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<div class="moz-cite-prefix">Le 18/09/2017 Ã 20:54, Samuel Gougeon a
écrit :<br>
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<blockquote type="cite"
cite="mid:03a6b389-083a-0dad-b6ab-b8a9a29ecf44@free.fr">
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<p> Now comes the issue:<br>
In (A), the relative difference 1/2^53 is too small (< %eps)
to be recorded and to change the number. OK.<br>
Since 1 / (2^53 +2) is even smaller than 1 / (2^53), it should
nor make a difference. Yet, it does:</p>
<p><tt>--> (2^53 + 2^1) + 1 == (2^53 + 2^1)</tt><tt><br>
</tt><tt> ans =</tt><tt><br>
</tt><tt>Â F</tt><tt><br>
</tt><br>
How is this possible ??!</p>
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<font size="+1">Â <tt> Â Â Â </tt><tt>Hi all, </tt><tt><br>
</tt><tt><br>
</tt><tt>Â Â no issue here but</tt> <tt>s</tt><tt>imply the round
to even rule. Every real number<br>
 should be approximated by the nearest floating point number
but in case of<br>
 a number exactly between 2 successive floats the one with an
even end<br>
 digit win. You can see this rule as a mean to approximate
those ambiguous cases,<br>
 one in two down and one in two up, but has more subtle
features in it such<br>
 (it is explained in Knuth I). <br>
<br>
<br>
 OK that</tt><tt> </tt></font><tt><font size="+1">2^53 + 2^1
is exactly represented<br>
 (it is a double float number). When computing : <br>
<br>
</font></tt><tt><tt>Â Â Â Â Â Â Â Â Â x = (2^53 + 2^1) + 1<br>
<br>
 it is not a float but is exactly at the same distance between
the two floats :<br>
<br>
        </tt></tt><tt><tt>2^53 + 2^1  and </tt></tt><tt><tt>2^53
+ 2^2Â but this last one ends with a even digit (0)<br>
 so : </tt></tt><br>
<tt><tt><tt><tt>Â Â Â Â Â Â Â Â fl( (2^53 + 2^1) + 1) = </tt></tt></tt></tt><tt><tt><tt><tt><tt><tt>2^53
+ 2^2<br>
<br>
 hth<br>
 Bruno<br>
<br>
</tt></tt></tt></tt></tt></tt>
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