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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. $0$

### 1 comment

(B) is correct.

Range of  $n-bit$  $2's$  complement numbers is :  $-2^{n-1}$  to   $2^{n-1}-1$  (Refer: Range of integers in 2's complement)

For  $n=8$, the smallest number that can be represented is : $-2^{8-1}=-2^{7}=-128$

Option B is correct.

by
the range of 8 bit number is -128 to 127. so smallest integer of 8 bit number is -128.

hence option (B) is correct.
Option B is right

Range of  n−bit 2′s  complement numbers is :  −2^(n−1) to   (2^n−1)−1\

then -128 is Correct