For a signed 6 bit binary number:
Maximum
is
0b011111
= 0*(2)
5
+1*(2)
4
+1*(2)
3
+1*(2)
2
+1*(2)
1
+1*(2)
0
= 31
Why? Because the first bit is used to signify whether the number is negative or not. Therefore, a positive
number cannot use that bit. For a ‘n’ bit binary number, the maximum possible number is when the first bit is 0
and the rest are 1’s. Since the left most bit is 0, two complement is not needed to find the decimal
representation of the binary number.
Minimum
is
0b100000
Since we see that the number has a 1 in the most left digit and we know it is a signed number, to find the
value, we need to use two’s complement.
0b100000
Binary number
0b011111
Take the complement
+1
Add a 1 to the complement
0b100000
Which is 32 in decimal
But since we see that the original binary number has a 1 in the most left digit, the number must be negative
giving us 32.
For a ‘n’ bit binary, the minimum possible number is when the first bit is 1 and the rest are 0’s. The binary
number is given in two’s complement representation.
Relative to HW2 Problem 5
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '05
 Cheng
 #include, #define, 6 bit, Stephen Nestinger

Click to edit the document details