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+13 votes
2.5k views

The smallest integer that can be represented by an $8-bit$ number in $2's$ complement form is

  1. $-256$
  2. $-128$
  3.  $-127$
  4. $0$
asked in Digital Logic by Veteran (370k points)
edited by | 2.5k views

2 Answers

+25 votes
Best answer
Range of $2$'s compliment no = > $(- 2^{n-1} )$to $ + (2^{n-1} - 1)$

Here $n =$ No of bits $= 8$.

So minimum no $= -2^7 =$ (B) $-128$
answered by Boss (43.2k points)
edited by
0
If i consider first bit as sign bit.. we can store my number in rest 7 bits. so in that case 127 is the smallest number that can be stored. so why -127 isn't the answer?
+1

@Shamim Ahmed

the range of 2's complement numbers with 8 digits = -128 to +127

in which smallest number is -128.

+2 votes
answer - B
answered by Loyal (8.8k points)
Answer:

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