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The distributive law \(a(b-c)=a b-a c\) does not hold, in general, if the products are rounded off to a smaller number of digits. To show this in a specific case take \(a=0.22,\) \(b=3.19,\) and \(c=2.17 .\) After each multiplication round off the last digit.

Short Answer

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Question: Prove that the distributive law does not hold in general if the products are rounded off to a smaller number of digits, using the specific numbers \(a=0.22\), \(b=3.19\), and \(c=2.17\). In this case, we calculated \(a(b-c)\) and \((ab-ac)\) with rounding off, and found that they both approximately equal to \(0.22\). Although they are equal in this specific example, this does not disprove the distributive law in general. Rounding might lead to different results than expected, so it should be taken into consideration when working with rounded values.

Step by step solution

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01

Find \((b-c)\)

First, calculate \(b-c\) using the given values \(b=3.19\) and \(c=2.17\). $$(b-c) = 3.19 - 2.17 = 1.02$$
02

Calculate \(a(b-c)\) and round off

Next, multiply the result from Step 1 by \(a=0.22\) and round off the last digit. $$a(b-c) = 0.22 \cdot 1.02 = 0.2244$$ After rounding off the last digit: $$a(b-c) \approx 0.22$$
03

Calculate \(ab\) and \(ac\) and round off

Now, calculate \(ab\) and \(ac\) separately, and round off the last digit. For \(ab\), we have: $$ab = 0.22 \cdot 3.19 = 0.7018 $$ After rounding off the last digit: $$ab \approx 0.70 $$ For \(ac\), we have: $$ac = 0.22 \cdot 2.17 = 0.4774$$ After rounding off the last digit: $$ac \approx 0.48$$
04

Calculate \((ab-ac)\)

Now, subtract the rounded off values of \(ab\) and \(ac\). $$(ab-ac) = 0.70 - 0.48 = 0.22$$
05

Compare results

We have obtained two results: - \(a(b-c)\) rounded off gives us \(0.22\) - \((ab-ac)\) rounded off gives us \(0.22\) In this specific case, the distributive law appears to hold when rounding off the last digit. Even though the problem states that the distributive law "does not hold, in general", it means that this principle might not always be true when rounding is involved. This example does not disprove the distributive law in general, but it should be taken into consideration that rounding might lead to different results than expected.

Key Concepts

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

Distributive Law
The distributive law is a basic property of arithmetic that allows the multiplication of a number by a group of numbers added together to be done individually and then summed up. In mathematical terms, it can be expressed as \( a(b+c) = ab + ac \). This law simplifies the multiplication process and is used extensively in algebra for expanding expressions.

When we apply the distributive law in calculations, we expect the outcomes to be consistent. However, when dealing with numbers that require rounding, especially in the context of limited arithmetic precision, the distributive law may not always hold true. In the given exercise, after rounding, the products \( ab \) and \( ac \) separately and then subtracting \( ac \) from \( ab \) may not yield the same result as multiplying \( a \) by the difference \( b-c \) due to loss of information from rounding. This is an important consideration in numerical analysis and computational mathematics, where tiny errors can accumulate and lead to significant inaccuracies.

To mitigate rounding errors in practical applications, it's crucial to maintain as many significant digits as possible until the final result and to understand that the distributive law should be applied carefully within the context of floating-point arithmetic.
Significant Digits
Significant digits, also known as significant figures, are the digits in a number that carry meaning contributing to its precision. These include all non-zero digits, any zeros between significant digits, and trailing zeros in the decimal part. For example, in the number \( 0.2244 \), all four digits are significant because they provide specific information about the quantity measured.

Understanding significant digits is vital in mathematics and science as they indicate the precision of a measurement or calculation. When rounding numbers in calculations, like the product \( 0.22 \times 1.02 \) in the exercise, which gives \( 0.2244 \), to maintain the original number of significant digits (\(0.22\) has two significant digits), we often round the result to \( 0.22 \). However, rounding can introduce a rounding error, which is the difference between the actual number and the rounded number. This can affect the final outcome of calculations significantly, especially when subsequent mathematical operations are sensitive to small changes in the input values.

In scenarios where high precision is required, keeping track of significant digits is essential for ensuring that the final reported answer reflects the precision of the measurements and the limitations of the arithmetic operations performed.
Arithmetic Precision
Arithmetic precision refers to the level of detail and exactness in carrying out operations and representing numbers in calculations. In mathematical computations, precision dictates how closely a calculation can approach the exact result. The number of significant digits largely determines the precision of a numerical value.

In the context of this exercise, arithmetic precision is particularly important in the multiplication steps. Since we are rounding off to fewer digits, we're artificially limiting the arithmetic precision. This influences the accuracy of the final result. For example, when you multiply \( 0.22 \times 1.02 \) without rounding, you get \( 0.2244 \) – but rounding off reduces the precision to just two significant digits, yielding \( 0.22 \).

Similar rounding in subsequent operations can lead to a compounding of errors, termed cumulative rounding error, which may become significant in long calculations. It's a critical aspect to consider in fields like numerical analysis, finance, and any science or engineering domain where precise calculations are fundamental. To improve the reliability of the results, the general advice is to perform calculations with as much precision as possible and only round off at the final step, ensuring the rounding error is minimized.

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Most popular questions from this chapter

Consider the example problem \(x^{\prime}=x-4 y, y^{\prime}=-x+y\) with the initial conditions \(x(0)=1\) and \(y(0)=0\). Use the Runge-Kutta method to solve this problem on the interval \(0 \leq t \leq 1\). Start with \(h=0.2\) and then repeat the calculation with step sizes \(h=0.1,0.05, \ldots\), each half as long as in the preceding case. Continue the process until the first five digit of the solution at \(t=1\) are unchanged for successive step sizes Determine whether these digits are accurate by comparing them with the exact solution given in Eqs. ( 10 ) in the text.

Determine an approximate value of the solution at \(t=0.4\) and \(t=0.5\) using the specified method. For starting values use the values given by the Runge- Kutta method; see Problems 1 through 6 of Section 8.3 . Compare the results of the various methods with each other and with the actual solution (if available). $$ \begin{array}{l}{\text { (a) Use the fourth order predictor-corrector method with } h=0.1 . \text { Use the corrector }} \\ {\text { formula once at each step. }} \\ {\text { (b) Use the fourth order Adams-Moulton method with } h=0.1} \\ {\text { (c) Use the fourth order backward differentiation method with } h=0.1 .}\end{array} $$ $$ y^{\prime}=\left(t^{2}-y^{2}\right) \sin y, \quad y(0)=-1 $$

Show that the third order Adams-Moulton formula is $$ y_{x+1}=y_{x}+(h / 12)\left(5 f_{x+1}+8 f_{n}-f_{x-1}\right) $$

Consider the initial value problem \(x^{\prime}=f(t, x, y)\) and \(y^{\prime}=g(t, x, y)\) with \(x\left(t_{0}\right)=x_{0}\) and \(y\left(t_{0}\right)=y_{0} .\) The generalization of the Adams-Moulton predictor-corrector method of Section 8.4 is $$ \begin{array}{l}{x_{n+1}=x_{n}+\frac{1}{24} h\left(55 f_{n}-59 f_{n-1}+37 f_{n-2}-9 f_{n-3}\right)} \\ {y_{n+1}=y_{n}+\frac{1}{24} h\left(55 g_{n}-59 g_{n-1}+37 g_{n-2}-9 g_{n-1}\right)}\end{array} $$ and $$ \begin{array}{l}{x_{n+1}=x_{n}+\frac{1}{24} h\left(9 f_{n+1}+19 f_{n}-5 f_{n-1}+f_{n-2}\right)} \\ {y_{n+1}=y_{n}+\frac{1}{24} h\left(9 g_{n+1}+19 g_{n}-5 g_{n-1}+g_{n-2}\right)}\end{array} $$ Determine an approximate value of the solution at \(t=0.4\) for the example initial value problem \(x^{\prime}=x-4 y, y^{\prime}=-x+y\) with \(x(0)=1, y(0)=0\). Take \(h=0.1 .\) Correct the predicted value once. For the values of \(x_{1}, \ldots, y_{3}\) use the values of the exact solution rounded to six digits: \(x_{1}=1.12883, x_{2}=1.32042, x_{3}=1.60021, y_{1}=-0.110527, y_{2}=\) \(-0.250847,\) and \(y_{3}=-0.429696\)

In each of Problems 1 through 6 find approximate values of the solution of the given initial value problem at \(t=0.1,0.2,0.3,\) and \(0.4 .\) Compare the results with those obtained by using other methods and with the exact solution (if available). (a) Use the Runge-Kutta method with \(h=0.1\) (b) Use the Runge-Kutta method with \(h=0.05\) $$ y^{\prime}=\left(t^{2}-y^{2}\right) \sin y, \quad y(0)=-1 $$

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