1. Stelle die Dezimalzahl ´47 871_10´ in den Positionssystemen zur Basis ´2´, zur Basis ´9´ und zur Basis ´16´ dar. (Ziffern im Hexadezimalsystem: ´0, 1, 2‚ …, 9, A, B, …‚ F´)
  2. Gib die Zahlen ´1100 1110 0010_2´ und ´210 012_3´ im Dezimalsystem an. (Tipp: Horner-Schema)
  3. Gib die Dezimalzahlen ´42´ und ´-42´ in 8-Bit Zweierkomplementdarstellung an.
Approach

Teil 1

47871 / 2 = 23935 Rest: 1
23935 / 2 = 11967 Rest: 1
11967 / 2 = 5983 Rest: 1
 5983 / 2 = 2991 Rest: 1
 2991 / 2 = 1495 Rest: 1
 1495 / 2 =  747 Rest: 1
  747 / 2 =  373 Rest: 1
  373 / 2 =  186 Rest: 1
  186 / 2 =  93 Rest: 0
   93 / 2 =  46 Rest: 1
   46 / 2 =  23 Rest: 0
   23 / 2 =  11 Rest: 1
   11 / 2 =   5 Rest: 1
    5 / 2 =   2 Rest: 1
    2 / 2 =   1 Rest: 0
    1 / 2 =   0 Rest: 1

´47 871_10 = 1011 1010 1111 1111_2´

47871 : 9 = 5319 Rest: 0
 5319 : 9 =  591 Rest: 0
  591 : 9 =  65 Rest: 6
   65 : 9 =   7 Rest: 2
    7 : 9 =   0 Rest: 7

´47 871_10 = 72 600_9´

  1011 1010 1111 1111
=    B    A    F    F

´47 871_10 = BAFF_16´

Teil 2

      1 =    1
* 2 + 1 =    3
* 2          6
* 2         12
* 2 + 1 =   25
* 2 + 1 =   51
* 2 + 1 =  103
* 2        206
* 2        412
* 2        824
* 2 + 1 = 1649
* 2     = 3298

´1100 1110 0010_2 = 3 298_10´

      2 =   2
* 3 + 1 =   7
* 3        21
* 3        63
* 3 + 1 = 190
* 3 + 2 = 572

´210 012_3 = 572_10´

Teil 3

´42´: Binär: ´101010´ Zweierkomplement: ´00101010´

´-42´: Zweierkomplement: ´11010101 + 1 = 11010110´


Solution

  • Teil 1

    ´47 871_10 = 1011 1010 1111 1111_2´

    ´47 871_10 = 72 600_9´

    ´47 871_10 = BAFF_16´

    Teil 2

    ´1100 1110 0010_2 = 3 298_10´

    ´210 012_3 = 572_10´

    Teil 3

    ´42´: ´00101010´

    ´-42´: ´11010110´

  • URL:
  • Language:
  • Subjects: math
  • Type: Transform
  • Duration: 30min
  • Credits: 3
  • Difficulty: 0.3
  • Tags: hpi horner's method
  • Note:
    HPI, 2014-04-07, Mathe 2, Blatt 1, Aufgabe 3
  • Created By: adius
  • Created At:
    2014-07-25 20:17:37 UTC
  • Last Modified:
    2014-07-25 20:17:37 UTC