floating point product expansion equivalence

Question

In IEEE 754 floating point, it is possible that

a*(b-c) != a*b-a*c // a, b, c double

So expansion of a product is not guaranteed to be equa

Answer 1

Even excluding overflow and underflow, the results may differ.

.3 * (.3+.3+.3) is 0.269999999999999962252417162744677625596523284912109375 while .3*.3 + .3*.3 + .3*.3 is 0.270000000000000017763568394002504646778106689453125 (when both are evaluated with IEEE-754 rules and 64-bit binary floating-point).

Regarding the questions added in an update:

There are two questions, one of which asks whether the computed average of a set of numbers each not exceeding one can exceed one. As David Hammon points out, calculating the average of nine 1s as 1./9*1 + 1./9*1 + 1./9*1 + 1./9*1 + 1./9*1 + 1./9*1 + 1./9*1 + 1./9*1 + 1./9*1 produces 1.0000000000000002220446049250313080847263336181640625.

The other asks whether the computed average of a set can exceed all of the numbers in the set. This is a superset of the first question, but here is an example using a different calculation of the average:

Let b = 1./3 (which is 0.333333333333333314829616256247390992939472198486328125).

Then 1./10 * (b+b+b+b+b+b+b+b+b+b) is 0.33333333333333337034076748750521801412105560302734375, which is greater than b.