Approximations and Errors
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In mathematics and science, exact values are not always necessary. In many situations, a number can be replaced by a nearby value that is easier to use while still remaining sufficiently accurate.
- Rounding down occurs when the approximate value \(a\) is less than the exact value \(n\): $$ a < n $$
- Rounding up occurs when the approximate value \(a\) is greater than the exact value \(n\): $$ a > n $$
Any decimal number \(n\) can be approximated by rounding it either down or up.
How Rounding Works
Rounding follows a simple rule based on the first digit that is removed.
- If the first discarded digit is 5 or greater, increase the last retained digit by one.
- If the first discarded digit is less than 5, leave the last retained digit unchanged.
Why Round Numbers? Rounded values are easier to read, compare, and use in calculations. They allow us to simplify numerical data without significantly changing its meaning. However, every approximation introduces a small error, and understanding that error is essential when accuracy matters.
Practical Examples
Consider the decimal number
$$ n = 87.427 $$
Suppose we want to round it to one decimal place, that is, to the nearest tenth.
The first discarded digit is 2, and the last retained digit is 4.
$$ n = 87.4 \color{red}27 $$
Since 2 is less than 5, the last retained digit remains unchanged.
$$ a = 87.4 $$
Example 2
Using the same number:
$$ n = 87.427 $$
This time, we round it to two decimal places, that is, to the nearest hundredth.
The first discarded digit is 7, and the last retained digit is 2.
$$ n = 87.42 \color{red}7 $$
Since 7 is greater than or equal to 5, the last retained digit must be increased by one.
$$ a = 87.43 $$
In this case, the digit 2 becomes 3.
Absolute Error and Relative Error
Whenever a number is approximated, some information is lost. The difference between the exact value and the approximation is called the error.
Two measures are commonly used to quantify this error.
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Absolute Error
The absolute error \(e\) is the absolute difference between the exact value \(n\) and the approximation \(a\): $$ e = |n - a| $$Example: For the approximation \(87.427 \approx 87.4\), $$ e = |87.427 - 87.4| = 0.027 $$
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Relative Error
The relative error \(e_r\) compares the absolute error with the size of the approximation itself: $$ e_r = \frac{|n-a|}{|a|} = \frac{e}{|a|} $$
Example: For the same approximation, $$ e_r = \frac{0.027}{|87.4|} = 0.000308924485 $$ Expressed as a percentage, $$ e_r \approx 0.0309\% $$
Which Error Measure Is More Useful?
Absolute error tells us how much an approximation differs from the exact value, but it does not tell us whether that difference is large or small relative to the magnitude of the number.
Relative error provides this additional perspective and is therefore often a better measure of accuracy.
Example: Consider the approximations $$ 802 \asymp 800 $$ and $$ 82 \asymp 80 $$ Both have the same absolute error: $$ e = 2 $$ Yet the approximation 800 is considerably more accurate than 80 because an error of 2 represents a much smaller fraction of 800 than of 80.
Error Propagation
Errors introduced by approximations do not disappear. As calculations continue, they can accumulate and affect the final result.
In general, the more operations performed using rounded values, the larger the potential deviation from the exact answer.
The way errors accumulate depends on the type of operation:
- For addition and subtraction, the maximum absolute error is the sum of the absolute errors of the individual terms: $$ e = e_1 + e_2 $$
- For multiplication and division, the maximum relative error is the sum of the relative errors of the individual factors: $$ e_r = e_{r1} + e_{r2} $$
Example of Error Propagation
Consider the numbers
$$ n_1 = 12331 $$
$$ n_2 = 8168 $$
Rounded to the nearest hundred, they become
$$ a_1 = 12300 $$
$$ a_2 = 8200 $$
The corresponding absolute errors are
$$ e_1 = |12331 - 12300| = 31 $$
$$ e_2 = |8168 - 8200| = 32 $$
Note: The relative errors are $$ e_{r1} = \frac{31}{|12300|} = 0.0025203252 \asymp 0.25\% $$ $$ e_{r2} = \frac{32}{|8200|} = 0.00390243902 \asymp 0.39\% $$
Now add the rounded values:
$$ s = a_1 + a_2 $$
$$ s = 12300 + 8200 = 20500 $$
The maximum absolute error in the result is
$$ e_s = e_1 + e_2 $$
$$ e_s = 31 + 32 = 63 $$
Note: The relative error of the sum is $$ e_r = \frac{63}{|20500|} = 0.00307317073 \asymp 0.31\% $$
This means that the exact sum \(s_r\) must lie within the interval
$$ s - e_s < s_r < s + e_s $$
$$ 20500 - 63 < s_r < 20500 + 63 $$
$$ 20437 < s_r < 20563 $$
Only the leading digits that remain unchanged throughout the entire interval can be regarded as reliable.
Verification: The exact sum of the original numbers is $$ s_r = n_1 + n_2 $$ $$ s_r = 12331 + 8168 = 20499 $$
