As you have been requested to execute a 32-bit operation, it is important to note that the leftmost bit denotes the sign of the number for signed values. To better understand this concept, you can use the following thought exercise, which applies to any number of bits and may help clarify any confusion you have.

Solution 1:

A group of 8 bits has the capacity to hold a maximum of 256 numerical values.

The range that can be stored for signed integers with a maximum of 8 bits will vary depending on the system and language. However, for most systems, this range will encompass all 256 (2^8) numbers, as indicated by

{-128,127}

and

-128...0...127

.

Solution 2:

The sign bit on the left in the standard notation would make 10000000 the most negative number, while 01111111 would represent the most positive number.

To avoid repetition, attempt the subsequent thought experiment that is applicable to any quantity of bits. For the purpose of brevity, only three bits will be utilized in this illustration.

```
-4 -3 -2 -1 0 1 2 3 4 5 6 7 .
unsigned 000 001 010 011 100 101 110 111
signed 100 101 110 111 000 001 010 011
```

Keep in mind that the order of the eight 3-bit codes is consistent, with only the “wraparound point” varying. Additionally, the positive numbers have identical bit representations in both forms.

Solution 3:

The decimal integer values that can be represented by a signed 8-bit cell range from -128 to 127.

Transforming them into binary using 2’s complement:

```
127 = 01111111
-128 = 10000000
```

Below, you will find details regarding the 2’s complement system.

An 8-bit cell that is unsigned has a range of values from 00000000 to 11111111, which in decimal is equivalent to 0 through 255.