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The equation has been solved for v, and you are now ready to substitute can be no more accurate than the values being treated. Because the
quantities and perform the needed operations (see the example measurement had only three significant figures (two certain, one
problem in chapter 1 on page 13 for information on this topic). uncertain), the answer can have only three significant figures. The
2
area is correctly expressed as 11.9 cm .
There are a few simple rules that will help you determine how
A.3 SIGNIFICANT FIGURES many significant figures are contained in a reported measurement:
1. All digits reported as a direct result of a measurement are
The numerical value of any measurement will always contain
some uncertainty. Suppose, for example, that you are measur- significant.
ing one side of a square piece of paper as shown in Figure A.1. 2. Zero is significant when it occurs between nonzero digits.
According to the enlarged scale, you could say that the paper For example, 607 has three significant figures, and the zero
is about 3.5 cm wide and you would be correct. This measure- is one of the significant figures.
ment, however, would be unsatisfactory for many purposes. It 3. In figures reported as larger than the digit 1, the digit 0 is
does not approach the true value of the length and contains too not significant when it follows a nonzero digit to indicate
much uncertainty. It seems clear that the paper width is larger place. For example, in a report that “23,000 people attended
than 3.4 cm but narrower than 3.5 cm. But how much larger the rock concert,” the digits 2 and 3 are significant but the
than 3.4 cm? You cannot be certain if the paper is 3.44, 3.45, zeros are not significant. In this situation, the 23 is the
or 3.46 cm wide. As your best estimate, you might say that the measured part of the figure, and the three zeros tell you
paper is 3.45 cm wide. Everyone would agree that you can be an estimate of how many attended the concert, that is,
certain about the first two numbers (3.4), and they should be 23 thousand. If the figure is a measurement rather than an
recorded. The last number (0.05) has been estimated and is not estimate, then it is written with a decimal point after the
certain. The two certain numbers, together with one uncertain last zero to indicate that the zeros are significant. Thus
number, represent the greatest accuracy possible with the ruler 23,000 has two significant figures (2 and 3), but 23,000.
being used. The paper is said to be 3.45 cm wide. has five significant figures. The figure 23,000 means “about
A signifi cant fi gure is a number that is believed to be correct 23 thousand,” but 23,000. means 23,000. and not 22,999
with some uncertainty only in the last digit. The value of the width or 23,001.
of the paper, 3.45 cm, represents three significant figures. As you 4. In figures reported as smaller than the digit 1, zeros after a
can see, the number of significant figures can be determined by decimal point that come before nonzero digits are not
the degree of accuracy of the measuring instrument being used. significant and serve only as place holders. For example,
But suppose you need to calculate the area of the paper. You would 0.0023 has two significant figures: 2 and 3. Zeros alone
multiply 3.45 cm × 3.45 cm, and the product for the area would be after a decimal point or zeros after a nonzero digit indicate
2
11.9025 cm . This is a greater precision than you were able to ob- a measurement, however, so these zeros are significant.
tain with your measuring instrument. The result of a calculation The figure 0.00230, for example, has three significant
figures since the 230 means 230 and not 229 or 231.
Likewise, the figure 3.000 cm has four significant figures
because the presence of the three zeros means that the
measurement was actually 3.000 and not 2.999 or 3.001.
2 3
MULTIPLICATION AND DIVISION
When you multiply or divide measurement figures, the answer
may have no more significant figures than the least number of
significant figures in the figures being multiplied or divided. This
simply means that an answer can be no more accurate than the
least accurate measurement entering into the calculation, and
that you cannot improve the accuracy of a measurement by doing
a calculation. For example, in multiplying 54.2 mi/h × 4.0 h to
find out the total distance traveled, the first figure (54.2) has three
significant figures, but the second (4.0) has only two significant
figures. The answer can contain only two significant figures since
this is the weakest number of those involved in the calculation.
The correct answer is therefore 220 mi, not 216.8 mi. This may
seem strange since multiplying the two numbers together gives the
answer of 216.8 mi. This answer, however, means a greater accu-
racy than is possible, and the accuracy cannot be improved over
the weakest number involved in the calculation. Since the weakest
FIGURE A.1 How wide is this sheet of paper? Write your answer number (4.0) has only two significant figures the answer must also
before reading the text ——— . have only two significant figures, which is 220 mi.

A-3 APPENDIX A Mathematical Review 625