I think the other replies are overcomplicating this.
+ 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.
"a+b+c" doesn't describe a unique evaluation order. You need some parentheses to disambiguate which changes are due to associativity vs commutativity. a+(b+c)=(c+b)+a should be true of floating point numbers, due to commutativity. a+(b+c)=(a+b)+c may fail due to the lack of associativity.
IEEE 754 doesn't (usually) distinguish between different NaN encodings for the purposes of semantics--if the result is a NaN, it doesn't specify which NaN the result is. Most hardware vendors implement a form of NaN propagation: when both inputs are NaN, one of the operands is returned, for example, always the left NaN is returned if both are NaN.
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.
For those to be equal you need both associativity and commutativity.
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.
We’re in very nitpicky terminology weeds here (and I’m not the person you’re replying to), but my understanding is “commutative” is specifically about reordering operands of one binary op (4+3 == 3+4), while “associative” is about reordering a longer chain of the same operation (1+2+3 == 1+3+2).
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.
It is not a nit it is fundamental, a•b•c is associativity, specifically operator associativity.
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.
IEEE 754 floating-point addition and multiplication are commutative in practice, even if there are exceptions with NaNs etc..
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.
