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Chapter 4: Problem 6

Give the 8-bit sign/magnitude representation of each of the following decimal values: a. \(+71\) b. \(-1\) c. \(-81\)

Short Answer

Expert verified
+71: 01000111; -1: 10000001; -81: Cannot be represented in 8-bit sign/magnitude.

Step by step solution

01

Understanding Sign/Magnitude Notation

Sign/magnitude notation represents numbers using one bit for the sign and the remaining bits for the magnitude. The first bit indicates the sign (0 for positive, 1 for negative), and the remaining bits represent the magnitude (absolute value) of the number.
02

Converting +71 to 8-bit Sign/Magnitude

The decimal number +71 is positive, so the sign bit is 0. To convert 71 into binary, start by dividing by powers of 2: \[71_{10} = 1000111_2\] This requires 7 bits. Pad with a leading zero to make it 8 bits: \[+71 = 01000111\].
03

Converting -1 to 8-bit Sign/Magnitude

The decimal number -1 is negative, so the sign bit is 1. The magnitude of -1 is 1, and its binary representation is \[0000001_2\]. Thus, the 8-bit sign/magnitude representation is: \[-1 = 10000001\].
04

Attempting to Convert -81 to 8-bit Sign/Magnitude

The decimal number -81 is negative, so the sign bit is 1. The magnitude of -81 is 81, and its binary representation is \[1010001_2\]. However, since this already uses 7 bits, adding the sign bit results in 9 bits, which does not fit in an 8-bit format. Therefore, -81 cannot be represented in 8-bit sign/magnitude notation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binary Conversion
Binary conversion is a fundamental concept for understanding how computers process and store numbers. To convert a decimal number to binary, you must break it down into sums of powers of two. This is because the binary system is base-2, meaning it only uses the digits 0 and 1. For example, to convert the decimal number 71 into binary:
  • Start with the highest power of 2 that fits into 71, which is 64, equivalent to \(2^6\), and subtract 64 from 71, leaving 7.
  • Next, the largest power of 2 that fits into 7 is 4 (\(2^2\)), leaving 3.
  • Then, the largest power of 2 for 3 is 2 (\(2^1\)), leaving 1.
  • Finally, 1 is \(2^0\), completing the conversion.
Putting these together, 71 is represented by 1s in the positions of \(2^6\), \(2^2\), \(2^1\), and \(2^0\), resulting in the binary number \(1000111_2\). This process ensures that numbers can be expressed with the binary digits 0 and 1, essential for digital communication.
8-bit Representation
The 8-bit representation refers to using exactly 8 binary digits to represent a number. This is a common format in computing because it fits well into a byte, which is 8 bits. In 8-bit representation, the leftmost bit often serves as the sign bit in systems that use sign/magnitude or similar notations. Here's how it works:
  • One bit is dedicated to indicating the sign of the number. A '0' indicates a positive number, while a '1' indicates a negative number.
  • The remaining 7 bits are used for the magnitude or size of the number.
To maintain an 8-bit format:
  • If the binary representation of the number requires less than 8 bits, add zeros at the beginning, known as padding, to meet the 8-bit requirement.
  • This ensures a uniform number length, crucial for operations in digital systems.
For example, the positive decimal number 71 is represented as \(01000111\) to adhere to the 8-bit format, including the sign bit.
Negative Numbers in Binary
Handling negative numbers in binary can be tricky. Sign/magnitude representation is one way to express them, though it has limitations. In sign/magnitude:
  • The first bit of an 8-bit number represents the sign (0 for positive, 1 for negative).
  • The remaining bits reflect the absolute value of the number.
For example, the number -1 is represented as:
  • The sign bit is 1 because the number is negative.
  • The binary form of the number 1 is \(0000001_2\).
  • Combining these gives \(10000001\) in 8-bit sign/magnitude format.
There is a drawback when converting larger negative numbers. For instance, -81 can't be represented in 8-bit sign/magnitude because its absolute value 81 needs more than the 7 bits reserved for magnitude. This brings attention to other methods like two's complement, which better handle the range limits for negative numbers in binary.

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Most popular questions from this chapter

Build a majority-rules circuit. This is a circuit that has three inputs and one output. The value of its output is 1 if and only if two or more of its inputs are 1; otherwise, the output of the circuit is 0 . For example, if the three inputs are \(0,1,1\), your circuit should output a 1. If its three inputs are \(0,1,0\), it should output a 0 . This circuit is frequently used in faulttolerant computing-environments where a computer must keep working correctly no matter what, for example as on a deep-space vehicle where making repairs is impossible. In these conditions, we might choose to put three computers on board and have all three do every computation; if two or more of the systems produce the same answer, we accept it. Thus, one of the machines could fail and the system would still work properly.

In Exercises 10 and 11, we used 16 bits to represent decimal numbers, allocating 10 bits for the mantissa and 6 bits for the exponent. What would be the impact on our representation if we still used 16 bits for each number but instead allocated 12 bits for the mantissa and 4 bits for the exponent?

Assume that we use 10 bits to represent signed integers, using \(\operatorname{sign} /\) magnitude notation. What are the largest (in absolute value) positive and negative numbers that can be represented on our system?

Assume that \(a=5, b=2\), and \(c=3\). What problem do you encounter when attempting to evaluate the following Boolean expression? \((a=1) \mathrm{AND}(b=2)\) OR \((c=3)\) How can this problem be solved?

Assume that you tried to store the signed integer value \(-200\) using an 8-bit sign/ magnitude representation. What happened? What type of error does this represent?
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