Approximate length in centimeters
Suppose that in a measurement of any object we find the number 3.8 cm as a value. For this result, we are presenting two significant figures, where 3 represents the correct figure and 8, the doubtful figure. It is also common to find significant figures with numerous decimal places, as is the case with pi ( π ). In this case, we have to pay more attention to correctly perform some basic operations, such as division, multiplication, subtraction and addition. Below we see the correct procedures to perform the fundamental operations. Let’s see then:
Let’s start with: Addition and Subtraction
For the fundamental operations of addition and subtraction, it is first necessary to round off the values of significant figures in order to equalize the decimal places. Let’s see the example below:
Let’s add three different measurements: 27.132 m, 115.18 m and 43.39 m.
The sum operation in the figure above can be expressed as follows:
S = 27.1 m+115.2 m+43.4 m
S = 185.7 m
After carrying out all the calculations, we choose as a reference the number that has the fewest decimal places. For subtraction operations we must follow the same reasoning as for addition, but following certain rules.
Multiplication and Division
Multiplication and division operations are solved normally, that is, by multiplying and dividing all numbers without rounding. The difference is that the result obtained in the division or multiplication must contain the same number of significant figures as the factor that has the fewest significant figures. Let’s look at an example:
Let’s multiply the following values: 2.083 m by 0.817 m.
S = (2.083 m).(0.817 m) = 1.70 m 2
The result obtained in the above multiplication must be rounded to three significant figures, which correspond to the number of significant figures of the factor 0.817 m. Therefore, we must round the result, giving as an answer 1.70 m 2 .
Let’s look at another example: let’s divide the following values: 6.832 cm and 2.365 cm.
S=(6.832 cm)÷(2.36 cm)