In Excel A1=1.2675-1.2664=.0011
B1=1.2664-1.2653=.0011
Ask A1=B1, result is False; why?

hi Zeno,

try to use round to see it as true
the results of the formula a

a1 = 0,0011000000000001
b1 = 0,0010999999999998

so they are not equal.

use =round((1.2675-1.2664),4)
and =round((1.2664-1.2653),4)

hth
regards from Brazil
Marcelo

"Zeno" escreveu:

In Excel A1=1.2675-1.2664=.0011
B1=1.2664-1.2653=.0011
Ask A1=B1, result is False; why?

If
A1=1.2675-1.2664=.0011 = FALSE
B1=1.2664-1.2653=.0011 = FALSE
A1=B1 = [FALSE = FALSE] = TRUE
if
A1=1.2675-1.2664
B1=1.2664-1.2653
A1=B1 = [.0011 = .0011] = FALSE



"Zeno" wrote:

In Excel A1=1.2675-1.2664=.0011

Ask A1=B1, result is False; why?

"Zeno" wrote in message
...
In Excel A1=1.2675-1.2664=.0011
B1=1.2664-1.2653=.0011
Ask A1=B1, result is False; why?

Because these numbers can't be represented exactly in binary.

http://www.cpearson.com/excel/rounding.htm
--
David Biddulph

The math (and logic) is exactly right, given unavoidable approximations to
the original numbers.

Most terminating decimal fractions (including .2675 .2774 .2653 and .0011)
are nonterminating binary fractions that can only be approximated, much as
1/3 can only be approximated as a decimal number. You are seeing the binary
equivalent of
2/3 - 1/3 = 0.6667 - 0.3333 = 0.3334 < 0.3333

As documented in Excel's Help for "Excel specifications and limits" subtopic
"Calculation specifications", the approximations used may not be exact beyond
15 digits. You cancelled 3 of those 15 digits in the subtraction, revealing
residue of the original approximations.

You can use the D2D function at
http://groups.google.com/group/micro...06871cf92f8465
To see the actual decimal values of the approximations involved.

Jerry

"Zeno" wrote:

In Excel A1=1.2675-1.2664=.0011
B1=1.2664-1.2653=.0011
Ask A1=B1, result is False; why?