3 ms·
For (1): Comparing sums of logarithms is pretty easy, since you can rewrite it as comparing products of integers. So not all sums of irrationals are hard. For
by thomasahle 5y ago
For (1): Comparing sums of logarithms is pretty easy, since you can rewrite it as comparing products of integers. So not all sums of irrationals are hard.
For (2): Allowing inputs to arbitrary precision presumably means they are arbitrarily long bit strings? But if you measure complexity in terms of the total number of input bits, summing long bit strings is very easy.