how can I create floating point?

[mineiro]

How can I create a floating point from a binary point of view? I'm talking in intergers point of view.
I have tried a lot, but some numbers needs more bits while others not. Why?
Why I cant represent all fractions into this sense?
I was reading some documents from doctors and academic, but this does not enter into my mind.
To be easy, how much is 1/??

[HSE]

Mineiro:

¿what is a point of view?

You want to represent fraccions with integers. ¿Are You well? 

[jimg]

Can you write 1/3 exactly as a decimal?   No, it is 1.333333333.....

Similarly, binary cannot represent all fractions.

[mineiro]

I'm fine sir HSE. I like to reproduce my question.
How much bits I need to represent a reason, division?
As an example: the number 4,5,6,7 in binary have 3 bits only. If i try 1/4,1/5, ... how many bits I need to represent that division? Why is not 3 bits plus one? Why some divisions need more than N bits?
This sounds easy in decimal, but lets take pi=3.1415. If pi ends like that instead of infinite, I know that I just need translate that decimal to binary which I know how to do, but how to reach exactly instead of aproaching number?

[mineiro]

Can you write 1/3 exactly as a decimal?   No, it is 1.333333333.....

Similarly, binary cannot represent all fractions.

Yes, 0.333... something like this. But how many numbers 3 is necessary? I know is infinite, but to precision of 1/3 we dont need too much because number 3 fits in 2 bits.

[HSE]

I'm fine sir HSE.

Fantastic 

If you are creating a system of representation the rules are what you want.
If you define that those bits means "inverse" 1/7 requiere 3 bits, but you can't represent 1/8.

[jj2007]

If i try 1/4,1/5, ... how many bits I need to represent that division? Why is not 3 bits plus one?

You can represent a division like 3/7 with 8 bits:

Code Select Expand

include \masm32\include\masm32rt.inc
.code
ThreeBySeven db 7*16+3
SevenByThree db 3*16+7
start:
  movzx eax, ThreeBySeven
  mov edx, eax
  and eax, 7
  sar edx, 4
  push eax
  fild dword ptr [esp]
  push edx
  fild dword ptr [esp]
  fdiv
  fstp REAL8 ptr [esp]
  invoke crt_printf, chr$("3/7=%0.18f", 13, 10), REAL8 ptr [esp]

  movzx eax, SevenByThree
  mov edx, eax
  and eax, 7
  sar edx, 4
  push eax
  fild dword ptr [esp]
  push edx
  fild dword ptr [esp]
  fdiv
  fstp REAL8 ptr [esp]
  invoke crt_printf, chr$("7/3=%0.18f", 13, 10), REAL8 ptr [esp]
  inkey
  exit
end start

The possible range is 1/7 ... 7/1. If you allow 16 bits, you can even use negative numbers, as in -66/77 - see attached source:

Code Select Expand

3/7=    0.428571283684245130
7/3=    7.000000955304130900
-66/77= -0.86842095837193090

[mineiro]

The number 4 and 5 in binary is 100 and 101, so they have a size of 3 bits.

1/4 = 0.25
1/5 = 0.2

transforming that decimal floating point to binary floating point.
1/4 = 0.25   ==>   0.5   1.0   ==>0.01
1/5 = 0.2   ==>   0.4   0.8   1.6   1.2   0.4   ...   ==>0.00110011...

1/4 need 2 binary digits in fractional part to be represented and result reach remainder zero, or end.
1/5 need 4 binary digits in fractional part to be represented until it continues repeating and result never reach remainder zero.

The question is: Why? Why 1/4 give as result 2 bits and 1/5 give as result 4 bits.
How can I know how many binary digits is necessary to represent a fraction until result is zero or a periodic tithe?
Why 1/4 fits in itself and 1/5 do not fit in itself?

So, if I have in decimal the fraction 100/123, how many binary digits is necessary to represent that fraction?

------while I was writing you posted sir jj2007. But without using floating point instructions, only interger instructions, well, interger math.

You can represent a division like 3/7 with 8 bits:

Yes, but why you know that the result fit in 8 bits? Whats the minimum size?
Thanks.

[jj2007]

Whats the minimum size?

It's different for any number. Try using this in my example, and put an int 3 between fdiv and fstp. Then check the number in ST(0).

HundredBy123   dw 100+123*256

[mineiro]

Thank you for your patience but I continue without understanding.

How much bits I need to represent 1/N, where N is a interger random number with sizeof 3 bits?
So can be 1/1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7.

Minimum size is 1 bit, because 1/1 = 1. So, how many bits is necessary to hold the result of that division where N is random?

[jj2007]

See below. What is the practical application of your question?

Code Select Expand

1 1/1
2 1/2
2 1/3
3 1/4
3 1/5
3 1/6
3 1/7
4 1/8
4 1/9
4 1/10
4 1/11
4 1/12
4 1/13
4 1/14
4 1/15

[mineiro]

I'm not able to represent that as 1/4 as an example, I need represent that as binary floating point number. That's why I talked about pi.
Not all numbers that I'm dealing are just 1/N, they can be X/Y (really huge numbers), so using pi as example, I can have 3 as interger part and that infinite as remainder.
Well, ok, inverting scenario, whats 1/pi? Assume that pi = 3,1415. How many binary digits is necessary to represent this arithmetic?

The pratical application of this is to me conquer the world.
Again, thank you.

[jj2007]

1/PI=0.31830988618379067153776752674503... and you need an infinite number of bits to represent that number.

[JonasS]

PI is not a good example, it is not an algebraic number, can not be put in fractional form.

[jj2007]

Right, Jonas. But 3/7 is a good example, and it can be exactly "represented" with 8 bits=1 byte, as shown above. The problem is what to do with the result... using the FPU with max 19 digits of precision?