decimal. The same process is followed with the exception that the base in the summation is changed to 2. Converting from binary to decimal can be accomplished quickly in your head due to the fact that the bit values in the products are either 1 or 0. That means any bit that is a 0 has no impact on the outcome and any bit that is a 1 simply yields the weight of its position. Example 2.2 shows the step-by-step process converting a binary number to decimal.

Example 2.2

Converting binary to decimal

2.2.1.2 Octal to Decimal

When converting from octal to decimal, the same process is followed with the exception that the base in the weight is changed to 8. Example 2.3 shows an example of converting an octal number to decimal.

Example 2.3

Converting octal to decimal

2.2.1.3 Hexadecimal to Decimal

Let’s convert 1AB.EF₁₆ to decimal. The same process is followed with the exception that the base is changed to 16. When performing the conversion, the decimal equivalent of the numerals A–F needs to be used. Example 2.4 shows the step-by-step process converting a hexadecimal number to decimal.

Example 2.4

Converting hexadecimal to decimal

In some cases, it is desired to specify a level of accuracy for the conversion in order to bound the number of fractional digits in the final result. For example, if the conversion in Example 2.4 was stated as “convert 1AB.EF₁₆ to decimal with a fractional accuracy of 2 digits,” the final result would be 427.93₁₀. How rounding is handled can also be specified with the two options being with or without rounding. In the case where the conversion is performed with rounding, additional fractional digits may need to be computed to determine if the least significant digit of the new decimal fraction needs to be altered. For example, let’s say the conversion in Example 2.4 is stated as “convert 1AB.EF₁₆ to decimal with a fractional accuracy of 4 digits with rounding.” In this case, the final result would be 427.9336₁₀. Notice how rounding was applied to the digit in position p = −3 changing it from a 5 to a 6 based on the value in position p = −4. Now let’s say the conversion in Example 2.4 is stated as “convert 1AB.EF₁₆ to decimal with a fractional accuracy of 4 digits without rounding.” In this case, the final result would be 427.9335₁₀. Notice how without rounding simply drops all of the digits beyond the specified level of accuracy.

2.2.2 Converting from Decimal

The process of converting from decimal to another base consists of two separate algorithms. There is one algorithm for converting the whole number portion of the number and another algorithm for converting the fractional portion of the number. The process for converting the whole number portion is to divide the decimal number by the base of the system you wish to convert to. The division will result in a quotient and a whole number remainder. The remainder is recorded as the least significant numeral in the converted number. The resulting quotient is then divided again by the base, which results in a new quotient and new remainder. The remainder is recorded as the next higher order numeral in the new number. This process is repeated until a quotient of 0 is achieved. At that point the conversion is complete. The remainders will always be within the numeral set of the base being converted to.

The process for converting the fractional portion is to multiply just the fractional component of the number by the base. This will result in a product that contains a whole number and a fraction. The whole number is recorded as the most significant digit of the new converted number. The new fractional portion is then multiplied again by the base with the whole number portion being recorded as the next lower order numeral. This process is repeated until the product yields a fractional component equal to zero or the desired level of accuracy has been achieved. The level of accuracy is specified by the number of numerals in the new converted number. For example, the conversion would be stated as “convert this decimal number to binary with a fractional accuracy of 4 bits.” This means the final result would only have 4 bits in the fraction. In cases where the conversion does not yield exactly 4 fractional bits, there are two approaches that can be used. The first is to have no rounding, which means the conversion simply stops at the desired accuracy. The second is to apply rounding, which means additional bits beyond the desired accuracy are computed in order to determine whether the least significant bit reported.

2.2.2.1 Decimal to Binary

Let’s convert 11.375₁₀ to binary. Example 2.5 shows the step-by-step process converting a decimal number to binary.

Example 2.5

Converting decimal to binary

In many binary conversions to binary, the number of fractional bits that result from the conversion is more than needed. In this case, rounding is applied to limit the fractional accuracy. The simplest rounding approach for binary numbers is to continue the conversion for one more bit beyond the desired fractional accuracy. If the next bit is a 0, then you leave the fractional component of the number as is. If the next bit is a 1, you round the least significant bit of your number up. Often this rounding will result in a cascade of roundings from the LSB to the MSB. As an example, let’s say that the conversion in Example 2.5 was specified to have a fractional accuracy of 2 bits. If the bit in position p = −3 was a 0 (which it is not, but let’s just say it is for the sake of this example), then the number would be left as is, and the final converted number would be 1011.01₂. However, if the bit in position p = −3 was a 1 (as it actually is in Example 2.5), then we would need to apply rounding. We would start with the bit in position p = −2. Since it is a 1, we would round that up to a 0, but we would need to apply the overflow of this rounding to the next higher order bit in position p = −1. That would then cause the value of p = −1 to go from a 0 to a 1. The final result of the conversion with rounding would be 1011.10₂.

2.2.2.2 Decimal to Octal

Let’s convert 10.4₁₀ to octal with an accuracy of 4 fractional digits. When converting the fractional component of the number, the algorithm is continued until 4 digits worth of fractional numerals has been achieved. Once the accuracy has been achieved, the conversion is finished even though a product with a zero fractional value has not been obtained. Example 2.6 shows the step-by-step process converting a decimal number to octal with a fractional accuracy of 4 digits.

Example 2.6

Converting decimal to octal

Rounding of octal digits uses a similar approach as when rounding decimal numbers, with the exception that the middle of the range of the numbers lies between digits 3₈ and 4₈. This means that any number to be rounded that is 4₈ or greater will be rounded up. Numbers that are 3₈ or less will be rounded down, which means the fractional component of the converted number is left as in.

2.2.2.3 Decimal to Hexadecimal

Let’s convert 254.655₁₀ to hexadecimal with an accuracy of 3 fractional digits. When doing this conversion, all of the divisions and multiplications are done using decimal. If the results end up between 10₁₀ and 15₁₀, then the decimal numbers are substituted with their hex symbol equivalent (i.e., A to F). Example 2.7 shows the step-by-step process of converting a decimal number to hex with a fractional accuracy of 3 digits.

Example 2.7

Converting decimal to hexadecimal

Rounding of hexadecimal digits uses a similar approach as when rounding decimal numbers, with the exception that the middle of the range of the numbers lies between digits 7₁₆ and 8₁₆. This means that any number to be rounded that is 8₁₆ or greater will be rounded up. Numbers that are 7₁₆ or less will be rounded down, which means the fractional component of the converted number is left as in.

2.2.3 Converting Between 2ⁿ Bases

Converting between 2ⁿ bases (e.g., 2, 4, 8, 16, etc.) takes advantage of the direct mapping that each of these bases has back to binary. Base 8 numbers take exactly 3 binary bits to represent all 8 symbols (i.e., 0₈ = 000₂, 7₈ = 111₂). Base 16 numbers take exactly 4 binary bits to represent all 16 symbols (i.e., 0₁₆ = 0000₂, F₁₆ = 1111₂).

When converting from binary to any other 2ⁿ base, the whole number bits are grouped into the appropriate-sized sets starting from the radix point and working left. If the final leftmost grouping does not have enough symbols, it is simply padded on the left with leading 0’s. Each of these groups is then directly substituted with their 2ⁿ base symbol. The fractional number bits are also grouped into the appropriate-sized sets starting from the radix point, but this time working right. Again, if the final rightmost grouping does not have enough symbols, it is simply padded on the right with trailing 0’s. Each of these groups is then directly substituted with their 2ⁿ base symbol.

2.2.3.1 Binary to Octal

Example 2.8 shows the step-by-step process of converting a binary number to octal.

Example 2.8

Converting binary to octal

2.2.3.2 Binary to Hexadecimal

Example 2.9 shows the step-by-step process of converting a binary number to hexadecimal.

Example 2.9

Converting binary to hexadecimal

2.2.3.3 Octal to Binary

When converting to binary from any 2ⁿ base, each of the symbols in the originating number are replaced with the appropriate-sized number of bits. An octal symbol will be replaced with 3 binary bits, while a hexadecimal symbol will be replaced with 4 binary bits. Any leading or trailing 0’s can be removed from the converted number once complete. Example 2.10 shows the step-by-step process of converting an octal number to binary.

Example 2.10

Converting Octal to Binary

2.2.3.4 Hexadecimal to Binary

Example 2.11 shows the step-by-step process of converting a hexadecimal number to binary.

Example 2.11

Converting hexadecimal to binary

2.2.3.5 Octal to Hexadecimal

When converting between 2ⁿ bases (excluding binary), the number is first converted into binary and then converted from binary into the final 2ⁿ base using the algorithms described before. Example 2.12 shows the step-by-step process of converting an octal number to hexadecimal.

Example 2.12

Converting Octal to Hexadecimal

2.2.3.6 Hexadecimal to Octal

Example 2.13 shows the step-by-step process of converting a hexadecimal number to octal.

Example 2.13

Converting Hexadecimal to Octal

Concept Check

CC2.2 A “googol” is the term for the decimal number 1e100. When written out manually this number is a 1 with 100 zeros after it (e.g., 10,000,000,000,000,000,000,000, 000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000). This term is more commonly associated with the search engine company Google, which uses a different spelling but is pronounced the same. How many bits does it take to represent a googol in binary?

• A) 100 bits   B) 256 bits   C) 332 bits   D) 333 bits

2.3 Binary Arithmetic

2.3.1 Addition (Carries)

Binary addition is a straightforward process that mirrors the approach we have learned for longhand decimal addition. The two numbers (or terms) to be added are aligned at the radix point and addition begins at the least significant bit. If the sum of the least significant position yields a value with two bits (e.g., 10₂), then the least significant bit is recorded, and the most significant bit is carried to the next higher position. The sum of the next higher position is then performed including the potential carry bit from the prior addition. This process continues from the least significant position to the most significant position. Example 2.14 shows how addition is performed on two individual bits.

Example 2.14

Single-bit binary addition

When performing binary addition, the width of the inputs and output is fixed (i.e., n-bits). Carries that exist within the n-bits are treated in the normal fashion of including them in the next higher position sum; however, if the highest position summation produces a carry, this is a uniquely named event. This event is called a carry out, or the sum is said to generate a carry. The reason this type of event is given special terminology is because in real circuitry, the number of bits of the inputs and output is fixed in hardware, and the carry out is typically handled by a separate circuit. Example 2.15 shows this process when adding two 4-bit numbers.

Example 2.15

Multiple-bit binary addition

The largest decimal sum that can result from the addition of two binary numbers is given by 2·(2ⁿ−1). For example, two 8-bit numbers to be added could both represent their highest decimal value of (2ⁿ−1) or 255₁₀ (i.e., 1111 1111₂). The sum of this number would result in 510₁₀ or (11111 1110₂). Notice that the largest sum achievable would only require one additional bit. This means that a single carry bit is sufficient to handle all possible magnitudes for binary addition.

2.3.2 Subtraction (Borrows)

Binary subtraction also mirrors longhand decimal subtraction. In subtraction, the formal terms for the two numbers being operated on are minuend and subtrahend. The subtrahend is subtracted from the minuend to find the difference. In longhand subtraction, the minuend is the top number, and the subtrahend is the bottom number. For a given position if the minuend is less than the subtrahend, it needs to borrow from the next higher order position to produce a difference that is positive. If the next higher position does not have a value that can be borrowed from (i.e., 0), then it in turn needs to borrow from the next higher position and so forth. Example 2.16 shows how subtraction is performed on two individual bits.

Example 2.16

Single-bit binary subtraction

As with binary addition, binary subtraction is accomplished on fixed widths of inputs and output (i.e., n-bits). The minuend and subtrahend are aligned at the radix point, and subtraction begins at the least significant bit position. Borrows are used as necessary as the subtractions move from the least significant position to the most significant position. If the most significant position requires a borrow, this is a uniquely named event. This event is called a borrow in or the subtraction is said to require a borrow. Again, the reason this event is uniquely named is because in real circuitry, the number of bits of the input and output is fixed in hardware, and the borrow in is typically handled by a separate circuit. Example 2.17 shows this process when subtracting two 4-bit numbers.

Example 2.17

Multiple-bit binary subtraction

Notice that if the minuend is less than the subtrahend, then the difference will be negative. At this point, we need a way to handle negative numbers.

Concept Check

CC2.3 If an 8-bit computer system can only perform unsigned addition on 8-bit inputs and produce an 8-bit sum, how is it possible for this computer to perform addition on numbers that are larger than what can be represented with 8-bits (e.g., 1,000₁₀ + 1,000₁₀ = 2,000₁₀)?

• A) There are multiple 8-bit adders in a computer to handle large numbers.

• B) The result is simply rounded to the nearest 8-bit number.

• C) The computer returns an error and requires smaller numbers to be entered.

• D) The computer keeps track of the carry out and uses it in a subsequent 8-bit addition, which enables larger numbers to be handled.

2.4 Unsigned and Signed Numbers

All of the number systems presented in the prior sections were positive. We need to also have a mechanism to indicate negative numbers. When looking at negative numbers, we only focus on the mapping between decimal and binary since octal and hexadecimal are used as just another representation of a binary number. In decimal, we are able to use the negative sign in front of a number to indicate it is negative (e.g., −34₁₀). In binary, this notation works fine for writing numbers on paper (e.g., −1010₂), but we need a mechanism that can be implemented using real circuitry. In a real digital circuit, the circuits can only deal with 0’s and 1’s. There is no “- “in a digital circuit. Since we only have 0’s and 1’s in the hardware, we use a bit to represent whether a number is positive or negative. This is referred to as the sign bit. If a binary number is not going to have any negative values, then it is called an unsigned number, and it can only represent positive numbers. If a binary number is going to allow negative numbers, it is called a signed number. It is important to always keep track of the type of number we are using as the same bit values can represent very different numbers depending on the coding mechanism that is being used.

2.4.1 Unsigned Numbers

An unsigned number is one that does not allow negative numbers. When talking about this type of code, the number of bits is fixed and stated up front. We use the variable n to represent the number of bits in the number. For example, if we had an 8-bit number, we would say, “This is an 8-bit, unsigned number.”

The number of unique codes in an unsigned number is given by 2ⁿ. For example, if we had an 8-bit number, we would have 2⁸ or 256 unique codes (e.g., 0000 0000₂ to 1111 1111₂).

The range of an unsigned number refers to the decimal values that the binary code can represent. If we use the notation N_unsigned to represent any possible value that an n-bit, unsigned number can take on, the range would be defined as: 0 < N_unsigned < (2ⁿ − 1).

$$ \mathbf{Range}\ \mathbf{of}\ \mathbf{an}\ \mathbf{UNSIGNED}\ \mathbf{number}\mathbf{\Rightarrow}\mathbf{0}\le {\mathbf{N}}_{\mathbf{unsigned}}\le \left({\mathbf{2}}^{\mathbf{n}}\hbox{--} \mathbf{1}\right) $$

For example, if we had an unsigned number with n = 4, it could take on a range of values from +0₁₀ (0000₂) to +15₁₀ (1111₂). Notice that while this number has 16 unique possible codes, the highest decimal value it can represent is 15₁₀. This is because one of the unique codes represents 0₁₀. This is the reason that the highest decimal value that can be represented is given by (2ⁿ−1). Example 2.18 shows this process for a 16-bit number.

Example 2.18

Finding the range of an unsigned number

2.4.2 Signed Numbers

Signed numbers are able to represent both positive and negative numbers. The most significant bit of these numbers is always the sign bit, which represents whether the number is positive or negative. The sign bit is defined to be a 0 if the number is positive and 1 if the number is negative. When using signed numbers, the number of bits is fixed so that the sign bit is always in the same position. There are a variety of ways to encode negative numbers using a sign bit. The encoding method used exclusively in modern computers is called two’s complement. There are two other encoding techniques called signed magnitude and one’s complement that are rarely used but are studied to motivate the power of two’s complement. When talking about a signed number, the number of bits and the type of encoding is always stated. For example, we would say, “This is an 8-bit, two’s complement number.”

2.4.2.1 Signed Magnitude

Signed Magnitude is the simplest way to encode a negative number. In this approach, the most significant bit (i.e., leftmost bit) of the binary number is considered the sign bit (0 = positive, 1 = negative). The rest of the bits to the right of the sign bit represent the magnitude or absolute value of the number. As an example of this approach, let’s look at the decimal values that a 4-bit, signed magnitude number can take on. These are shown in Example 2.19.

Example 2.19

Decimal values that a 4-bit, signed magnitude code can represent

There are drawbacks of signed magnitude encoding that are apparent from this example. First, the value of 0₁₀ has two signed magnitude codes (0000₂ and 1000₂). This is an inefficient use of the available codes and leads to complexity when building arithmetic circuitry since it must account for two codes representing the same number.

The second drawback is that addition using the negative numbers does not directly map to how decimal addition works. For example, in decimal if we added (−5) + (1), the result would be −4. In signed magnitude, adding these numbers using a traditional adder would produce (−5) + (1) = (−6). This is because the traditional addition would take place on the magnitude portion of the number. A 5₁₀ is represented with 101₂. Adding 1 to this number would result in the next higher binary code 110₂ or 6₁₀. Since the sign portion is separate, the addition is performed on |5|, thus yielding 6. Once the sign bit is included, the resulting number is −6. It is certainly possible to build an addition circuit that works on signed magnitude numbers, but it is more complex than a traditional adder because it must perform a different addition operation for the negative numbers versus the positive numbers. It is advantageous to have a single adder that works across the entire set of numbers.

Due to the duplicate codes for 0, the range of decimal numbers that signed magnitude can represent is reduced by 1 compared to unsigned encoding. For an n-bit number, there are 2ⁿ unique binary codes available, but only 2ⁿ−1 can be used to represent unique decimal numbers. If we use the notation N_SM to represent any possible value that an n-bit, signed magnitude number can take on, the range would be defined as:

$$ \mathbf{Range}\ \mathbf{of}\ \mathbf{a}\ \mathbf{SIGNED}\ \mathbf{MAGNITUDE}\ \mathbf{number}\mathbf{\Rightarrow }-\left({\mathbf{2}}^{\mathbf{n}\hbox{-} \mathbf{1}}-\mathbf{1}\right)\mathbf{\le}{\mathbf{N}}_{\mathbf{SM}}\mathbf{\le}+\left({\mathbf{2}}^{\mathbf{n}\hbox{-} \mathbf{1}}-\mathbf{1}\right) $$

Example 2.20 shows how to use this expression to find the range of decimal values that an 8-bit, signed magnitude code can represent.

Example 2.20

Finding the range of a signed magnitude number

The process to determine the decimal value from a signed magnitude binary code involves treating the sign bit separately from the rest of the code. The sign bit provides the polarity of the decimal number (0 = positive, 1 = negative). The remaining bits in the code are treated as unsigned numbers and converted to decimal using the standard conversion procedure described in the prior sections. This conversion yields the magnitude of the decimal number. The final decimal value is found by applying the sign. Example 2.21 shows an example of this process.

Example 2.21

Finding the decimal value of a signed magnitude number

2.4.2.2 One’s Complement

One’s complement is another simple way to encode negative numbers. In this approach, the negative number is obtained by taking its positive equivalent and flipping all of the 1’s to 0’s and 0’s to 1’s. This procedure of flipping the bits is called a complement (notice the two e’s). In this way, the most significant bit of the number is still the sign bit (0 = positive, 1 = negative). The rest of the bits represent the value of the number, but in this encoding scheme, the negative number values are less intuitive. As an example of this approach, let’s look at the decimal values that a 4-bit, one’s complement number can take on. These are shown in Example 2.22.

Example 2.22

Decimal values that a 4-bit, one’s complement code can represent

Again, we notice that there are two different codes for 0₁₀ (0000₂ and 1111₂). This is a drawback of one’s complement because it reduces the possible range of numbers that can be represented from 2ⁿ to (2ⁿ−1) and requires arithmetic operations that take into account the gap in the number system. There are advantages of one’s complement, however. First, the numbers are ordered such that traditional addition works on both positive and negative numbers (excluding the double 0 gap). Taking the example of (−5) + (1) again, in one’s complement the result yields −4, just as in a traditional decimal system. Notice in one’s complement, −5₁₀ is represented with 1010₂. Adding 1 to this entire binary code would result in the next higher binary code 1011₂ or − 4₁₀ from the above table. This makes addition circuitry less complicated, but still not as simple as if the double 0 gap was eliminated. Another advantage of one’s complement is that as the numbers are incremented beyond the largest value in the set, they roll over and start counting at the lowest number. For example, if you increment the number 0111₂ (7₁₀), it goes to the next higher binary code 1000₂, which is −7₁₀. The ability to have the numbers roll over is a useful feature for computer systems.

If we use the notation N_1comp to represent any possible value that an n-bit, one’s complement number can take on, the range is defined as:

$$ \mathbf{Range}\ \mathbf{of}\ \mathbf{a}\ {\mathbf{ONE}}^{'}\mathbf{S}\ \mathbf{COMPLEMENT}\ \mathbf{number}\mathbf{\Rightarrow }-\left({\mathbf{2}}^{\mathbf{n}\hbox{-} \mathbf{1}}-\mathbf{1}\right)\mathbf{\le}{\mathbf{N}}_{\mathbf{1}`\mathbf{s}\ \mathbf{comp}}\mathbf{\le}+\left({\mathbf{2}}^{\mathbf{n}\hbox{-} \mathbf{1}}-\mathbf{1}\right) $$

Example 2.23 shows how to use this expression to find the range of decimal values that a 24-bit, one’s complement code can represent.

Example 2.23

Finding the range of a one’s complement number

The process of finding the decimal value of a one’s complement number involves first identifying whether the number is positive or negative by looking at the sign bit. If the number is positive (i.e., the sign bit is 0), then the number is treated as an unsigned code and is converted to decimal using the standard conversion procedure described in prior sections. If the number is negative (i.e., the sign bit is 1), then the number sign is recorded separately, and the code is complemented in order to convert it to its positive magnitude equivalent. This new positive number is then converted to decimal using the standard conversion procedure. As the final step, the sign is applied. Example 2.24 shows an example of this process.

Example 2.24

Finding the decimal value of a one’s complement number

2.4.2.3 Two’s Complement

Two’s complement is an encoding scheme that addresses the double 0 issue in signed magnitude and one’s complement representations. In this approach, the negative number is obtained by subtracting its positive equivalent from 2ⁿ. This is identical to performing a complement on the positive equivalent and then adding one. If a carry is generated, it is discarded. This procedure is called taking the two’s complement of a number. The procedure of complementing each bit and adding one is the most common technique to perform a two’s complement. In this way, the most significant bit of the number is still the sign bit (0 = positive, 1 = negative), but all of the negative numbers are in essence shifted up so that the double 0 gap is eliminated. Taking the two’s complement of a positive number will give its negative counterpart and vice versa. Let’s look at the decimal values that a 4-bit, two’s complement number can take on. These are shown in Example 2.25.

Example 2.25

Decimal values that a 4-bit, two’s complement code can represent

There are many advantages of two’s complement encoding. First, there is no double 0 gap, which means that all possible 2ⁿ unique codes that can exist in an n-bit number are used. This gives the largest possible range of numbers that can be represented. Another advantage of two’s complement is that addition with negative numbers works exactly the same as decimal. In our example of (−5) + (1), the result (−4). Arithmetic circuitry can be built to mimic the way our decimal arithmetic works without the need to consider the double 0 gap. Finally, the rollover characteristic is preserved from one’s complement. Incrementing +7 by +1 will result in −8.

If we use the notation N_2comp to represent any possible value that an n-bit, two’s complement number can take on, the range is defined as:

$$ \mathbf{Range}\ \mathbf{of}\ \mathbf{a}\ {\mathbf{TWO}}^{'}\mathbf{S}\ \mathbf{COMPLEMENT}\ \mathbf{number}\mathbf{\Rightarrow }-\left({\mathbf{2}}^{\mathbf{n}\hbox{-} \mathbf{1}}\right)\mathbf{\le}{\mathbf{N}}_{\mathbf{2}'\mathbf{s}\ \mathbf{comp}}\mathbf{\le}+\left({\mathbf{2}}^{\mathbf{n}\hbox{-} \mathbf{1}}-\mathbf{1}\right) $$

Example 2.26 shows how to use this expression to find the range of decimal values that a 32-bit, two’s complement code can represent.

Example 2.26

Finding the range of a two’s complement number

The process of finding the decimal value of a two’s complement number involves first identifying whether the number is positive or negative by looking at the sign bit. If the number is positive (i.e., the sign bit is 0), then the number is treated as an unsigned code and is converted to decimal using the standard conversion procedure described in prior sections. If the number is negative (i.e., the sign bit is 1), then the number sign is recorded separately, and a two’s complement is performed on the code in order to convert it to its positive magnitude equivalent. This new positive number is then converted to decimal using the standard conversion procedure. The final step is to apply the sign. Example 2.27 shows an example of this process.

Example 2.27

Finding the decimal value of a two’s complement number

To convert a decimal number into its two’s complement code, the range is first checked to determine whether the number can be represented with the allocated number of bits. The next step is to convert the decimal number into unsigned binary. The final step is to apply the sign bit. If the original decimal number was positive, then the conversion is complete. If the original decimal number was negative, then the two’s complement is taken on the unsigned binary code to find its negative equivalent. Example 2.28 shows this procedure when converting −99₁₀ to its 8-bit, two’s complement code.

Example 2.28

Finding the two’s complement code of a decimal number

2.4.2.4 Arithmetic with Two’s Complement

Two’s complement has a variety of arithmetic advantages. First, the operations of addition, subtraction, and multiplication are handled exactly the same as when using unsigned numbers. This means that duplicate circuitry is not needed in a system that uses both number types. Second, the ability to convert a number from positive to its negative representation by performing a two’s complement means that an adder circuit can be used for subtraction. For example, if we wanted to perform the subtraction 13₁₀ – 4₁₀ = 9₁₀, this is the same as performing 13₁₀ + (−4₁₀) = 9₁₀. This allows us to use a single adder circuit to perform both addition and subtraction as long as we have the ability to take the two’s complement of a number. Creating a circuit to perform two’s complement can be simpler and faster than building a separate subtraction circuit, so this approach can sometimes be advantageous.

There are specific rules for performing two’s complement arithmetic that must be followed to ensure proper results. First, any carry or borrow that is generated is ignored. The second rule that must be followed is to always check if two’s complement overflow occurred. Two’s complement overflow refers to when the result of the operation falls outside of the range of values that can be represented by the number of bits being used. For example, if you are performing 8-bit, two’s complement addition, the range of decimal values that can be represented is −128₁₀ to +127₁₀. Having two input terms of 127₁₀ (0111 1111₂) is perfectly legal because they can be represented by the 8-bits of the two’s complement number; however, the summation of 127₁₀ + 127₁₀ = 254₁₀ (1111 1110₂). This number does not fit within the range of values that can be represented and is actually the two’s complement code for −2₁₀, which is obviously incorrect. Two’s complement overflow occurs if any of the following occurs:

• The sum of like signs results in an answer with opposite sign

    (i.e., positive + positive = negative or negative + negative = positive)

• The subtraction of a positive number from a negative number results in a positive number

    (i.e., negative − positive = positive)

• The subtraction of a negative number from a positive number results in a negative number

    (i.e., positive − negative = negative)

Computer systems that use two’s complement have a dedicated logic circuit that monitors for any of these situations and lets the operator know that overflow has occurred. These circuits are straightforward since they simply monitor the sign bits of the input and output codes. Example 2.29 shows how to use two’s complement in order to perform subtraction using an addition operation.

Example 2.29

Two’s complement addition

Concept Check

CC2.4 A 4-bit, two’s complement number has 16 unique codes and can represent decimal numbers between −8₁₀ and +7₁₀. If the number of unique codes is even, why is it that the range of integers it can represent is not symmetrical about zero?

• A) One of the positive codes is used to represent zero. This prevents the highest positive number from reaching +8₁₀ and being symmetrical.

• B) It is asymmetrical because the system allows the numbers to roll over.

• C) It isn’t asymmetrical if zero is considered a positive integer. That way there are eight positive numbers and eight negatives numbers.

• D) It is asymmetrical because there are duplicate codes for 0.

Summary

• The base, or radix, of a number system refers to the number of unique symbols within its set. The definition of a number system includes both the symbols used and the relative values of each symbol within the set.

• The most common number systems are base 10 (decimal), base 2 (binary), and base 16 (hexadecimal). Base 10 is used because it is how the human brain has been trained to treat numbers. Base 2 is used because the two values are easily represented using electrical switches. Base 16 is a convenient way to describe large groups of bits.

• A positional number system allows larger (or smaller) numbers to be represented beyond the values within the original symbol set. This is accomplished by having each position within a number have a different weight.

• There are specific algorithms that are used to convert any base to or from decimal. There are also algorithms to convert between number systems that contain a power-of-two symbols (e.g., binary to hexadecimal and hexadecimal to binary).

• Binary arithmetic is performed on a fixed width of bits (n). When an n-bit addition results in a sum that cannot fit within n-bits, it generates a carry out bit. In an n-bit subtraction, if the minuend is smaller than the subtrahend, a borrow in can be used to complete the operation.

• Binary codes can represent both unsigned and signed numbers. For an arbitrary n-bit binary code, it is important to know the encoding technique and the range of values that can be represented.

• Signed numbers use the most significant position to represent whether the number is negative (0 = positive, 1 = negative). The width of a signed number is always fixed.

• Two’s complement is the most common encoding technique for signed numbers. It has an advantage that there are no duplicate codes for zero and that the encoding approach provides a monotonic progression of codes from the most negative number that can be represented to the most positive. This allows addition and subtraction to work the same on two’s complement numbers as it does on unsigned numbers.

• When performing arithmetic using two’s complement codes, the carry bit is ignored.

• When performing arithmetic using two’s complement codes, if the result lies outside of the range that can be represented it is called two’s complement overflow. Two’s complement overflow can be determined by looking at the sign bits of the input arguments and the sign bit of the result.

Exercise Problems

Section 2.1: Positional Number Systems

• 2.1.1 What is the radix of the binary number system?

• 2.1.2 What is the radix of the decimal number system?

• 2.1.3 What is the radix of the hexadecimal number system?

• 2.1.4 What is the radix of the octal number system?

• 2.1.5 What is the radix of a number system with base 3?

• 2.1.6 For the number 261.367, what position (p) is the number 2 in?

• 2.1.7 For the number 261.367, what position (p) is the number leftmost 6 in?

• 2.1.8 For the number 261.367, what position (p) is the number 1 in?

• 2.1.9 For the number 261.367, what position (p) is the number 3 in?

• 2.1.10 For the number 261.367, what position (p) is the number rightmost 6 in?

• 2.1.11 For the number 261.367, what position (p) is the number 7 in?

• 2.1.12 What is the name of the number system containing 10₂?

• 2.1.13 What is the name of the number system containing 10₁₀?

• 2.1.14 What is the name of the number system containing 10₁₆?

• 2.1.15 What is the name of the number system containing 10₈?

• 2.1.16 Which of the four number systems covered in this chapter (i.e., binary, decimal, hexadecimal, and octal) could the number 22 be part of? Give all that are possible.

• 2.1.17 Which of the four number systems covered in this chapter (i.e., binary, decimal, hexadecimal, and octal) could the number 99 be part of? Give all that are possible.

• 2.1.18 Which of the four number systems covered in this chapter (i.e., binary, decimal, hexadecimal, and octal) could the number 1F be part of? Give all that are possible.

• 2.1.19 Which of the four number systems covered in this chapter (i.e., binary, decimal, hexadecimal, and octal) could the number 88 be part of? Give all that are possible.

• 2.1.20 Which symbols could be used in all of the four number systems covered in this chapter (i.e., binary, decimal, hexadecimal, and octal)?

• 2.1.21 What is the only symbol that could be used in every number system from base 1 to base ∞?

Section 2.2: Base Conversions

• 2.2.1 If the number 101.111 has a radix of 2, what is the weight of the position containing the leftmost 1?

• 2.2.2 If the number 101.111 has a radix of 2, what is the weight of the position containing the bit 0?

• 2.2.3 If the number 101.111 has a radix of 2, what is the weight of the position containing the 1 immediately to the left of the radix point?

• 2.2.4 If the number 101.111 has a radix of 2, what is the weight of the position containing the 1 immediately to the right of the radix point?

• 2.2.5 If the number 101.111 has a radix of 2, what is the weight of the position containing the 1 that second to the right of the radix point?

• 2.2.6 If the number 101.111 has a radix of 2, what is the weight of the position containing the rightmost 1?

• 2.2.7 If the number 261.367 has a radix of 10, what is the weight of the position containing the numeral 2?

• 2.2.8 If the number 261.367 has a radix of 10, what is the weight of the position containing the leftmost 6?

• 2.2.9 If the number 261.367 has a radix of 10, what is the weight of the position containing the numeral 1?

• 2.2.10 If the number 261.367 has a radix of 10, what is the weight of the position containing the numeral 3?

• 2.2.11 If the number 261.367 has a radix of 10, what is the weight of the position containing the rightmost 6?

• 2.2.12 If the number 261.367 has a radix of 10, what is the weight of the position containing the numeral 7?

• 2.2.13 If the number 261.367 has a radix of 16, what is the weight of the position containing the numeral 2?

• 2.2.14 If the number 261.367 has a radix of 16, what is the weight of the position containing the leftmost 6?

• 2.2.15 If the number 261.367 has a radix of 16, what is the weight of the position containing the numeral 1?

• 2.2.16 If the number 261.367 has a radix of 16, what is the weight of the position containing the numeral 3?

• 2.2.17 If the number 261.367 has a radix of 16, what is the weight of the position containing the rightmost 6?

• 2.2.18 If the number 261.367 has a radix of 16, what is the weight of the position containing the numeral 7?

• 2.2.19 If the number 261.367 has a radix of 8, what is the weight of the position containing the numeral 2?

• 2.2.20 If the number 261.367 has a radix of 8, what is the weight of the position containing the leftmost 6?

• 2.2.21 If the number 261.367 has a radix of 8, what is the weight of the position containing the numeral 1?

• 2.2.22 If the number 261.367 has a radix of 8, what is the weight of the position containing the numeral 3?

• 2.2.23 If the number 261.367 has a radix of 8, what is the weight of the position containing the rightmost 6?

• 2.2.24 If the number 261.367 has a radix of 8, what is the weight of the position containing the numeral 7?

• 2.2.25 Convert 11 1111₂ to decimal. Treat all numbers as unsigned.

• 2.2.26 Convert 10 1001₂ to decimal. Treat all numbers as unsigned.

• 2.2.27 Convert 1100 1100₂ to decimal. Treat all numbers as unsigned.

• 2.2.28 Convert 1001 1001₂ to decimal. Treat all numbers as unsigned.

• 2.2.29 Convert 0.1111₂ to decimal. Provide the full answer without limiting its accuracy or rounding. Treat all numbers as unsigned.

• 2.2.30 Convert 0.1111₂ to decimal with a fractional accuracy of 2 digits without rounding. Treat all numbers as unsigned.

• 2.2.31 Convert 0.1111₂ to decimal with a fractional accuracy of 2 digits with rounding. Treat all numbers as unsigned.

• 2.2.32 Convert 11.01₂ to decimal. Provide the full answer without limiting its accuracy or rounding. Treat all numbers as unsigned.

• 2.2.33 Convert 11.01₂ to decimal with a fractional accuracy of 1 digit without rounding. Treat all numbers as unsigned.

• 2.2.34 Convert 11.01₂ to decimal with a fractional accuracy of 1 digit with rounding. Treat all numbers as unsigned.

• 2.2.35 Convert 1001.1001₂ to decimal. Provide the full answer without limiting its accuracy or rounding. Treat all numbers as unsigned.

• 2.2.36 Convert 1001.1001₂ to decimal with a fractional accuracy of 3 digits without rounding. Treat all numbers as unsigned.

• 2.2.37 Convert 1001.1001₂ to decimal with a fractional accuracy of 3 digits with rounding. Treat all numbers as unsigned.

• 2.2.38 Convert 1100.1101₂ to decimal. Provide the full answer without limiting its accuracy or rounding. Treat all numbers as unsigned.

• 2.2.39 Convert 1100.1101₂ to decimal with a fractional accuracy of 3 digits without rounding. Treat all numbers as unsigned.

• 2.2.40 Convert 1100.1101₂ to decimal with a fractional accuracy of 3 digits with rounding. Treat all numbers as unsigned.

• 2.2.41 Convert 72₈ to decimal. Treat all numbers as unsigned.

• 2.2.42 Convert 777₈ to decimal. Treat all numbers as unsigned.

• 2.2.43 Convert 123₈ to decimal. Treat all numbers as unsigned.

• 2.2.44 Convert 7654₈ to decimal. Treat all numbers as unsigned

• 2.2.45 Convert 0.777₈ to decimal. Provide the full answer without limiting its accuracy or rounding. Treat all numbers as unsigned.

• 2.2.46 Convert 0.777₈ to decimal with a fractional accuracy of 2 digits without rounding. Treat all numbers as unsigned.

• 2.2.47 Convert 0.777₈ to decimal with a fractional accuracy of 2 digits with rounding. Treat all numbers as unsigned.

• 2.2.48 Convert 12.57₈ to decimal. Provide the full answer without limiting its accuracy or rounding. Treat all numbers as unsigned.

• 2.2.49 Convert 12.57₈ to decimal with a fractional accuracy of 4 digits without rounding. Treat all numbers as unsigned.

• 2.2.50 Convert 12.57₈ to decimal with a fractional accuracy of 4 digits with rounding. Treat all numbers as unsigned.

• 2.2.51 Convert 123.123₈ to decimal. Provide the full answer without limiting its accuracy or rounding. Treat all numbers as unsigned.

• 2.2.52 Convert 123.123₈ to decimal with a fractional accuracy of 1 digit without rounding. Treat all numbers as unsigned.

• 2.2.53 Convert 123.123₈ to decimal with a fractional accuracy of 1 digit with rounding. Treat all numbers as unsigned.

• 2.2.54 Convert 7654.7654₈ to decimal. Provide the full answer without limiting its accuracy or rounding. Treat all numbers as unsigned.

• 2.2.55 Convert 7654.7654₈ to decimal with a fractional accuracy of 4 digits without rounding. Treat all numbers as unsigned.

• 2.2.56 Convert 7654.7654₈ to decimal with a fractional accuracy of 4 digits with rounding. Treat all numbers as unsigned.

• 2.2.57 Convert F3₁₆ to decimal. Treat all numbers as unsigned.

• 2.2.58 Convert FFF₁₆ to decimal. Treat all numbers as unsigned.

• 2.2.59 Convert FACE₁₆ to decimal. Treat all numbers as unsigned.

• 2.2.60 Convert BEEF FEED₁₆ to decimal. Treat all numbers as unsigned.

• 2.2.61 Convert 0.FF₁₆ to decimal. Provide the full answer without limiting its accuracy or rounding. Treat all numbers as unsigned.

• 2.2.62 Convert 0.FF₁₆ to decimal with a fractional accuracy of 4 digits without rounding. Treat all numbers as unsigned.

• 2.2.63 Convert 0.FF₁₆ to decimal with a fractional accuracy of 4 digits with rounding. Treat all numbers as unsigned.

• 2.2.64 Convert EE.0F₁₆ to decimal. Provide the full answer without limiting its accuracy or rounding. Treat all numbers as unsigned.

• 2.2.65 Convert EE.0F ₁₆ to decimal with a fractional accuracy of 4 digits without rounding. Treat all numbers as unsigned.

• 2.2.66 Convert EE.0F ₁₆ to decimal with a fractional accuracy of 4 digits with rounding. Treat all numbers as unsigned.

• 2.2.67 Convert 15B.CEF₁₆ to decimal. Provide the full answer without limiting its accuracy or rounding. Treat all numbers as unsigned.

• 2.2.68 Convert 15B.CEF₁₆ to decimal with a fractional accuracy of 2 digits without rounding. Treat all numbers as unsigned.

• 2.2.69 Convert 15B.CEF₁₆ to decimal with a fractional accuracy of 2 digits with rounding. Treat all numbers as unsigned.

• 2.2.70 Convert 1ACE.E1F₁₆ to decimal. Provide the full answer without limiting its accuracy or rounding. Treat all numbers as unsigned.

• 2.2.71 Convert 1ACE.E1F₁₆ to decimal with a fractional accuracy of 4 digits without rounding. Treat all numbers as unsigned.

• 2.2.72 Convert 1ACE.E1F₁₆ to decimal with a fractional accuracy of 4 digits with rounding. Treat all numbers as unsigned.

• 2.2.73 Convert 67₁₀ to binary. Treat all numbers as unsigned.

• 2.2.74 Convert 100₁₀ to binary. Treat all numbers as unsigned.

• 2.2.75 Convert 999₁₀ to binary. Treat all numbers as unsigned.

• 2.2.76 Convert 1111₁₀ to binary. Treat all numbers as unsigned.

• 2.2.77 Convert 0.875₁₀ to binary. Provide the full answer without limiting its accuracy or rounding. Treat all numbers as unsigned.

• 2.2.78 Convert 0.875₁₀ to binary with a fractional accuracy of 2 bits without rounding. Treat all numbers as unsigned.

• 2.2.79 Convert 0.875₁₀ to binary with a fractional accuracy of 2 bits with rounding. Treat all numbers as unsigned.

• 2.2.80 Convert 1.4375₁₀ to binary. Provide the full answer without limiting its accuracy or rounding. Treat all numbers as unsigned.

• 2.2.81 Convert 1.4375₁₀ to binary with a fractional accuracy of 3 bits without rounding. Treat all numbers as unsigned.

• 2.2.82 Convert 1.4375₁₀ to binary with a fractional accuracy of 3 bits with rounding. Treat all numbers as unsigned.

• 2.2.83 Convert 31.65625₁₀ to binary. Provide the full answer without limiting its accuracy or rounding. Treat all numbers as unsigned.

• 2.2.84 Convert 31.65625₁₀ to binary with a fractional accuracy of 3 bits without rounding. Treat all numbers as unsigned.

• 2.2.85 Convert 31.65625₁₀ to binary with a fractional accuracy of 3 bits with rounding. Treat all numbers as unsigned.

• 2.2.86 Convert 252.987₁₀ to binary. Provide the full answer without limiting its accuracy or rounding. Treat all numbers as unsigned.

• 2.2.87 Convert 252.987₁₀ to binary with a fractional accuracy of 4 bits without rounding. Treat all numbers as unsigned.

• 2.2.88 Convert 252.987₁₀ to binary with a fractional accuracy of 4 bits with rounding. Treat all numbers as unsigned.

• 2.2.89 Convert 67₁₀ to octal. Treat all numbers as unsigned.

• 2.2.90 Convert 101₁₀ to octal. Treat all numbers as unsigned.

• 2.2.91 Convert 777₁₀ to octal. Treat all numbers as unsigned.

• 2.2.92 Convert 7654₁₀ to octal. Treat all numbers as unsigned.

• 2.2.93 Convert 0.1875 ₁₀ to octal. Provide the full answer without limiting its accuracy or rounding. Treat all numbers as unsigned.

• 2.2.94 Convert 0.1875₁₀ to octal with a fractional accuracy of 1 digit without rounding. Treat all numbers as unsigned.

• 2.2.95 Convert 0.1875₁₀ to octal with a fractional accuracy of 1 digit with rounding. Treat all numbers as unsigned.

• 2.2.96 Convert 4.5625₁₀ to octal. Provide the full answer without limiting its accuracy or rounding. Treat all numbers as unsigned.

• 2.2.97 Convert 4.5625₁₀ to octal with a fractional accuracy of 1 digit without rounding. Treat all numbers as unsigned.

• 2.2.98 Convert 4.5625₁₀ to octal with a fractional accuracy of 1 digit with rounding. Treat all numbers as unsigned.

• 2.2.99 Convert 77.15625₁₀ to octal. Provide the full answer without limiting its accuracy or rounding. Treat all numbers as unsigned.

• 2.2.100 Convert 77.15625₁₀ to octal with a fractional accuracy of 1 digit without rounding. Treat all numbers as unsigned.

• 2.2.101 Convert 77.15625₁₀ to octal with a fractional accuracy of 1 digit with rounding. Treat all numbers as unsigned.

• 2.2.102 Convert 22.2890625₁₀ to octal. Provide the full answer without limiting its accuracy or rounding. Treat all numbers as unsigned.

• 2.2.103 Convert 22.2890625₁₀ to octal with a fractional accuracy of 2 digits without rounding. Treat all numbers as unsigned.

• 2.2.104 Convert 22.2890625₁₀ to octal with a fractional accuracy of 2 digits with rounding. Treat all numbers as unsigned.

• 2.2.105 Convert 67₁₀ to hexadecimal. Treat all numbers as unsigned.

• 2.2.106 Convert 100₁₀ to hexadecimal. Treat all numbers as unsigned.

• 2.2.107 Convert 999₁₀ to hexadecimal. Treat all numbers as unsigned.

• 2.2.108 Convert 6789₁₀ to hexadecimal. Treat all numbers as unsigned.

• 2.2.109 Convert 0.109375₁₀ to hexadecimal. Provide the full answer without limiting its accuracy or rounding. Treat all numbers as unsigned.

• 2.2.110 Convert 0.109375₁₀ to hexadecimal with a fractional accuracy of 1 digit without rounding. Treat all numbers as unsigned.

• 2.2.111 Convert 0.109375₁₀ to hexadecimal with a fractional accuracy of 1 digit with rounding. Treat all numbers as unsigned.

• 2.2.112 Convert 10.6640625₁₀ to hexadecimal. Provide the full answer without limiting its accuracy or rounding. Treat all numbers as unsigned.

• 2.2.113 Convert 10.6640625₁₀ to hexadecimal with a fractional accuracy of 1 digit without rounding. Treat all numbers as unsigned.

• 2.2.114 Convert 10.6640625₁₀ to hexadecimal with a fractional accuracy of 1 digit with rounding. Treat all numbers as unsigned.

• 2.2.115 Convert 186.66796875₁₀ to hexadecimal. Provide the full answer without limiting its accuracy or rounding. Treat all numbers as unsigned.

• 2.2.116 Convert 186.66796875₁₀ to hexadecimal with a fractional accuracy of 2 digits without rounding. Treat all numbers as unsigned.

• 2.2.117 Convert 186.66796875₁₀ to hexadecimal with a fractional accuracy of 2 digits with rounding. Treat all numbers as unsigned.

• 2.2.118 Convert 57005.74560546875₁₀ to hexadecimal. Provide the full answer without limiting its accuracy or rounding. Treat all numbers as unsigned.

• 2.2.119 Convert 57005.74560546875₁₀ to hexadecimal with a fractional accuracy of 2 digits without rounding. Treat all numbers as unsigned.

• 2.2.120 Convert 57005.74560546875₁₀ to hexadecimal with a fractional accuracy of 2 digits with rounding. Treat all numbers as unsigned.

• 2.2.121 Convert 111110₂ to octal. Treat all numbers as unsigned.

• 2.2.122 Convert 10 1010.01₂ to octal. Treat all numbers as unsigned.

• 2.2.123 Convert 1010 1010.0101₂ to octal. Treat all numbers as unsigned.

• 2.2.124 Convert 1 0000 1111₂ to octal. Treat all numbers as unsigned.

• 2.2.125 Convert 11 1110₂ to hexadecimal. Treat all numbers as unsigned.

• 2.2.126 Convert 10 1010.01₂ to hexadecimal. Treat all numbers as unsigned.

• 2.2.127 Convert 1010 1010.0101₂ to hexadecimal. Treat all numbers as unsigned.

• 2.2.128 Convert 1 0000 1111.011₂ to hexadecimal. Treat all numbers as unsigned.

• 2.2.129 Convert 77₈ to binary. Treat all numbers as unsigned.

• 2.2.130 Convert 77.7₈ to binary. Treat all numbers as unsigned.

• 2.2.131 Convert 123.4₈ to binary. Treat all numbers as unsigned.

• 2.2.132 Convert 261.367₈ to binary. Treat all numbers as unsigned.

• 2.2.133 Convert AB₁₆ to binary. Treat all numbers as unsigned.

• 2.2.134 Convert F.A₁₆ to binary. Treat all numbers as unsigned.

• 2.2.135 Convert AB.CD₁₆ to binary. Treat all numbers as unsigned.

• 2.2.136 Convert 261.367₁₆ to binary. Treat all numbers as unsigned.

• 2.2.137 Convert 66₈ to hexadecimal. Treat all numbers as unsigned.

• 2.2.138 Convert 66.7₈ to hexadecimal. Treat all numbers as unsigned.

• 2.2.139 Convert 261.367₈ to hexadecimal. Treat all numbers as unsigned.

• 2.2.140 Convert 1234.5678₈ to hexadecimal. Treat all numbers as unsigned.

• 2.2.141 Convert AB₁₆ to octal. Treat all numbers as unsigned.

• 2.2.142 Convert AB.D₁₆ to octal. Treat all numbers as unsigned.

• 2.2.143 Convert ABC.DE₁₆ to octal. Treat all numbers as unsigned.

• 2.2.144 Convert BABE.FACE₁₆ to octal. Treat all numbers as unsigned.

Section 2.3: Binary Arithmetic

• 2.3.1 Compute 11₂ + 01₂ by hand. Treat all numbers as unsigned. Provide the 2-bit sum and indicate whether a carry out occurred.

• 2.3.2 Compute 1010₂ + 1011₂ by hand. Treat all numbers as unsigned. Provide the 4-bit sum and indicate whether a carry out occurred.

• 2.3.3 Compute 1111 1111₂ + 0000 0001₂ by hand. Treat all numbers as unsigned. Provide the 8-bit sum and indicate whether a carry out occurred.

• 2.3.4 Compute 1010.1010₂ + 1011.1011₂ by hand. Treat all numbers as unsigned. Provide the 8-bit sum and indicate whether a carry out occurred.

• 2.3.5 Compute 1111 1111.1011₂ + 0000 0001.1100₂ by hand. Treat all numbers as unsigned. Provide the 12-bit sum and indicate whether a carry out occurred.

• 2.3.6 Compute 10₂₋01₂ by hand. Treat all numbers as unsigned. Provide the 2-bit difference and indicate whether a borrow in occurred.

• 2.3.7 Compute 1010₂₋1011₂ by hand. Treat all numbers as unsigned. Provide the 4-bit difference and indicate whether a borrow in occurred.

• 2.3.8 Compute 1111 1111₂₋0000 0001₂ by hand. Treat all numbers as unsigned. Provide the 8-bit difference and indicate whether a borrow in occurred.

• 2.3.9 Compute 1010.1010₂–1011.1011₂ by hand. Treat all numbers as unsigned. Provide the 8-bit difference and indicate whether a borrow in occurred.

• 2.3.10 Compute 1111 1111.1011₂₋0000 0001.1100₂ by hand. Treat all numbers as unsigned. Provide the 12-bit difference and indicate whether a borrow in occurred.

Section 2.4: Unsigned and Signed Numbers

• 2.4.1 What range of decimal numbers can be represented by 8-bit, two’s complement numbers?

• 2.4.2 What range of decimal numbers can be represented by 16-bit, two’s complement numbers?

• 2.4.3 What range of decimal numbers can be represented by 32-bit, two’s complement numbers?

• 2.4.4 What range of decimal numbers can be represented by 64-bit, two’s complement numbers?

• 2.4.5 What is the 8-bit, two’s complement code for +88₁₀?

• 2.4.6 What is the 8-bit, two’s complement code for −88₁₀?

• 2.4.7 What is the 8-bit, two’s complement code for −128₁₀?

• 2.4.8 What is the 8-bit, two’s complement code for −1₁₀?

• 2.4.9 What is the decimal value of the 4-bit, two’s complement code 0010₂?

• 2.4.10 What is the decimal value of the 4-bit, two’s complement code 1010₂?

• 2.4.11 What is the decimal value of the 8-bit, two’s complement code 0111 1110₂?

• 2.4.12 What is the decimal value of the 8-bit, two’s complement code 1111 1110₂?

• 2.4.13 Compute 1110₂ + 1011₂ by hand. Treat all numbers as 4-bit, two’s complement codes. Provide the 4-bit sum and indicate whether two’s complement overflow occurred.

• 2.4.14 Compute 1101 1111₂ + 0000 0001₂ by hand. Treat all numbers as 8-bit, two’s complement codes. Provide the 8-bit sum and indicate whether two’s complement overflow occurred.

• 2.4.15 Compute 1010.1010₂ + 1000.1011₂ by hand. Treat all numbers as 8-bit, two’s complement codes. Provide the 8-bit sum and indicate whether two’s complement overflow occurred.

• 2.4.16 Compute 1110 1011.1001₂ + 0010 0001.1101₂ by hand. Treat all numbers as 12-bit, two’s complement codes. Provide the 12-bit sum and indicate whether two’s complement overflow occurred.

• 2.4.17 Compute 4₁₀ – 5₁₀ using 4-bit two’s complement addition. You will need to first convert each number into its 4-bit two’s complement code and then perform binary addition (i.e., 4₁₀ + (−5₁₀)). Provide the 4-bit result and indicate whether two’s complement overflow occurred. Check your work by converting the 4-bit result back to decimal.

• 2.4.18 Compute 7₁₀ – 7₁₀ using 4-bit two’s complement addition. You will need to first convert each decimal number into its 4-bit two’s complement code and then perform binary addition (i.e., 7₁₀ + (−7₁₀)). Provide the 4-bit result and indicate whether two’s complement overflow occurred. Check your work by converting the 4-bit result back to decimal.

• 2.4.19 Compute 7₁₀ + 1₁₀ using 4-bit two’s complement addition. You will need to first convert each decimal number into its 4-bit two’s complement code and then perform binary addition. Provide the 4-bit result and indicate whether two’s complement overflow occurred. Check your work by converting the 4-bit result back to decimal.

• 2.4.20 Compute 64₁₀ – 100₁₀ using 8-bit two’s complement addition. You will need to first convert each number into its 8-bit two’s complement code and then perform binary addition (i.e, 64₁₀ + (−100₁₀)). Provide the 8-bit result and indicate whether two’s complement overflow occurred. Check your work by converting the 8-bit result back to decimal.

• 2.4.21 Compute (−99)₁₀–11₁₀ using 8-bit two’s complement addition. You will need to first convert each decimal number into its 8-bit two’s complement code and then perform binary addition (i.e., (−99₁₀) + (−11₁₀)). Provide the 8-bit result and indicate whether two’s complement overflow occurred. Check your work by converting the 8-bit result back to decimal.

• 2.4.22 Compute 50₁₀ + 100₁₀ using 8-bit two’s complement addition. You will need to first convert each decimal number into its 8-bit two’s complement code and then perform binary addition. Provide the 8-bit result and indicate whether two’s complement overflow occurred. Check your work by converting the 8-bit result back to decimal.