This doesn’t solve the general problem that binary floating point cannot exactly represent decimal fractions.

The first example

  new Double('0.3').sub(new Double('0.1')).toNumber()

does the equivalent of

  float f = (float)(0.3 - 0.1);

which happens to produce a ‘float32’ that prints as “0.2”, but for ‘float64’ vs ‘float128’.

I expect that adding a sufficient number of zeroes, as in

  double f = (float128)0.00003 - (float128)0.00001);

(actual number of zeroes is too low here, to prevent wrapping on phone screens) will surface the problem again.

The fastest way are probably rational numbers, as they are natively available in many functional languages or Perl6. That is, 0.1 is represented as the integer tupls (1,10) instead of a floating point number.