https://bugs.documentfoundation.org/show_bug.cgi?id=165738

            Bug ID: 165738
           Summary: Round inside functions
           Product: LibreOffice
           Version: 25.2.1.2 release
          Hardware: All
                OS: Windows (All)
            Status: UNCONFIRMED
          Severity: normal
          Priority: medium
         Component: LibreOffice
          Assignee: [email protected]
          Reporter: [email protected]

Description:
I encountered some unexpected behavior in LibreOffice—using numbers in
functions and arithmetic operations rounds them after a certain decimal place.
For example, the result of the multiplication =0.6 * 3 will be 1.8 in the cell
and in the file, which differs from the result of the calculation in Python,
1.7999999999999998. Additionally, if I use large numbers inside formulas, the
result will be rounded. For instance, the result of the comparison
=CLEAN(1.7999999999999998) = CLEAN(0.6 * 3) will be TRUE, and the formula will
change to =CLEAN(1.8) = CLEAN(0.6 * 3). However, =CLEAN(1.79999999999998) =
CLEAN(0.6 * 3) will return FALSE and will not alter the formula. Is this
behavior a bug, or is rounding after a certain decimal place supposed to happen
in arithmetic operations and functions?

Steps to Reproduce:
1. Enter used formulas in any cell

Actual Results:
=1.7999999999999991 = 1.7999999999999999 - TRUE
=CLEAN(1.7999999999999991) - 1.8
=CLEAN(1.799999999999991) - 1.79999999999999

Expected Results:
=1.7999999999999991 = 1.7999999999999999 - FALSE
=CLEAN(1.7999999999999991) - 1.7999999999999991
=CLEAN(1.799999999999991) - 1.799999999999991


Reproducible: Always


User Profile Reset: No

Additional Info:
Used formulas:
=1.7999999999999991 = 1.7999999999999999
=0.6*3
=10.01-10
=1234567890 + 0.0123456789
=CLEAN(1.7999999999999991)
=CLEAN(1.799999999999991)
=CLEAN(1.79999999999991)
=CLEAN(0.6*3)
=TRUNC(0.6*3,15)
=TRUNC(1.7999999999999998, 15)
=TRUNC(1.799999999999998, 15)
=TRUNC(1.79999999999998, 15)
=CLEAN(1.7999999999999998) = CLEAN(0.6*3)
=CLEAN(1.79999999999998) = CLEAN(0.6*3)

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