Question #119200
compute the value of 410-78 using 8-bit sign magnitude in binary
1
Expert's answer
2020-06-07T17:35:32-0400

An 8-bit binary number can have a value ranging from 0(00000000)20 (00000000)_2 to 255(11111111)2255 (11111111)_2.

78 = 64+8+4+2=(01001110)264+8+4+2 = (01001110)_2.

Hence, 8-bit sign magnitude in binary of -78 is 78=(01110010)2-78 = (01110010)_2 .

410=255+155=(11111111)2+(10011011)2410 = 255 + 155 = (11111111)_2 + (10011011)_2 .

41078=255+15578=(11111111)2+(10011011)2+(01110010)2410-78 = 255 +155-78 = (11111111)_2 + (10011011)_2 + (01110010)_2 .

Sum will be like that

  • 0 + 0 = 0
  • 1 + 0 = 1
  • 1 + 1 = 10 (binary for 2)
  • 1 + 1 + 1 = 11 (binary for 3)

Hence Sum is

      (11111111)2      (10011011)2=(110011010)2\ \ \ \ \ \ (11111111)_2 \\ \ \ \ \ \ \ (10011011)_2 \\ = (110011010)_2

So, 255+155=(110011010)2255+155 = (110011010)_2.

Hence,

41078=(110011010)2+(01110010)2=    (110011010)2  +(01110010)2=(101001100)2=332410 - 78 = (110011010)_2 + (01110010)_2 = \\ \ \ \ \ (110011010)_2 \\ \ \ + (01110010)_2 \\ = (101001100)_2 = 332


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