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    <p>This has to do with how computers store floating point numbers. A detailed description of this is <a href="http://docs.sun.com/source/806-3568/ncg_goldberg.html" rel="nofollow noreferrer">here</a>. However, for your case, the quick solution is to check not the printed representation of <code>point.x</code> but if <code>point.x</code> is equal to <code>4.7</code>. So...</p> <pre><code>&gt;&gt;&gt; point = Point(4.7, 8.2) &gt;&gt;&gt; point.x == 4.7 True </code></pre> <p>Or better:</p> <pre><code>&gt;&gt;&gt; point = Point(4.7, 8.2) &gt;&gt;&gt; eps = 2**-53 #get epsilon for standard double precision number &gt;&gt;&gt; -eps &lt;= point.x - 4.7 &lt;= eps True </code></pre> <p>Where <code>eps</code> is the maximum value for rounding errors in floating-point arithmetic. For details on epsilon, see <a href="http://en.wikipedia.org/wiki/Machine_epsilon" rel="nofollow noreferrer">here</a>.</p> <p><strong>EDIT:</strong> <code>-eps &lt;= point.x - 4.7 &lt;= eps</code> is equivalent to <code>abs(point.x - 4.7) &lt;= eps</code>. I only add this because not everyone is familiar with Python's chaining of comparison operators.</p> <p><strong>EDIT 2:</strong> Since you mentioned numpy, numpy has a method to get the eps without calculating it yourself. Use <code>eps = numpy.finfo(float).eps</code> instead of <code>2**-53</code> if you're using numpy. Note that the numpy epsilon is for some reason bigger than it should be and is equal to <code>2**-52</code> rather than <code>2**-53</code>. I have no idea why this is.</p>
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    1. POPitfalls of number values in Python, "How deep?"
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    1. USChinmay Kanchi
    1. USChinmay Kanchi
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    1. COMachine epsilon is a bound for **relative** error. You can't use it as you did, as the absolute error will be larger for values farther away from zero. In this specific case, `point.x - 4.7` will always give exactly 0 anyway.
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