Chapter 3 Number Systems and Boolean Algebra 3.1.2 Positional notation n 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. 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一位数的值为2-或05。一般来说, 十进制使用的计数法可以用于任何其他数制。 《什第机专出美语 31
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 3.1.2 Positional notation 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). 例3-1给出了一个二进制数110.101(通常写成10.1012),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 6625(4+2+0.540.125)。注意,这个和的整数部分是由224加21(2)构成, 之所以没有用201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) 什算机专些英语 3-12
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 3.1.3 Conversion to Decimal 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 1x64 plus2×8plus5×1plus7×.125. 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12578(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,12578的值是十进制的85.875,即1×642×8+5×1+7× 0.125。 计算机专业英语 3-13
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 164 plus 2 8 plus 51 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 3.1.3 Conversion to Decimal Notice that the weight of the position to the left of the units position is 8. This is 8 times I. 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 178, or 125. The next position is 125/8, or 015625, which can also be written as 1/64 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51264×8)。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0125/8,即0.015625,也可以写成164。 3-14
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(18)。再前一位的权是64(88)。如果存 在更前一位，则其权将是512(648)。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 3.1.3 Conversion to Decimal H 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 6x16 plus 10(A)x1. 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 计算机经常使用十六进制。例3-2给出了一个十六进制数 6ACH(H表示十六进制),以及它的权。它的各位数值之和是 10675,即106。整数部分用6×16加10(4)×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 《什第机美语 3-15
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 616 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616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