Even making calculations only by the Excel submitted figures are not correct
from the 6 decimal place, and I updated with Office (last SP), which can be?

Other e-mail for contact:

It's no good just saying vaguely: "... are not correct"
Tell us what formula you used, what numbers were the inputs to that formula,
what answer you got, and what answer you expected.
--
David Biddulph

"Claudio" wrote in message
...
Even making calculations only by the Excel submitted figures are not
correct
from the 6 decimal place, and I updated with Office (last SP), which can
be?

Other e-mail for contact:

Suppose cell A1 = 1,0198424825731 (thirteen decimal places after comma) and
cell B1 = 2136,613241 (six decimal places after comma) the result is
presented in Excel = 2179,008952 (only six decimal places after comma) and
should be presented the following results = 2179,00895199999.

After a six decimal places is presented 000000 (zeros)

"David Biddulph" wrote:

It's no good just saying vaguely: "... are not correct"
Tell us what formula you used, what numbers were the inputs to that formula,
what answer you got, and what answer you expected.
--
David Biddulph

"Claudio" wrote in message
...
Even making calculations only by the Excel submitted figures are not
correct
from the 6 decimal place, and I updated with Office (last SP), which can
be?

Other e-mail for contact:

This problem also occurs in Excel 2007.

"Claudio" wrote:

Suppose cell A1 = 1,0198424825731 (thirteen decimal places after comma) and
cell B1 = 2136,613241 (six decimal places after comma) the result is
presented in Excel = 2179,008952 (only six decimal places after comma) and
should be presented the following results = 2179,00895199999.

After a six decimal places is presented 000000 (zeros)

"David Biddulph" wrote:

It's no good just saying vaguely: "... are not correct"
Tell us what formula you used, what numbers were the inputs to that formula,
what answer you got, and what answer you expected.
--
David Biddulph

"Claudio" wrote in message
...
Even making calculations only by the Excel submitted figures are not
correct
from the 6 decimal place, and I updated with Office (last SP), which can
be?

Other e-mail for contact:

Actually Excel produces

2137,63308348257

as opposed to

2137,6330834825731

which is due to the limit of precision available in Excel.

Claudio wrote:

This problem also occurs in Excel 2007.

"Claudio" wrote:


Suppose cell A1 = 1,0198424825731 (thirteen decimal places after comma) and
cell B1 = 2136,613241 (six decimal places after comma) the result is
presented in Excel = 2179,008952 (only six decimal places after comma) and
should be presented the following results = 2179,00895199999.

After a six decimal places is presented 000000 (zeros)

"David Biddulph" wrote:


It's no good just saying vaguely: "... are not correct"
Tell us what formula you used, what numbers were the inputs to that formula,
what answer you got, and what answer you expected.
--
David Biddulph

"Claudio" wrote in message
...

Even making calculations only by the Excel submitted figures are not
correct
from the 6 decimal place, and I updated with Office (last SP), which can
be?

Other e-mail for contact:

'Not that this is happening with me after the sixth decimal place has only
zeros. How to reach us below after the sixth decimal place?

"Bob I" wrote:

Actually Excel produces

2137,63308348257

as opposed to

2137,6330834825731

which is due to the limit of precision available in Excel.

Claudio wrote:

This problem also occurs in Excel 2007.

"Claudio" wrote:


Suppose cell A1 = 1,0198424825731 (thirteen decimal places after comma) and
cell B1 = 2136,613241 (six decimal places after comma) the result is
presented in Excel = 2179,008952 (only six decimal places after comma) and
should be presented the following results = 2179,00895199999.

After a six decimal places is presented 000000 (zeros)

"David Biddulph" wrote:


It's no good just saying vaguely: "... are not correct"
Tell us what formula you used, what numbers were the inputs to that formula,
what answer you got, and what answer you expected.
--
David Biddulph

"Claudio" wrote in message
...

Even making calculations only by the Excel submitted figures are not
correct
from the 6 decimal place, and I updated with Office (last SP), which can
be?

Other e-mail for contact:

What calculation is giving you 2137.63308348257 ?
And where do you get 2137.6330834825731 ?
Doesn't 1.0198424825731 * 2136.613241 come to something a lot closer to the
OP's 2179.008952 ?
--
David Biddulph

"Bob I" wrote in message
...
Actually Excel produces

2137,63308348257

as opposed to

2137,6330834825731

which is due to the limit of precision available in Excel.

Claudio wrote:

This problem also occurs in Excel 2007.

"Claudio" wrote:


Suppose cell A1 = 1,0198424825731 (thirteen decimal places after comma)
and cell B1 = 2136,613241 (six decimal places after comma) the result is
presented in Excel = 2179,008952 (only six decimal places after comma)
and should be presented the following results = 2179,00895199999.

After a six decimal places is presented 000000 (zeros)

"David Biddulph" wrote:


It's no good just saying vaguely: "... are not correct"
Tell us what formula you used, what numbers were the inputs to that
formula, what answer you got, and what answer you expected.
--
David Biddulph

"Claudio" wrote in message
...

Even making calculations only by the Excel submitted figures are not
correct
from the 6 decimal place, and I updated with Office (last SP), which
can be?

Other e-mail for contact:

You must have missed my earlier reply where I said that Excel's precision is
15 significant figures.

If the answer to a higher precision is 2179.0089519999972104171, then when
you calculate it to 15 significant figures the correct answer is
2179.00895200000, not 2179.00895199999.
--
David Biddulph

"Claudio" wrote in message
...
Suppose cell A1 = 1,0198424825731 (thirteen decimal places after comma)
and
cell B1 = 2136,613241 (six decimal places after comma) the result is
presented in Excel = 2179,008952 (only six decimal places after comma) and
should be presented the following results = 2179,00895199999.

After a six decimal places is presented 000000 (zeros)

"David Biddulph" wrote:

It's no good just saying vaguely: "... are not correct"
Tell us what formula you used, what numbers were the inputs to that
formula,
what answer you got, and what answer you expected.
--
David Biddulph

"Claudio" wrote in message
...
Even making calculations only by the Excel submitted figures are not
correct
from the 6 decimal place, and I updated with Office (last SP), which
can
be?

Other e-mail for contact:

OK I understand now about the 15 decimal digits.

So there isn´t a way (function, macro, programming) to obtain the number
2179.00895199999 instead of the number 2179.00895200000?

"David Biddulph" wrote:

You must have missed my earlier reply where I said that Excel's precision is
15 significant figures.

If the answer to a higher precision is 2179.0089519999972104171, then when
you calculate it to 15 significant figures the correct answer is
2179.00895200000, not 2179.00895199999.
--
David Biddulph

"Claudio" wrote in message
...
Suppose cell A1 = 1,0198424825731 (thirteen decimal places after comma)
and
cell B1 = 2136,613241 (six decimal places after comma) the result is
presented in Excel = 2179,008952 (only six decimal places after comma) and
should be presented the following results = 2179,00895199999.

After a six decimal places is presented 000000 (zeros)

"David Biddulph" wrote:

It's no good just saying vaguely: "... are not correct"
Tell us what formula you used, what numbers were the inputs to that
formula,
what answer you got, and what answer you expected.
--
David Biddulph

"Claudio" wrote in message
...
Even making calculations only by the Excel submitted figures are not
correct
from the 6 decimal place, and I updated with Office (last SP), which
can
be?

Other e-mail for contact:

If you were looking at a number of digits smaller than 15, and you don't
want to round to the closest value at the specified number of digits, which
Excel does by default, but instead you wanted to round down or truncate, you
may wish to look at the ROUNDDOWN, FLOOR, or TRUNCATE functions. Excel
help will tell you about them.

However in this case you don't have the number 2179.0089519999972104171 in
the first place, so I think you're out of luck trying to do it all in one
go. If you want to work to more than 15 digits, don't use Excel.

There might be a complicated workaround in splitting your input numbers into
most significant and least significant parts, manipulating those separately,
deciding where you needed to do your truncation in the least significant
part, and then gluing together the answers.
In this case, as your input numbers are greater than 1 but have non-integer
parts, it is as simple as
=INT(A1)*INT(B1)+TRUNC(INT(A1)*MOD(B1,1)+INT(B1)*M OD(A1,1)+MOD(A1,1)*MOD(B1,1),15-LOG(A1*B1))
but in a more general case it would be more complicated.
--
David Biddulph

"Claudio" wrote in message
...
OK I understand now about the 15 decimal digits.

So there isn´t a way (function, macro, programming) to obtain the number
2179.00895199999 instead of the number 2179.00895200000?

"David Biddulph" wrote:

You must have missed my earlier reply where I said that Excel's precision
is
15 significant figures.

If the answer to a higher precision is 2179.0089519999972104171, then
when
you calculate it to 15 significant figures the correct answer is
2179.00895200000, not 2179.00895199999.
--
David Biddulph

"Claudio" wrote in message
...
Suppose cell A1 = 1,0198424825731 (thirteen decimal places after comma)
and
cell B1 = 2136,613241 (six decimal places after comma) the result is
presented in Excel = 2179,008952 (only six decimal places after comma)
and
should be presented the following results = 2179,00895199999.

After a six decimal places is presented 000000 (zeros)

"David Biddulph" wrote:

It's no good just saying vaguely: "... are not correct"
Tell us what formula you used, what numbers were the inputs to that
formula,
what answer you got, and what answer you expected.
--
David Biddulph

"Claudio" wrote in message
...
Even making calculations only by the Excel submitted figures are not
correct
from the 6 decimal place, and I updated with Office (last SP), which
can
be?

Other e-mail for contact:

On Thu, 7 Aug 2008 04:34:01 -0700, Claudio
wrote:

OK I understand now about the 15 decimal digits.

So there isn´t a way (function, macro, programming) to obtain the number
2179.00895199999 instead of the number 2179.00895200000?

You can use the CDec data type in VBA and truncate the resultant string.

====================
Function DecMult(n1 As Variant, n2 As Variant) As Variant
DecMult = CStr(CDec(n1) * CDec(n2))
End Function
======================

-- 2179.0089519999972104171

Or you could use the Xnumbers add-in from
http://digilander.libero.it/foxes/SoftwareDownload.htm

which affords high-precision math routines for Excel.
--ron