$n1 = 0;
$n2 = 0;
for (1..1000) {
   # 0.1 is periodic in binary.
   # 0.1 requires an infinite number of bits to represent.
   $n1 += 0.1;

   # 0.5 = 2^(-1)
   # 0.5 requires one bit to represent.
   $n2 += 0.5;

   printf("%17.13f %17.13f\n", $n1, $n2);
}

##</code><code>##

...
 99.6999999999986 498.5000000000000
 99.7999999999986 499.0000000000000
 99.8999999999986 499.5000000000000
 99.9999999999986 500.0000000000000