2 votes 2 votes The smallest integer that can be represented by an $8$-bit number in $2$'s complement form is : $-256$ $-128$ $-127$ $0$ Digital Logic nielit2017dec-assistanta digital-logic number-system + – admin asked Mar 31, 2020 • recategorized Aug 24, 2020 by Lakshman Bhaiya admin 1.1k views answer comment Share Follow See 1 comment See all 1 1 comment reply mohit rathore commented Aug 26, 2020 reply Follow Share (B) is correct. 0 votes 0 votes Please log in or register to add a comment.
4 votes 4 votes 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. haralk10 answered Apr 1, 2020 haralk10 comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes the range of 8 bit number is -128 to 127. so smallest integer of 8 bit number is -128. hence option (B) is correct. mohit rathore answered Aug 26, 2020 mohit rathore comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes 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 Himanshu Kumar Gupta answered Sep 20, 2020 • edited Sep 20, 2020 by Himanshu Kumar Gupta Himanshu Kumar Gupta comment Share Follow See all 0 reply Please log in or register to add a comment.