Electronics - Arithmetic Operations and Circuits - Discussion

Discussion Forum : Arithmetic Operations and Circuits - General Questions (Q.No. 13)
The range of an 8-bit two's complement word is from:
+12810 to –12810
–12810 to +12710
+12810 to –12710
+12710 to –12710
Answer: Option
No answer description is available. Let's discuss.
3 comments Page 1 of 1.

Lokesha M said:   7 years ago
Here in 8 bit, 2's complement MSB is used for sign so remaining 7 bits are used.

So, that the range from 01111111(+127) to 111111111(-128) because the value 00000000 is considered as positive value.

Arif Akhondi said:   5 years ago
In 2'compliment the MSB indicates the sign if 1 then +ive if 0 then -ive. So the range will be from +128(11111111) to -127(0111111).

Pankaj said:   8 years ago
But why? Please explain this answer.

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