Microsoft Office

Making Your Worksheets Error-Free 32
You can instruct Excel to use the displayed values by checking the Set Precision As Displayed check box on
the Advanced section of the Excel Options dialog box. (Choose Office ➪ Excel Options to display this dialog
box.)
Be very careful with the Set Precision As Displayed option. This option also affects normal values
(nonformulas) that have been entered into cells. For example, if a cell contains the value
4.68 and is displayed with no decimal places (that is, 5), checking the Precision As Displayed check box converts
4.68 to 5.00. This change is permanent, and you can’t restore the original value if you later uncheck the
Set Precision As Displayed check box. A better approach is to use Excel’s ROUND function to round off the
values to the desired number of decimal places.
CAUTION
Floating point number errors
Computers, by their very nature, don’t have infinite precision. Excel stores numbers in binary format by
using eight bytes, which can handle numbers with 15-digit accuracy. Some numbers can’t be expressed precisely
by using eight bytes, so the number stores as an approximation.
To demonstrate how this lack of precision may cause problems, enter the following formula into cell A1:
=(5.1-5.2)+1
The result should be 0.9. However, if you format the cell to display 15 decimal places, you discover that
Excel calculates the formula with a result of 0.899999999999999. This result occurs because the operation
in parentheses is performed first, and this intermediate result stores in binary format by using an approximation.
The formula then adds 1 to this value, and the approximation error is propagated to the final result.
In many cases, this type of error doesn’t present a problem. However, if you need to test the result of that
formula by using a logical operator, it may present a problem. For example, the following formula (which
assumes that the previous formula is in cell A1) returns False:
=A1=.9
One solution to this type of error is to use Excel’s ROUND function. The following formula, for example,
returns True because the comparison is made by using the value in A1 rounded to one decimal place.
=ROUND(A1,1)=0.9
Here’s another example of a “precision” problem. Try entering the following formula:
=(1.333-1.233)-(1.334-1.234)
This formula should return 0, but it actually returns –2.220446E-16 (a number very close to zero).
If that formula is in cell A1, the following formula returns Not Zero.
=IF(A1=0,”Zero”,”Not Zero”)
One way to handle these “very close to zero” rounding errors is to use a formula like this:
=IF(ABS(A1)