Question

Consider the following toy model for a normalized floating-point representation in base \(2: x=(-1)^{s}\left(1 . a_{2} a_{3}\right)_{2} \times 2^{e}\) where \(-1 \leq e \leq 1 .\) Find all positive machine numbers (there are 12 of them) that can be represented in this model. Convert the numbers to base 10 , and then carefully plot them on the number line, by hand, and comment on how the numbers are spaced.

Short answer

Answer: In a toy model of a floating-point representation system, the spacing of positive machine numbers is not uniform, and it increases as the magnitude of numbers increases. Consequently, the relative precision of the representation decreases with increasing magnitude.

Step-by-step solution

Identify all possible combinations for \(s, a_{2}, a_{3}\) and \(e\)

There are only two possible values for the sign \(s\), i.e., 0 (positive) or 1 (negative). However, since we are only looking for positive machine numbers, we can take \(s = 0\). For the two fractional bits \(a_{2}\) and \(a_{3}\), we have two possible values each, 0 and 1. Thus, we have \(2^2 = 4\) possible combinations for \(a_{2}\) and \(a_{3}\). Since \(-1 \leq e \leq 1\), we have three possible values for \(e\): -1, 0, and 1. Combining all possibilities, we have a total of \(4 \times 3 = 12\) positive machine numbers.

Calculate the decimal values of each combination

Now that we have the possible values for \(s, a_{2}, a_{3}\), and \(e\), we can compute the decimal value for each combination. For each combination, we can substitute the values for \(s, a_{2}, a_{3}\), and \(e\) in the floating-point formula $x=(-1)^{s}\left(1 . a_{2} a_{3}\right)_{2} \times 2^{e}$ and obtain the decimal value. Recall that \(s=0\) which means \((-1)^s = 1\), so we can simplify the equation to \(x=(1 . a_{2} a_{3})_{2} \times 2^{e}\). We will compute the decimal values for each combination and list them below. | \(a_{2}\) | \(a_{3}\) | \((1 . a_{2} a_{3})_{2}\) | \(e\) | Decimal Value (\(x\)) | |---|---|---------------------|-----|-----| | 0 | 0 | \((1.00)_{2}\) | -1 | 1/2 | | 0 | 1 | \((1.01)_{2}\) | -1 | 5/8 | | 1 | 0 | \((1.10)_{2}\) | -1 | 3/4 | | 1 | 1 | \((1.11)_{2}\) | -1 | 7/8 | | \(a_{2}\) | \(a_{3}\) | \((1 . a_{2} a_{3})_{2}\) | \(e\) | Decimal Value (\(x\)) | |---|---|---------------------|-----|-----| | 0 | 0 | \((1.00)_{2}\) | 0 | 1 | | 0 | 1 | \((1.01)_{2}\) | 0 | 5/4 | | 1 | 0 | \((1.10)_{2}\) | 0 | 3/2 | | 1 | 1 | \((1.11)_{2}\) | 0 | 7/4 | | \(a_{2}\) | \(a_{3}\) | \((1 . a_{2} a_{3})_{2}\) | \(e\) | Decimal Value (\(x\)) | |---|---|---------------------|-----|-----| | 0 | 0 | \((1.00)_{2}\) | 1 | 2 | | 0 | 1 | \((1.01)_{2}\) | 1 | 5/2 | | 1 | 0 | \((1.10)_{2}\) | 1 | 3 | | 1 | 1 | \((1.11)_{2}\) | 1 | 7/2 |

Plot the decimal values on a number line

Now that we have found all the decimal values for the 12 positive machine numbers, we can plot them on a number line. ``` 0 1/2 1 3/2 2 5/2 7/2 |------|-------------|----------|------|---|---| ``` Keep in mind that this is a rough sketch of the number line, and the actual spacing should be more accurately measured using a proper scale.

Comment on the spacing between the numbers

Looking at the number line, we can see that the positive machine numbers have different spacing. The spacing between the numbers is not uniform. As the numbers get larger, the spacing between them also increases. This is a characteristic of floating-point representation systems - as the magnitude of numbers increases, the relative precision decreases.

Key concepts

Machine Numbers in Floating-Point

Understanding machine numbers in floating-point representation is key to grasping many computational and numerical analysis concepts. Floating-point representation allows computers to approximate real numbers within a certain precision, depending on the size of the mantissa and exponent.

The toy model from our exercise is a simplified version of the standard floating-point system used in computers, which is often based on the IEEE 754 standard. This model provides a visual understanding of how numbers are stored in a computer. Essentially, we define a machine number as the combination of a sign bit, mantissa (or significand), and exponent, conforming to a specific format which, in our model, is normalized and in base 2.

Base Conversion

Base conversion is the process of converting numbers from one base to another. In computer science, base 2, also called binary, is fundamental because it represents the off (0) and on (1) states of transistors in digital circuits.

To convert from binary to decimal, as required in step two of the exercise, you multiply each bit by its corresponding power of two according to its position and then sum all the values obtained. For instance, the binary number \(1.01)_2\) converts to \(1 \times 2^0 + 0 \times 2^{-1} + 1 \times 2^{-2} = 1 + 0 + 0.25 = 1.25\) in decimal, an important procedure for visualizing how floating-point numbers are spread on the real number line.

Number Line Plotting

Plotting numbers on a number line is an instructive way to visualize their distribution, relative sizes, and how they might span across an interval or several ranges. In the final step of the solution, the task was to represent the converted decimal values on a number line.

In our floating-point context, plotting these machine numbers provides insight into an important feature of digital representation of numbers: as values get larger, the spacing between representable numbers becomes larger too. This illustrates the concept of precision in floating-point arithmetic—the larger the number, the fewer the number of digits after the decimal point that can be precisely represented.

Floating-Point Arithmetic

Lastly, let's delve into floating-point arithmetic, which involves operations on numbers represented in floating-point format. This type of arithmetic is vital in ensuring that calculations with large or small scales can be handled accurately without the restrictions of fixed-point representation.

Floating-point arithmetic comes with its own set of challenges, including round-off errors, overflow, and underflow. Round-off error occurs because of the limited number of bits available to represent the mantissa and exponent. The exercise highlights the idea that as numbers become larger, the relative error due to this limitation increases as well. It's therefore important to understand the precision of the arithmetic involved when representing and manipulating numbers in a floating-point system.