This sounds like the smart thing to do

It's my preferred option, but there's a view that 17 digits is just too much clutter and only gives one a level of precision that one neither generally wants nor can comprehend.
And I think that's how choices of 12 or 15 might have come about.
So ... having print() output a "thereabouts" value is not universally condemned - given that one can always see the extra decimal digits by doing printf "%.16e", $double

However, whenever I've tried to get the full 17 decimal digits by using perl6's printf() the last 2 digits are always "0" : 
C:\>perl6 -e "printf '%.16e\n', 1.7320508075688772e0;"
1.7320508075688800e+00

[download]
(The perl6 developers have been recently informed of this.)

What’s your take on the second-to-last example, where $a < $b, and yet Num($a) > Num($b)?

I speculate that there's a bug in the conversion of rational values to doubles: 
C:\>perl6 -e "say Rat(0.7777777777777777777770) > Rat(0.77777777777777
+77777771);"
False

C:\>perl6 -e "say Num(Rat(0.7777777777777777777770)) > Num(Rat(0.77777
+77777777777777771));"
True

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It is known that the way that perl6 assigns doubles can result in weird discrepancies and, given that the 2 rationals are so close together in value, I would not be surprised if that method of assigning the doubles is the culprit.
Note that if you do: 
C:\>perl6 -e "say Num(0.7777777777777777777770e0) == Num(0.77777777777
+77777777771e0);"
True

[download]
then you avoid the "Rat to Num" conversion and get the correct result.
(The two values are deemed equal because, of course, they both equate to the same double.)

And I believe that last one-liner can equivalently be written as: 
C:\>perl6 -e "say 0.7777777777777777777770e0 == 0.77777777777777777777
+71e0;"
True

[download]

Cheers,
Rob

In reply to Re^3: Perl6 discoveries — floating-point by syphilis
in thread Perl6 discoveries — floating-point by Grimy

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