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Calculations Involving Significant Figures

Date: 02/22/2002 at 20:34:42
From: Cassandra Rideout
Subject: Calculations involving Significant Digits

Hi!  This is a great project you have here, but on with my question!
My physics class is having a fair amount of trouble with significant 
digits. We understand what significant digits are and their purpose. 
We have laboured over the rules for rounding and for multiplication 
and addition, and so on. However, our BIG problem lies with lengthy 
calculations that require several intermediate steps before the final 
result is obtained. In other words, we don't know (even our teacher!) 
what to do when we have to add and multiply in the same problem.  

For example: 3.95 x 1.15 + 2.7503 / 8.49. Do you multiply the first 
two numbers (3.95 x 1.15) and determine the correct number of 
significant figures (4.59), which you then add to (0.324)? Or do we 
not round or use significant figures (is there a difference between 
these two terms?) until the very end? That is, add the raw data 
(4.5425 and 0.3239458...) and then use the addition significant figure 

Likewise, when determining slope m = (11.2cm - 10.2cm) / (200g - 0g) 
do you say that 1.0 / 200 = .0050 cm/g, or do you look at your initial 
numbers (least amount of significant figures would be 3 because masses 
are constants and contain an infinite number of significant figures) 
and say that the answer is 0.00500 cm/g?  

I hope that you can help me (in fact our whole class). Thanks in 

Date: 02/22/2002 at 23:39:54
From: Doctor Peterson
Subject: Re: Calculations involving Significant Digits

Hi, Cassandra.

It sounds as if you are forgetting the fact that significant figures 
are only appropriate in multiplication, not in addition. This page 
discusses that:

   Decimal Places and Significant Figures   

So you really have to think about the precision at each step, since 
each step has a different effect, some via significant figures, others 
via decimal places. But you don't want to round to the appropriate 
precision at each step, because then you would be introducing error. 
Significant figures are only a rule of thumb, and rounding too early 
can allow errors in the dropped digits to affect the result. In these 
days of calculators you don't have to round anything until the end; in 
the old days you would have been advised to keep at least a digit or 
two extra until the end. (Calculators do that too - they have more 
precision internally than they show, for exactly this reason.)

So let's take your examples:

    3.95 * 1.15 + 2.7503 / 8.49

You will do all the arithmetic with as much precision as your 
calculator allows, giving 4.5425 + 0.3239..., and then add them to get 
4.8664.... Now you look at the precision. The product has three 
significant figures, giving you hundredths, and the quotient has three 
as well, this time giving thousandths; the sum is accurate only to the 
hundredths, so you use 4.87. This happens to have three significant 
figures as well, but that is not necessarily going to happen.

    (11.2 cm - 10.2 cm) / (200 g - 0 g)

Here you first subtract, giving tenths in the dividend and units in 
the divisor, so you have 1.0 / 200. Since the dividend has only two 
significant figures (and I'm going to assume the divisor has three, 
though it could be taken as one), your answer should have two: 
0.0050 cm/g.

I'm not sure why you said that masses are constants and have 
infinitely many significant figures; if they are measured, they have 
finite precision. Only _defined_ constants (like 2 when you are 
doubling something) are treated as exact. But the 0 probably is exact. 
I can't tell without knowing the source of these numbers.

I hope this helps!

- Doctor Peterson, The Math Forum   
Associated Topics:
Elementary Place Value
High School Physics/Chemistry

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