How many decimal places can a cell display?

How many decimal places can be displayed in a cell?  I'm running a brute 
force VBA procedure of finding fractions that will approximate pi to as many 
decimal places as Excel will display, but I don't know how many decimal 
places Excel will display accurately.  Anybody know?  I guess this is also a 
matter of how many decimal places VBA will calculate accurately as well.

Sub PiFractions()
Dim dividend As Integer, divisor As Integer, quotient As Double
Dim rowpointer As Byte

rowpointer = 1

For dividend = 22 To 10000
  For divisor = 7 To dividend \ 3
    quotient = dividend / divisor
    If quotient > 3.14159 And quotient < 3.1416 Then
      Cells(rowpointer, 1) = dividend
      Cells(rowpointer, 2) = divisor
      Cells(rowpointer, 3) = quotient
      rowpointer = rowpointer + 1
    End If
  Next
Next

End Sub 


0
yeah (16)
4/2/2006 9:30:04 PM
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You can only get 15 digit precision in Excel.

Pete

0
pashurst (2576)
4/2/2006 9:35:07 PM
Thanks.  I'm getting 14 now with this little procedure.

"Pete_UK" <pashurst@auditel.net> wrote in message 
news:1144013707.000566.119580@e56g2000cwe.googlegroups.com...
> You can only get 15 digit precision in Excel.
>
> Pete
> 


0
yeah (16)
4/2/2006 9:53:36 PM
>... VBA procedure of finding fractions that will approximate pi to as many 
>decimal places as Excel will display,

Hi.  At 15 digits, I believe the minimum fraction for Pi is:

=80143857/25510582

As a side note, the limit in vba is:
Num = 428224593349304#
Den = 136308121570117#

Debug.Print CDec(Num) / Den
' 3.1415926535897932384626433833

-- 
HTH.  :>)
Dana DeLouis
Windows XP, Office 2003


"Spaz" <yeah@right.com> wrote in message 
news:svKdndamzu_E363ZRVn-tQ@comcast.com...
> How many decimal places can be displayed in a cell?  I'm running a brute 
> force VBA procedure of finding fractions that will approximate pi to as 
> many decimal places as Excel will display, but I don't know how many 
> decimal places Excel will display accurately.  Anybody know?  I guess this 
> is also a matter of how many decimal places VBA will calculate accurately 
> as well.
>
> Sub PiFractions()
> Dim dividend As Integer, divisor As Integer, quotient As Double
> Dim rowpointer As Byte
>
> rowpointer = 1
>
> For dividend = 22 To 10000
>  For divisor = 7 To dividend \ 3
>    quotient = dividend / divisor
>    If quotient > 3.14159 And quotient < 3.1416 Then
>      Cells(rowpointer, 1) = dividend
>      Cells(rowpointer, 2) = divisor
>      Cells(rowpointer, 3) = quotient
>      rowpointer = rowpointer + 1
>    End If
>  Next
> Next
>
> End Sub
> 


0
ddelouis (141)
4/2/2006 10:18:32 PM
Here is one that shows 15 digits

? application.pi()
 3.14159265358979

or in the worksheet
=pi()

and format the cell to show 14 decimals.

-- 
Regards,
Tom Ogilvy

"Spaz" <yeah@right.com> wrote in message
news:3uadncd2utpC2q3ZnZ2dnUVZ_vednZ2d@comcast.com...
> Thanks.  I'm getting 14 now with this little procedure.
>
> "Pete_UK" <pashurst@auditel.net> wrote in message
> news:1144013707.000566.119580@e56g2000cwe.googlegroups.com...
> > You can only get 15 digit precision in Excel.
> >
> > Pete
> >
>
>


0
twogilvy (1078)
4/2/2006 10:45:25 PM
Excel's numeric display limit is on significant figures, not decimal places.  
Excel (like almost all software) follows the IEEE standard for double 
precision binary representation of numbers.
  http://www.cpearson.com/excel/rounding.htm
In particular, all 15 digit and most 16 digit integers can be exactly 
represented.  But rather than explain why some 16 digit numbers unavoidably 
change value from what you enter, MS chose to display only 15 digits (See 
Help for "specifications").
and It requires 17 decimal digits to uniquely specify a double precision 
binary number, and An exact conversion from binary to decimal of a floating 
point number may require many more than 17 decimal digits
  http://groups.google.com/group/microsoft.public.excel/msg/b106871cf92f8465

If you want to write a routine that will handle more precision than Excel 
natively gives, you might find the VBA code at that last link instructive.  
There are some Excel add-ins like
  http://digilander.libero.it/foxes/index.htm
  http://precisioncalc.com/
that already implement higher precision.

Also there are commercial packages like Maple, Mathematica,
MacSyma and open source packages like Maxima 
  http://maxima.sourceforge.net/
that implement algebraic math and user-specified numeric precision.

Jerry

"Spaz" wrote:

> How many decimal places can be displayed in a cell?  I'm running a brute 
> force VBA procedure of finding fractions that will approximate pi to as many 
> decimal places as Excel will display, but I don't know how many decimal 
> places Excel will display accurately.  Anybody know?  I guess this is also a 
> matter of how many decimal places VBA will calculate accurately as well.
> 
> Sub PiFractions()
> Dim dividend As Integer, divisor As Integer, quotient As Double
> Dim rowpointer As Byte
> 
> rowpointer = 1
> 
> For dividend = 22 To 10000
>   For divisor = 7 To dividend \ 3
>     quotient = dividend / divisor
>     If quotient > 3.14159 And quotient < 3.1416 Then
>       Cells(rowpointer, 1) = dividend
>       Cells(rowpointer, 2) = divisor
>       Cells(rowpointer, 3) = quotient
>       rowpointer = rowpointer + 1
>     End If
>   Next
> Next
> 
> End Sub 
> 
> 
> 
0
post_a_reply (1395)
4/2/2006 10:55:02 PM
If you want to do a program loop, this is one of a few ways to get a jump 
start...

Sub Demo()
    Dim s As String
    s = WorksheetFunction.Rept("?", 16)
    s = s & "/" & s

    Range("A1").FormulaR1C1 = "=PI()"
    Range("A1").NumberFormat = s
    Debug.Print Range("A1").Text
End Sub

  5419351/1725033

As you can see, the fraction format can get close(~14), but not quite...:>(
-- 
HTH.  :>)
Dana DeLouis
Windows XP, Office 2003


"Dana DeLouis" <ddelouis@bellsouth.net> wrote in message 
news:OvEepNqVGHA.4336@TK2MSFTNGP14.phx.gbl...
> >... VBA procedure of finding fractions that will approximate pi to as 
> >many decimal places as Excel will display,
>
> Hi.  At 15 digits, I believe the minimum fraction for Pi is:
>
> =80143857/25510582
>
> As a side note, the limit in vba is:
> Num = 428224593349304#
> Den = 136308121570117#
>
> Debug.Print CDec(Num) / Den
> ' 3.1415926535897932384626433833
>
> -- 
> HTH.  :>)
> Dana DeLouis
> Windows XP, Office 2003
>
>
> "Spaz" <yeah@right.com> wrote in message 
> news:svKdndamzu_E363ZRVn-tQ@comcast.com...
>> How many decimal places can be displayed in a cell?  I'm running a brute 
>> force VBA procedure of finding fractions that will approximate pi to as 
>> many decimal places as Excel will display, but I don't know how many 
>> decimal places Excel will display accurately.  Anybody know?  I guess 
>> this is also a matter of how many decimal places VBA will calculate 
>> accurately as well.
>>
>> Sub PiFractions()
>> Dim dividend As Integer, divisor As Integer, quotient As Double
>> Dim rowpointer As Byte
>>
>> rowpointer = 1
>>
>> For dividend = 22 To 10000
>>  For divisor = 7 To dividend \ 3
>>    quotient = dividend / divisor
>>    If quotient > 3.14159 And quotient < 3.1416 Then
>>      Cells(rowpointer, 1) = dividend
>>      Cells(rowpointer, 2) = divisor
>>      Cells(rowpointer, 3) = quotient
>>      rowpointer = rowpointer + 1
>>    End If
>>  Next
>> Next
>>
>> End Sub
>>
>
> 


0
ddelouis (141)
4/3/2006 1:22:24 AM
Wow, that's some crazy code.  Thanks!

"Dana DeLouis" <ddelouis@bellsouth.net> wrote in message 
news:uQz2X0rVGHA.4772@TK2MSFTNGP14.phx.gbl...
> If you want to do a program loop, this is one of a few ways to get a jump 
> start...
>
> Sub Demo()
>    Dim s As String
>    s = WorksheetFunction.Rept("?", 16)
>    s = s & "/" & s
>
>    Range("A1").FormulaR1C1 = "=PI()"
>    Range("A1").NumberFormat = s
>    Debug.Print Range("A1").Text
> End Sub
>
>  5419351/1725033
>
> As you can see, the fraction format can get close(~14), but not 
> quite...:>(
> -- 
> HTH.  :>)
> Dana DeLouis
> Windows XP, Office 2003
>
>
> "Dana DeLouis" <ddelouis@bellsouth.net> wrote in message 
> news:OvEepNqVGHA.4336@TK2MSFTNGP14.phx.gbl...
>> >... VBA procedure of finding fractions that will approximate pi to as 
>> >many decimal places as Excel will display,
>>
>> Hi.  At 15 digits, I believe the minimum fraction for Pi is:
>>
>> =80143857/25510582
>>
>> As a side note, the limit in vba is:
>> Num = 428224593349304#
>> Den = 136308121570117#
>>
>> Debug.Print CDec(Num) / Den
>> ' 3.1415926535897932384626433833
>>
>> -- 
>> HTH.  :>)
>> Dana DeLouis
>> Windows XP, Office 2003
>>
>>
>> "Spaz" <yeah@right.com> wrote in message 
>> news:svKdndamzu_E363ZRVn-tQ@comcast.com...
>>> How many decimal places can be displayed in a cell?  I'm running a brute 
>>> force VBA procedure of finding fractions that will approximate pi to as 
>>> many decimal places as Excel will display, but I don't know how many 
>>> decimal places Excel will display accurately.  Anybody know?  I guess 
>>> this is also a matter of how many decimal places VBA will calculate 
>>> accurately as well.
>>>
>>> Sub PiFractions()
>>> Dim dividend As Integer, divisor As Integer, quotient As Double
>>> Dim rowpointer As Byte
>>>
>>> rowpointer = 1
>>>
>>> For dividend = 22 To 10000
>>>  For divisor = 7 To dividend \ 3
>>>    quotient = dividend / divisor
>>>    If quotient > 3.14159 And quotient < 3.1416 Then
>>>      Cells(rowpointer, 1) = dividend
>>>      Cells(rowpointer, 2) = divisor
>>>      Cells(rowpointer, 3) = quotient
>>>      rowpointer = rowpointer + 1
>>>    End If
>>>  Next
>>> Next
>>>
>>> End Sub
>>>
>>
>>
>
> 


0
yeah (16)
4/3/2006 5:31:42 AM
Just for interest, in xl97 it returned:  

355/113

-- 
Regards,
Tom Ogilvy


"Dana DeLouis" wrote:

> If you want to do a program loop, this is one of a few ways to get a jump 
> start...
> 
> Sub Demo()
>     Dim s As String
>     s = WorksheetFunction.Rept("?", 16)
>     s = s & "/" & s
> 
>     Range("A1").FormulaR1C1 = "=PI()"
>     Range("A1").NumberFormat = s
>     Debug.Print Range("A1").Text
> End Sub
> 
>   5419351/1725033
> 
> As you can see, the fraction format can get close(~14), but not quite...:>(
> -- 
> HTH.  :>)
> Dana DeLouis
> Windows XP, Office 2003
> 
> 
> "Dana DeLouis" <ddelouis@bellsouth.net> wrote in message 
> news:OvEepNqVGHA.4336@TK2MSFTNGP14.phx.gbl...
> > >... VBA procedure of finding fractions that will approximate pi to as 
> > >many decimal places as Excel will display,
> >
> > Hi.  At 15 digits, I believe the minimum fraction for Pi is:
> >
> > =80143857/25510582
> >
> > As a side note, the limit in vba is:
> > Num = 428224593349304#
> > Den = 136308121570117#
> >
> > Debug.Print CDec(Num) / Den
> > ' 3.1415926535897932384626433833
> >
> > -- 
> > HTH.  :>)
> > Dana DeLouis
> > Windows XP, Office 2003
> >
> >
> > "Spaz" <yeah@right.com> wrote in message 
> > news:svKdndamzu_E363ZRVn-tQ@comcast.com...
> >> How many decimal places can be displayed in a cell?  I'm running a brute 
> >> force VBA procedure of finding fractions that will approximate pi to as 
> >> many decimal places as Excel will display, but I don't know how many 
> >> decimal places Excel will display accurately.  Anybody know?  I guess 
> >> this is also a matter of how many decimal places VBA will calculate 
> >> accurately as well.
> >>
> >> Sub PiFractions()
> >> Dim dividend As Integer, divisor As Integer, quotient As Double
> >> Dim rowpointer As Byte
> >>
> >> rowpointer = 1
> >>
> >> For dividend = 22 To 10000
> >>  For divisor = 7 To dividend \ 3
> >>    quotient = dividend / divisor
> >>    If quotient > 3.14159 And quotient < 3.1416 Then
> >>      Cells(rowpointer, 1) = dividend
> >>      Cells(rowpointer, 2) = divisor
> >>      Cells(rowpointer, 3) = quotient
> >>      rowpointer = rowpointer + 1
> >>    End If
> >>  Next
> >> Next
> >>
> >> End Sub
> >>
> >
> > 
> 
> 
> 
0
TomOgilvy (26)
4/3/2006 12:37:04 PM
While better than xl97 (as Tom showed), formatting as a fraction is still not 
entirely reliable when you request many digits.  The DP (IEEE double 
precision) approximation to Pi is exactly
  884279719003555/281474976710656
which has a 15 digit denominator.  However, you get the same value as the DP 
approximation to
  245850922/78256779
which only has an 8 digit denominator.

Jerry

"Dana DeLouis" wrote:

> If you want to do a program loop, this is one of a few ways to get a jump 
> start...
> 
> Sub Demo()
>     Dim s As String
>     s = WorksheetFunction.Rept("?", 16)
>     s = s & "/" & s
> 
>     Range("A1").FormulaR1C1 = "=PI()"
>     Range("A1").NumberFormat = s
>     Debug.Print Range("A1").Text
> End Sub
> 
>   5419351/1725033
> 
> As you can see, the fraction format can get close(~14), but not quite...:>(
> -- 
> HTH.  :>)
> Dana DeLouis
> Windows XP, Office 2003
> 
> 
> "Dana DeLouis" <ddelouis@bellsouth.net> wrote in message 
> news:OvEepNqVGHA.4336@TK2MSFTNGP14.phx.gbl...
> > >... VBA procedure of finding fractions that will approximate pi to as 
> > >many decimal places as Excel will display,
> >
> > Hi.  At 15 digits, I believe the minimum fraction for Pi is:
> >
> > =80143857/25510582
> >
> > As a side note, the limit in vba is:
> > Num = 428224593349304#
> > Den = 136308121570117#
> >
> > Debug.Print CDec(Num) / Den
> > ' 3.1415926535897932384626433833
> >
> > -- 
> > HTH.  :>)
> > Dana DeLouis
> > Windows XP, Office 2003
> >
> >
> > "Spaz" <yeah@right.com> wrote in message 
> > news:svKdndamzu_E363ZRVn-tQ@comcast.com...
> >> How many decimal places can be displayed in a cell?  I'm running a brute 
> >> force VBA procedure of finding fractions that will approximate pi to as 
> >> many decimal places as Excel will display, but I don't know how many 
> >> decimal places Excel will display accurately.  Anybody know?  I guess 
> >> this is also a matter of how many decimal places VBA will calculate 
> >> accurately as well.
> >>
> >> Sub PiFractions()
> >> Dim dividend As Integer, divisor As Integer, quotient As Double
> >> Dim rowpointer As Byte
> >>
> >> rowpointer = 1
> >>
> >> For dividend = 22 To 10000
> >>  For divisor = 7 To dividend \ 3
> >>    quotient = dividend / divisor
> >>    If quotient > 3.14159 And quotient < 3.1416 Then
> >>      Cells(rowpointer, 1) = dividend
> >>      Cells(rowpointer, 2) = divisor
> >>      Cells(rowpointer, 3) = quotient
> >>      rowpointer = rowpointer + 1
> >>    End If
> >>  Next
> >> Next
> >>
> >> End Sub
> >>
> >
> > 
> 
> 
> 
0
post_a_reply (1395)
4/3/2006 7:33:01 PM
Reply:

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I have just upgraded from Money 2004 STD (Australia). All my account numbers in the 'accounts list' dialog box display mainly as 'xxxxx. My account numbers are alphnumeric. They are correct when I open the individual accounts. Does anyone have a work around or know of a fix for this bug? Rgds - Pete This is nothing more than stupid knee jerk pseudo-security. R-click and go to the account settings and it will look correct. "Peter" <anonymous@discussions.microsoft.com> wrote in message news:1b3e01c4ac1c$02961af0$a601280a@phx.gbl... >I have just upgraded ...

how do I get the if function to return a blank cell, not 0?
I am trying to create a chart from a series that contains data for each month. The series is calculated on other worksheets and copied to the worksheet containing the chart. I would like to have the cells for the months that have not been updated yet (now is January, there are 0' in all cells for Feb-Dec) to be blank (to create gaps in the chart) not 0's. Can this be done? Unfortunately, what you want, and what many of us have requested but doesn't exist, is a worksheet function like BLANK() or NULL(). The best we can do is use NA() in a chart's data source, which is...

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All of a sudden, I can't open jpg attachments to email in my Outlook 2000. I have reinstalled Outlook, but with no change. If you can solve this problem, I'll dance at your wedding. Harlan HH <hhhague@comcast.net> wrote: > All of a sudden, I can't open jpg attachments to email in my Outlook > 2000. What happens when you try? Use Windows Explorer's Tools>Folder Options menu to reestablish the association between JPEG files and whatever application you want to use to open them. -- Brian Tillman I have just discovered that I can look at attachment in a rou...

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I have a clicking sound which I can't stop, I have gone through all the System sounds from Control Panel and done the test for every sound and it isn't there. I've even tried turning off all sounds, but it still keeps on clicking! It's driving me nuts! effdee wrote: > I have a clicking sound which I can't stop, I have gone through all the > System sounds from Control Panel and done the test for every sound and > it isn't there. > > I've even tried turning off all sounds, but it still keeps on clicking! > > It's driving me ...

Can you set up a Word document to accept present tense only?
I was told that you could set a document up using Microsoft Word 2007 to only accept present tense. Is this true? If so, how do you do it? Word 2007 has a "Find all word forms" option in the Find and Replace dialog. This, among other things, will let it search for all verb forms (among other things). That might be what the person was talking about. For example, with this option enabled, if I search for "have", it will also match "having". If I search for "is", it will also match are, be, was, and were (for example). But, there's nothin...

work out how many trucks came in between the hours of 9am and 3pm
hi there l am after a formula which will calculate how many trucks came in from the hours of 9 am and 3 pm on any day. the main information is in sheet 1 and l want the value to return to sheet 2. How do l do this? and what formula is best to use. A:A time in time out transport coompany amount of pallets 1-Apr 4:00 5:26 LEOCATAS WJT982 BRIAN B 68 TATURA 656223 T15.02 68 1 1-Apr 4:06 5:45 LEOCATAS UMV991 DAVE B 68 TATURA 656224 T13.02 68 1 1-Apr 5:08 6:00 HUNTERS QZW243 RODNEY B 68 TATURA 656228 T17.17 68 1 1-Apr 5:50 6:25 LEOCATA...

How to slot cell values into pre-defined ranges
I would like to slot cell values starting from 500 upto 75000 in ranges (bins) like 500-999, 1000-1499, 1500to 1999 etc. How do i do this? I have 15 bins. pls help Not sure what you want here. Do you want to count the number of times values within a range occur? -- Ian -- "KDD" <KDD@discussions.microsoft.com> wrote in message news:766017CE-A55E-42FF-AD8D-9E74E48CD568@microsoft.com... >I would like to slot cell values starting from 500 upto 75000 in ranges > (bins) like 500-999, 1000-1499, 1500to 1999 etc. How do i do this? I have > 15 > bins. > > pls...