 <p>You still only have about 16 significant decimal digits; this means that only the first 16 or so digits of the output are correct; and all of the remaining digits are likely wildly incorrect, and are not actually stored as part of the double-precision number. This is why adding 1 to the number doesn't do anything -- it's a change to one of the <em>insignificant</em> digits that isn't actually recorded.</p>