If you want arbitrary precision rational arithmetic, you can use a FatRat

I didn't know about FatRats (yes, I know very little about perl6) - so I had a bit of a play (on rakudo-star-2017.07) and encountered confusing results: 
> my $x = 1.111111111111111111111.FatRat; my $y = 1.111111111111111111
+111.Rat; $x - $y
0
> $x == $y
True
> say $x
1.111111111111111111111
> say $y
1.11111111111111111604544
>

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On the bases that $x-$y==0 and $x==$y one is led to believe that $x and $y are exactly equivalent.
Yet, say() presents us with different values.

Are the 2 rationals equivalent ?
If so, then why does say() output different values ?
If not, then why do both $x-$y==0 and $x==$y evaluate as "True" ?

Interestingly, 1.11111111111111111604544 is the value of the double 1.1111111111111111 (16 decimal places) rounded to 23 decimals: 
C:\>perl -le "printf '%.22e\n', 1.1111111111111111;"
1.1111111111111111604544e+000\n

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Perhaps this ties in with: 
> my $x = 1.111111111111111111111.FatRat; my $y = 1.1111111111111111.N
+um; $x - $y
0
> $x == $y
True

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How does one coerce perl6 into displaying the actual numerator and denominator of these rationals ?

Cheers,
Rob

In reply to Re^3: Small Perl 6 discoveries II, Rats by syphilis
in thread [Perl6] Small discoveries I, __DATA__ by holli

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