Apparently 5.2 MOD 4.6 = 0 but why? [resolved]

[Matt_T_hat]

can anyone tell me why 5.2 MOD 4.6 = 0 is returned by access accross some 55 PCs all over the UK?

[si_the_geek]

Mod is for Integer arithmetic.. it doesnt work with floating point numbers

an aternative method is:

5.2-(4.6*(5.2\4.6)) = 0.600000000000001

[Matt_T_hat]

Originally posted by si_the_geek
an aternative method is:

5.2-(4.6*(5.2\4.6)) = 0.600000000000001

So if I understand correctly

Hmm...

I would code something...

I understand the reason for the MOD not working now

How does... never mind

[Matt_T_hat]

Originally posted by si_the_geek
5.2-(4.6*(5.2\4.6)) = 0.600000000000001

no it doesn't

[si_the_geek]

no it doesn't

yes it does... notice \ rather than /
( \ returns an integer only, which is what we want )

so:
5.2-(4.6*(5.2\4.6))
=> 5.2-(4.6*(1))
=> 5.2-(4.6)
=> 0.6 (usual floating point problem gives the extra dp's)

How does your alternative work if the numbers A and B are unknown and you want to find if A is and exact multiple of B?

if you mean the equivalent of:
If (A MOD B = 0) Then ...

that is:
If (A-(B*(A\B)) = 0) Then ...


this Mod in a function:

Function MyMod(NumberA, NumberB)
  MyMod = NumberA-(NumberB*(NumberA\NumberB))  
End Function
 
'usage:
If MyMod(A,B) = 0 Then ...

admittedly you'll want to cut off the extra decimal places, but you get the idea

[Matt_T_hat]

Originally posted by si_the_geek
admittedly you'll want to cut off the extra decimal places, but you get the idea

I do indeed get the idea... I was not aware / and \ functioned differently in VB in fact I was unaware of '\' as a math function.

One last Q if I may "...cut off the extra decimal places..." ?

not that it matters... thankyou I tried you code snippit and wow - like a dream! Man that's cool.

You rock.

[si_the_geek]

One last Q if I may "...cut off the extra decimal places..." ?

that returns 0.600000000000001 rather than 0.6 as it theoretically should (due to the problems of storing floating point numbers on a binary machine).

I think you need to use Val or Format to get it right...

just found this thread: http://www.vbforums.com/showthread.p...hreadid=193491

and it seems CSng() works!

so you should use:

MyMod = CSng(NumberA-(NumberB*(NumberA\NumberB)))

[DiGiTaIErRoR]

Alternatively, you could multiply the values by 10^x, then divide the result by 10^x.

[Matt_T_hat]

Originally posted by DiGiTaIErRoR
Alternatively, you could multiply the values by 10^x, then divide the result by 10^x.

Yes I understand there is a general Access bug where by you should do n=(n/100)*100 (I am told) the good news is that si_the_geek's method did the job but now we have DiGiTaIErRoR's as well it seems like there truely is more than one way to skin a cat...

[evolver]

Originally posted by Matt_T_hat
can anyone tell me why 5.2 MOD 4.6 = 0 is returned by access accross some 55 PCs all over the UK?

I wipped out my thrustworthy windows calculator
5.2 mod 4.6 = 0.6

What is mod?
It is the remainder of // after division:
8 mod 4 yields 0 because 2 * 4 = 8 remain = 0
8.2 mod 4 yields 0.2 because 2 * 4 = 8 remain = 0.2

[si_the_geek]

Originally posted by evolver
What is mod?
It is the remainder of // after division

yep

[evolver]

Thank you!
No problemo

[RAEsquivelC]

And, calling up Excel, and
entering 5.2 in A1, and
entering 4.6 in B1, and using the equation
=MOD(A1,B1) in C1, I get 0.6 in C1. So, Excel DOES know how to work out this kind of arithmatic!

[evolver]

Originally posted by RAEsquivelC
And, calling up Excel, and
entering 5.2 in A1, and
entering 4.6 in B1, and using the equation
=MOD(A1,B1) in C1, I get 0.6 in C1. So, Excel DOES know how to work out this kind of arithmatic!

I sincerely hope so

[RAEsquivelC]

Well, Access aparently does NOT!

[evolver]

Very strange indeed...

5.5 Mod 4.2 yields 0