MathWorld disagrees, and so does http://wombat.doc.ic.ac.uk/foldoc/foldoc.cgi?linear+space , and every other reference I could turn up. So do I. I've never once encountered any context where "linear space" and "vector space" were not complete synonyms.
I'm certainly willing to have my opinion...
Sure, no problem, first here's the original problem, in morbid detail, just to be clear:
\begin{align*}
(f+g)(s) &= f(s) + g(s) \quad \text{definition of addition of \textbf{functions}} \\
& = g(s) + f(s) \quad \text{addition of \textbf{numbers} is commutative} \\
& = (g+f)(s) \quad...
I'm telling you to assume cummutativity of the real numbers under addition, which is not what you are trying to prove. You are trying to prove commutativity of a functionspace under an "addition" rule defined by (f+g)(s)=f(s)+g(s). That is something completely different from real numbers being...
You make it more complicated than it is. Adding another f(s) as you have done is completely unneccessary. Commutativity in the function space follws directly from the commutativity of the real numbers under addition. Nothing else is needed.
For instance, you want to show that (f+g)(s) = (g+f)(s). By the definition, (f+g)(s) = f(s) + g(s), and (g+f)(s) = g(s) + f(s). But f(s) and g(s) are just numbers, so (assuming the standard properties of real numbers) g(s) + f(s) = f(s) + g(s). From which the desired result obtains.
f(s) and g(s) are just numbers. You should invoke the comutivity and associativity of operations on real numbers, along with the definition you were given.
What? Why is any of that well said?
1) I have never heard of a crocodile eating a baby; I've also never heard of someone holding a baby at arms length from a crocodile maw, either.
2) Where did anyone say anywhere that he should not have kids? No one ever said that; they just said he...
Does anyone have any physics they want to talk about here?
Edit: I mean really, the questions you ask -- the ultimate "why" sort of questions -- belong to the realm of philosophy. Why is a photon evidently massless? Who knows, that's just the way our universe operates. No experiment will...
A matter of opinion, surely. It looks to me like you posted your question, with your answer already in mind ("I know more than I have let on.") just so that you could pounce on a well-meaning respondent. I call that a troll.
In fact I don't; I was being facetious.
A few people come to mind.