Is floating point math broken?

+2400 votes
asked Feb 25, 2009 by cato-johnston

Consider the following code:

0.1 + 0.2 == 0.3 -> false
0.1 + 0.2 -> 0.30000000000000004

Why do these inaccuracies happen?

25 Answers

+377 votes
answered Feb 25, 2009 by chris-jester-young

When you convert .1 or 1/10 to base 2 (binary) you get a repeating pattern after the decimal point, just like trying to represent 1/3 in base 10. The value is not exact, and therefore you can't do exact math with it using normal floating point methods.

+84 votes
answered Feb 25, 2009 by mark-ransom

My answer is quite long, so I've split it into three sections. Since the question is about floating point mathematics, I've put the emphasis on what the machine actually does. I've also made it specific to double (64 bit) precision, but the argument applies equally to any floating point arithmetic.


An IEEE 754 double-precision binary floating-point format (binary64) number represents a number of the form

value = (-1)^s * (1.m51m50...m2m1m0)2 * 2e-1023

in 64 bits:

  • The first bit is the sign bit: 1 if the number is negative, 0 otherwise1.
  • The next 11 bits are the exponent, which is offset by 1023. In other words, after reading the exponent bits from a double-precision number, 1023 must be subtracted to obtain the power of two.
  • The remaining 52 bits are the significand (or mantissa). In the mantissa, an 'implied' 1. is always2 omitted since the most significant bit of any binary value is 1.

1 - IEEE 754 allows for the concept of a signed zero - +0 and -0 are treated differently: 1 / (+0) is positive infinity; 1 / (-0) is negative infinity. For zero values, the mantissa and exponent bits are all zero. Note: zero values (+0 and -0) are explicitly not classed as denormal2.

2 - This is not the case for denormal numbers, which have an offset exponent of zero (and an implied 0.). The range of denormal double precision numbers is dmin ≤ |x| ≤ dmax, where dmin (the smallest representable nonzero number) is 2-1023 - 51 (≈ 4.94 * 10-324) and dmax (the largest denormal number, for which the mantissa consists entirely of 1s) is 2-1023 + 1 - 2-1023 - 51 (≈ 2.225 * 10-308).

Turning a double precision number to binary

Many online converters exist to convert a double precision floating point number to binary (e.g. at, but here is some sample C# code to obtain the IEEE 754 representation for a double precision number (I separate the three parts with colons (:):

public static string BinaryRepresentation(double value)
{ long valueInLongType = BitConverter.DoubleToInt64Bits(value); string bits = Convert.ToString(valueInLongType, 2); string leadingZeros = new string('0', 64 - bits.Length); string binaryRepresentation = leadingZeros + bits; string sign = binaryRepresentation[0].ToString(); string exponent = binaryRepresentation.Substring(1, 11); string mantissa = binaryRepresentation.Substring(12); return string.Format("{0}:{1}:{2}", sign, exponent, mantissa);

Getting to the point: the original question

(Skip to the bottom for the TL;DR version)

Cato Johnston (the question asker) asked why 0.1 + 0.2 != 0.3.

Written in binary (with colons separating the three parts), the IEEE 754 representations of the values are:

0.1 => 0:01111111011:1001100110011001100110011001100110011001100110011010
0.2 => 0:01111111100:1001100110011001100110011001100110011001100110011010

Note that the mantissa is composed of recurring digits of 0011. This is key to why there is any error to the calculations - 0.1, 0.2 and 0.3 cannot be represented in binary precisely in a finite number of binary bits any more than 1/9, 1/3 or 1/7 can be represented precisely in decimal digits.

Converting the exponents to decimal, removing the offset, and re-adding the implied 1 (in square brackets), 0.1 and 0.2 are:

0.1 = 2^-4 * [1].1001100110011001100110011001100110011001100110011010
0.2 = 2^-3 * [1].1001100110011001100110011001100110011001100110011010

To add two numbers, the exponent needs to be the same, i.e.:

0.1 = 2^-3 * 0.1100110011001100110011001100110011001100110011001101(0)
0.2 = 2^-3 * 1.1001100110011001100110011001100110011001100110011010
sum = 2^-3 * 10.0110011001100110011001100110011001100110011001100111

Since the sum is not of the form 2n * 1.{bbb} we increase the exponent by one and shift the decimal (binary) point to get:

sum = 2^-2 * 1.0011001100110011001100110011001100110011001100110011(1)

There are now 53 bits in the mantissa (the 53rd is in square brackets in the line above). The default rounding mode for IEEE 754 is 'Round to Nearest' - i.e. if a number x falls between two values a and b, the value where the least significant bit is zero is chosen.

a = 2^-2 * 1.0011001100110011001100110011001100110011001100110011
x = 2^-2 * 1.0011001100110011001100110011001100110011001100110011(1)
b = 2^-2 * 1.0011001100110011001100110011001100110011001100110100

Note that a and b differ only in the last bit; ...0011 + 1 = ...0100. In this case, the value with the least significant bit of zero is b, so the sum is:

sum = 2^-2 * 1.0011001100110011001100110011001100110011001100110100


Writing 0.1 + 0.2 in a IEEE 754 binary representation (with colons separating the three parts) and comparing it to 0.3, this is (I've put the distinct bits in square brackets):

0.1 + 0.2 => 0:01111111101:0011001100110011001100110011001100110011001100110[100]
0.3 => 0:01111111101:0011001100110011001100110011001100110011001100110[011]

Converted back to decimal, these values are:

0.1 + 0.2 => 0.300000000000000044408920985006...
0.3 => 0.299999999999999988897769753748...

The difference is exactly 2-54, which is ~5.5511151231258 × 10-17 - insignificant (for many applications) when compared to the original values.

Comparing the last few bits of a floating point number is inherently dangerous, as anyone who reads the famous "What Every Computer Scientist Should Know About Floating-Point Arithmetic" (which covers all the major parts of this answer) will know.

Most calculators use additional guard digits to get around this problem, which is how 0.1 + 0.2 would give 0.3: the final few bits are rounded.

+1810 votes
answered Feb 25, 2009 by kernelpanik

Binary floating point math is like this. In most programming languages, it is based on the IEEE 754 standard. JavaScript uses 64-bit floating point representation, which is the same as Java's double. The crux of the problem is that numbers are represented in this format as a whole number times a power of two; rational numbers (such as 0.1, which is 1/10) whose denominator is not a power of two cannot be exactly represented.

For 0.1 in the standard binary64 format, the representation can be written exactly as

  • 0.1000000000000000055511151231257827021181583404541015625 in decimal, or
  • 0x1.999999999999ap-4 in C99 hexfloat notation.

In contrast, the rational number 0.1, which is 1/10, can be written exactly as

  • 0.1 in decimal, or
  • 0x1.99999999999999...p-4 in an analogue of C99 hexfloat notation, where the ... represents an unending sequence of 9's.

The constants 0.2 and 0.3 in your program will also be approximations to their true values. It happens that the closest double to 0.2 is larger than the rational number 0.2 but that the closest double to 0.3 is smaller than the rational number 0.3. The sum of 0.1 and 0.2 winds up being larger than the rational number 0.3 and hence disagreeing with the constant in your code.

A fairly comprehensive treatment of floating-point arithmetic issues is What Every Computer Scientist Should Know About Floating-Point Arithmetic. For an easier-to-digest explanation, see

+244 votes
answered Apr 9, 2010 by devin-jeanpierre

Most answers here address this question in very dry, technical terms. I'd like to address this in terms that normal human beings can understand.

Imagine that you are trying to slice up pizzas. You have a robotic pizza cutter that can cut pizza slices exactly in half. It can halve a whole pizza, or it can halve an existing slice, but in any case, the halving is always exact.

That pizza cutter has very fine movements, and if you start with a whole pizza, then halve that, and continue halving the smallest slice each time, you can do the halving 53 times before the slice is too small for even its high-precision abilities. At that point, you can no longer halve that very thin slice, but must either include or exclude it as is.

Now, how would you piece all the slices in such a way that would add up to one-tenth (0.1) or one-fifth (0.2) of a pizza? Really think about it, and try working it out. You can even try to use a real pizza, if you have a mythical precision pizza cutter at hand. :-)

Most experienced programmers, of course, know the real answer, which is that there is no way to piece together an exact tenth or fifth of the pizza using those slices, no matter how finely you slice them. You can do a pretty good approximation, and if you add up the approximation of 0.1 with the approximation of 0.2, you get a pretty good approximation of 0.3, but it's still just that, an approximation.

For double-precision numbers (which is the precision that allows you to halve your pizza 53 times), the numbers immediately less and greater than 0.1 are 0.09999999999999999167332731531132594682276248931884765625 and 0.1000000000000000055511151231257827021181583404541015625. The latter is quite a bit closer to 0.1 than the former, so a numeric parser will, given an input of 0.1, favour the latter.

(The difference between those two numbers is the "smallest slice" that we must decide to either include, which introduces an upward bias, or exclude, which introduces a downward bias. The technical term for that smallest slice is an ulp.)

In the case of 0.2, the numbers are all the same, just scaled up by a factor of 2. Again, we favour the value that's slightly higher than 0.2.

Notice that in both cases, the approximations for 0.1 and 0.2 have a slight upward bias. If we add enough of these biases in, they will push the number further and further away from what we want, and in fact, in the case of 0.1 + 0.2, the bias is high enough that the resulting number is no longer the closest number to 0.3.

In particular, 0.1 + 0.2 is really 0.1000000000000000055511151231257827021181583404541015625 + 0.200000000000000011102230246251565404236316680908203125 = 0.3000000000000000444089209850062616169452667236328125, whereas the number closest to 0.3 is actually 0.299999999999999988897769753748434595763683319091796875.

P.S. Some programming languages also provide pizza cutters that can split slices into exact tenths. Although such pizza cutters are uncommon, if you do have access to one, you should use it when it's important to be able to get exactly one-tenth or one-fifth of a slice.

(Originally posted on Quora.)

+52 votes
answered Dec 26, 2011 by brett-daniel

Floating point numbers stored in the computer consist of two parts, an integer and an exponent that the base is taken to and multiplied by the integer part.

If the computer were working in base 10, 0.1 would be 1 x 10⁻¹, 0.2 would be 2 x 10⁻¹, and 0.3 would be 3 x 10⁻¹. Integer math is easy and exact, so adding 0.1 + 0.2 will obviously result in 0.3.

Computers don't usually work in base 10, they work in base 2. You can still get exact results for some values, for example 0.5 is 1 x 2⁻¹ and 0.25 is 1 x 2⁻², and adding them results in 3 x 2⁻², or 0.75. Exactly.

The problem comes with numbers that can be represented exactly in base 10, but not in base 2. Those numbers need to be rounded to their closest equivalent. Assuming the very common IEEE 64-bit floating point format, the closest number to 0.1 is 3602879701896397 x 2⁻⁵⁵, and the closest number to 0.2 is 7205759403792794 x 2⁻⁵⁵; adding them together results in 10808639105689191 x 2⁻⁵⁵, or an exact decimal value of 0.3000000000000000444089209850062616169452667236328125. Floating point numbers are generally rounded for display.

+25 votes
answered Aug 1, 2012 by digitalross

Some statistics related to this famous double precision question.

When adding all values (a + b) using a step of 0.1 (from 0.1 to 100) we have ~15% chance of precision error. Note that the error could result in slightly bigger or smaller values. Here are some examples:

0.1 + 0.2 = 0.30000000000000004 (BIGGER)
0.1 + 0.7 = 0.7999999999999999 (SMALLER)
1.7 + 1.9 = 3.5999999999999996 (SMALLER)
1.7 + 2.2 = 3.9000000000000004 (BIGGER)
3.2 + 3.6 = 6.800000000000001 (BIGGER)
3.2 + 4.4 = 7.6000000000000005 (BIGGER)

When subtracting all values (a - b where a > b) using a step of 0.1 (from 100 to 0.1) we have ~34% chance of precision error. Here are some examples:

0.6 - 0.2 = 0.39999999999999997 (SMALLER)
0.5 - 0.4 = 0.09999999999999998 (SMALLER)
2.1 - 0.2 = 1.9000000000000001 (BIGGER)
2.0 - 1.9 = 0.10000000000000009 (BIGGER)
100 - 99.9 = 0.09999999999999432 (SMALLER)
100 - 99.8 = 0.20000000000000284 (BIGGER)

*15% and 34% are indeed huge, so always use BigDecimal when precision is of big importance. With 2 decimal digits (step 0.01) the situation worsens a bit more (18% and 36%).

+23 votes
answered Oct 14, 2013 by workoverflow

No, not broken, but most decimal fractions must be approximated


Floating point arithmetic is exact, unfortunately, it doesn't match up well with our usual base-10 number representation, so it turns out we are often giving it input that is slightly off from what we wrote.

Even simple numbers like 0.01, 0.02, 0.03, 0.04 ... 0.24 are not representable exactly as binary fractions. If you count up 0.01, .02, .03 ..., not until you get to 0.25 will you get the first fraction representable in base2. If you tried that using FP, your 0.01 would have been slightly off, so the only way to add 25 of them up to a nice exact 0.25 would have required a long chain of causality involving guard bits and rounding. It's hard to predict so we throw up our hands and say "FP is inexact", but that's not really true.

We constantly give the FP hardware something that seems simple in base 10 but is a repeating fraction in base 2.

How did this happen?

When we write in decimal, every fraction (specifically, every terminating decimal) is a rational number of the form

          a / (2n x 5m)

In binary, we only get the 2n term, that is:

           a / 2n

So in decimal, we can't represent 1/3. Because base 10 includes 2 as a prime factor, every number we can write as a binary fraction also can be written as a base 10 fraction. However, hardly anything we write as a base10 fraction is representable in binary. In the range from 0.01, 0.02, 0.03 ... 0.99, only three numbers can be represented in our FP format: 0.25, 0.50, and 0.75, because they are 1/4, 1/2, and 3/4, all numbers with a prime factor using only the 2n term.

In base10 we can't represent 1/3. But in binary, we can't do 1/10 or 1/3.

So while every binary fraction can be written in decimal, the reverse is not true. And in fact most decimal fractions repeat in binary.

Dealing with it

Developers are usually instructed to do < epsilon comparisons, better advice might be to round to integral values (in the C library: round() and roundf(), i.e., stay in the FP format) and then compare. Rounding to a specific decimal fraction length solves most problems with output.

Also, on real number-crunching problems (the problems that FP was invented for on early, frightfully expensive computers) the physical constants of the universe and all other measurements are only known to a relatively small number of significant figures, so the entire problem space was "inexact" anyway. FP "accuracy" isn't a problem in this kind of application.

The whole issue really arises when people try to use FP for bean counting. It does work for that, but only if you stick to integral values, which kind of defeats the point of using it. This is why we have all those decimal fraction software libraries.

I love the Pizza answer by Chris, because it describes the actual problem, not just the usual handwaving about "inaccuracy". If FP were simply "inaccurate", we could fix that and would have done it decades ago. The reason we haven't is because the FP format is compact and fast and it's the best way to crunch a lot of numbers. Also, it's a legacy from the space age and arms race and early attempts to solve big problems with very slow computers using small memory systems. (Sometimes, individual magnetic cores for 1-bit storage, but that's another story.)


If you are just counting beans at a bank, software solutions that use decimal string representations in the first place work perfectly well. But you can't do quantum chromodynamics or aerodynamics that way.

+43 votes
answered Oct 5, 2014 by justineo

Floating point rounding error. From What Every Computer Scientist Should Know About Floating-Point Arithmetic:

Squeezing infinitely many real numbers into a finite number of bits requires an approximate representation. Although there are infinitely many integers, in most programs the result of integer computations can be stored in 32 bits. In contrast, given any fixed number of bits, most calculations with real numbers will produce quantities that cannot be exactly represented using that many bits. Therefore the result of a floating-point calculation must often be rounded in order to fit back into its finite representation. This rounding error is the characteristic feature of floating-point computation.

+516 votes
answered Nov 20, 2014 by joel-coehoorn

A Hardware Designer's Perspective

I believe I should add a hardware designer’s perspective to this since I design and build floating point hardware. Knowing the origin of the error may help in understanding what is happening in the software, and ultimately, I hope this helps explain the reasons for why floating point errors happen and seem to accumulate over time.

1. Overview

From an engineering perspective, most floating point operations will have some element of error since the hardware that does the floating point computations is only required to have an error of less than one half of one unit in the last place. Therefore, much hardware will stop at a precision that's only necessary to yield an error of less than one half of one unit in the last place for a single operation which is especially problematic in floating point division. What constitutes a single operation depends upon how many operands the unit takes. For most, it is two, but some units take 3 or more operands. Because of this, there is no guarantee that repeated operations will result in a desirable error since the errors add up over time.

2. Standards

Most processors follow the IEEE-754 standard but some use denormalized, or different standards . For example, there is a denormalized mode in IEEE-754 which allows representation of very small floating point numbers at the expense of precision. The following, however, will cover the normalized mode of IEEE-754 which is the typical mode of operation.

In the IEEE-754 standard, hardware designers are allowed any value of error/epsilon as long as it's less than one half of one unit in the last place, and the result only has to be less than one half of one unit in the last place for one operation. This explains why when there are repeated operations, the errors add up. For IEEE-754 double precision, this is the 54th bit, since 53 bits are used to represent the numeric part (normalized), also called the mantissa, of the floating point number (e.g. the 5.3 in 5.3e5). The next sections go into more detail on the causes of hardware error on various floating point operations.

3. Cause of Rounding Error in Division

The main cause of the error in floating point division is the division algorithms used to calculate the quotient. Most computer systems calculate division using multiplication by an inverse, mainly in Z=X/Y, Z = X * (1/Y). A division is computed iteratively i.e. each cycle computes some bits of the quotient until the desired precision is reached, which for IEEE-754 is anything with an error of less than one unit in the last place. The table of reciprocals of Y (1/Y) is known as the quotient selection table (QST) in the slow division, and the size in bits of the quotient selection table is usually the width of the radix, or a number of bits of the quotient computed in each iteration, plus a few guard bits. For the IEEE-754 standard, double precision (64-bit), it would be the size of the radix of the divider, plus a few guard bits k, where k>=2. So for example, a typical Quotient Selection Table for a divider that computes 2 bits of the quotient at a time (radix 4) would be 2+2= 4 bits (plus a few optional bits).

3.1 Division Rounding Error: Approximation of Reciprocal

What reciprocals are in the quotient selection table depend on the division method: slow division such as SRT division, or fast division such as Goldschmidt division; each entry is modified according to the division algorithm in an attempt to yield the lowest possible error. In any case, though, all reciprocals are approximations of the actual reciprocal and introduce some element of error. Both slow division and fast division methods calculate the quotient iteratively, i.e. some number of bits of the quotient are calculated each step, then the result is subtracted from the dividend, and the divider repeats the steps until the error is less than one half of one unit in the last place. Slow division methods calculate a fixed number of digits of the quotient in each step and are usually less expensive to build, and fast division methods calculate a variable number of digits per step and are usually more expensive to build. The most important part of the division methods is that most of them rely upon repeated multiplication by an approximation of a reciprocal, so they are prone to error.

4. Rounding Errors in Other Operations: Truncation

Another cause of the rounding errors in all operations are the different modes of truncation of the final answer that IEEE-754 allows. There's truncate, round-towards-zero, round-to-nearest (default), round-down, and round-up. All methods introduce an element of error of less than one unit in the last place for a single operation. Over time and repeated operations, truncation also adds cumulatively to the resultant error. This truncation error is especially problematic in exponentiation, which involves some form of repeated multiplication.

5. Repeated Operations

Since the hardware that does the floating point calculations only needs to yield a result with an error of less than one half of one unit in the last place for a single operation, the error will grow over repeated operations if not watched. This is the reason that in computations that require a bounded error, mathematicians use methods such as using the round-to-nearest even digit in the last place of IEEE-754, because, over time, the errors are more likely to cancel each other out, and Interval Arithmetic combined with variations of the IEEE 754 rounding modes to predict rounding errors, and correct them. Because of its low relative error compared to other rounding modes, round to nearest even digit (in the last place), is the default rounding mode of IEEE-754.

Note that the default rounding mode, round-to-nearest even digit in the last place, guarantees an error of less than one half of one unit in the last place for one operation. Using the truncation, round-up, and round down alone may result in an error that is greater than one half of one unit in the last place, but less than one unit in the last place, so these modes are not recommended unless they are used in Interval Arithmetic.

6. Summary

In short, the fundamental reason for the errors in floating point operations is a combination of the truncation in hardware, and the truncation of a reciprocal in the case of division. Since the IEEE-754 standard only requires an error of less than one half of one unit in the last place for a single operation, the floating point errors over repeated operations will add up unless corrected.

+31 votes
answered Jan 3, 2015 by bruziuz

My workaround:

function add(a, b, precision) { var x = Math.pow(10, precision || 2); return (Math.round(a * x) + Math.round(b * x)) / x;

precision refers to the number of digits you want to preserve after the decimal point during addition.

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