Related
Here is the subtraction
First number
Decimal 3.0000002
Hexadecimal 0x4040001
Binary: Sign[0], Exponent[1000_0000], Mantissa[100_0000_0000_0000_0000_0001]
substract second number:
Decimal 3.000000
Hexadecimal 0x4040000
Binary: Sign[0], Exponent[1000_0000], Mantissa[100_0000_0000_0000_0000_0000]
==========================================
At this situation, the exponent is already same, we just need to substract the mantissa. We know in IEEE754, there is a hiding bit 1 in front of mantissa. Therefore, the result mantissa should be:
Mantissa_1[1100_0000_0000_0000_0000_0001] - Mantissa_2[1100_0000_0000_0000_0000_0000]
which equal to
Mantissa_Rst = [0000_0000_0000_0000_0000_0001]
But this number is not normalized, Because of the first hiding bit is not 1. Thus we shift the Mantissa_Rst right 23 times, and the exponent minuses 23 at the same time.
Then we have the result value
Hexadecimal 0x4040000
Binary: Sign[0], Exponent[0110_1000], Mantissa[000_0000_0000_0000_0000_0000].
32 bits total, no rounding needed.
Notice that in the mantissa region, there still is a hidden 1.
If my calculations were correct, then converting result to decimal number is 0.00000023841858, comparing with the real result 0.0000002, I still think that is not very precise.
So the question is, are my calculations wrong? or actually this is a real situation and happens all the time in computer?
The inaccuracy already starts with your input. 3.0000002 is a fraction with a prime factor of five in the denominator, so its "decimal" expansion in base 2 is periodic. No amount of mantissa bits will suffice to represent it exactly. The float you give actually has the value 3.0000002384185791015625 (this is exact). Yes, this happens all the time.
Don't despair, though! Base ten has the same problem (for example 1/3). It isn't a problem. Well, it is for some people, but luckily there are other number types available for their needs. Floating point numbers have many advantages, and slight rounding error is irrelevant for many applications, for example when not even your inputs are perfectly accurate measurements of what you're interested in (a lot of scientific computing and simulation). Also remember that 64-bit floats also exist. Additionally, the error is bounded: With the best possible rounding, your result will be within 0.5 units in the last place removed from the infinite-precision result. For a 32-bit float of the magnitude as your example, this is approximately 2^-25, or 3 * 10^-8. This gets worse and worse as you do additional operations that have to round, but with careful numeric analysis and the right algorithms, you can get a lot of milage out of them.
Whenever x/2 ≤ y ≤ 2x, the calculation x - y is exact which means there is no rounding error whatsoever. That is also the case in your example.
You just made the wrong assumption that you could have a floating point number that is equal to 3.0000002. You can't. The type "float" can only ever represent integers less than 2^24, multiplied by a power of two. 3.0000002 is not such a number, therefore it is rounded to the nearest floating point number, which is closer to 3.00000023841858. Subtracting 3 calculates the difference exactly and gives a result close to 0.00000023841858.
Why do some numbers lose accuracy when stored as floating point numbers?
For example, the decimal number 9.2 can be expressed exactly as a ratio of two decimal integers (92/10), both of which can be expressed exactly in binary (0b1011100/0b1010). However, the same ratio stored as a floating point number is never exactly equal to 9.2:
32-bit "single precision" float: 9.19999980926513671875
64-bit "double precision" float: 9.199999999999999289457264239899814128875732421875
How can such an apparently simple number be "too big" to express in 64 bits of memory?
In most programming languages, floating point numbers are represented a lot like scientific notation: with an exponent and a mantissa (also called the significand). A very simple number, say 9.2, is actually this fraction:
5179139571476070 * 2 -49
Where the exponent is -49 and the mantissa is 5179139571476070. The reason it is impossible to represent some decimal numbers this way is that both the exponent and the mantissa must be integers. In other words, all floats must be an integer multiplied by an integer power of 2.
9.2 may be simply 92/10, but 10 cannot be expressed as 2n if n is limited to integer values.
Seeing the Data
First, a few functions to see the components that make a 32- and 64-bit float. Gloss over these if you only care about the output (example in Python):
def float_to_bin_parts(number, bits=64):
if bits == 32: # single precision
int_pack = 'I'
float_pack = 'f'
exponent_bits = 8
mantissa_bits = 23
exponent_bias = 127
elif bits == 64: # double precision. all python floats are this
int_pack = 'Q'
float_pack = 'd'
exponent_bits = 11
mantissa_bits = 52
exponent_bias = 1023
else:
raise ValueError, 'bits argument must be 32 or 64'
bin_iter = iter(bin(struct.unpack(int_pack, struct.pack(float_pack, number))[0])[2:].rjust(bits, '0'))
return [''.join(islice(bin_iter, x)) for x in (1, exponent_bits, mantissa_bits)]
There's a lot of complexity behind that function, and it'd be quite the tangent to explain, but if you're interested, the important resource for our purposes is the struct module.
Python's float is a 64-bit, double-precision number. In other languages such as C, C++, Java and C#, double-precision has a separate type double, which is often implemented as 64 bits.
When we call that function with our example, 9.2, here's what we get:
>>> float_to_bin_parts(9.2)
['0', '10000000010', '0010011001100110011001100110011001100110011001100110']
Interpreting the Data
You'll see I've split the return value into three components. These components are:
Sign
Exponent
Mantissa (also called Significand, or Fraction)
Sign
The sign is stored in the first component as a single bit. It's easy to explain: 0 means the float is a positive number; 1 means it's negative. Because 9.2 is positive, our sign value is 0.
Exponent
The exponent is stored in the middle component as 11 bits. In our case, 0b10000000010. In decimal, that represents the value 1026. A quirk of this component is that you must subtract a number equal to 2(# of bits) - 1 - 1 to get the true exponent; in our case, that means subtracting 0b1111111111 (decimal number 1023) to get the true exponent, 0b00000000011 (decimal number 3).
Mantissa
The mantissa is stored in the third component as 52 bits. However, there's a quirk to this component as well. To understand this quirk, consider a number in scientific notation, like this:
6.0221413x1023
The mantissa would be the 6.0221413. Recall that the mantissa in scientific notation always begins with a single non-zero digit. The same holds true for binary, except that binary only has two digits: 0 and 1. So the binary mantissa always starts with 1! When a float is stored, the 1 at the front of the binary mantissa is omitted to save space; we have to place it back at the front of our third element to get the true mantissa:
1.0010011001100110011001100110011001100110011001100110
This involves more than just a simple addition, because the bits stored in our third component actually represent the fractional part of the mantissa, to the right of the radix point.
When dealing with decimal numbers, we "move the decimal point" by multiplying or dividing by powers of 10. In binary, we can do the same thing by multiplying or dividing by powers of 2. Since our third element has 52 bits, we divide it by 252 to move it 52 places to the right:
0.0010011001100110011001100110011001100110011001100110
In decimal notation, that's the same as dividing 675539944105574 by 4503599627370496 to get 0.1499999999999999. (This is one example of a ratio that can be expressed exactly in binary, but only approximately in decimal; for more detail, see: 675539944105574 / 4503599627370496.)
Now that we've transformed the third component into a fractional number, adding 1 gives the true mantissa.
Recapping the Components
Sign (first component): 0 for positive, 1 for negative
Exponent (middle component): Subtract 2(# of bits) - 1 - 1 to get the true exponent
Mantissa (last component): Divide by 2(# of bits) and add 1 to get the true mantissa
Calculating the Number
Putting all three parts together, we're given this binary number:
1.0010011001100110011001100110011001100110011001100110 x 1011
Which we can then convert from binary to decimal:
1.1499999999999999 x 23 (inexact!)
And multiply to reveal the final representation of the number we started with (9.2) after being stored as a floating point value:
9.1999999999999993
Representing as a Fraction
9.2
Now that we've built the number, it's possible to reconstruct it into a simple fraction:
1.0010011001100110011001100110011001100110011001100110 x 1011
Shift mantissa to a whole number:
10010011001100110011001100110011001100110011001100110 x 1011-110100
Convert to decimal:
5179139571476070 x 23-52
Subtract the exponent:
5179139571476070 x 2-49
Turn negative exponent into division:
5179139571476070 / 249
Multiply exponent:
5179139571476070 / 562949953421312
Which equals:
9.1999999999999993
9.5
>>> float_to_bin_parts(9.5)
['0', '10000000010', '0011000000000000000000000000000000000000000000000000']
Already you can see the mantissa is only 4 digits followed by a whole lot of zeroes. But let's go through the paces.
Assemble the binary scientific notation:
1.0011 x 1011
Shift the decimal point:
10011 x 1011-100
Subtract the exponent:
10011 x 10-1
Binary to decimal:
19 x 2-1
Negative exponent to division:
19 / 21
Multiply exponent:
19 / 2
Equals:
9.5
Further reading
The Floating-Point Guide: What Every Programmer Should Know About Floating-Point Arithmetic, or, Why don’t my numbers add up? (floating-point-gui.de)
What Every Computer Scientist Should Know About Floating-Point Arithmetic (Goldberg 1991)
IEEE Double-precision floating-point format (Wikipedia)
Floating Point Arithmetic: Issues and Limitations (docs.python.org)
Floating Point Binary
This isn't a full answer (mhlester already covered a lot of good ground I won't duplicate), but I would like to stress how much the representation of a number depends on the base you are working in.
Consider the fraction 2/3
In good-ol' base 10, we typically write it out as something like
0.666...
0.666
0.667
When we look at those representations, we tend to associate each of them with the fraction 2/3, even though only the first representation is mathematically equal to the fraction. The second and third representations/approximations have an error on the order of 0.001, which is actually much worse than the error between 9.2 and 9.1999999999999993. In fact, the second representation isn't even rounded correctly! Nevertheless, we don't have a problem with 0.666 as an approximation of the number 2/3, so we shouldn't really have a problem with how 9.2 is approximated in most programs. (Yes, in some programs it matters.)
Number bases
So here's where number bases are crucial. If we were trying to represent 2/3 in base 3, then
(2/3)10 = 0.23
In other words, we have an exact, finite representation for the same number by switching bases! The take-away is that even though you can convert any number to any base, all rational numbers have exact finite representations in some bases but not in others.
To drive this point home, let's look at 1/2. It might surprise you that even though this perfectly simple number has an exact representation in base 10 and 2, it requires a repeating representation in base 3.
(1/2)10 = 0.510 = 0.12 = 0.1111...3
Why are floating point numbers inaccurate?
Because often-times, they are approximating rationals that cannot be represented finitely in base 2 (the digits repeat), and in general they are approximating real (possibly irrational) numbers which may not be representable in finitely many digits in any base.
While all of the other answers are good there is still one thing missing:
It is impossible to represent irrational numbers (e.g. π, sqrt(2), log(3), etc.) precisely!
And that actually is why they are called irrational. No amount of bit storage in the world would be enough to hold even one of them. Only symbolic arithmetic is able to preserve their precision.
Although if you would limit your math needs to rational numbers only the problem of precision becomes manageable. You would need to store a pair of (possibly very big) integers a and b to hold the number represented by the fraction a/b. All your arithmetic would have to be done on fractions just like in highschool math (e.g. a/b * c/d = ac/bd).
But of course you would still run into the same kind of trouble when pi, sqrt, log, sin, etc. are involved.
TL;DR
For hardware accelerated arithmetic only a limited amount of rational numbers can be represented. Every not-representable number is approximated. Some numbers (i.e. irrational) can never be represented no matter the system.
There are infinitely many real numbers (so many that you can't enumerate them), and there are infinitely many rational numbers (it is possible to enumerate them).
The floating-point representation is a finite one (like anything in a computer) so unavoidably many many many numbers are impossible to represent. In particular, 64 bits only allow you to distinguish among only 18,446,744,073,709,551,616 different values (which is nothing compared to infinity). With the standard convention, 9.2 is not one of them. Those that can are of the form m.2^e for some integers m and e.
You might come up with a different numeration system, 10 based for instance, where 9.2 would have an exact representation. But other numbers, say 1/3, would still be impossible to represent.
Also note that double-precision floating-points numbers are extremely accurate. They can represent any number in a very wide range with as much as 15 exact digits. For daily life computations, 4 or 5 digits are more than enough. You will never really need those 15, unless you want to count every millisecond of your lifetime.
Why can we not represent 9.2 in binary floating point?
Floating point numbers are (simplifying slightly) a positional numbering system with a restricted number of digits and a movable radix point.
A fraction can only be expressed exactly using a finite number of digits in a positional numbering system if the prime factors of the denominator (when the fraction is expressed in it's lowest terms) are factors of the base.
The prime factors of 10 are 5 and 2, so in base 10 we can represent any fraction of the form a/(2b5c).
On the other hand the only prime factor of 2 is 2, so in base 2 we can only represent fractions of the form a/(2b)
Why do computers use this representation?
Because it's a simple format to work with and it is sufficiently accurate for most purposes. Basically the same reason scientists use "scientific notation" and round their results to a reasonable number of digits at each step.
It would certainly be possible to define a fraction format, with (for example) a 32-bit numerator and a 32-bit denominator. It would be able to represent numbers that IEEE double precision floating point could not, but equally there would be many numbers that can be represented in double precision floating point that could not be represented in such a fixed-size fraction format.
However the big problem is that such a format is a pain to do calculations on. For two reasons.
If you want to have exactly one representation of each number then after each calculation you need to reduce the fraction to it's lowest terms. That means that for every operation you basically need to do a greatest common divisor calculation.
If after your calculation you end up with an unrepresentable result because the numerator or denominator you need to find the closest representable result. This is non-trivil.
Some Languages do offer fraction types, but usually they do it in combination with arbitary precision, this avoids needing to worry about approximating fractions but it creates it's own problem, when a number passes through a large number of calculation steps the size of the denominator and hence the storage needed for the fraction can explode.
Some languages also offer decimal floating point types, these are mainly used in scenarios where it is imporant that the results the computer gets match pre-existing rounding rules that were written with humans in mind (chiefly financial calculations). These are slightly more difficult to work with than binary floating point, but the biggest problem is that most computers don't offer hardware support for them.
There have been several questions posted to SO about floating-point representation. For example, the decimal number 0.1 doesn't have an exact binary representation, so it's dangerous to use the == operator to compare it to another floating-point number. I understand the principles behind floating-point representation.
What I don't understand is why, from a mathematical perspective, are the numbers to the right of the decimal point any more "special" that the ones to the left?
For example, the number 61.0 has an exact binary representation because the integral portion of any number is always exact. But the number 6.10 is not exact. All I did was move the decimal one place and suddenly I've gone from Exactopia to Inexactville. Mathematically, there should be no intrinsic difference between the two numbers -- they're just numbers.
By contrast, if I move the decimal one place in the other direction to produce the number 610, I'm still in Exactopia. I can keep going in that direction (6100, 610000000, 610000000000000) and they're still exact, exact, exact. But as soon as the decimal crosses some threshold, the numbers are no longer exact.
What's going on?
Edit: to clarify, I want to stay away from discussion about industry-standard representations, such as IEEE, and stick with what I believe is the mathematically "pure" way. In base 10, the positional values are:
... 1000 100 10 1 1/10 1/100 ...
In binary, they would be:
... 8 4 2 1 1/2 1/4 1/8 ...
There are also no arbitrary limits placed on these numbers. The positions increase indefinitely to the left and to the right.
Decimal numbers can be represented exactly, if you have enough space - just not by floating binary point numbers. If you use a floating decimal point type (e.g. System.Decimal in .NET) then plenty of values which can't be represented exactly in binary floating point can be exactly represented.
Let's look at it another way - in base 10 which you're likely to be comfortable with, you can't express 1/3 exactly. It's 0.3333333... (recurring). The reason you can't represent 0.1 as a binary floating point number is for exactly the same reason. You can represent 3, and 9, and 27 exactly - but not 1/3, 1/9 or 1/27.
The problem is that 3 is a prime number which isn't a factor of 10. That's not an issue when you want to multiply a number by 3: you can always multiply by an integer without running into problems. But when you divide by a number which is prime and isn't a factor of your base, you can run into trouble (and will do so if you try to divide 1 by that number).
Although 0.1 is usually used as the simplest example of an exact decimal number which can't be represented exactly in binary floating point, arguably 0.2 is a simpler example as it's 1/5 - and 5 is the prime that causes problems between decimal and binary.
Side note to deal with the problem of finite representations:
Some floating decimal point types have a fixed size like System.Decimal others like java.math.BigDecimal are "arbitrarily large" - but they'll hit a limit at some point, whether it's system memory or the theoretical maximum size of an array. This is an entirely separate point to the main one of this answer, however. Even if you had a genuinely arbitrarily large number of bits to play with, you still couldn't represent decimal 0.1 exactly in a floating binary point representation. Compare that with the other way round: given an arbitrary number of decimal digits, you can exactly represent any number which is exactly representable as a floating binary point.
For example, the number 61.0 has an exact binary representation because the integral portion of any number is always exact. But the number 6.10 is not exact. All I did was move the decimal one place and suddenly I've gone from Exactopia to Inexactville. Mathematically, there should be no intrinsic difference between the two numbers -- they're just numbers.
Let's step away for a moment from the particulars of bases 10 and 2. Let's ask - in base b, what numbers have terminating representations, and what numbers don't? A moment's thought tells us that a number x has a terminating b-representation if and only if there exists an integer n such that x b^n is an integer.
So, for example, x = 11/500 has a terminating 10-representation, because we can pick n = 3 and then x b^n = 22, an integer. However x = 1/3 does not, because whatever n we pick we will not be able to get rid of the 3.
This second example prompts us to think about factors, and we can see that for any rational x = p/q (assumed to be in lowest terms), we can answer the question by comparing the prime factorisations of b and q. If q has any prime factors not in the prime factorisation of b, we will never be able to find a suitable n to get rid of these factors.
Thus for base 10, any p/q where q has prime factors other than 2 or 5 will not have a terminating representation.
So now going back to bases 10 and 2, we see that any rational with a terminating 10-representation will be of the form p/q exactly when q has only 2s and 5s in its prime factorisation; and that same number will have a terminating 2-representatiion exactly when q has only 2s in its prime factorisation.
But one of these cases is a subset of the other! Whenever
q has only 2s in its prime factorisation
it obviously is also true that
q has only 2s and 5s in its prime factorisation
or, put another way, whenever p/q has a terminating 2-representation, p/q has a terminating 10-representation. The converse however does not hold - whenever q has a 5 in its prime factorisation, it will have a terminating 10-representation , but not a terminating 2-representation. This is the 0.1 example mentioned by other answers.
So there we have the answer to your question - because the prime factors of 2 are a subset of the prime factors of 10, all 2-terminating numbers are 10-terminating numbers, but not vice versa. It's not about 61 versus 6.1 - it's about 10 versus 2.
As a closing note, if by some quirk people used (say) base 17 but our computers used base 5, your intuition would never have been led astray by this - there would be no (non-zero, non-integer) numbers which terminated in both cases!
The root (mathematical) reason is that when you are dealing with integers, they are countably infinite.
Which means, even though there are an infinite amount of them, we could "count out" all of the items in the sequence, without skipping any. That means if we want to get the item in the 610000000000000th position in the list, we can figure it out via a formula.
However, real numbers are uncountably infinite. You can't say "give me the real number at position 610000000000000" and get back an answer. The reason is because, even between 0 and 1, there are an infinite number of values, when you are considering floating-point values. The same holds true for any two floating point numbers.
More info:
http://en.wikipedia.org/wiki/Countable_set
http://en.wikipedia.org/wiki/Uncountable_set
Update:
My apologies, I appear to have misinterpreted the question. My response is about why we cannot represent every real value, I hadn't realized that floating point was automatically classified as rational.
To repeat what I said in my comment to Mr. Skeet: we can represent 1/3, 1/9, 1/27, or any rational in decimal notation. We do it by adding an extra symbol. For example, a line over the digits that repeat in the decimal expansion of the number. What we need to represent decimal numbers as a sequence of binary numbers are 1) a sequence of binary numbers, 2) a radix point, and 3) some other symbol to indicate the repeating part of the sequence.
Hehner's quote notation is a way of doing this. He uses a quote symbol to represent the repeating part of the sequence. The article: http://www.cs.toronto.edu/~hehner/ratno.pdf and the Wikipedia entry: http://en.wikipedia.org/wiki/Quote_notation.
There's nothing that says we can't add a symbol to our representation system, so we can represent decimal rationals exactly using binary quote notation, and vice versa.
BCD - Binary-coded Decimal - representations are exact. They are not very space-efficient, but that's a trade-off you have to make for accuracy in this case.
This is a good question.
All your question is based on "how do we represent a number?"
ALL the numbers can be represented with decimal representation or with binary (2's complement) representation. All of them !!
BUT some (most of them) require infinite number of elements ("0" or "1" for the binary position, or "0", "1" to "9" for the decimal representation).
Like 1/3 in decimal representation (1/3 = 0.3333333... <- with an infinite number of "3")
Like 0.1 in binary ( 0.1 = 0.00011001100110011.... <- with an infinite number of "0011")
Everything is in that concept. Since your computer can only consider finite set of digits (decimal or binary), only some numbers can be exactly represented in your computer...
And as said Jon, 3 is a prime number which isn't a factor of 10, so 1/3 cannot be represented with a finite number of elements in base 10.
Even with arithmetic with arbitrary precision, the numbering position system in base 2 is not able to fully describe 6.1, although it can represent 61.
For 6.1, we must use another representation (like decimal representation, or IEEE 854 that allows base 2 or base 10 for the representation of floating-point values)
If you make a big enough number with floating point (as it can do exponents), then you'll end up with inexactness in front of the decimal point, too. So I don't think your question is entirely valid because the premise is wrong; it's not the case that shifting by 10 will always create more precision, because at some point the floating point number will have to use exponents to represent the largeness of the number and will lose some precision that way as well.
It's the same reason you cannot represent 1/3 exactly in base 10, you need to say 0.33333(3). In binary it is the same type of problem but just occurs for different set of numbers.
(Note: I'll append 'b' to indicate binary numbers here. All other numbers are given in decimal)
One way to think about things is in terms of something like scientific notation. We're used to seeing numbers expressed in scientific notation like, 6.022141 * 10^23. Floating point numbers are stored internally using a similar format - mantissa and exponent, but using powers of two instead of ten.
Your 61.0 could be rewritten as 1.90625 * 2^5, or 1.11101b * 2^101b with the mantissa and exponents. To multiply that by ten and (move the decimal point), we can do:
(1.90625 * 2^5) * (1.25 * 2^3) = (2.3828125 * 2^8) = (1.19140625 * 2^9)
or in with the mantissa and exponents in binary:
(1.11101b * 2^101b) * (1.01b * 2^11b) = (10.0110001b * 2^1000b) = (1.00110001b * 2^1001b)
Note what we did there to multiply the numbers. We multiplied the mantissas and added the exponents. Then, since the mantissa ended greater than two, we normalized the result by bumping the exponent. It's just like when we adjust the exponent after doing an operation on numbers in decimal scientific notation. In each case, the values that we worked with had a finite representation in binary, and so the values output by the basic multiplication and addition operations also produced values with a finite representation.
Now, consider how we'd divide 61 by 10. We'd start by dividing the mantissas, 1.90625 and 1.25. In decimal, this gives 1.525, a nice short number. But what is this if we convert it to binary? We'll do it the usual way -- subtracting out the largest power of two whenever possible, just like converting integer decimals to binary, but we'll use negative powers of two:
1.525 - 1*2^0 --> 1
0.525 - 1*2^-1 --> 1
0.025 - 0*2^-2 --> 0
0.025 - 0*2^-3 --> 0
0.025 - 0*2^-4 --> 0
0.025 - 0*2^-5 --> 0
0.025 - 1*2^-6 --> 1
0.009375 - 1*2^-7 --> 1
0.0015625 - 0*2^-8 --> 0
0.0015625 - 0*2^-9 --> 0
0.0015625 - 1*2^-10 --> 1
0.0005859375 - 1*2^-11 --> 1
0.00009765625...
Uh oh. Now we're in trouble. It turns out that 1.90625 / 1.25 = 1.525, is a repeating fraction when expressed in binary: 1.11101b / 1.01b = 1.10000110011...b Our machines only have so many bits to hold that mantissa and so they'll just round the fraction and assume zeroes beyond a certain point. The error you see when you divide 61 by 10 is the difference between:
1.100001100110011001100110011001100110011...b * 2^10b
and, say:
1.100001100110011001100110b * 2^10b
It's this rounding of the mantissa that leads to the loss of precision that we associate with floating point values. Even when the mantissa can be expressed exactly (e.g., when just adding two numbers), we can still get numeric loss if the mantissa needs too many digits to fit after normalizing the exponent.
We actually do this sort of thing all the time when we round decimal numbers to a manageable size and just give the first few digits of it. Because we express the result in decimal it feels natural. But if we rounded a decimal and then converted it to a different base, it'd look just as ugly as the decimals we get due to floating point rounding.
I'm surprised no one has stated this yet: use continued fractions. Any rational number can be represented finitely in binary this way.
Some examples:
1/3 (0.3333...)
0; 3
5/9 (0.5555...)
0; 1, 1, 4
10/43 (0.232558139534883720930...)
0; 4, 3, 3
9093/18478 (0.49209871198181621387596060179673...)
0; 2, 31, 7, 8, 5
From here, there are a variety of known ways to store a sequence of integers in memory.
In addition to storing your number with perfect accuracy, continued fractions also have some other benefits, such as best rational approximation. If you decide to terminate the sequence of numbers in a continued fraction early, the remaining digits (when recombined to a fraction) will give you the best possible fraction. This is how approximations to pi are found:
Pi's continued fraction:
3; 7, 15, 1, 292 ...
Terminating the sequence at 1, this gives the fraction:
355/113
which is an excellent rational approximation.
In the equation
2^x = y ;
x = log(y) / log(2)
Hence, I was just wondering if we could have a logarithmic base system for binary like,
2^1, 2^0, 2^(log(1/2) / log(2)), 2^(log(1/4) / log(2)), 2^(log(1/8) / log(2)),2^(log(1/16) / log(2)) ........
That might be able to solve the problem, so if you wanted to write something like 32.41 in binary, that would be
2^5 + 2^(log(0.4) / log(2)) + 2^(log(0.01) / log(2))
Or
2^5 + 2^(log(0.41) / log(2))
The problem is that you do not really know whether the number actually is exactly 61.0 . Consider this:
float a = 60;
float b = 0.1;
float c = a + b * 10;
What is the value of c? It is not exactly 61, because b is not really .1 because .1 does not have an exact binary representation.
The number 61.0 does indeed have an exact floating-point operation—but that's not true for all integers. If you wrote a loop that added one to both a double-precision floating point number and a 64-bit integer, eventually you'd reach a point where the 64-bit integer perfectly represents a number, but the floating point doesn't—because there aren't enough significant bits.
It's just much easier to reach the point of approximation on the right side of the decimal point. If you started writing out all the numbers in binary floating point, it'd make more sense.
Another way of thinking about it is that when you note that 61.0 is perfectly representable in base 10, and shifting the decimal point around doesn't change that, you're performing multiplication by powers of ten (10^1, 10^-1). In floating point, multiplying by powers of two does not affect the precision of the number. Try taking 61.0 and dividing it by three repeatedly for an illustration of how a perfectly precise number can lose its precise representation.
There's a threshold because the meaning of the digit has gone from integer to non-integer. To represent 61, you have 6*10^1 + 1*10^0; 10^1 and 10^0 are both integers. 6.1 is 6*10^0 + 1*10^-1, but 10^-1 is 1/10, which is definitely not an integer. That's how you end up in Inexactville.
A parallel can be made of fractions and whole numbers. Some fractions eg 1/7 cannot be represented in decimal form without lots and lots of decimals. Because floating point is binary based the special cases change but the same sort of accuracy problems present themselves.
There are an infinite number of rational numbers, and a finite number of bits with which to represent them. See http://en.wikipedia.org/wiki/Floating_point#Accuracy_problems.
you know integer numbers right? each bit represent 2^n
2^4=16
2^3=8
2^2=4
2^1=2
2^0=1
well its the same for floating point(with some distinctions) but the bits represent 2^-n
2^-1=1/2=0.5
2^-2=1/(2*2)=0.25
2^-3=0.125
2^-4=0.0625
Floating point binary representation:
sign Exponent Fraction(i think invisible 1 is appended to the fraction )
B11 B10 B9 B8 B7 B6 B5 B4 B3 B2 B1 B0
The high scoring answer above nailed it.
First you were mixing base 2 and base 10 in your question, then when you put a number on the right side that is not divisible into the base you get problems. Like 1/3 in decimal because 3 doesnt go into a power of 10 or 1/5 in binary which doesnt go into a power of 2.
Another comment though NEVER use equal with floating point numbers, period. Even if it is an exact representation there are some numbers in some floating point systems that can be accurately represented in more than one way (IEEE is bad about this, it is a horrible floating point spec to start with, so expect headaches). No different here 1/3 is not EQUAL to the number on your calculator 0.3333333, no matter how many 3's there are to the right of the decimal point. It is or can be close enough but is not equal. so you would expect something like 2*1/3 to not equal 2/3 depending on the rounding. Never use equal with floating point.
As we have been discussing, in floating point arithmetic, the decimal 0.1 cannot be perfectly represented in binary.
Floating point and integer representations provide grids or lattices for the numbers represented. As arithmetic is done, the results fall off the grid and have to be put back onto the grid by rounding. Example is 1/10 on a binary grid.
If we use binary coded decimal representation as one gentleman suggested, would we be able to keep numbers on the grid?
For a simple answer: The computer doesn't have infinite memory to store fraction (after representing the decimal number as the form of scientific notation). According to IEEE 754 standard for double-precision floating-point numbers, we only have a limit of 53 bits to store fraction.
For more info: http://mathcenter.oxford.emory.edu/site/cs170/ieee754/
I will not bother to repeat what the other 20 answers have already summarized, so I will just answer briefly:
The answer in your content:
Why can't base two numbers represent certain ratios exactly?
For the same reason that decimals are insufficient to represent certain ratios, namely, irreducible fractions with denominators containing prime factors other than two or five which will always have an indefinite string in at least the mantissa of its decimal expansion.
Why can't decimal numbers be represented exactly in binary?
This question at face value is based on a misconception regarding values themselves. No number system is sufficient to represent any quantity or ratio in a manner that the thing itself tells you that it is both a quantity, and at the same time also gives the interpretation in and of itself about the intrinsic value of the representation. As such, all quantitative representations, and models in general, are symbolic and can only be understood a posteriori, namely, after one has been taught how to read and interpret these numbers.
Since models are subjective things that are true insofar as they reflect reality, we do not strictly need to interpret a binary string as sums of negative and positive powers of two. Instead, one may observe that we can create an arbitrary set of symbols that use base two or any other base to represent any number or ratio exactly. Just consider that we can refer to all of infinity using a single word and even a single symbol without "showing infinity" itself.
As an example, I am designing a binary encoding for mixed numbers so that I can have more precision and accuracy than an IEEE 754 float. At the time of writing this, the idea is to have a sign bit, a reciprocal bit, a certain number of bits for a scalar to determine how much to "magnify" the fractional portion, and then the remaining bits are divided evenly between the integer portion of a mixed number, and the latter a fixed-point number which, if the reciprocal bit is set, should be interpreted as one divided by that number. This has the benefit of allowing me to represent numbers with infinite decimal expansions by using their reciprocals which do have terminating decimal expansions, or alternatively, as a fraction directly, potentially as an approximation, depending on my needs.
You can't represent 0.1 exactly in binary for the same reason you can't measure 0.1 inch using a conventional English ruler.
English rulers, like binary fractions, are all about halves. You can measure half an inch, or a quarter of an inch (which is of course half of a half), or an eighth, or a sixteenth, etc.
If you want to measure a tenth of an inch, though, you're out of luck. It's less than an eighth of an inch, but more than a sixteenth. If you try to get more exact, you find that it's a little more than 3/32, but a little less than 7/64. I've never seen an actual ruler that had gradations finer than 64ths, but if you do the math, you'll find that 1/10 is less than 13/128, and it's more than 25/256, and it's more than 51/512. You can keep going finer and finer, to 1024ths and 2048ths and 4096ths and 8192nds, but you will never find an exact marking, even on an infinitely-fine base-2 ruler, that exactly corresponds to 1/10, or 0.1.
You will find something interesting, though. Let's look at all the approximations I've listed, and for each one, record explicitly whether 0.1 is less or greater:
fraction
decimal
0.1 is...
as 0/1
1/2
0.5
less
0
1/4
0.25
less
0
1/8
0.125
less
0
1/16
0.0625
greater
1
3/32
0.09375
greater
1
7/64
0.109375
less
0
13/128
0.1015625
less
0
25/256
0.09765625
greater
1
51/512
0.099609375
greater
1
103/1024
0.1005859375
less
0
205/2048
0.10009765625
less
0
409/4096
0.099853515625
greater
1
819/8192
0.0999755859375
greater
1
Now, if you read down the last column, you get 0001100110011. It's no coincidence that the infinitely-repeating binary fraction for 1/10 is 0.0001100110011...
c++ pow(2,1000) is normaly to big for double, but it's working. why?
So I've been learning C++ for couple weeks but the datatypes are still confusing me.
One small minor thing first: the code that 0xbadc0de posted in the other thread is not working for me.
First of all pow(2,1000) gives me this more than once instance of overloaded function "pow" matches the argument list.
I fixed it by changing pow(2,1000) -> pow(2.0,1000)
Seems fine, i run it and get this:
http://i.stack.imgur.com/bbRat.png
Instead of
10715086071862673209484250490600018105614048117055336074437503883703510511249361224931983788156958581275946729175531468251871452856923140435984577574698574803934567774824230985421074605062371141877954182153046474983581941267398767559165543946077062914571196477686542167660429831652624386837205668069376
it is missing a lot of the values, what might be cause that?
But now for the real problem.
I'm wondering how can 302 digits long number fit a double (8 bytes)?
0xFFFFFFFFFFFFFFFF = 18446744073709551616 so how can the number be larger than that?
I think it has something to do with the floating point number encoding stuff.
Also what is the largest number that can possibly be stored in 8 bytes if it's not 0xFFFFFFFFFFFFFFFF?
Eight bytes contain 64 bits of information, so you can store 2^64 ~ 10^20 unique items using those bits. Those items can easily be interpreted as the integers from 0 to 2^64 - 1. So you cannot store 302 decimal digits in 8 bytes; most numbers between 0 and 10^303 - 1 cannot be so represented.
Floating point numbers can hold approximations to numbers with 302 decimal digits; this is because they store the mantissa and exponent separately. Numbers in this representation store a certain number of significant digits (15-16 for doubles, if I recall correctly) and an exponent (which can go into the hundreds, of memory serves). However, if a decimal is X bytes long, then it can only distinguish between 2^(8X) different values... unlikely enough for exactly representing integers with 302 decimal digits.
To represent such numbers, you must use many more bits: about 1000, actually, or 125 bytes.
It's called 'floating point' for a reason. The datatype contains a number in the standard sense, and an exponent which says where the decimal point belongs. That's why pow(2.0, 1000) works, and it's why you see a lot of zeroes. A floating point (or double, which is just a bigger floating point) number contains a fixed number of digits of precision. All the remaining digits end up being zero. Try pow(2.0, -1000) and you'll see the same situation in reverse.
The number of decimal digits of precision in a float (32 bits) is about 7, and for a double (64 bits) it's about 16 decimal digits.
Most systems nowadays use IEEE floating point, and I just linked to a really good description of it. Also, the article on the specific standard IEEE 754-1985 gives a detailed description of the bit layouts of various sizes of floating point number.
2.0 ^ 1000 mathematically will have a decimal (non-floating) output. IEEE floating point numbers, and in your case doubles (as the pow function takes in doubles and outputs a double) have 52 bits of the 64 bit representation allocated to the mantissa. If you do the math, 2^52 = 4,503,599,627,370,496. Because a floating point number can represent positive and negative numbers, really the integer representation will be ~ 2^51 = 2,251,799,813,685,248. Notice there are 16 digits. there are 16 quality (non-zero) digits in the output you see.
Essentially the pow function is going to perform the exponentiation, but once the exponentiation moves past ~2^51, it is going to begin losing precision. Ultimately it will hold precision for the top ~16 decimal digits, but all other digits right will be un-guaranteed.
Thus it is a floating point precision / rounding problem.
If you were strictly in unsigned integer land, the number would overflow after (2^64 - 1) = 18,446,744,073,709,551,616. What overflowing means, is that you would never actually see the number go ANY HIGHER than the one provided, infact I beleive the answer would be 0 from this operation. Once the answer goes beyond 2^64, the result register would be zero, and any multiply afterwords would be 0 * 2, which would always result in 0. I would have to try it.
The exact answer (as you show) can be obtained using a standard computer using a multi-precision libary. What these do is to emulate a larger bit computer by concatenating multiple of the smaller data types, and use algorithms to convert and print on the fly. Mathematica is one example of a math engine that implements an arbitrary precision math calculation library.
Floating point types can cover a much larger range than integer types of the same size, but with less precision.
They represent a number as:
a sign bit s to indicate positive or negative;
a mantissa m, a value between 1 and 2, giving a certain number of bits of precision;
an exponent e to indicate the scale of the number.
The value itself is calculated as m * pow(2,e), negated if the sign bit is set.
A standard double has a 53-bit mantissa, which gives about 16 decimal digits of precision.
So, if you need to represent an integer with more than (say) 64 bits of precision, then neither a 64-bit integer nor a 64-bit floating-point type will work. You will need either a large integer type, with as many bits as necessary to represent the values you're using, or (depending on the problem you're solving) some other representation such as a prime factorisation. No such type is available in standard C++, so you'll need to make your own.
If you want to calculate the range of the digits that can be hold by some bytes, it should be (2^(64bits - 1bit)) to (2^(64bits - 1bit) - 1).
Because the left most digit of the variable is for representing sign (+ and -).
So the range for negative side of the number should be : (2^(64bits - 1bit))
and the range for positive side of the number should be : (2^(64bits - 1bit) - 1)
there is -1 for the positive range because of 0(to avoid reputation of counting 0 for each side).
For example if we are calculating 64bits, the range should be ==> approximately [-9.223372e+18] to [9.223372e+18]
I mean, for example, I have the following number encoded in IEEE-754 single precision:
"0100 0001 1011 1110 1100 1100 1100 1100" (approximately 23.85 in decimal)
The binary number above is stored in literal string.
The question is, how can I convert this string into IEEE-754 double precision representation(somewhat like the following one, but the value is not the same), WITHOUT losing precision?
"0100 0000 0011 0111 1101 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1010"
which is the same number encoded in IEEE-754 double precision.
I have tried using the following algorithm to convert the first string back to decimal number first, but it loses precision.
num in decimal = (sign) * (1 + frac * 2^(-23)) * 2^(exp - 127)
I'm using Qt C++ Framework on Windows platform.
EDIT: I must apologize maybe I didn't get the question clearly expressed.
What I mean is that I don't know the true value 23.85, I only got the first string and I want to convert it to double precision representation without precision loss.
Well: keep the sign bit, rewrite the exponent (minus old bias, plus new bias), and pad the mantissa with zeros on the right...
(As #Mark says, you have to treat some special cases separately, namely when the biased exponent is either zero or max.)
IEEE-754 (and floating point in general) cannot represent periodic binary decimals with full precision. Not even when they, in fact, are rational numbers with relatively small integer numerator and denominator. Some languages provide a rational type that may do it (they are the languages that also support unbounded precision integers).
As a consequence those two numbers you posted are NOT the same number.
They in fact are:
10111.11011001100110011000000000000000000000000000000000000000 ...
10111.11011001100110011001100110011001100110011001101000000000 ...
where ... represent an infinite sequence of 0s.
Stephen Canon in a comment above gives you the corresponding decimal values (did not check them, but I have no reason to doubt he got them right).
Therefore the conversion you want to do cannot be done as the single precision number does not have the information you would need (you have NO WAY to know if the number is in fact periodic or simply looks like being because there happens to be a repetition).
First of all, +1 for identifying the input in binary.
Second, that number does not represent 23.85, but slightly less. If you flip its last binary digit from 0 to 1, the number will still not accurately represent 23.85, but slightly more. Those differences cannot be adequately captured in a float, but they can be approximately captured in a double.
Third, what you think you are losing is called accuracy, not precision. The precision of the number always grows by conversion from single precision to double precision, while the accuracy can never improve by a conversion (your inaccurate number remains inaccurate, but the additional precision makes it more obvious).
I recommend converting to a float or rounding or adding a very small value just before displaying (or logging) the number, because visual appearance is what you really lost by increasing the precision.
Resist the temptation to round right after the cast and to use the rounded value in subsequent computation - this is especially risky in loops. While this might appear to correct the issue in the debugger, the accummulated additional inaccuracies could distort the end result even more.
It might be easiest to convert the string into an actual float, convert that to a double, and convert it back to a string.
Binary floating points cannot, in general, represent decimal fraction values exactly. The conversion from a decimal fractional value to a binary floating point (see "Bellerophon" in "How to Read Floating-Point Numbers Accurately" by William D.Clinger) and from a binary floating point back to a decimal value (see "Dragon4" in "How to Print Floating-Point Numbers Accurately" by Guy L.Steele Jr. and Jon L.White) yield the expected results because one converts a decimal number to the closest representable binary floating point and the other controls the error to know which decimal value it came from (both algorithms are improved on and made more practical in David Gay's dtoa.c. The algorithms are the basis for restoring std::numeric_limits<T>::digits10 decimal digits (except, potentially, trailing zeros) from a floating point value stored in type T.
Unfortunately, expanding a float to a double wrecks havoc on the value: Trying to format the new number will in many cases not yield the decimal original because the float padded with zeros is different from the closest double Bellerophon would create and, thus, Dragon4 expects. There are basically two approaches which work reasonably well, however:
As someone suggested convert the float to a string and this string into a double. This isn't particularly efficient but can be proven to produce the correct results (assuming a correct implementation of the not entirely trivial algorithms, of course).
Assuming your value is in a reasonable range, you can multiply it by a power of 10 such that the least significant decimal digit is non-zero, convert this number to an integer, this integer to a double, and finally divide the resulting double by the original power of 10. I don't have a proof that this yields the correct number but for the range of value I'm interested in and which I want to store accurately in a float, this works.
One reasonable approach to avoid this entirely issue is to use decimal floating point values as described for C++ in the Decimal TR in the first place. Unfortunately, these are not, yet, part of the standard but I have submitted a proposal to the C++ standardization committee to get this changed.