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Why does the floating-point value of 4*0.1 look nice in Python 3 but 3*0.1 doesn't?

I know that most decimals don't have an exact floating point representation (Is floating point math broken?).
But I don't see why 4*0.1 is printed nicely as 0.4, but 3*0.1 isn't, when
both values actually have ugly decimal representations:
>>> 3*0.1
0.30000000000000004
>>> 4*0.1
0.4
>>> from decimal import Decimal
>>> Decimal(3*0.1)
Decimal('0.3000000000000000444089209850062616169452667236328125')
>>> Decimal(4*0.1)
Decimal('0.40000000000000002220446049250313080847263336181640625')

The simple answer is because 3*0.1 != 0.3 due to quantization (roundoff) error (whereas 4*0.1 == 0.4 because multiplying by a power of two is usually an "exact" operation). Python tries to find the shortest string that would round to the desired value, so it can display 4*0.1 as 0.4 as these are equal, but it cannot display 3*0.1 as 0.3 because these are not equal.
You can use the .hex method in Python to view the internal representation of a number (basically, the exact binary floating point value, rather than the base-10 approximation). This can help to explain what's going on under the hood.
>>> (0.1).hex()
'0x1.999999999999ap-4'
>>> (0.3).hex()
'0x1.3333333333333p-2'
>>> (0.1*3).hex()
'0x1.3333333333334p-2'
>>> (0.4).hex()
'0x1.999999999999ap-2'
>>> (0.1*4).hex()
'0x1.999999999999ap-2'
0.1 is 0x1.999999999999a times 2^-4. The "a" at the end means the digit 10 - in other words, 0.1 in binary floating point is very slightly larger than the "exact" value of 0.1 (because the final 0x0.99 is rounded up to 0x0.a). When you multiply this by 4, a power of two, the exponent shifts up (from 2^-4 to 2^-2) but the number is otherwise unchanged, so 4*0.1 == 0.4.
However, when you multiply by 3, the tiny little difference between 0x0.99 and 0x0.a0 (0x0.07) magnifies into a 0x0.15 error, which shows up as a one-digit error in the last position. This causes 0.1*3 to be very slightly larger than the rounded value of 0.3.
Python 3's float repr is designed to be round-trippable, that is, the value shown should be exactly convertible into the original value (float(repr(f)) == f for all floats f). Therefore, it cannot display 0.3 and 0.1*3 exactly the same way, or the two different numbers would end up the same after round-tripping. Consequently, Python 3's repr engine chooses to display one with a slight apparent error.

repr (and str in Python 3) will put out as many digits as required to make the value unambiguous. In this case the result of the multiplication 3*0.1 isn't the closest value to 0.3 (0x1.3333333333333p-2 in hex), it's actually one LSB higher (0x1.3333333333334p-2) so it needs more digits to distinguish it from 0.3.
On the other hand, the multiplication 4*0.1 does get the closest value to 0.4 (0x1.999999999999ap-2 in hex), so it doesn't need any additional digits.
You can verify this quite easily:
>>> 3*0.1 == 0.3
False
>>> 4*0.1 == 0.4
True
I used hex notation above because it's nice and compact and shows the bit difference between the two values. You can do this yourself using e.g. (3*0.1).hex(). If you'd rather see them in all their decimal glory, here you go:
>>> Decimal(3*0.1)
Decimal('0.3000000000000000444089209850062616169452667236328125')
>>> Decimal(0.3)
Decimal('0.299999999999999988897769753748434595763683319091796875')
>>> Decimal(4*0.1)
Decimal('0.40000000000000002220446049250313080847263336181640625')
>>> Decimal(0.4)
Decimal('0.40000000000000002220446049250313080847263336181640625')

Here's a simplified conclusion from other answers.
If you check a float on Python's command line or print it, it goes through function repr which creates its string representation.
Starting with version 3.2, Python's str and repr use a complex rounding scheme, which prefers
nice-looking decimals if possible, but uses more digits where
necessary to guarantee bijective (one-to-one) mapping between floats
and their string representations.
This scheme guarantees that value of repr(float(s)) looks nice for simple
decimals, even if they can't be
represented precisely as floats (eg. when s = "0.1").
At the same time it guarantees that float(repr(x)) == x holds for every float x

Not really specific to Python's implementation but should apply to any float to decimal string functions.
A floating point number is essentially a binary number, but in scientific notation with a fixed limit of significant figures.
The inverse of any number that has a prime number factor that is not shared with the base will always result in a recurring dot point representation. For example 1/7 has a prime factor, 7, that is not shared with 10, and therefore has a recurring decimal representation, and the same is true for 1/10 with prime factors 2 and 5, the latter not being shared with 2; this means that 0.1 cannot be exactly represented by a finite number of bits after the dot point.
Since 0.1 has no exact representation, a function that converts the approximation to a decimal point string will usually try to approximate certain values so that they don't get unintuitive results like 0.1000000000004121.
Since the floating point is in scientific notation, any multiplication by a power of the base only affects the exponent part of the number. For example 1.231e+2 * 100 = 1.231e+4 for decimal notation, and likewise, 1.00101010e11 * 100 = 1.00101010e101 in binary notation. If I multiply by a non-power of the base, the significant digits will also be affected. For example 1.2e1 * 3 = 3.6e1
Depending on the algorithm used, it may try to guess common decimals based on the significant figures only. Both 0.1 and 0.4 have the same significant figures in binary, because their floats are essentially truncations of (8/5)(2^-4) and (8/5)(2^-6) respectively. If the algorithm identifies the 8/5 sigfig pattern as the decimal 1.6, then it will work on 0.1, 0.2, 0.4, 0.8, etc. It may also have magic sigfig patterns for other combinations, such as the float 3 divided by float 10 and other magic patterns statistically likely to be formed by division by 10.
In the case of 3*0.1, the last few significant figures will likely be different from dividing a float 3 by float 10, causing the algorithm to fail to recognize the magic number for the 0.3 constant depending on its tolerance for precision loss.
Edit:
https://docs.python.org/3.1/tutorial/floatingpoint.html
Interestingly, there are many different decimal numbers that share the same nearest approximate binary fraction. For example, the numbers 0.1 and 0.10000000000000001 and 0.1000000000000000055511151231257827021181583404541015625 are all approximated by 3602879701896397 / 2 ** 55. Since all of these decimal values share the same approximation, any one of them could be displayed while still preserving the invariant eval(repr(x)) == x.
There is no tolerance for precision loss, if float x (0.3) is not exactly equal to float y (0.1*3), then repr(x) is not exactly equal to repr(y).

Division by 3 in Python

I am new to Python and while experimenting with operators, I came across this:
>>> 7.0 / 3
2.3333333333333335
Shouldn't the result be 2.3333333333333333 or maybe 2.3333333333333334. Why is it rounding the number in such a way?
Also, with regard to floor division in Python 2.7 my results were:
>>> 5 / 2
2
>>> 5 // 2
2
>>> 5.0 / 2
2.5
>>> 5.0 // 2
2.0
So my observation is that floor division returns the integer quotient even in case of floating numbers, while normal division return the decimal value. Is this true?

Take a look at this 0.30000000000000004.com
Your language isn't broken, it's doing floating point math. Computers can only natively store integers, so they need some way of representing decimal numbers. This representation comes with some degree of inaccuracy. That's why, more often than not, .1 + .2 != .3.

Shouldn't the result be 2.3333333333333333 or maybe 2.3333333333333334. Why is it rounding the number in such a way?
The key is the number is being rounded twice.
The first rounding is part of the division operation, rounding the number to the nearest double-precision floating point value. This is a binary operation not a decimal one.
The second rounding is part of converting the floating point number to a decimal representation for display. It is possible to represent the exact value of any binary fraction in decimal, but it is usually not desirable as in most applications doing so will simply result in many digits of false-precision. Python instead outputs the shortest decimal approximation that will round-trip to the correct floating point value.
We can better see what is going on by using the Fraction and Decimal types, unlike converting directly to a string converting a floating point number to a Fraction or Decimal will give the exact value. We can also use the Fraction type to determine the error in our calculation.
>>> from fractions import Fraction
>>> from decimal import Decimal
>>> 7.0 / 3
2.3333333333333335
>>> Decimal(7.0 / 3)
Decimal('2.333333333333333481363069950020872056484222412109375')
>>> Fraction(7.0 / 3)
Fraction(5254199565265579, 2251799813685248)
>>> Fraction(7,3) - Fraction(7.0 / 3)
Fraction(-1, 6755399441055744)
The conversion via type Decimal shows us the exact value of the floating point number and demonstrates the many digits of false-precision that typically result from exact conversion of a floating point value to decimal.
The conversion to a Fraction is also interesting, the denominator is 2251799813685248 which is equivalent to 251. This makes perfect sense, a Double precision floating point has 53 effective bits of mantissa and we need two of those for the integral part of the result leaving 51 for the fractional part.
The error in our floating point calculation is 1/6755399441055744 or ⅓ * 2-51. This error is less than half our precision step of 2-51 so the answer was indeed correctly rounded to a double precision floating point value.

numpy float64 1.0 is not equal to 1.0 [duplicate]

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.

Convert to Float without Rounding Decimal Places

I have a list and it contains a certain number '5.74536541' in it which I convert to a float.
I am printing it out in Python 3 using ("%0.2f" % (variable)) but it always prints out 5.75 instead of 5.74.
I know you're thinking who cares, but it is for a currency converter program and I don't want the currencies to round up/down but to be exact.
How can I keep it from rounding but also keep the 2 decimal places?

You shouldn't use floating point numbers for currency, due to rounding errors like you mentioned.
Your best bet is to use a fixed-precision decimal where you also have full control over how rounding and truncation works. From the docs:
>>> from decimal import *
>>> getcontext()
Context(prec=28, rounding=ROUND_HALF_EVEN, Emin=-999999999, Emax=999999999,
capitals=1, flags=[], traps=[Overflow, DivisionByZero,
InvalidOperation])
>>> getcontext().prec = 6
>>> Decimal('3.0')
Decimal('3.0')
>>> Decimal('3.1415926535')
Decimal('3.1415926535')
>>> Decimal('3.1415926535') + Decimal('2.7182818285')
Decimal('5.85987')
>>> getcontext().rounding = ROUND_UP
>>> Decimal('3.1415926535') + Decimal('2.7182818285')
Decimal('5.85988')
You should represent all currency-based values internally as Decimals with a high precision (the standard level of precision should be fine in your case - just leave the prec alone!). If you want to print a nicely formatted dollars and cents value to the user, using the locale module is a straightforward way to do this.
Be careful when printing as you will have to quantize the Decimal down to the correct number of places for display or the rounding will not be based on your Decimal context! You should only perform the quantize step for final display or for a single, final value - all intermediate steps should use high-precision Decimals to make any operations as accurate as possible.
>>> from decimal import *
>>> import locale
>>> locale.setlocale(locale.LC_ALL, '')
'en_AU.UTF-8'
>>> getcontext().rounding = ROUND_DOWN
>>> TWOPLACES = Decimal(10) ** -2
>>> var = Decimal('5.74536541')
Decimal('5.74536541')
>>> var.quantize(TWOPLACES)
Decimal('5.74')
>>> locale.currency(var.quantize(TWOPLACES))
'$5.74'

If you're dealing with currency and accuracy matters, don't use float, use decimal.

Take away the number mod 0.01
i.e.
rounded = number - (number % 0.01)
then print it the same as before.
This said, rounding down is not more accurate. Are you trying the old steal money from a bank by exploiting rounding errors scheme?

Floating point values are known as "useful approximations". Whatever you do to a floating point number—round it, truncate it, whatever—if the result is a floating point value, you don't get to decide how many digits to the right of the decimal point it has.
Never use floating point values for currency. See pydoc decimal, for example. Python's decimal module supports decimal fixed point and decimal floating point arithmetic.
Python docs warn about rounding floats.
Note The behavior of round() for floats can be surprising: for
example, round(2.675, 2) gives 2.67 instead of the expected 2.68. This
is not a bug: it’s a result of the fact that most decimal fractions
can’t be represented exactly as a float.
If you're not careful, you'll be misled by the value that appears at the interpreter prompt.
Python only prints a decimal approximation to the true decimal value
of the binary approximation stored by the machine.
And
It’s important to realize that this is, in a real sense, an illusion:
the value in the machine is not exactly 1/10, you’re simply rounding
the display of the true machine value. This fact becomes apparent as
soon as you try to do arithmetic with these values

If the number is a string then truncate the string to only 2 characters after the decimal and then convert it to a float.
Otherwise multiply it with 10^n where n is the number of digits after the decimal and then divide your float by 10^n.

Why is Python's Decimal function defaulting to 54 places?

After inputting
from decimal import *
getcontext().prec = 6
Decimal (1) / Decimal (7)
I get the value
Decimal('0.142857')
However if I enter Decimal (1.0/7) I get
Decimal('0.142857142857142849212692681248881854116916656494140625')

The 1.0 / 7 computes a binary floating point number to 17 digits of precision. This happens before the Decimal constructor sees it:
>>> d = 1.0 / 7
>>> type(d)
<type 'float'>
>>> d.as_integer_ratio()
(2573485501354569, 18014398509481984)
The binary fraction, 2573485501354569 / 18014398509481984 is as close as binary floating point can get using 53 bits of precision. It is not exactly 1/7th, but it's pretty close.
The Decimal constructor then converts the binary fraction to as many places as necessary to get an exact decimal equivalent. The result you're are seeing is what you get when you evaluate 2573485501354569 / 18014398509481984 exactly:
>>> from decimal import Decimal, getcontext
>>> getcontext().prec = 100
>>> Decimal(2573485501354569) / Decimal(18014398509481984)
Decimal('0.142857142857142849212692681248881854116916656494140625')
Learning point 1: Binary floating point computes binary fractions to 53 bits of precision. The result is rounded if necessary.
Learning point 2: The Decimal constructor converts binary floating point numbers to decimals losslessly (no rounding). This tends to result in many more digits of precision than you might expect (See the 6th question in the Decimal FAQ).
Learning point 3: The decimal module is designed to treat all numbers as being exact. Only the results of computations get rounded to the context precision. The binary floating point input is converted to decimal exactly and context precision isn't applied until you do a computation with the number (See the final question and answer in the Decimal FAQ for details).
Executive summary: Don't do binary floating point division before handing the numbers to the decimal module. Let it do the work to your desired precision.
Hope this helps :-)