Nitpick, but IEEE float operations are commutative (when relevant and appropriate). Associative and distributive they indeed are not.
a+b+c != c+b+a
That’s why you need techniques like Kahan summation.
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!
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.
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.
+ 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.
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.