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Excel's Formula - sisman

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586<br />

Part V: Miscellaneous <strong>Formula</strong> Techniques<br />

Here’s another example of a “precision” problem. Try entering the following formula:<br />

=(1.333–1.233)–(1.334–1.234)<br />

This formula should return 0, but it actually returns –2.220446E–16 (a number very close to zero).<br />

If that formula were in cell A1, the following formula would return the string “Not Zero”.<br />

=IF(A1=0,”Zero”,”Not Zero”)<br />

One way to handle these very-close-to-zero rounding errors is to use a formula like this:<br />

=IF(ABS(A1)

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