Computer English
Chapter 3 Number Systems and
Boolean Algebra
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-2
Key points:
useful terms and definitions
of Number system and Boolean
Algbra
Difficult points:
Conversion of the Number
Systems and Boolean Algbra
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-3
Requirements:
1,Concepts of Number System and their conversion
2,Boolean Algebra
3,Moore’s Law
4,科技英语中数学公式的读法
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-4
New Words & Expressions:
hexadecimal adj.十六进制的 ; n.十六进制 radix n.根,基数
octal adj.八进制的 ; n.八进制 alphabet n.字母表
fractional adj.分数的,小数的 whole number n.整数
remainder n.余数 significant figure n.有效数字
quotient n.商 algorithm n.算法
complement n,补码，余角 carry n.进位
3.1 Number Systems
Abbreviations:
Binary-coded hexadecimal (BCH) 二进制编码的十六进制
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-5
The use of the microprocessor requires a working knowledge
of binary,decimal,and hexadecimal numbering systems,This
section provides a background for those who are unfamiliar
with number systems,Conversions between decimal and
binary,decimal and hexadecimal,and binary and hexadecimal
are described.
3.1 Number Systems
使用微处理器需要掌握二进制、十进制和十六进制数制系统
的基本知识，本节为那些不熟悉数制系统的读者提供这方面
的背景知识。说明了十进制与二进制之间、十进制与十六进
制之间，及二进制与十六进制之间的转换。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-6
Before numbers are converted from one number base to another,
the digits of a number system must be understood,Early in our
education,we learned that a decimal,or base 10,number was
constructed with 10 digits,0 through 9,The first digit in any
numbering system is always a zero,For example,a base 8 (octal)
number contains 8 digits,0 through 7; a base 2 (binary) number
contains 2 digits,0 and 1.
3.1.1 Digits
将数从 —种数制向另一种数制转换之前，必须了解数的计数系统。在早期
教育中，我们已学习了十进制数，或以 10为基的数，它由 10个数字组成,0
到 9。任何计数制的第一个数字总是零，这种规则适用于任何其他数制。例
如，以 8为基的数 (八进制 )包含 8个数字,0到 7，而以 2为基的数 (二进制 )包
含 2个数字,0和 l。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-7
If the base of a number exceeds 10,the additional digits use the
letters of the alphabet,beginning with an A,For example,a base
12 number contains 12 digits,0 through 9,followed by A for 10
and B for 11,Note that a base 10 number does not contain a 10
digit,just as a base 8 number does net contain an 8 digit,The
most common numbering systems used with computers are
decimal,binary,and hexadecimal (base 16),(Many years ago
octal numbers were popular.) Each system is described and used
in this section of the chapter.
3.1.1 Digits
如果基数大于 10，其余数字用从 A开始的字母表示，例如，以 12为基的数包含 12个数
字,0到 9，之后用 A代表 10,B代表 11。注意，以 10为基的数不包含数字 10，如同以 8
为基的数不包括数字 8一样。计算机中最通用的计数制是十进制、二进制、八进制和
十六进制 (基为 16)。每种计数制都将在本节中进行说明和应用。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-8
Once the digits of a number system are understood,larger
numbers are constructed by using positional notation,In grade
school,we learned that the position to the left of the units
position was the tens position,the position to the left of the tens
position was the hundreds position,and so forth,(An example
is the decimal number 132,This number has 1 hundred,3 tens,
and 2 units.) What probably was not learned was the
exponential value of each position,The units position has a
weight of 100 or 1; the tens position has weight of 101,or 10;
and the hundreds position has a weight of 102,or 100.
3.1.2 Positional Notation
一旦我们理解了计数制的数字后，就可用位计数法构造更大的数值。在小学时我
们都学过个位的左边一位是十位，十位左边一位是百位，以此类推 (例如十进制数
132,这个数字有 —个百，三个十和两个一 )。或许我们没有学过每个位的指数值：
个位的权为 l00，即 1；十位的权为 101或 10；而百位的权为 102或 l00。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-9
The exponential powers of the positions are critical for
understanding numbers in other numbering systems,The
position to the left of the radix (number base) point,called a
decimal point only in the decimal system,is always the units
position in any number system,For example,the position to
the left of the binary point is always 20 or 1; the position to the
left of the octal point is 80 or 1,In any case,any number raised
to its zero power is always 1,or the units position.
3.1.2 Positional Notation
位的指数幂在理解其他计数制中的数时是个关键。基数小数
点，在十进制中称为十进制小数点，其左边的位在任何数制
中都是个位。例如，二进制小数点左边的位是 20或 1。而八进
制小数点左边的位是 80或 1。在任何情况下，任何数的零次幂
总是 1，或 1个单位。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-10
The position to the left of the units position is always the
number base raised to the first power; in a decimal system,
this is l01,or l0,In a binary system,it is 21,or 2; and in an
octal system it is 81,or 8,Therefore,an 11 decimal has a
different value from an 11 binary,The 1l decimal is composed
of 1 ten plus 1 unit and has a value of 11 units; while the
binary number 11 is composed of 1 two plus 1 unit,for a value
of 3 decimal units,The 11 octal has a value of 9 units.
3.1.2 Positional Notation
个位左边的位总是基数的 1次幂，在十进制系统中是 101，或
10；在二进制中是 21，或 2；而在八进制中是 81，或 8。因此，
十进制的 11与二进制的 11具有不同的数值。十进制 11表示 —
个 10加上一个 1，其值为 11；二进制 11表示 —个 2加上 —个 1，
其值为 3；八进制 11的值为 9。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-11
In the decimal system,positions to the right of the decimal
point have negative powers,The first digit to the right of the
decimal point has a value of 10-1,or 0.1,In the binary system,
the first digit to the right of the binary point has a value of 2-1,
or 0.5,In general,the principles that apply to decimal
numbers also apply to numbers in any other number system.
3.1.2 Positional Notation
在十进制系统中，对于十进制小数点右边的位，它的幂为负
数。十进制小数点右边第一位数的值为 10-1，或 0.1。在二进
制中,二进制小数点右边第 —位数的值为 2-1或 0.5。一般来说，
十进制使用的计数法可以用于任何其他数制。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-12
Example 3-1 shows a 110.101 in binary (often written as
110.1012),It also shows the power and weight or value of each
digit position,To convert a binary number to decimal,add the
weights of each digit to form its decimal equivalent,The
110.1012 is equivalent to a 6.625 in decimal (4+2+0.5+ 0.125).
Notice that this is the sum of 22 (or 4) plus 21 (or 2),but 20 (or
1) is not added because there are no digits under this position.
The fraction part is composed of 2-1 (0.5) plus 2-3 (or.125),but
there is no digit under the 2-2 (or,25).
3.1.2 Positional Notation
例 3-1给出了一个二进制数 110.101(通常写成 110.1012)，也给出了这个数每
个位的幂、权和值。为了把二进制数转换为十进制，将每位数字的权相加，
就得到了它的等效十进制值。二进制 110.101等于十进制的
6.625(4+2+0.5+0.125)。注意，这个和的整数部分是由 22(4)加 21(2)构成，
之所以没有用 20(1)是因为这个位的数为零。小数部分由 2-1(0.5),加 2-
3(0.125)构成，但是没有用 2-2(0.25)。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-13
The prior examples have shown that to convert from any
number base to decimal,determine the weights or values of
each position of the number,and then sum the weights to form
the decimal equivalent,Suppose that a 125.78 octal is
converted to decimal,To accomplish this conversion,first write
down the weights of each position of the number,This appears
in Example 3-2,The value of 125.78 is 85.875 decimal,or 1?64
plus 2? 8 plus 5?1 plus 7?,125.
3.1.3 Conversion to Decimal
前面的例子说明了将任何其他基数的数转换为十进制数时，
十进制数的值取决于该数每个位上的权或值，它们的和就是
等效的十进制数值。假定要将 125.78(八进制 )转换为十进制。
为了完成这个转换，首先写出该数每一位数的权，如例 3-2所
示,125.78的值是十进制的 85.875，即 1?64+2? 8+5 ? 1+7?
0.125。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-14
Notice that the weight of the position to the left of the units position
is 8,This is 8 times 1,Then notice that the weight of the next position
is 64,or 8 times 8,If another position existed,it would be 64 times 8,
or 51 2,To find the weight of the next higher-order position,
multiply the weight of the current position by the number base (or 8,
in this example),To calculate the weights of position to the right of
the radix point,divide by the number base,In the octal system,the
position immediately to the fight of the octal point is 1/8,or,125,The
next position is,125/8,or,015625,which can also be written as 1/64.
3.1.3 Conversion to Decimal
注意，该数个位左边那位的权是 8(1?8)。再前一位的权是 64(8?8)。如果存
在更前一位，则其权将是 512(64?8)。将当前位的权乘上基数，就可得到
更高一位的权 (本例中是乘 8)。而计算小数点右边那些位的权，需要用基
数去除。在八进制中，紧跟八进制小数点右边的那位的权是 1/8，即 0.125。
下一位是 0.125/8，即 0.015625，也可以写成 1/64。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-15
Hexadecimal numbers are often used with computers,A
6A.CH (H for hexadecimal) is illustrated with its weights in
Example 3-3,The sum of its digits is 106.75,or 106,The whole
number part is represented with 6?16 plus 10 (A)?1,The
fraction part is 12 (C) as a numerator and 16 (16-1) as the
denominator,or 12/16,which is reduced to 3/4.
3.1.3 Conversion to Decimal
计算机经常使用十六进制。例 3-2给出了一个十六进制数
6A.CH(H表示十六进制 )，以及它的权。它的各位数值之和是
106.75，即 106。整数部分用 6?16加 10(A) ?1表示；分数部分
用 12(C)作为分子,16作为分母 (16-1)，或表示为 12/16，化简
得 3/4。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-16
Conversions from decimal to other number systems are more
difficult to accomplish than conversion to decimal,To convert
the whole number portion of a number to decimal,divide by
the radix,To convert the fractional portion,multiply by the
radix.
3.1.4 Conversion From Decimal
由十进制转换成其他进制比由其他进制转换成十进制困难。
转换十进制整数部分时，要用基数去除，转换分数部分时，
要用基数去乘它们。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-17
Whole Number Conversion from Decimal,To convert a decimal
whole number to another number system,divide by the radix
and save the remainders as significant digits of the result,An
algorithm for this conversion as is follows:
1,Divide the decimal number by the radix (number base).
2,Save the remainder (first remainder is the least significant
digit),
3,Repeat steps 1 and 2 until the quotient is zero.
3.1.4 Conversion From Decimal
转换十进制整数部分 将十进制整数转换成其他数制时，要用基数去除，
并且保存余数，作为结果的有效数字。这种转换的算法如下：
1,用基数除十进制数。
2,保存余数 (最先得到的余数是最低有效位数字 )。
3,重复步骤 l和 2，直到商为零。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-18
Converting from a Decimal Fraction,Conversion from decimal
fraction to another number base is accomplished with
multiplication by the radix,For example,to convert a decimal
fraction into binary,multiply by 2,After the multiplication,
the whole number portion of the result is saved as a significant
digit of the result,and the fractional remainder is again multiplied
by the radix,When the fraction remainder is zero,multiplication
ends,Note that some numbers are never-ending,That is,a zero is
never a remainder,An algorithm for conversion from a decimal
fraction is as follows
3.1.4 Conversion From Decimal
转换十进制小数部分 转换 10进制小数部分是用基数乘来完成的。例如，
要将十进制 小数转换成二进制，要用 2乘。乘法之后，乘积的整数部分
保存起来作为结果的一个有效位，剩余的小数再用基数 2去乘。当剩余的
小数部分为 0时，乘法结束。有些数可能永远不会结束，即余数总不为 0。
转换十进制小数部分的算法如下：
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-19
1,Multiply the decimal fraction by the radix (number base).
2,Save the whole number portion of the result (even if zero) as
a digit,Note that the first result is written immediately to the
fight of the radix point.
3,Repeat steps 1 and 2,using the fractional part of step 2 until
the fractional part of step 2 is zero.
3.1.4 Conversion From Decimal
1,用基数乘十进制小数。
2,保存结果的整数部分（即使是零）作为一位数。注意，第
一个得到的结果写在紧挨着小数点的右边。
3,用步骤 2的小数部分重复步骤 l和 2，直到步骤 2的小数部分
是零。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-20
Binary-coded hexadecimal (BCH) is used to represent hexadecimal
data in binary code,A binary-coded hexadecimal number is a
hexadecimal number written so that each digit is represented by a 4-
bit binary number,The values for the BCH digits appear in Table 3--
1.
Hexadecimal numbers are represented in BCH code by converting
each digit to BCH code,with a space between each coded digit,
3.1.5 Binary-Coded Hexadecimal
二进制编码的十六进制 (BCH)是用二进制编码表示的十六进
制数据，二进制编码的十六进制数是将十六进制数的每一位
都用 4位二进制数表示。表 3-1给出了 BCH数的值。
用 BCH表示十六进制数时，将每个十六进制数字都转换成
BCH码，并且每个数位之间用空格分开。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-21
The purpose of BCH code is to allow a binary version of a
hexadecimal number to be written in a form that can easily be
converted between BCH and hexadecimal,Example 3-8 shows
a BCH coded number converted back to hexadecimal code.
3.1.5 Binary-Coded Hexadecimal
BCH码的目的在于能将十六进制数以二进制的形式写出，使
BCH与十六进制之间转换很容易。例 3-8表示如何将 BCH代码
数据转换为十六进制码。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-22
At times,data are stored in complement form to represent
negative numbers,There are two systems that are used to
represent negative data,radix and radix -1 complements,The
earliest system was the radix -1 complement,in which each
digit of the number is subtracted from the radix -1 to generate
the radix -1 complement to represent a negative number.
3.1.6 Complements
有时，数据以补码的形式存储，以便表示负数。有两种表示
负数的方式：补码和反码 (基数减 l的补 )，最早的方式是反码。
为了得到负数的反码表示，用基数 -1减去该数的每一个数位
上的数字。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-23
Example 3-9 shows how the 8-bit binary number 01001100 is
one's (radix -1) complemented to represent it as a negative value.
Notice that each digit of the number is subtracted from one to
generate the radix -1 (one's) complement,In this example,the
negative of 01001100 is 10110011,The same technique can be
applied to any number system,as illustrated in Example 3-10,in
which the fifteen's (radix -l) complement of a 5CD hexadecimal is
computed by subtracting each digit from a fifteen.
3.1.6 Complements
例 3-9表示了如何将 8位二进制数 01001100对 l取补 (基数减 1的
补 )，以便表示成 —个负数。注意，用 1减去该数的每一位数
字，以便生成反码。在此例中,01001100的负数是 10110011。
同样的技术可适用于任何数制。如例 3-10所示，十六进制数
5CD的反码是从 15(基 -1)中减去它的每一位数字得到的。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-24
Today,the radix -1 complement is not used by itself; it is used
as a step for finding the radix complement,The radix
complement is used to represent negative numbers in modern
computer systems,(The radix -1 complement was used in the
early days of computer technology.) The main problem with
the radix -1 complement is that a negative or a positive zero
exists; in the radix complement system,only a positive zero
can exist.
3.1.6 Complements
如今，反码已不单独使用，而作为求补码的一个步骤使用，
补码是当代计算机系统表示负数的方法 (反码用于早期的计算
技术中 )。反码的主要问题是它存在负零或者正零，而补码系
统中只能存在正零。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-25
To form the radix complement,first find the radix -1
complement,and then add a one to the result,Example 3-11
shows how the number 0100 1000 is converted to a negative
value by two's (radix) complementing it.
3.1.6 Complements
为得到补码，先求反码，然后将 1加到结果上。例 3-11表示了
如何通过对 2(基为 2)取补的方式，将数 01001000转换成负数。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-26
To prove that a 0100 1000 is the inverse (negative) of a 1011 0111,add
the two together to form an 8-digit result,The ninth digit is dropped and
the result is zero because a 0l00100 is a positive 72,while a 1011 0111 is
a negative 72,The same technique applied to any number system.
Example 3-12 shows how the inverse of a 345 hexadecimal is found by
first fifteen's complementing the number,and then by adding one to the
result to form the sixteen’s complement,As before,if the original 3-digit
number 345 is added to the inverse of CBB,the result is a 3-digit 000.
As before,the fourth bit (carry) is dropped,This proves that 345 is the
inverse of CBB.
3.1.6 Complements
为验证 0100 1000是 1011 1000的反 (负数 )，将两者相加得到一个 8位
结果。去掉第 9位数字，结果是零。因为 0100 1000是正数 72，而
1011 0111是负数 72。同样的枝术可用于任何数制。例 3-12表示如何
求十六进制数 345的负数，首先求该数 15的补，然后将 1加到结果上，
得到 16的补，同前面类似，如把原来的 3位数 345加上其负数 CBB，
则结果是 3位 000，第 4位 (进位 )被丢掉。这证明了 345是 CBB的反。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-27
The concept of a Boolean algebra was first proposed by the
English mathematician George Boole in 1847,Since that time,
Boole’s original conception has been extensively developed and
refined by algebraists and logicians,The relationships among
Boolean algebra,set algebra,logic,and binary arithmetic have
given Boolean algebras a central role in the development of
electronic digital computers.
3.2 Boolean Algebra
布尔代数的概念最初是由英国数学家 George Boole于 1847年提
出来的，从那时起，代数学家和逻辑学家们更广泛地发展了
Boole最初的概念，并使之更加精练。由于布尔代数、集合代
数、逻辑学和二进制算术之间的内在联系，使得布尔代数的
理论在电子计算机的发展中起到举足轻重的作用。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-28
The most intuitive development of Boolean algebras arises from the concept of a
set algebra,Let S={a,b,c} and T={a,b,c,d,e} be two sets consisting of three
and five elements,respectively,We say that S is a subset of T,since every
element of S (namely,a,b,and c) belongs to T,Since T has five elements,there
are 25 subsets of T,for we may choose any individual element to be included or
omitted from a subset,Note that these 32 subsets include T itself and the empty
set,which contains no elements at all,If T contains all elements of concern,it is
called the universal set,Given a subset of T,such as S,we may define the
complement of S with respect to a universal set T to consist of precisely those
elements of T which are not included in the given subset.
3.2 Boolean Algebra
布尔代数最直觉的发展产生于集合代数的概念。设 S={a,b,c}和 T={a,b,c,d,e}
分别为两个含有三个和五个元素的集合。由于 S中的每一个元素（ a,b,c）
都属于 T，所以我们说 S是 T的一个子集。由于 T有五个元素，因而 T共有
25个子集，这是因为我们可以选择任何一个元素使其包含于某个子集中或
从该子集中删除。应该注意到这 32个子集中包含 T本身和空集（空集即不
含任何元素的集合）。如果 T包含了所讨论的所有元素，则称之为全集。
给定 T的一个子集，例如子集 S，我们可以定义一个关于全集 T的 S的补集。
其中正好包含那些不在子集 S中而在 T中的元素。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-29
Thus,S as above defined has its complement (with respect to T),
The union of any two sets (subsets of a given set) consists of those
elements that are in one or the other or in both given sets; the
intersection of two sets consists of those elements that are in both
given sets,We use the symbol ∪ to denote the union,and ∩ to
denote the intersection of two sets,For example,if B={b,d,e},
then B∪ S={a,b,c,d,e},and B∩S={b}.
3.2 Boolean Algebra
于是，如上定义的集合 S就有一个它的补集（相对于集合 T）。
任何两个集合（已给定集合的若干子集）的并集包含了出现
于这两个子集中某一个集合或同时出现于这两个集合中的所
有元素；两个集合的交集包含了同时出现于这两个集合中的
元素。我们用符号,∪,来表示两个集合的“并（运算）”，
用,∩” 来表示两个集合的“交（运算）”。例如，如果
B={b,d,e},那么,B∪ S={a,b,c,d,e},B∩S={b}。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-30
While other set operations may be defined,the operations of
complementation union and intersection are of primary interest to
us,A Boolean algebra is a finite or infinite set of elements together
with three operations— negation,addition,and multiplication—
that correspond to the set operations of complementation,union,
and intersection,respectively,Among the elements of a Boolean
algebra are two distinguished elements,0,corresponding to the
empty set; and 1,corresponding to the universal set.
3.2 Boolean Algebra
虽然我们可以定义其他一些集合运算，但求补、并和交运算
是我们最感兴趣的三个集合运算。一个布尔代数就是一个有
限集或无限集，以及建立在该有限集或无限集上的三种运
算 ——否定、加或乘，这三个运算分别对应于集合的求补、
并和交运算。在布尔代数的元素中有两个特殊的元素,0，对
应于空集； 1，对应于全集。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-31
For any given element of a Boolean algebra,there is a unique
complement a' with the property that a+a'=1 and aa'=0,Boolean
addition and multiplication are associative and commutative,as are
ordinary addition and multiplication,but otherwise have somewhat
different properties,The principal properties are given in Table 3-2,
where a,b,and c are any elements of a Boolean algebra.
3.2 Boolean Algebra
对于一个布尔代数中任意给定元素 a，都有一个唯一的补 a?，
它满足 a+a?=1和 aa?=0。布尔加和布尔乘与普通的加和乘一样，
满足结合律和交换律，但除此之外含有一些不太相同的特性。
其主要特性由表 3-2给出，其中 a,b和 c是一个布尔代数中的任
意元素。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-32
3.2 Boolean Algebra
?Table 3-2
Distributivity 分配律 a(b+c)=ab+aca+(bc)=(a+b)(a+c)
Idempotency 同一律 a+a=aaa=a
Absorption laws 吸收律 a+ab=aa(a+b)=a
DeMorgan’s laws德 ?摩根定理 (a+b)'=a'b'(ab)'=a'+b'
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-33
Since a finite set of n elements has exactly 2n subsets,and it can
be shown that the finite Boolean algebras are precisely the finite
set algebras,each finite Boolean algebra consists of exactly 2n
elements for some integer n,For example,the set algebra for the
set T defined above corresponds to a Boolean algebra of 32
elements.
3.2 Boolean Algebra
由于 n个元素的有限集有且只有 2n个子集，而且很显然有限布
尔代数一定是有限集合代数，所以对某个整数 n而言，每个有
限布尔代数也有且只有 2n个元素。例如，上文定义的集合 T的
集合代数就对应一个有 32个元素的布尔代数。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-34
While it is possible to use a different symbol to denote each
element of a Boolean algebra,it is often more useful to
represent the 2n elements of a finite Boolean algebra by binary
vectors having n components,With such a representation the
operations of the Boolean algebra are accomplished
componentwise by considering each component as an
independent two-element Boolean algebra,This corresponds to
representing subsets of a finite set by binary vectors.
3.2 Boolean Algebra
虽然我们可以用不同的符号来表示布尔代数中的每一个元素，
但最常用的方法是用一个有 n个分量的二进制向量来表示一个
有限布尔代数的 2n个元素。用这样一种表示方法，布尔代数
的所有运算都以分量的形式完成，而每一个分量被认为是一
个独立的二值布尔代数。这种做法对应于用二进制向量来表
示一个有限集的子集。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-35
For example,since the set T has five elements,we may represent its subsets
by five-component binary vectors,each component denoting an element of
the set T,A numeral l in the i-th component of the vector denotes the
inclusion of the i-th element of that particular subset; a 0 denotes its
exclusion,Thus,the subset S={a,b,c} has the binary vector representation
{1,1,1,0,0},The set operations become Boolean operations on the
components of the vectors,This representation of sets,and the
correspondence to Boolean or logical operations,is very useful in
information retrieval,Because of it,sets of document and query
characteristics may be easily and rapidly matched.
3.2 Boolean Algebra
例如，由于集合 T有 5个元素，所以我们可以用 5个分量的二进制向量表示
它的子集，其中每一个分量表示集合 T的一个元素。向量中的第 i个分量为
数字 1表示集合 T的第 i个元素在某一特定子集中，用数字 0表示不在某一特
定子集中。于是，子集 S={a,b,c}可用二进制向量表示为 {1,1,1,0,0}。
集合运算变成了向量分量上的布尔运算。集合的这种表示方法以及相应的
布尔或逻辑运算，对于 信息检索 是非常有用的。由于这一原因，文件的集
合和查询特性可以很容易而迅速地得到匹配。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-36
数学公式的读法 (Pronunciation of mathematical expressions)
1 逻辑 (Logic)
there exists
for all
p? q p implies q / if p,then q
p?q p if and only if q
p is equivalent to q
p and q are equivalent
?
?
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-37
2 集合 (Sets)
x?A x belongs to A / x is an element (or a member) of A
x?A x does not belong to A / x is not an element (or a member)
of A
A?B A is contained in B / A is a subset of B
A?B A contains B / B is a subset of A
A?B A cap B / A meet B/ A intersection B
A?B A cup B/ A join B / A union B
B/A A minus B/the difference between A and B
A× B A cross B / the Cartesian product of A and B(A与 B的笛卡
尔积 )
数学公式的读法 (Pronunciation of mathematical expressions)
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-38
数学公式的读法 (Pronunciation of mathematical expressions)
3 实数 (Real numbers)
x+1 x plus one
x-1 x minus one
x± 1 x plus or minus one
xy xy / x multiplied by y
(x-y)(x+y) x minus y,x plus y
x over y
= the equals sign
x=5 x equals 5 / x is equal to 5
x≠5 x (is) not equal to 5
x≡y x is equivalent to (or identical with) y
y
x
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-39
3 实数 (Real numbers)
x>y x is greater than y
x≥y x is greater than or equal to y
x< y x is less than y
x?y x is less than or equal to y
0<x<1 zero is less than x is less than 1
0?x?1 zero is less than or equal to x is less than or equal to 1
|x| mod x / modulus x
x2 x squared / x (raised) to the power 2
x3 x cubed
x4 x to the fourth / x to the power four
xn x to the nth / x to the power n
x-n x to the (power) minus n
数学公式的读法 (Pronunciation of mathematical expressions)
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-40
数学公式的读法 (Pronunciation of mathematical expressions)
3 实数 (Real numbers)
n! n factorial
(x+y)2 x plus y all squared
xi xi / x subscript i / x suffix i / x sub i
the sum from i equals one to n ai / the sum as i runs
from 1 to n of the ai
x over y all squared
x hat x bar x tilde
x? x x~
?
?
n
i
ia
1
2
???
?
???
?
y
x
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-41
数学公式的读法 (Pronunciation of mathematical expressions)
4 线性代数 (Linear algebra)
||A|| the norm (or modulus) of x
OA / vector OA
OA / the length of the segment OA
AT A transpose / the transpose of A
A-1 A inverse / the inverse of A
OA
OA
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-42
数学公式的读法 (Pronunciation of mathematical expressions)
5 函数 (Functions)
f(x) fx / f of x / the function f of x
f, S→T a function f from S to T
x maps to y / x is sent (or mapped) to y
f?(x) f prime x / f dash x / the (rst) derivative of f with
respect to x
f?(x) f double–prime x / f double–dash x / the second
derivative of f with respect to x
f??(x) f triple–prime x / f triple–dash x / the third
derivative of f with respect to x
f(4) four x / the fourth derivative of f with respect to x
ln y log y to the base e / log to the base e of y / natural
log (of) y
yx?
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-43
数学公式的读法 (Pronunciation of mathematical expressions)
5 函数 (Functions)
the partial (derivative) of f with respect to x1
the second partial (derivative) of f with respect to x1
the integral from zero to infinity
the limit as x approaches zero from above
1x
f
?
?
2
1
2
x
f
?
?
??0
0lim??x
Chapter 3 Number Systems and
Boolean Algebra
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-2
Key points:
useful terms and definitions
of Number system and Boolean
Algbra
Difficult points:
Conversion of the Number
Systems and Boolean Algbra
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-3
Requirements:
1,Concepts of Number System and their conversion
2,Boolean Algebra
3,Moore’s Law
4,科技英语中数学公式的读法
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-4
New Words & Expressions:
hexadecimal adj.十六进制的 ; n.十六进制 radix n.根,基数
octal adj.八进制的 ; n.八进制 alphabet n.字母表
fractional adj.分数的,小数的 whole number n.整数
remainder n.余数 significant figure n.有效数字
quotient n.商 algorithm n.算法
complement n,补码，余角 carry n.进位
3.1 Number Systems
Abbreviations:
Binary-coded hexadecimal (BCH) 二进制编码的十六进制
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-5
The use of the microprocessor requires a working knowledge
of binary,decimal,and hexadecimal numbering systems,This
section provides a background for those who are unfamiliar
with number systems,Conversions between decimal and
binary,decimal and hexadecimal,and binary and hexadecimal
are described.
3.1 Number Systems
使用微处理器需要掌握二进制、十进制和十六进制数制系统
的基本知识，本节为那些不熟悉数制系统的读者提供这方面
的背景知识。说明了十进制与二进制之间、十进制与十六进
制之间，及二进制与十六进制之间的转换。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-6
Before numbers are converted from one number base to another,
the digits of a number system must be understood,Early in our
education,we learned that a decimal,or base 10,number was
constructed with 10 digits,0 through 9,The first digit in any
numbering system is always a zero,For example,a base 8 (octal)
number contains 8 digits,0 through 7; a base 2 (binary) number
contains 2 digits,0 and 1.
3.1.1 Digits
将数从 —种数制向另一种数制转换之前，必须了解数的计数系统。在早期
教育中，我们已学习了十进制数，或以 10为基的数，它由 10个数字组成,0
到 9。任何计数制的第一个数字总是零，这种规则适用于任何其他数制。例
如，以 8为基的数 (八进制 )包含 8个数字,0到 7，而以 2为基的数 (二进制 )包
含 2个数字,0和 l。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-7
If the base of a number exceeds 10,the additional digits use the
letters of the alphabet,beginning with an A,For example,a base
12 number contains 12 digits,0 through 9,followed by A for 10
and B for 11,Note that a base 10 number does not contain a 10
digit,just as a base 8 number does net contain an 8 digit,The
most common numbering systems used with computers are
decimal,binary,and hexadecimal (base 16),(Many years ago
octal numbers were popular.) Each system is described and used
in this section of the chapter.
3.1.1 Digits
如果基数大于 10，其余数字用从 A开始的字母表示，例如，以 12为基的数包含 12个数
字,0到 9，之后用 A代表 10,B代表 11。注意，以 10为基的数不包含数字 10，如同以 8
为基的数不包括数字 8一样。计算机中最通用的计数制是十进制、二进制、八进制和
十六进制 (基为 16)。每种计数制都将在本节中进行说明和应用。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-8
Once the digits of a number system are understood,larger
numbers are constructed by using positional notation,In grade
school,we learned that the position to the left of the units
position was the tens position,the position to the left of the tens
position was the hundreds position,and so forth,(An example
is the decimal number 132,This number has 1 hundred,3 tens,
and 2 units.) What probably was not learned was the
exponential value of each position,The units position has a
weight of 100 or 1; the tens position has weight of 101,or 10;
and the hundreds position has a weight of 102,or 100.
3.1.2 Positional Notation
一旦我们理解了计数制的数字后，就可用位计数法构造更大的数值。在小学时我
们都学过个位的左边一位是十位，十位左边一位是百位，以此类推 (例如十进制数
132,这个数字有 —个百，三个十和两个一 )。或许我们没有学过每个位的指数值：
个位的权为 l00，即 1；十位的权为 101或 10；而百位的权为 102或 l00。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-9
The exponential powers of the positions are critical for
understanding numbers in other numbering systems,The
position to the left of the radix (number base) point,called a
decimal point only in the decimal system,is always the units
position in any number system,For example,the position to
the left of the binary point is always 20 or 1; the position to the
left of the octal point is 80 or 1,In any case,any number raised
to its zero power is always 1,or the units position.
3.1.2 Positional Notation
位的指数幂在理解其他计数制中的数时是个关键。基数小数
点，在十进制中称为十进制小数点，其左边的位在任何数制
中都是个位。例如，二进制小数点左边的位是 20或 1。而八进
制小数点左边的位是 80或 1。在任何情况下，任何数的零次幂
总是 1，或 1个单位。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-10
The position to the left of the units position is always the
number base raised to the first power; in a decimal system,
this is l01,or l0,In a binary system,it is 21,or 2; and in an
octal system it is 81,or 8,Therefore,an 11 decimal has a
different value from an 11 binary,The 1l decimal is composed
of 1 ten plus 1 unit and has a value of 11 units; while the
binary number 11 is composed of 1 two plus 1 unit,for a value
of 3 decimal units,The 11 octal has a value of 9 units.
3.1.2 Positional Notation
个位左边的位总是基数的 1次幂，在十进制系统中是 101，或
10；在二进制中是 21，或 2；而在八进制中是 81，或 8。因此，
十进制的 11与二进制的 11具有不同的数值。十进制 11表示 —
个 10加上一个 1，其值为 11；二进制 11表示 —个 2加上 —个 1，
其值为 3；八进制 11的值为 9。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-11
In the decimal system,positions to the right of the decimal
point have negative powers,The first digit to the right of the
decimal point has a value of 10-1,or 0.1,In the binary system,
the first digit to the right of the binary point has a value of 2-1,
or 0.5,In general,the principles that apply to decimal
numbers also apply to numbers in any other number system.
3.1.2 Positional Notation
在十进制系统中，对于十进制小数点右边的位，它的幂为负
数。十进制小数点右边第一位数的值为 10-1，或 0.1。在二进
制中,二进制小数点右边第 —位数的值为 2-1或 0.5。一般来说，
十进制使用的计数法可以用于任何其他数制。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-12
Example 3-1 shows a 110.101 in binary (often written as
110.1012),It also shows the power and weight or value of each
digit position,To convert a binary number to decimal,add the
weights of each digit to form its decimal equivalent,The
110.1012 is equivalent to a 6.625 in decimal (4+2+0.5+ 0.125).
Notice that this is the sum of 22 (or 4) plus 21 (or 2),but 20 (or
1) is not added because there are no digits under this position.
The fraction part is composed of 2-1 (0.5) plus 2-3 (or.125),but
there is no digit under the 2-2 (or,25).
3.1.2 Positional Notation
例 3-1给出了一个二进制数 110.101(通常写成 110.1012)，也给出了这个数每
个位的幂、权和值。为了把二进制数转换为十进制，将每位数字的权相加，
就得到了它的等效十进制值。二进制 110.101等于十进制的
6.625(4+2+0.5+0.125)。注意，这个和的整数部分是由 22(4)加 21(2)构成，
之所以没有用 20(1)是因为这个位的数为零。小数部分由 2-1(0.5),加 2-
3(0.125)构成，但是没有用 2-2(0.25)。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-13
The prior examples have shown that to convert from any
number base to decimal,determine the weights or values of
each position of the number,and then sum the weights to form
the decimal equivalent,Suppose that a 125.78 octal is
converted to decimal,To accomplish this conversion,first write
down the weights of each position of the number,This appears
in Example 3-2,The value of 125.78 is 85.875 decimal,or 1?64
plus 2? 8 plus 5?1 plus 7?,125.
3.1.3 Conversion to Decimal
前面的例子说明了将任何其他基数的数转换为十进制数时，
十进制数的值取决于该数每个位上的权或值，它们的和就是
等效的十进制数值。假定要将 125.78(八进制 )转换为十进制。
为了完成这个转换，首先写出该数每一位数的权，如例 3-2所
示,125.78的值是十进制的 85.875，即 1?64+2? 8+5 ? 1+7?
0.125。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-14
Notice that the weight of the position to the left of the units position
is 8,This is 8 times 1,Then notice that the weight of the next position
is 64,or 8 times 8,If another position existed,it would be 64 times 8,
or 51 2,To find the weight of the next higher-order position,
multiply the weight of the current position by the number base (or 8,
in this example),To calculate the weights of position to the right of
the radix point,divide by the number base,In the octal system,the
position immediately to the fight of the octal point is 1/8,or,125,The
next position is,125/8,or,015625,which can also be written as 1/64.
3.1.3 Conversion to Decimal
注意，该数个位左边那位的权是 8(1?8)。再前一位的权是 64(8?8)。如果存
在更前一位，则其权将是 512(64?8)。将当前位的权乘上基数，就可得到
更高一位的权 (本例中是乘 8)。而计算小数点右边那些位的权，需要用基
数去除。在八进制中，紧跟八进制小数点右边的那位的权是 1/8，即 0.125。
下一位是 0.125/8，即 0.015625，也可以写成 1/64。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-15
Hexadecimal numbers are often used with computers,A
6A.CH (H for hexadecimal) is illustrated with its weights in
Example 3-3,The sum of its digits is 106.75,or 106,The whole
number part is represented with 6?16 plus 10 (A)?1,The
fraction part is 12 (C) as a numerator and 16 (16-1) as the
denominator,or 12/16,which is reduced to 3/4.
3.1.3 Conversion to Decimal
计算机经常使用十六进制。例 3-2给出了一个十六进制数
6A.CH(H表示十六进制 )，以及它的权。它的各位数值之和是
106.75，即 106。整数部分用 6?16加 10(A) ?1表示；分数部分
用 12(C)作为分子,16作为分母 (16-1)，或表示为 12/16，化简
得 3/4。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-16
Conversions from decimal to other number systems are more
difficult to accomplish than conversion to decimal,To convert
the whole number portion of a number to decimal,divide by
the radix,To convert the fractional portion,multiply by the
radix.
3.1.4 Conversion From Decimal
由十进制转换成其他进制比由其他进制转换成十进制困难。
转换十进制整数部分时，要用基数去除，转换分数部分时，
要用基数去乘它们。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-17
Whole Number Conversion from Decimal,To convert a decimal
whole number to another number system,divide by the radix
and save the remainders as significant digits of the result,An
algorithm for this conversion as is follows:
1,Divide the decimal number by the radix (number base).
2,Save the remainder (first remainder is the least significant
digit),
3,Repeat steps 1 and 2 until the quotient is zero.
3.1.4 Conversion From Decimal
转换十进制整数部分 将十进制整数转换成其他数制时，要用基数去除，
并且保存余数，作为结果的有效数字。这种转换的算法如下：
1,用基数除十进制数。
2,保存余数 (最先得到的余数是最低有效位数字 )。
3,重复步骤 l和 2，直到商为零。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-18
Converting from a Decimal Fraction,Conversion from decimal
fraction to another number base is accomplished with
multiplication by the radix,For example,to convert a decimal
fraction into binary,multiply by 2,After the multiplication,
the whole number portion of the result is saved as a significant
digit of the result,and the fractional remainder is again multiplied
by the radix,When the fraction remainder is zero,multiplication
ends,Note that some numbers are never-ending,That is,a zero is
never a remainder,An algorithm for conversion from a decimal
fraction is as follows
3.1.4 Conversion From Decimal
转换十进制小数部分 转换 10进制小数部分是用基数乘来完成的。例如，
要将十进制 小数转换成二进制，要用 2乘。乘法之后，乘积的整数部分
保存起来作为结果的一个有效位，剩余的小数再用基数 2去乘。当剩余的
小数部分为 0时，乘法结束。有些数可能永远不会结束，即余数总不为 0。
转换十进制小数部分的算法如下：
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-19
1,Multiply the decimal fraction by the radix (number base).
2,Save the whole number portion of the result (even if zero) as
a digit,Note that the first result is written immediately to the
fight of the radix point.
3,Repeat steps 1 and 2,using the fractional part of step 2 until
the fractional part of step 2 is zero.
3.1.4 Conversion From Decimal
1,用基数乘十进制小数。
2,保存结果的整数部分（即使是零）作为一位数。注意，第
一个得到的结果写在紧挨着小数点的右边。
3,用步骤 2的小数部分重复步骤 l和 2，直到步骤 2的小数部分
是零。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-20
Binary-coded hexadecimal (BCH) is used to represent hexadecimal
data in binary code,A binary-coded hexadecimal number is a
hexadecimal number written so that each digit is represented by a 4-
bit binary number,The values for the BCH digits appear in Table 3--
1.
Hexadecimal numbers are represented in BCH code by converting
each digit to BCH code,with a space between each coded digit,
3.1.5 Binary-Coded Hexadecimal
二进制编码的十六进制 (BCH)是用二进制编码表示的十六进
制数据，二进制编码的十六进制数是将十六进制数的每一位
都用 4位二进制数表示。表 3-1给出了 BCH数的值。
用 BCH表示十六进制数时，将每个十六进制数字都转换成
BCH码，并且每个数位之间用空格分开。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-21
The purpose of BCH code is to allow a binary version of a
hexadecimal number to be written in a form that can easily be
converted between BCH and hexadecimal,Example 3-8 shows
a BCH coded number converted back to hexadecimal code.
3.1.5 Binary-Coded Hexadecimal
BCH码的目的在于能将十六进制数以二进制的形式写出，使
BCH与十六进制之间转换很容易。例 3-8表示如何将 BCH代码
数据转换为十六进制码。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-22
At times,data are stored in complement form to represent
negative numbers,There are two systems that are used to
represent negative data,radix and radix -1 complements,The
earliest system was the radix -1 complement,in which each
digit of the number is subtracted from the radix -1 to generate
the radix -1 complement to represent a negative number.
3.1.6 Complements
有时，数据以补码的形式存储，以便表示负数。有两种表示
负数的方式：补码和反码 (基数减 l的补 )，最早的方式是反码。
为了得到负数的反码表示，用基数 -1减去该数的每一个数位
上的数字。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-23
Example 3-9 shows how the 8-bit binary number 01001100 is
one's (radix -1) complemented to represent it as a negative value.
Notice that each digit of the number is subtracted from one to
generate the radix -1 (one's) complement,In this example,the
negative of 01001100 is 10110011,The same technique can be
applied to any number system,as illustrated in Example 3-10,in
which the fifteen's (radix -l) complement of a 5CD hexadecimal is
computed by subtracting each digit from a fifteen.
3.1.6 Complements
例 3-9表示了如何将 8位二进制数 01001100对 l取补 (基数减 1的
补 )，以便表示成 —个负数。注意，用 1减去该数的每一位数
字，以便生成反码。在此例中,01001100的负数是 10110011。
同样的技术可适用于任何数制。如例 3-10所示，十六进制数
5CD的反码是从 15(基 -1)中减去它的每一位数字得到的。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-24
Today,the radix -1 complement is not used by itself; it is used
as a step for finding the radix complement,The radix
complement is used to represent negative numbers in modern
computer systems,(The radix -1 complement was used in the
early days of computer technology.) The main problem with
the radix -1 complement is that a negative or a positive zero
exists; in the radix complement system,only a positive zero
can exist.
3.1.6 Complements
如今，反码已不单独使用，而作为求补码的一个步骤使用，
补码是当代计算机系统表示负数的方法 (反码用于早期的计算
技术中 )。反码的主要问题是它存在负零或者正零，而补码系
统中只能存在正零。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-25
To form the radix complement,first find the radix -1
complement,and then add a one to the result,Example 3-11
shows how the number 0100 1000 is converted to a negative
value by two's (radix) complementing it.
3.1.6 Complements
为得到补码，先求反码，然后将 1加到结果上。例 3-11表示了
如何通过对 2(基为 2)取补的方式，将数 01001000转换成负数。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-26
To prove that a 0100 1000 is the inverse (negative) of a 1011 0111,add
the two together to form an 8-digit result,The ninth digit is dropped and
the result is zero because a 0l00100 is a positive 72,while a 1011 0111 is
a negative 72,The same technique applied to any number system.
Example 3-12 shows how the inverse of a 345 hexadecimal is found by
first fifteen's complementing the number,and then by adding one to the
result to form the sixteen’s complement,As before,if the original 3-digit
number 345 is added to the inverse of CBB,the result is a 3-digit 000.
As before,the fourth bit (carry) is dropped,This proves that 345 is the
inverse of CBB.
3.1.6 Complements
为验证 0100 1000是 1011 1000的反 (负数 )，将两者相加得到一个 8位
结果。去掉第 9位数字，结果是零。因为 0100 1000是正数 72，而
1011 0111是负数 72。同样的枝术可用于任何数制。例 3-12表示如何
求十六进制数 345的负数，首先求该数 15的补，然后将 1加到结果上，
得到 16的补，同前面类似，如把原来的 3位数 345加上其负数 CBB，
则结果是 3位 000，第 4位 (进位 )被丢掉。这证明了 345是 CBB的反。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-27
The concept of a Boolean algebra was first proposed by the
English mathematician George Boole in 1847,Since that time,
Boole’s original conception has been extensively developed and
refined by algebraists and logicians,The relationships among
Boolean algebra,set algebra,logic,and binary arithmetic have
given Boolean algebras a central role in the development of
electronic digital computers.
3.2 Boolean Algebra
布尔代数的概念最初是由英国数学家 George Boole于 1847年提
出来的，从那时起，代数学家和逻辑学家们更广泛地发展了
Boole最初的概念，并使之更加精练。由于布尔代数、集合代
数、逻辑学和二进制算术之间的内在联系，使得布尔代数的
理论在电子计算机的发展中起到举足轻重的作用。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-28
The most intuitive development of Boolean algebras arises from the concept of a
set algebra,Let S={a,b,c} and T={a,b,c,d,e} be two sets consisting of three
and five elements,respectively,We say that S is a subset of T,since every
element of S (namely,a,b,and c) belongs to T,Since T has five elements,there
are 25 subsets of T,for we may choose any individual element to be included or
omitted from a subset,Note that these 32 subsets include T itself and the empty
set,which contains no elements at all,If T contains all elements of concern,it is
called the universal set,Given a subset of T,such as S,we may define the
complement of S with respect to a universal set T to consist of precisely those
elements of T which are not included in the given subset.
3.2 Boolean Algebra
布尔代数最直觉的发展产生于集合代数的概念。设 S={a,b,c}和 T={a,b,c,d,e}
分别为两个含有三个和五个元素的集合。由于 S中的每一个元素（ a,b,c）
都属于 T，所以我们说 S是 T的一个子集。由于 T有五个元素，因而 T共有
25个子集，这是因为我们可以选择任何一个元素使其包含于某个子集中或
从该子集中删除。应该注意到这 32个子集中包含 T本身和空集（空集即不
含任何元素的集合）。如果 T包含了所讨论的所有元素，则称之为全集。
给定 T的一个子集，例如子集 S，我们可以定义一个关于全集 T的 S的补集。
其中正好包含那些不在子集 S中而在 T中的元素。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-29
Thus,S as above defined has its complement (with respect to T),
The union of any two sets (subsets of a given set) consists of those
elements that are in one or the other or in both given sets; the
intersection of two sets consists of those elements that are in both
given sets,We use the symbol ∪ to denote the union,and ∩ to
denote the intersection of two sets,For example,if B={b,d,e},
then B∪ S={a,b,c,d,e},and B∩S={b}.
3.2 Boolean Algebra
于是，如上定义的集合 S就有一个它的补集（相对于集合 T）。
任何两个集合（已给定集合的若干子集）的并集包含了出现
于这两个子集中某一个集合或同时出现于这两个集合中的所
有元素；两个集合的交集包含了同时出现于这两个集合中的
元素。我们用符号,∪,来表示两个集合的“并（运算）”，
用,∩” 来表示两个集合的“交（运算）”。例如，如果
B={b,d,e},那么,B∪ S={a,b,c,d,e},B∩S={b}。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-30
While other set operations may be defined,the operations of
complementation union and intersection are of primary interest to
us,A Boolean algebra is a finite or infinite set of elements together
with three operations— negation,addition,and multiplication—
that correspond to the set operations of complementation,union,
and intersection,respectively,Among the elements of a Boolean
algebra are two distinguished elements,0,corresponding to the
empty set; and 1,corresponding to the universal set.
3.2 Boolean Algebra
虽然我们可以定义其他一些集合运算，但求补、并和交运算
是我们最感兴趣的三个集合运算。一个布尔代数就是一个有
限集或无限集，以及建立在该有限集或无限集上的三种运
算 ——否定、加或乘，这三个运算分别对应于集合的求补、
并和交运算。在布尔代数的元素中有两个特殊的元素,0，对
应于空集； 1，对应于全集。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-31
For any given element of a Boolean algebra,there is a unique
complement a' with the property that a+a'=1 and aa'=0,Boolean
addition and multiplication are associative and commutative,as are
ordinary addition and multiplication,but otherwise have somewhat
different properties,The principal properties are given in Table 3-2,
where a,b,and c are any elements of a Boolean algebra.
3.2 Boolean Algebra
对于一个布尔代数中任意给定元素 a，都有一个唯一的补 a?，
它满足 a+a?=1和 aa?=0。布尔加和布尔乘与普通的加和乘一样，
满足结合律和交换律，但除此之外含有一些不太相同的特性。
其主要特性由表 3-2给出，其中 a,b和 c是一个布尔代数中的任
意元素。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-32
3.2 Boolean Algebra
?Table 3-2
Distributivity 分配律 a(b+c)=ab+aca+(bc)=(a+b)(a+c)
Idempotency 同一律 a+a=aaa=a
Absorption laws 吸收律 a+ab=aa(a+b)=a
DeMorgan’s laws德 ?摩根定理 (a+b)'=a'b'(ab)'=a'+b'
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-33
Since a finite set of n elements has exactly 2n subsets,and it can
be shown that the finite Boolean algebras are precisely the finite
set algebras,each finite Boolean algebra consists of exactly 2n
elements for some integer n,For example,the set algebra for the
set T defined above corresponds to a Boolean algebra of 32
elements.
3.2 Boolean Algebra
由于 n个元素的有限集有且只有 2n个子集，而且很显然有限布
尔代数一定是有限集合代数，所以对某个整数 n而言，每个有
限布尔代数也有且只有 2n个元素。例如，上文定义的集合 T的
集合代数就对应一个有 32个元素的布尔代数。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-34
While it is possible to use a different symbol to denote each
element of a Boolean algebra,it is often more useful to
represent the 2n elements of a finite Boolean algebra by binary
vectors having n components,With such a representation the
operations of the Boolean algebra are accomplished
componentwise by considering each component as an
independent two-element Boolean algebra,This corresponds to
representing subsets of a finite set by binary vectors.
3.2 Boolean Algebra
虽然我们可以用不同的符号来表示布尔代数中的每一个元素，
但最常用的方法是用一个有 n个分量的二进制向量来表示一个
有限布尔代数的 2n个元素。用这样一种表示方法，布尔代数
的所有运算都以分量的形式完成，而每一个分量被认为是一
个独立的二值布尔代数。这种做法对应于用二进制向量来表
示一个有限集的子集。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-35
For example,since the set T has five elements,we may represent its subsets
by five-component binary vectors,each component denoting an element of
the set T,A numeral l in the i-th component of the vector denotes the
inclusion of the i-th element of that particular subset; a 0 denotes its
exclusion,Thus,the subset S={a,b,c} has the binary vector representation
{1,1,1,0,0},The set operations become Boolean operations on the
components of the vectors,This representation of sets,and the
correspondence to Boolean or logical operations,is very useful in
information retrieval,Because of it,sets of document and query
characteristics may be easily and rapidly matched.
3.2 Boolean Algebra
例如，由于集合 T有 5个元素，所以我们可以用 5个分量的二进制向量表示
它的子集，其中每一个分量表示集合 T的一个元素。向量中的第 i个分量为
数字 1表示集合 T的第 i个元素在某一特定子集中，用数字 0表示不在某一特
定子集中。于是，子集 S={a,b,c}可用二进制向量表示为 {1,1,1,0,0}。
集合运算变成了向量分量上的布尔运算。集合的这种表示方法以及相应的
布尔或逻辑运算，对于 信息检索 是非常有用的。由于这一原因，文件的集
合和查询特性可以很容易而迅速地得到匹配。
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-36
数学公式的读法 (Pronunciation of mathematical expressions)
1 逻辑 (Logic)
there exists
for all
p? q p implies q / if p,then q
p?q p if and only if q
p is equivalent to q
p and q are equivalent
?
?
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-37
2 集合 (Sets)
x?A x belongs to A / x is an element (or a member) of A
x?A x does not belong to A / x is not an element (or a member)
of A
A?B A is contained in B / A is a subset of B
A?B A contains B / B is a subset of A
A?B A cap B / A meet B/ A intersection B
A?B A cup B/ A join B / A union B
B/A A minus B/the difference between A and B
A× B A cross B / the Cartesian product of A and B(A与 B的笛卡
尔积 )
数学公式的读法 (Pronunciation of mathematical expressions)
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-38
数学公式的读法 (Pronunciation of mathematical expressions)
3 实数 (Real numbers)
x+1 x plus one
x-1 x minus one
x± 1 x plus or minus one
xy xy / x multiplied by y
(x-y)(x+y) x minus y,x plus y
x over y
= the equals sign
x=5 x equals 5 / x is equal to 5
x≠5 x (is) not equal to 5
x≡y x is equivalent to (or identical with) y
y
x
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-39
3 实数 (Real numbers)
x>y x is greater than y
x≥y x is greater than or equal to y
x< y x is less than y
x?y x is less than or equal to y
0<x<1 zero is less than x is less than 1
0?x?1 zero is less than or equal to x is less than or equal to 1
|x| mod x / modulus x
x2 x squared / x (raised) to the power 2
x3 x cubed
x4 x to the fourth / x to the power four
xn x to the nth / x to the power n
x-n x to the (power) minus n
数学公式的读法 (Pronunciation of mathematical expressions)
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-40
数学公式的读法 (Pronunciation of mathematical expressions)
3 实数 (Real numbers)
n! n factorial
(x+y)2 x plus y all squared
xi xi / x subscript i / x suffix i / x sub i
the sum from i equals one to n ai / the sum as i runs
from 1 to n of the ai
x over y all squared
x hat x bar x tilde
x? x x~
?
?
n
i
ia
1
2
???
?
???
?
y
x
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-41
数学公式的读法 (Pronunciation of mathematical expressions)
4 线性代数 (Linear algebra)
||A|| the norm (or modulus) of x
OA / vector OA
OA / the length of the segment OA
AT A transpose / the transpose of A
A-1 A inverse / the inverse of A
OA
OA
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-42
数学公式的读法 (Pronunciation of mathematical expressions)
5 函数 (Functions)
f(x) fx / f of x / the function f of x
f, S→T a function f from S to T
x maps to y / x is sent (or mapped) to y
f?(x) f prime x / f dash x / the (rst) derivative of f with
respect to x
f?(x) f double–prime x / f double–dash x / the second
derivative of f with respect to x
f??(x) f triple–prime x / f triple–dash x / the third
derivative of f with respect to x
f(4) four x / the fourth derivative of f with respect to x
ln y log y to the base e / log to the base e of y / natural
log (of) y
yx?
Chapter 3 Number Systems and Boolean Algebra
计算机专业英语 3-43
数学公式的读法 (Pronunciation of mathematical expressions)
5 函数 (Functions)
the partial (derivative) of f with respect to x1
the second partial (derivative) of f with respect to x1
the integral from zero to infinity
the limit as x approaches zero from above
1x
f
?
?
2
1
2
x
f
?
?
??0
0lim??x
