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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. $0$
edited | 2.6k views

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) $-128$
edited by
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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.

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