But is there more to it than that? As mathematicians, we know how to evaluate expressions involving lots of brackets. In fact we all learned how to prioritize brackets over the other mathematical operations when we were at school. But somewhere implicit in this is a confidence that we know how to evaluate expressions involving nested brackets and that these expressions are unambiguous.
But how do we know that evaluating expressions with nested brackets is unambiguous? Are we assuming here that composition of functions is associative? First of all, how else could we define the function. Given , set , and then set. Then the function is the function , where is defined in terms of as above. How can this possibly help? Set , set and set. There is no question of associativity in interpreting an expression formed from brackets, and therefore, as far as I can see no possibility of circularity.
Indeed, a similar complication of brackets is easily obtained in basic algebra. Like Like. Just noticed the date of posting of the original article. It was a personal response to the issue, and one which made me feel a bit happier about it! Some might find my comments helpful, others irrelevant. Let me ask: why do you think that there is no issue involved in evaluating nested brackets?
Presumably the general case is discussed elsewhere? Brackets are a means of writing unambiguous instructions to carry out multiple binary operations or function evaluations of course. Associativity is about whether two different bracketed expressions are in fact equal as a consequence of the properties of the particular binary operation.
There is not even the scope for circularity. I quite see that there is an issue to think about as to whether bracketed expressions do define a calculation to be done unambiguously, and doubtless somebody has written on this although it strikes me as obvious that they do.
And, as I said before, if we are in any doubt about this then we should sort it out far earlier since it would affect every reasonably complicated calculation we ever do.
I certainly have not experienced a failure to understand brackets as being an issue with UG maths teaching — although of course they are often used or more often, not used rather carelessly. But students will certainly find the associativity of function composition statement and proof difficult, but not, I would claim, because of imagining an ambiguity with brackets.
I would suggest that the issue is probably in understanding the statement and the need for a proof. One sees this at a more elementary level when students have difficulty reasoning about an abstract function about which only some properties are known, compared with a concrete one which can be evaluated on a calculator. The mathematics required to explain what it actually means to evaluate a mathematical expression is standard enough, but involves an inductive construction which I suspect would be somewhat tricky for a typical first-year mathematics undergraduate.
So I doubt that this would be covered rigorously before students have to meet composition of functions. That means that students are taking certain things on trust at this point. What I hope my post does is to give a short cut for those who, like me, might otherwise need to look at the full details involved in the inductive construction, and check that there is no hidden use of the associativity of composition of functions there. I suspect that I am, and will remain, in a vanishingly small minority in having had any concerns at all!
I have certainly never come across a student with this concern, and I may never meet one. Somewhat related to my concerns here is a more advanced fact.
If a Banach space is reflexive then so is its topological dual. Some of my students asked me whether the following argument was acceptable:. That attempted argument is definitely dubious!
It is this type of concern that brought me back to wondering whether there is an issue. Perhaps there really is nothing to it. If you are not worried then please feel free to ignore this post. Part of the point of the post is to convince the rare concerned reader someone like myself that there really is no circularity after all here. Thanks for the post. I was confusing myself with this today. I also thought I would be the only one with an issue but apparently not.
Which may be of some help. Take functions to be defined by their source, target and graph. Ie, ordered pairs with elements from given sets. Which is sort of weird:. Maybe I should just reproduce the set theoretic proof of associativity.
Recall that fg x is the unique element y such that x,y in fg. So really the question is how best to define function composition and how best to prove associativity. This will lose a lot in other places though.
Seems the answer is no. We have not shown otherwise. How should we prove associativity? Personally I prefer the proof straight from the set theoretic definition. This really helped me with something that I was doing. Nor did I like some of the other alternatives out there. At least, when maps are supposed to be defined on the whole domain as is the case when talking of surjectivity or bijectivity. Associativity is almost obvious. Perhaps it's EVEN easier clearer? We can view a relation R A,B as a collection of paths so that each path has a unique initial point in the set A and a unique terminal point in the set B.
In the special case of a relation a path is determined by its endpoints, but this is NOT a necessary assumption. Precisely the same input is needed to manufacture a typical arrow in each side of the previous equation.
Perhaps we have at least gotten to the heart of the matter, and established a precise sense in which 3 formally different things are the same. It is tempting to say that in the special case of functions and relations, that the 3 formaly different things are precisely the same, but even here perhaps we should be careful, since we are perhaps relying on canonical isomorphisms among products of finite sets:.
AxB xC is canonically isomorphic to Ax BxC which in turn is canonically isomorphic to the collection of all ordered triples a,b,c from the respective factors.
It is also tempting to say that in practice, no confusion arises, if our biggest crime is conflating canonically isomorphic things. However the vector a,b,c reminds me of a purported proof of a VERY famous conjecture in number theory, which according to my extremely pedestrian grasp of the opinion of experts, is perhaps problematic, due in part the conflation of: thing versus thing up to isomorphism or perhaps isomorphic copy of thing.
Attempted summary of the post at hand: The candidate isomorphism from the types x y z and x y z is in fact an isomorphism, rendering a sense in which x y z is unambiguous. I am unsatisfied with the post at hand. It would be fun to have a long thoughtful discussion leading to the agreement that indeed all the matters at are trivially obvious, obviously trivial, and that nowhere has there been any sleight of hand after all. Sign up to join this community.
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Some people write left-to-right, but then the argument should also be written to the left of the function. Add a comment. Active Oldest Votes. Otherwise, there is no way for the definition to work. Chaining functions in calculus uses different language and conventions.
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