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<section id="floating-point-arithmetic-issues-and-limitations">
<span id="tut-fp-issues"></span><h1><span class="section-number">15. </span>Floating-Point Arithmetic: Issues and Limitations<a class="headerlink" href="#floating-point-arithmetic-issues-and-limitations" title="Link to this heading"></a></h1>
<p>Floating-point numbers are represented in computer hardware as base 2 (binary)
fractions. For example, the <strong>decimal</strong> fraction <code class="docutils literal notranslate"><span class="pre">0.625</span></code>
has value 6/10 + 2/100 + 5/1000, and in the same way the <strong>binary</strong> fraction <code class="docutils literal notranslate"><span class="pre">0.101</span></code>
has value 1/2 + 0/4 + 1/8. These two fractions have identical values, the only
real difference being that the first is written in base 10 fractional notation,
and the second in base 2.</p>
<p>Unfortunately, most decimal fractions cannot be represented exactly as binary
fractions. A consequence is that, in general, the decimal floating-point
numbers you enter are only approximated by the binary floating-point numbers
actually stored in the machine.</p>
<p>The problem is easier to understand at first in base 10. Consider the fraction
1/3. You can approximate that as a base 10 fraction:</p>
<div class="highlight-python3 notranslate"><div class="highlight"><pre><span></span><span class="mf">0.3</span>
</pre></div>
</div>
<p>or, better,</p>
<div class="highlight-python3 notranslate"><div class="highlight"><pre><span></span><span class="mf">0.33</span>
</pre></div>
</div>
<p>or, better,</p>
<div class="highlight-python3 notranslate"><div class="highlight"><pre><span></span><span class="mf">0.333</span>
</pre></div>
</div>
<p>and so on. No matter how many digits you’re willing to write down, the result
will never be exactly 1/3, but will be an increasingly better approximation of
1/3.</p>
<p>In the same way, no matter how many base 2 digits you’re willing to use, the
decimal value 0.1 cannot be represented exactly as a base 2 fraction. In base
2, 1/10 is the infinitely repeating fraction</p>
<div class="highlight-python3 notranslate"><div class="highlight"><pre><span></span><span class="mf">0.0001100110011001100110011001100110011001100110011</span><span class="o">...</span>
</pre></div>
</div>
<p>Stop at any finite number of bits, and you get an approximation. On most
machines today, floats are approximated using a binary fraction with
the numerator using the first 53 bits starting with the most significant bit and
with the denominator as a power of two. In the case of 1/10, the binary fraction
is <code class="docutils literal notranslate"><span class="pre">3602879701896397</span> <span class="pre">/</span> <span class="pre">2</span> <span class="pre">**</span> <span class="pre">55</span></code> which is close to but not exactly
equal to the true value of 1/10.</p>
<p>Many users are not aware of the approximation because of the way values are
displayed. Python only prints a decimal approximation to the true decimal
value of the binary approximation stored by the machine. On most machines, if
Python were to print the true decimal value of the binary approximation stored
for 0.1, it would have to display:</p>
<div class="highlight-python3 notranslate"><div class="highlight"><pre><span></span><span class="gp">&gt;&gt;&gt; </span><span class="mf">0.1</span>
<span class="go">0.1000000000000000055511151231257827021181583404541015625</span>
</pre></div>
</div>
<p>That is more digits than most people find useful, so Python keeps the number
of digits manageable by displaying a rounded value instead:</p>
<div class="highlight-pycon notranslate"><div class="highlight"><pre><span></span><span class="gp">&gt;&gt;&gt; </span><span class="mi">1</span> <span class="o">/</span> <span class="mi">10</span>
<span class="go">0.1</span>
</pre></div>
</div>
<p>Just remember, even though the printed result looks like the exact value
of 1/10, the actual stored value is the nearest representable binary fraction.</p>
<p>Interestingly, there are many different decimal numbers that share the same
nearest approximate binary fraction. For example, the numbers <code class="docutils literal notranslate"><span class="pre">0.1</span></code> and
<code class="docutils literal notranslate"><span class="pre">0.10000000000000001</span></code> and
<code class="docutils literal notranslate"><span class="pre">0.1000000000000000055511151231257827021181583404541015625</span></code> are all
approximated by <code class="docutils literal notranslate"><span class="pre">3602879701896397</span> <span class="pre">/</span> <span class="pre">2</span> <span class="pre">**</span> <span class="pre">55</span></code>. Since all of these decimal
values share the same approximation, any one of them could be displayed
while still preserving the invariant <code class="docutils literal notranslate"><span class="pre">eval(repr(x))</span> <span class="pre">==</span> <span class="pre">x</span></code>.</p>
<p>Historically, the Python prompt and built-in <a class="reference internal" href="../library/functions.html#repr" title="repr"><code class="xref py py-func docutils literal notranslate"><span class="pre">repr()</span></code></a> function would choose
the one with 17 significant digits, <code class="docutils literal notranslate"><span class="pre">0.10000000000000001</span></code>. Starting with
Python 3.1, Python (on most systems) is now able to choose the shortest of
these and simply display <code class="docutils literal notranslate"><span class="pre">0.1</span></code>.</p>
<p>Note that this is in the very nature of binary floating point: this is not a bug
in Python, and it is not a bug in your code either. You’ll see the same kind of
thing in all languages that support your hardware’s floating-point arithmetic
(although some languages may not <em>display</em> the difference by default, or in all
output modes).</p>
<p>For more pleasant output, you may wish to use string formatting to produce a
limited number of significant digits:</p>
<div class="highlight-pycon notranslate"><div class="highlight"><pre><span></span><span class="gp">&gt;&gt;&gt; </span><span class="nb">format</span><span class="p">(</span><span class="n">math</span><span class="o">.</span><span class="n">pi</span><span class="p">,</span> <span class="s1">&#39;.12g&#39;</span><span class="p">)</span> <span class="c1"># give 12 significant digits</span>
<span class="go">&#39;3.14159265359&#39;</span>
<span class="gp">&gt;&gt;&gt; </span><span class="nb">format</span><span class="p">(</span><span class="n">math</span><span class="o">.</span><span class="n">pi</span><span class="p">,</span> <span class="s1">&#39;.2f&#39;</span><span class="p">)</span> <span class="c1"># give 2 digits after the point</span>
<span class="go">&#39;3.14&#39;</span>
<span class="gp">&gt;&gt;&gt; </span><span class="nb">repr</span><span class="p">(</span><span class="n">math</span><span class="o">.</span><span class="n">pi</span><span class="p">)</span>
<span class="go">&#39;3.141592653589793&#39;</span>
</pre></div>
</div>
<p>It’s important to realize that this is, in a real sense, an illusion: you’re
simply rounding the <em>display</em> of the true machine value.</p>
<p>One illusion may beget another. For example, since 0.1 is not exactly 1/10,
summing three values of 0.1 may not yield exactly 0.3, either:</p>
<div class="highlight-pycon notranslate"><div class="highlight"><pre><span></span><span class="gp">&gt;&gt;&gt; </span><span class="mf">0.1</span> <span class="o">+</span> <span class="mf">0.1</span> <span class="o">+</span> <span class="mf">0.1</span> <span class="o">==</span> <span class="mf">0.3</span>
<span class="go">False</span>
</pre></div>
</div>
<p>Also, since the 0.1 cannot get any closer to the exact value of 1/10 and
0.3 cannot get any closer to the exact value of 3/10, then pre-rounding with
<a class="reference internal" href="../library/functions.html#round" title="round"><code class="xref py py-func docutils literal notranslate"><span class="pre">round()</span></code></a> function cannot help:</p>
<div class="highlight-pycon notranslate"><div class="highlight"><pre><span></span><span class="gp">&gt;&gt;&gt; </span><span class="nb">round</span><span class="p">(</span><span class="mf">0.1</span><span class="p">,</span> <span class="mi">1</span><span class="p">)</span> <span class="o">+</span> <span class="nb">round</span><span class="p">(</span><span class="mf">0.1</span><span class="p">,</span> <span class="mi">1</span><span class="p">)</span> <span class="o">+</span> <span class="nb">round</span><span class="p">(</span><span class="mf">0.1</span><span class="p">,</span> <span class="mi">1</span><span class="p">)</span> <span class="o">==</span> <span class="nb">round</span><span class="p">(</span><span class="mf">0.3</span><span class="p">,</span> <span class="mi">1</span><span class="p">)</span>
<span class="go">False</span>
</pre></div>
</div>
<p>Though the numbers cannot be made closer to their intended exact values,
the <a class="reference internal" href="../library/math.html#math.isclose" title="math.isclose"><code class="xref py py-func docutils literal notranslate"><span class="pre">math.isclose()</span></code></a> function can be useful for comparing inexact values:</p>
<div class="highlight-pycon notranslate"><div class="highlight"><pre><span></span><span class="gp">&gt;&gt;&gt; </span><span class="n">math</span><span class="o">.</span><span class="n">isclose</span><span class="p">(</span><span class="mf">0.1</span> <span class="o">+</span> <span class="mf">0.1</span> <span class="o">+</span> <span class="mf">0.1</span><span class="p">,</span> <span class="mf">0.3</span><span class="p">)</span>
<span class="go">True</span>
</pre></div>
</div>
<p>Alternatively, the <a class="reference internal" href="../library/functions.html#round" title="round"><code class="xref py py-func docutils literal notranslate"><span class="pre">round()</span></code></a> function can be used to compare rough
approximations:</p>
<div class="highlight-pycon notranslate"><div class="highlight"><pre><span></span><span class="gp">&gt;&gt;&gt; </span><span class="nb">round</span><span class="p">(</span><span class="n">math</span><span class="o">.</span><span class="n">pi</span><span class="p">,</span> <span class="n">ndigits</span><span class="o">=</span><span class="mi">2</span><span class="p">)</span> <span class="o">==</span> <span class="nb">round</span><span class="p">(</span><span class="mi">22</span> <span class="o">/</span> <span class="mi">7</span><span class="p">,</span> <span class="n">ndigits</span><span class="o">=</span><span class="mi">2</span><span class="p">)</span>
<span class="go">True</span>
</pre></div>
</div>
<p>Binary floating-point arithmetic holds many surprises like this. The problem
with “0.1” is explained in precise detail below, in the “Representation Error”
section. See <a class="reference external" href="https://jvns.ca/blog/2023/01/13/examples-of-floating-point-problems/">Examples of Floating Point Problems</a> for
a pleasant summary of how binary floating point works and the kinds of
problems commonly encountered in practice. Also see
<a class="reference external" href="http://www.indowsway.com/floatingpoint.htm">The Perils of Floating Point</a>
for a more complete account of other common surprises.</p>
<p>As that says near the end, “there are no easy answers.” Still, don’t be unduly
wary of floating point! The errors in Python float operations are inherited
from the floating-point hardware, and on most machines are on the order of no
more than 1 part in 2**53 per operation. That’s more than adequate for most
tasks, but you do need to keep in mind that it’s not decimal arithmetic and
that every float operation can suffer a new rounding error.</p>
<p>While pathological cases do exist, for most casual use of floating-point
arithmetic you’ll see the result you expect in the end if you simply round the
display of your final results to the number of decimal digits you expect.
<a class="reference internal" href="../library/stdtypes.html#str" title="str"><code class="xref py py-func docutils literal notranslate"><span class="pre">str()</span></code></a> usually suffices, and for finer control see the <a class="reference internal" href="../library/stdtypes.html#str.format" title="str.format"><code class="xref py py-meth docutils literal notranslate"><span class="pre">str.format()</span></code></a>
method’s format specifiers in <a class="reference internal" href="../library/string.html#formatstrings"><span class="std std-ref">Format String Syntax</span></a>.</p>
<p>For use cases which require exact decimal representation, try using the
<a class="reference internal" href="../library/decimal.html#module-decimal" title="decimal: Implementation of the General Decimal Arithmetic Specification."><code class="xref py py-mod docutils literal notranslate"><span class="pre">decimal</span></code></a> module which implements decimal arithmetic suitable for
accounting applications and high-precision applications.</p>
<p>Another form of exact arithmetic is supported by the <a class="reference internal" href="../library/fractions.html#module-fractions" title="fractions: Rational numbers."><code class="xref py py-mod docutils literal notranslate"><span class="pre">fractions</span></code></a> module
which implements arithmetic based on rational numbers (so the numbers like
1/3 can be represented exactly).</p>
<p>If you are a heavy user of floating-point operations you should take a look
at the NumPy package and many other packages for mathematical and
statistical operations supplied by the SciPy project. See &lt;<a class="reference external" href="https://scipy.org">https://scipy.org</a>&gt;.</p>
<p>Python provides tools that may help on those rare occasions when you really
<em>do</em> want to know the exact value of a float. The
<a class="reference internal" href="../library/stdtypes.html#float.as_integer_ratio" title="float.as_integer_ratio"><code class="xref py py-meth docutils literal notranslate"><span class="pre">float.as_integer_ratio()</span></code></a> method expresses the value of a float as a
fraction:</p>
<div class="highlight-pycon notranslate"><div class="highlight"><pre><span></span><span class="gp">&gt;&gt;&gt; </span><span class="n">x</span> <span class="o">=</span> <span class="mf">3.14159</span>
<span class="gp">&gt;&gt;&gt; </span><span class="n">x</span><span class="o">.</span><span class="n">as_integer_ratio</span><span class="p">()</span>
<span class="go">(3537115888337719, 1125899906842624)</span>
</pre></div>
</div>
<p>Since the ratio is exact, it can be used to losslessly recreate the
original value:</p>
<div class="highlight-pycon notranslate"><div class="highlight"><pre><span></span><span class="gp">&gt;&gt;&gt; </span><span class="n">x</span> <span class="o">==</span> <span class="mi">3537115888337719</span> <span class="o">/</span> <span class="mi">1125899906842624</span>
<span class="go">True</span>
</pre></div>
</div>
<p>The <a class="reference internal" href="../library/stdtypes.html#float.hex" title="float.hex"><code class="xref py py-meth docutils literal notranslate"><span class="pre">float.hex()</span></code></a> method expresses a float in hexadecimal (base
16), again giving the exact value stored by your computer:</p>
<div class="highlight-pycon notranslate"><div class="highlight"><pre><span></span><span class="gp">&gt;&gt;&gt; </span><span class="n">x</span><span class="o">.</span><span class="n">hex</span><span class="p">()</span>
<span class="go">&#39;0x1.921f9f01b866ep+1&#39;</span>
</pre></div>
</div>
<p>This precise hexadecimal representation can be used to reconstruct
the float value exactly:</p>
<div class="highlight-pycon notranslate"><div class="highlight"><pre><span></span><span class="gp">&gt;&gt;&gt; </span><span class="n">x</span> <span class="o">==</span> <span class="nb">float</span><span class="o">.</span><span class="n">fromhex</span><span class="p">(</span><span class="s1">&#39;0x1.921f9f01b866ep+1&#39;</span><span class="p">)</span>
<span class="go">True</span>
</pre></div>
</div>
<p>Since the representation is exact, it is useful for reliably porting values
across different versions of Python (platform independence) and exchanging
data with other languages that support the same format (such as Java and C99).</p>
<p>Another helpful tool is the <a class="reference internal" href="../library/functions.html#sum" title="sum"><code class="xref py py-func docutils literal notranslate"><span class="pre">sum()</span></code></a> function which helps mitigate
loss-of-precision during summation. It uses extended precision for
intermediate rounding steps as values are added onto a running total.
That can make a difference in overall accuracy so that the errors do not
accumulate to the point where they affect the final total:</p>
<div class="highlight-pycon notranslate"><div class="highlight"><pre><span></span><span class="gp">&gt;&gt;&gt; </span><span class="mf">0.1</span> <span class="o">+</span> <span class="mf">0.1</span> <span class="o">+</span> <span class="mf">0.1</span> <span class="o">+</span> <span class="mf">0.1</span> <span class="o">+</span> <span class="mf">0.1</span> <span class="o">+</span> <span class="mf">0.1</span> <span class="o">+</span> <span class="mf">0.1</span> <span class="o">+</span> <span class="mf">0.1</span> <span class="o">+</span> <span class="mf">0.1</span> <span class="o">+</span> <span class="mf">0.1</span> <span class="o">==</span> <span class="mf">1.0</span>
<span class="go">False</span>
<span class="gp">&gt;&gt;&gt; </span><span class="nb">sum</span><span class="p">([</span><span class="mf">0.1</span><span class="p">]</span> <span class="o">*</span> <span class="mi">10</span><span class="p">)</span> <span class="o">==</span> <span class="mf">1.0</span>
<span class="go">True</span>
</pre></div>
</div>
<p>The <a class="reference internal" href="../library/math.html#math.fsum" title="math.fsum"><code class="xref py py-func docutils literal notranslate"><span class="pre">math.fsum()</span></code></a> goes further and tracks all of the “lost digits”
as values are added onto a running total so that the result has only a
single rounding. This is slower than <a class="reference internal" href="../library/functions.html#sum" title="sum"><code class="xref py py-func docutils literal notranslate"><span class="pre">sum()</span></code></a> but will be more
accurate in uncommon cases where large magnitude inputs mostly cancel
each other out leaving a final sum near zero:</p>
<div class="highlight-pycon notranslate"><div class="highlight"><pre><span></span><span class="gp">&gt;&gt;&gt; </span><span class="n">arr</span> <span class="o">=</span> <span class="p">[</span><span class="o">-</span><span class="mf">0.10430216751806065</span><span class="p">,</span> <span class="o">-</span><span class="mf">266310978.67179024</span><span class="p">,</span> <span class="mf">143401161448607.16</span><span class="p">,</span>
<span class="gp">... </span> <span class="o">-</span><span class="mf">143401161400469.7</span><span class="p">,</span> <span class="mf">266262841.31058735</span><span class="p">,</span> <span class="o">-</span><span class="mf">0.003244936839808227</span><span class="p">]</span>
<span class="gp">&gt;&gt;&gt; </span><span class="nb">float</span><span class="p">(</span><span class="nb">sum</span><span class="p">(</span><span class="nb">map</span><span class="p">(</span><span class="n">Fraction</span><span class="p">,</span> <span class="n">arr</span><span class="p">)))</span> <span class="c1"># Exact summation with single rounding</span>
<span class="go">8.042173697819788e-13</span>
<span class="gp">&gt;&gt;&gt; </span><span class="n">math</span><span class="o">.</span><span class="n">fsum</span><span class="p">(</span><span class="n">arr</span><span class="p">)</span> <span class="c1"># Single rounding</span>
<span class="go">8.042173697819788e-13</span>
<span class="gp">&gt;&gt;&gt; </span><span class="nb">sum</span><span class="p">(</span><span class="n">arr</span><span class="p">)</span> <span class="c1"># Multiple roundings in extended precision</span>
<span class="go">8.042178034628478e-13</span>
<span class="gp">&gt;&gt;&gt; </span><span class="n">total</span> <span class="o">=</span> <span class="mf">0.0</span>
<span class="gp">&gt;&gt;&gt; </span><span class="k">for</span> <span class="n">x</span> <span class="ow">in</span> <span class="n">arr</span><span class="p">:</span>
<span class="gp">... </span> <span class="n">total</span> <span class="o">+=</span> <span class="n">x</span> <span class="c1"># Multiple roundings in standard precision</span>
<span class="gp">...</span>
<span class="gp">&gt;&gt;&gt; </span><span class="n">total</span> <span class="c1"># Straight addition has no correct digits!</span>
<span class="go">-0.0051575902860057365</span>
</pre></div>
</div>
<section id="representation-error">
<span id="tut-fp-error"></span><h2><span class="section-number">15.1. </span>Representation Error<a class="headerlink" href="#representation-error" title="Link to this heading"></a></h2>
<p>This section explains the “0.1” example in detail, and shows how you can perform
an exact analysis of cases like this yourself. Basic familiarity with binary
floating-point representation is assumed.</p>
<p><em class="dfn">Representation error</em> refers to the fact that some (most, actually)
decimal fractions cannot be represented exactly as binary (base 2) fractions.
This is the chief reason why Python (or Perl, C, C++, Java, Fortran, and many
others) often won’t display the exact decimal number you expect.</p>
<p>Why is that? 1/10 is not exactly representable as a binary fraction. Since at
least 2000, almost all machines use IEEE 754 binary floating-point arithmetic,
and almost all platforms map Python floats to IEEE 754 binary64 “double
precision” values. IEEE 754 binary64 values contain 53 bits of precision, so
on input the computer strives to convert 0.1 to the closest fraction it can of
the form <em>J</em>/2**<em>N</em> where <em>J</em> is an integer containing exactly 53 bits.
Rewriting</p>
<div class="highlight-python3 notranslate"><div class="highlight"><pre><span></span><span class="mi">1</span> <span class="o">/</span> <span class="mi">10</span> <span class="o">~=</span> <span class="n">J</span> <span class="o">/</span> <span class="p">(</span><span class="mi">2</span><span class="o">**</span><span class="n">N</span><span class="p">)</span>
</pre></div>
</div>
<p>as</p>
<div class="highlight-python3 notranslate"><div class="highlight"><pre><span></span><span class="n">J</span> <span class="o">~=</span> <span class="mi">2</span><span class="o">**</span><span class="n">N</span> <span class="o">/</span> <span class="mi">10</span>
</pre></div>
</div>
<p>and recalling that <em>J</em> has exactly 53 bits (is <code class="docutils literal notranslate"><span class="pre">&gt;=</span> <span class="pre">2**52</span></code> but <code class="docutils literal notranslate"><span class="pre">&lt;</span> <span class="pre">2**53</span></code>),
the best value for <em>N</em> is 56:</p>
<div class="highlight-pycon notranslate"><div class="highlight"><pre><span></span><span class="gp">&gt;&gt;&gt; </span><span class="mi">2</span><span class="o">**</span><span class="mi">52</span> <span class="o">&lt;=</span> <span class="mi">2</span><span class="o">**</span><span class="mi">56</span> <span class="o">//</span> <span class="mi">10</span> <span class="o">&lt;</span> <span class="mi">2</span><span class="o">**</span><span class="mi">53</span>
<span class="go">True</span>
</pre></div>
</div>
<p>That is, 56 is the only value for <em>N</em> that leaves <em>J</em> with exactly 53 bits. The
best possible value for <em>J</em> is then that quotient rounded:</p>
<div class="highlight-pycon notranslate"><div class="highlight"><pre><span></span><span class="gp">&gt;&gt;&gt; </span><span class="n">q</span><span class="p">,</span> <span class="n">r</span> <span class="o">=</span> <span class="nb">divmod</span><span class="p">(</span><span class="mi">2</span><span class="o">**</span><span class="mi">56</span><span class="p">,</span> <span class="mi">10</span><span class="p">)</span>
<span class="gp">&gt;&gt;&gt; </span><span class="n">r</span>
<span class="go">6</span>
</pre></div>
</div>
<p>Since the remainder is more than half of 10, the best approximation is obtained
by rounding up:</p>
<div class="highlight-pycon notranslate"><div class="highlight"><pre><span></span><span class="gp">&gt;&gt;&gt; </span><span class="n">q</span><span class="o">+</span><span class="mi">1</span>
<span class="go">7205759403792794</span>
</pre></div>
</div>
<p>Therefore the best possible approximation to 1/10 in IEEE 754 double precision
is:</p>
<div class="highlight-python3 notranslate"><div class="highlight"><pre><span></span><span class="mi">7205759403792794</span> <span class="o">/</span> <span class="mi">2</span> <span class="o">**</span> <span class="mi">56</span>
</pre></div>
</div>
<p>Dividing both the numerator and denominator by two reduces the fraction to:</p>
<div class="highlight-python3 notranslate"><div class="highlight"><pre><span></span><span class="mi">3602879701896397</span> <span class="o">/</span> <span class="mi">2</span> <span class="o">**</span> <span class="mi">55</span>
</pre></div>
</div>
<p>Note that since we rounded up, this is actually a little bit larger than 1/10;
if we had not rounded up, the quotient would have been a little bit smaller than
1/10. But in no case can it be <em>exactly</em> 1/10!</p>
<p>So the computer never “sees” 1/10: what it sees is the exact fraction given
above, the best IEEE 754 double approximation it can get:</p>
<div class="highlight-pycon notranslate"><div class="highlight"><pre><span></span><span class="gp">&gt;&gt;&gt; </span><span class="mf">0.1</span> <span class="o">*</span> <span class="mi">2</span> <span class="o">**</span> <span class="mi">55</span>
<span class="go">3602879701896397.0</span>
</pre></div>
</div>
<p>If we multiply that fraction by 10**55, we can see the value out to
55 decimal digits:</p>
<div class="highlight-pycon notranslate"><div class="highlight"><pre><span></span><span class="gp">&gt;&gt;&gt; </span><span class="mi">3602879701896397</span> <span class="o">*</span> <span class="mi">10</span> <span class="o">**</span> <span class="mi">55</span> <span class="o">//</span> <span class="mi">2</span> <span class="o">**</span> <span class="mi">55</span>
<span class="go">1000000000000000055511151231257827021181583404541015625</span>
</pre></div>
</div>
<p>meaning that the exact number stored in the computer is equal to
the decimal value 0.1000000000000000055511151231257827021181583404541015625.
Instead of displaying the full decimal value, many languages (including
older versions of Python), round the result to 17 significant digits:</p>
<div class="highlight-pycon notranslate"><div class="highlight"><pre><span></span><span class="gp">&gt;&gt;&gt; </span><span class="nb">format</span><span class="p">(</span><span class="mf">0.1</span><span class="p">,</span> <span class="s1">&#39;.17f&#39;</span><span class="p">)</span>
<span class="go">&#39;0.10000000000000001&#39;</span>
</pre></div>
</div>
<p>The <a class="reference internal" href="../library/fractions.html#module-fractions" title="fractions: Rational numbers."><code class="xref py py-mod docutils literal notranslate"><span class="pre">fractions</span></code></a> and <a class="reference internal" href="../library/decimal.html#module-decimal" title="decimal: Implementation of the General Decimal Arithmetic Specification."><code class="xref py py-mod docutils literal notranslate"><span class="pre">decimal</span></code></a> modules make these calculations
easy:</p>
<div class="highlight-pycon notranslate"><div class="highlight"><pre><span></span><span class="gp">&gt;&gt;&gt; </span><span class="kn">from</span><span class="w"> </span><span class="nn">decimal</span><span class="w"> </span><span class="kn">import</span> <span class="n">Decimal</span>
<span class="gp">&gt;&gt;&gt; </span><span class="kn">from</span><span class="w"> </span><span class="nn">fractions</span><span class="w"> </span><span class="kn">import</span> <span class="n">Fraction</span>
<span class="gp">&gt;&gt;&gt; </span><span class="n">Fraction</span><span class="o">.</span><span class="n">from_float</span><span class="p">(</span><span class="mf">0.1</span><span class="p">)</span>
<span class="go">Fraction(3602879701896397, 36028797018963968)</span>
<span class="gp">&gt;&gt;&gt; </span><span class="p">(</span><span class="mf">0.1</span><span class="p">)</span><span class="o">.</span><span class="n">as_integer_ratio</span><span class="p">()</span>
<span class="go">(3602879701896397, 36028797018963968)</span>
<span class="gp">&gt;&gt;&gt; </span><span class="n">Decimal</span><span class="o">.</span><span class="n">from_float</span><span class="p">(</span><span class="mf">0.1</span><span class="p">)</span>
<span class="go">Decimal(&#39;0.1000000000000000055511151231257827021181583404541015625&#39;)</span>
<span class="gp">&gt;&gt;&gt; </span><span class="nb">format</span><span class="p">(</span><span class="n">Decimal</span><span class="o">.</span><span class="n">from_float</span><span class="p">(</span><span class="mf">0.1</span><span class="p">),</span> <span class="s1">&#39;.17&#39;</span><span class="p">)</span>
<span class="go">&#39;0.10000000000000001&#39;</span>
</pre></div>
</div>
</section>
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