Unless I’m having a brain fart it’s not commutative or you mean something by “relevant and appropriate” that I’m not understanding.
a+b+c != c+b+a
That’s why you need techniques like Kahan summation.
a+b+c != c+b+a
That’s why you need techniques like Kahan summation.
+ is a binary operation, and a+b+c can’t be interpreted without knowing whether one treats + as left-associative or right-associative. Let’s assume the former: a+b+c really means (a+b)+c.
If + is commutative, you can turn (a+b)+c into (b+a)+c or c+(a+b) or (commuting twice) c+(b+a).
But that last expression is not the same thing as (c+b)+a. Getting there requires associativity, and floating point addition is not associative.
IEEE 754 operations are nonassociative, but they are commutative (at least if you ignore the effect of NaN payloads).
As a side note: all compilers I'm aware of make almost no guarantees on preserving the value of NaN payloads, hence they consider floating-point operations to be fully commutative, and there's no general way to guarantee that they evaluate in exactly the order you specified.
You're supposed to do (a+b) to demonstrate the effect, because floating point subtraction that results in a number near zero is sensitive to rounding (worst case, a non-zero number gets you a zero number), which can introduce a huge error when a and b are very similar numbers.
The confusion between associativity and commutativity is the entire point of this thread!
Commutativity says that a*b = b*a, but that's not enough to allow arbitrary reordering. When you write a*b*c depending on whether * is left or right associative that either means a*(b*c) or (a*b)*c. If those are equal we say the operation is associative. You need both to allow arbitrary reordering. If an operation is only commutative you can turn a*(b*c) into a*(c*b) or (b*c)*a but there is no way to put a in the middle.
Edit: Wikipedia actually says associativity is definitionally about changing parens[0]. Mostly amounts to the same thing for standard arithmetic operators, but it’s an interesting distinction.
Rounding and eventual underflow in IEEE means an expression X•Y for any algebraic operation • produces, if finite, a result (X•Y)·( 1 + ß ) + µ where |µ| cannot exceed half the smallest gap between numbers in the destination’s format, and |ß| < 2^-N , and ß·µ = 0 . ( µ ≠ 0 only when Underflow occurs.)
And yes that is a binary relation only
a•b•c is really (a•b)•c assuming left operator associativity, one of the properties that IEEE doesn't have.
But remember that commutative is on the operations (+,x) which are binary operations, a+b=b+a and ab=ba, you can get accumulated rounding errors on iterated forms of those binary operations.