1.5 Measurement uncertainty, accuracy, and precision  (Page 3/11)

1. When we add or subtract numbers, we should round the result to the same number of decimal places as the number with the least number of decimal places (the least precise value in terms of addition and subtraction).
2. When we multiply or divide numbers, we should round the result to the same number of digits as the number with the least number of significant figures (the least precise value in terms of multiplication and division).
3. If the digit to be dropped (the one immediately to the right of the digit to be retained) is less than 5, we “round down” and leave the retained digit unchanged; if it is more than 5, we “round up” and increase the retained digit by 1; if the dropped digit is 5, we round up or down, whichever yields an even value for the retained digit. (The last part of this rule may strike you as a bit odd, but it’s based on reliable statistics and is aimed at avoiding any bias when dropping the digit “5,” since it is equally close to both possible values of the retained digit.)

The following examples illustrate the application of this rule in rounding a few different numbers to three significant figures:

• 0.028675 rounds “up” to 0.0287 (the dropped digit, 7, is greater than 5)
• 18.3384 rounds “down” to 18.3 (the dropped digit, 3, is less than 5)
• 6.8752 rounds “up” to 6.88 (the dropped digit is 5, and the retained digit is even)
• 92.85 rounds “down” to 92.8 (the dropped digit is 5, and the retained digit is even)

Let’s work through these rules with a few examples.

Addition and subtraction with significant figures

Rule: When we add or subtract numbers, we should round the result to the same number of decimal places as the number with the least number of decimal places (i.e., the least precise value in terms of addition and subtraction).

(a) Add 1.0023 g and 4.383 g.

(b) Subtract 421.23 g from 486 g.

Solution

(a) $\begin{array}{l}\\ \begin{array}{l}\hfill \\ \frac{\begin{array}{c}\mathrm{1.0023\; g}\\ \text{+ 4.383 g}\end{array}}{\mathrm{5.3853\; g}}\hfill \end{array}\end{array}$

Answer is 5.385 g (round to the thousandths place; three decimal places)

(b) $\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \end{array}\\ \frac{\begin{array}{l}\mathrm{486\; g}\hfill \\ -\mathrm{421.23\; g}\hfill \end{array}}{\mathrm{64.77\; g}}\end{array}$

Answer is 65 g (round to the ones place; no decimal places)

Check your learning

(a) Add 2.334 mL and 0.31 mL.

(b) Subtract 55.8752 m from 56.533 m.

Answer:

(a) 2.64 mL; (b) 0.658 m

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