**Please Note: **
This article is written for users of the following Microsoft Excel versions: 2007, 2010, 2013, 2016, 2019, and Excel in Office 365. If you are using an earlier version (Excel 2003 or earlier), *this tip may not work for you*. For a version of this tip written specifically for earlier versions of Excel, click here: Avoiding Rounding Errors in Formula Results.

Nick notes that the formula "=0.28*100-INT(0.28*100)=0" returns False even though it is obviously true. He believes the issue has to do with rounding and how the computer uses binary arithmetic, etc. Nick is using the formula as a part of a larger IF statement, and he assumes there are other rounding errors in Excel that can reach out and bite him. He wonders if there is a simple way to avoid formula pitfalls such as this.

Before looking at simple ways to avoid this type of problem, it is important to understand why the problem exists. As Nick notes, it really has to do with binary arithmetic and the fact that Excel rounds information. Behind the scenes, Excel always rounds information to 15 digits. Consider the calculation that Nick is working with:

=0.28*100-INT(0.28*100)=0

When Excel first calculates this, the precedence followed by Excel calculates it in this manner:

=(0.28*100-INT(0.28*100))=0

Note the extra set of parentheses. The result of everything to the left of that final equal sign is 3.55E-15, which means that you end up with this (longhand) formula:

=0.00000000000000355=0

This is obviously not true, which is why you get the value False returned. There are several ways to "fix" this situation. In this case, perhaps the easiest is to simply change your formula to remove the need to compare to zero:

=0.28*100=INT(0.28*100)

This formula returns True, as you would expect. This may not work for all your needs, however. So, a better rule of thumb to avoid problems is to never rely on Excel's rounding. You do this by implementing your own explicit rounding, as shown with these formulas:

=INT(0.28*100)-INT(0.28*100)=0 =ROUND(0.28*100,0)-INT(0.28*100)=0 =ROUND(0.28*100,5)-ROUND(0.28*100,5)=0

Note that you are, in these instances, not allowing Excel to perform calculations to its full precision since that can cause some unexpected results when you are comparing to a specific value, such as zero. Instead, you are forcing Excel to round the values—all the values you are working with—to whatever level of precision you need for your comparison.

Another approach is to not do the comparison to an exact value, like zero. Instead, allow for some "fudge factor" in the comparison, which allows for rounding issues. For instance, you may determine that you only care if the comparison is accurate to one one-hundredth of whatever units you are assuming. In that case, the original formula could be modified in this manner:

=0.28*100-INT(0.28*100)<0.01

This returns True, as one would expect.

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This tip (8145) applies to Microsoft Excel 2007, 2010, 2013, 2016, 2019, and Excel in Office 365. You can find a version of this tip for the older menu interface of Excel here: **Avoiding Rounding Errors in Formula Results**.

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2019-05-15 12:33:41

Dennis Costello

Specifically, the 0.28 was not represented exactly - it's accurate to 1 part in 2^55 (because there are 55 mantissa bits in the double-precision floating-point format. Multiplying that by 100 means the product could be off by as much as 100 parts in 2^55, which is approximately 1 part in 2^48. Subtracting 28 exactly from 28 +- 1/2^48 means we get a number that's somewhere between -1/2^48 and +1/2^48. The product turned out to be 28.0000000000000035527136788005 (1/2^48 larger than 28) - when you subtracted exactly 28 from that, the difference was far enough away from 0 that Excel wasn't willing to treat it as being equal to 0.

2019-01-02 13:20:27

Dave Bonin

Unfortunately, you cannot know whether the leftover portion will be positive or negative. Therefore, cover both bases by using the Absolute Value function:

= ABS( 0.28 * 100 - INT( 0.28 * 100 )) < 0.01

2018-12-31 22:12:53

Alex B

=Round(0.28*100-INT(0.28*100),5)=0

I have used 5 decimals here but anything up to 14 decimals worked in this case which seems consistent with the 15 precision limit.

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