When you tutor physics or chemistry, most calculations require management of significant figures (aka significant digits).  Today, we’ll look at multiplying and dividing.

We know from earlier posts (see the previous one here) why significant figures are needed and how to tell when zeros are significant. Now, we’ll discuss how to report the result of a calculation to the correct number of significant figures.

Remember that, “on sight”, all nonzero digits are significant. The rules for zeros, once again, are covered here. However, once you do a calculation, you must apply other rules to determine how many significant digits are in the answer.

With mutliplying and dividing, the method is simple:

1)  Decide which input number has the least number of significant figures.

2)  The least number from 1) is the proper number of significant figures for your answer.

Putting it a different way:  the correct number of significant figures for the answer is how many the input with the least of them has.

Consider an example:

75.04 x 0.407

We know that 75.04 has four significant digits, while 0.407 has 3.  Our answer must then have three significant digits.

Of course, as often happens, the calculator gives us more than the proper number of significant digits:

75.04 x 0.407 = 30.54128

We round our answer to three significant digits:

30.54128 ≈ 30.5

Just to review:  We were multiplying.  One input number had four significant digits and one had three.  Therefore, we rounded our answer to three significant digits.

Here’s another example:

0.5695 ÷ 1.7 = 0.335

We know that 0.5674 has four significant digits while 1.7 has only two.  Our answer, therefore, should have two significant digits:

0.335 ≈ 0.34

(Since the next digit is 5 or more, we round up to 0.34.)

The method explained in this post applies to multiplying and dividing with significant digits.  Adding and subtracting use a different method, which will be covered in a future post.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.