Minimum amount of digits in double

hello everyone,
i was doing quite some research on how many digits a double can hold.
it's clear, that this is limited by the amount of bits a double consists of.
the algorithm is too complicated for me to understand, though.
Depending on the number , a double is supposed to be able
to hold 15-19 digits.
For example the number "12345678901234,121" can be held by a double,
but "12345678901234,122" can't. Also "999999999999999999" (18 digits)
and "2220000000000000000000" (23 digits) are ok but .
I need a reliable statement (prooved by some article or book) for boss,
which amount of digits (fraction and digital) a double can hold as
a minimum (e.g. 12 digital and 5 fraction-digits). even after hours of search i wasn't able to come up with anything, so I hope someone can help me here.
here are some links on the topic:
http://www.trotek.ec-lyon.fr/~muller/cours/numeration/flp_ieee.html.en
http://www.particle.kth.se/~lindsey/JavaCourse/Book/Part1/Tech/Chapter02/floatingPt.html
http://forum.java.sun.com/thread.jspa?forumID=31&threadID=403027
many greetings
marc johnen, berlin

i was doing quite some research on how many digits a double can hold.Check out the API documentation for the Double.longBitsToDouble method.
It explainsexactly what the format of a double is all about. In binary, a double can
hold 52 binary digits; always. The notion of an eleven bit exponent implements
the 'floating point' idea. When you want to translate this notion to decimal, you
have to be prepared for certain fractions that cannot be exactly expressed as
a finite sum of negative piwers of 2. A safe estimation is <= 15 decimal digits for
all double values.
kind regards,
Jos

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