edited by
7,231 views

2 Answers

Best answer
40 votes
40 votes
Range of $2$'s compliment no $\Rightarrow (- 2^{n-1} )\;\text{to}\; + (2^{n-1} - 1)$

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

So minimum no $= -2^7 = -128$
edited by
2 votes
2 votes
Lets take the biggest number possible with 8 bits i.e., 1111 1111

Now since the question asked for 2's complement then MSB is reserved for sign bit therefore considering sign bit, biggest number possible with 8 bits is:

(+/-bit) 111 1111

Smallest number possible will be the reverse of this number:

(+/-bit) 000 0000

Now as per convention 1 is for -ve and 0 is for +ve therefore smallest number will be:

1 000 0000

Which comes out to be -128.
Answer:

Related questions

31 votes
31 votes
7 answers
1
Arjun asked Sep 24, 2014
9,389 views
Which one of the following expressions does NOT represent exclusive NOR of $x$ and $y$?$xy + x′ y′$$x\oplus y′$$x′\oplus y$$x′\oplus y′$
17 votes
17 votes
3 answers
2
21 votes
21 votes
2 answers
3
17 votes
17 votes
5 answers
4
Arjun asked Sep 24, 2014
5,600 views
What will be the maximum sum of $44, 42, 40, \dots$ ?$502$$504$$506$$500$