matt grime

If a function is self inverse then its graph is invariant under reflection in the line y=x. That IS the answer to your question. And for whoeever said that f(x)=x is the only self inverse function, can I suggest f(x)=-x, or f(x)=1-x, or how about f(x)=2-x. Spotting a pattern? What about all the graphs of these? They're all invariant under reflection in the line y=x.

Your arguement fails because you don't understand the meanings of any of the terms involved. With the usual identification of the power set with the set of all functions to {0,1}, then cardP(X) = 2^cardP{X}, which is extended to the transfinite case by anology.

You see that's where everyone but you disagrees. You don't get to change N. You can create whatever you care for, but it (probably) won't replace the naturals as we know them. We have a simple set of rules for the naturals, and there is no need to change them. You may create something new and different, but stop trying to sell them as "the natural numbers", because they aren't. Sometimes definitions do change over time, for many different reasons: some used to consider 1 a prime, now we do not. I doubt that you are going to change the definition of the naturals.

No, the inverse function is not 1/f. If f is its own inverse (NOT RECIPROCAL) and you know the graph of f inverse is the graph of f reflected in the line y=x, then surely you can see that the graph of f must be unchanged afer reflection in that line.

The domain is not the continuous thing, the function is. Riemann integration is the integration you are doing. Every continuous function on an interval [a,b] can be integrated. Sometimes that integral can be expressed as in general terms as a nice function like the thing you're integrating: eg integral of 2x is x^2 + k for some k. We say it is an elementary integral if we can find a function that differentiates to give the thing we are integrating (the integrand). Mostly that isn't true though. exp and log are inverse functions between R and R+.

Also, since every constructed model of the Real Numbers I can think of is done from taking the Naturals, adding inverses, localizing, and then completing or some other analytic process, we can assume that 2 "exists", and that base two expansions are valid. Clearly the smallest infinite cardinal is that of the Natural numbers, and we say that anything with cardinality equal to them has card aleph-0. So it "exists". If one allows the axiom of choice to be true then we can create a system of (small) transfinite numbers quite easily.

Because, in its domain (strictly positive Reals) it is continuous, and hence Riemann integrable.

I am not sure that I would say this approach is remotely sound. It is philosophically dubious, and mathematically unnecessary, and since at no point do you use the fact you are reducing mod 7 you should worry. the map n to 2n mod 10 isn't bijective for instance. The simple and better way is: suppose f(n)=f(m) then 2n=2m mod 7 ie 7 divides 2(n-m), as 2 and 7 are coprime, it must be that 7 divides n-m, ie n=m mod 7. Thus proving f is injective. f is surjective is now either trivial, by some larger result, or needs proving. Here is the larger result: if f is a function between two finite sets and f is injective, then f is surjective. proof: if f is not surjective, then there is some element not mapped to, hence, omitting this element, we can define a map that is injective from a set to one strictly smaller. by the pigeon hole principal this cannot be injective # so f was surjective. Or, we could say: Since 2 and 7 are coprime, there are integers a nd b such that 2a+7b=1. Let [n] be in Z_7, then 2an+7bn=1, ie 2an=n mod 7. n ws arbitrary hence the map is surjective.

Doron, you are once more moving the goal posts. The proofs hold still because they are not written about *your* objects. If you change the meaning of all the terms in some proposition, then of course it will almost certainly fail to be true, but you don't get to do that and then tell others they are wrong. All that remains is in *your* system (one of so far no demonstrable value, in fact the exact opposite since so many things are false) the statement is not true. Truth depends only on the context, there is no absolute truth in mathematics. As a simple example of why your observation might not be true anyway: hte natural numbers possess an arithmetic, and we may define primes in them. I do not at any point use these properties in an inductive proof : that doesn't mean the induction is invalid in the system in which I am operating.

"proofs by induction that are based only on the quantitative aspect of the Number concept" No they aren't. Proofs by induction are based upon the well-ordering principal, and do not *necessarily* make any reference to quantity, though in the natural numbers the well ordering statement is equivalent to the existence of a least element in any non-empty set, but it is not a necessary condition, merely a sufficient condition. That doesn't mean one cannot do induction without the natural numbers, nor does it mean that if we do not "examine" the "structure" of the natural numbers, whatever that may be, that proofs by induction fail to be true. Your "structure" concept has no bearing on proof by induction since all the properties of induction are only dependent on well-ordering. If you are about to redefine the concept of natural numbers so as they aren't well ordered, then you cannot talk about induction: it doesn't make it false since you are changing the premise on which it is founded and not using a well ordered set.

In what sense are you using "times" and how does this affect the Reals? What exceptions to the usual field axioms, which we still presume hold for all other defined operations, do you now need to inttoduce in order to give a complete decription of all of the newly valid operations on he reals? NB I don't care about any handy wavy heuristic "real life" arguments, it is only the mathematical consistency that counts, especially as there is little "real life" about infinity when treated as a number in this casual fashion.

Here is an example of how sums are usefully used in determinants. Let A be nxn, and let A(i) be the matrix obtained by setting all entries in the first row of A to be zero except for the i'th, for 1<=i<=n, then det(A)= det( sumA(i)) = sum det(A(i)) that is after all how we practically calculate the determinant by expanding about the first row.

There is no simple elementary formula, of the type that det(A+B)= f(det(A),det(B)) for some simple f. There is obviously a formula in terms of detA, detB and (some of) the entries in A and B, in fact there are obviously many such formulas. I don't know that any of them is easier (or less computationally intensive) to calculate than just working out det(A+B) the hard way purely in terms of the entries in A and B without using det(A) or det(B).

Then look up the class equation (of a group acting on a G-set) if you don't know it already, and the answer is immediate.

i think there's a mistake here does the question really ask exactly that? A cyclic group only ever has one generator, that is the definition of cyclic.

All orbits are a power of p, agreed? do you know what the class equation is? This is true even if omega were just a set, and it isn't necessary, though it may be true that the fixed points are in the centre of omega (I don't see it myself, in fact I'd go so far as to say it was trivially false: let G be the trivial group acting on any non-abelian group, then all points are fixed).

homEomorphisms, not homomorphisms

I have had many discussions with Doron, most of them have now been deleted from physicsforums.com, where he often hijacked threads, hence my post to which some people's respsonses have vanished. Whatever Shadmi Theory is, it is ill-presented and lacks any coherent definitions of the terms it introduces - the state it's in now constitutes a vast improvement over its original form. I am tired of trying to explain to Doron the misconceptions he has about mathematics and pointing out where his unfounded allegations about its paucity were wrong. Have you tried explaining why cantor's argument works to someone who doesn't even know what a bijection is? Explaining that there is no such thing as the axiom of infinity induction making mathematics internally inconsistent? Explaining basic logic theory to someone who thinks that if a proposition is equivalent to one of the axioms of the system that the proposition is undecidable? Tried explaining that the collatz conjecture isn't equivalent to the statement 'there is an infinite set'? I've done all those and it was painful; I don't intend to start again. See, already this thread is being h ijacked by it. Moderators feel free to delete my posts if you want to get this back on topic.

Call me an overzealous bodyguard of mathematics if you will, but let's not do any such thing and let's not hijack another thread with your own personal pet theory that no one else cares about. The question has been answered properly in mathematical terms which is the scope within which it was asked.

Ok, was hoping not to have to use this, but i can add up zero infinitely many times and get any number you care for. it's called measure theory. all these questions have been asked before, they have answers, they are understood, and they all are limited by exactly what they claim to be. you can extend the numbers to allow a limited arithmetic with infinity, however 0/0 and 0*inf are still not permitted in that system because there is no consistent or necessary way of assigning a MATHEMATICAL meaning to them. Nothing more nor less. there is no mystery or great con trick going on here. you don't make a right on a red light in New York, but you do in California, so if you're in california make that right, and when you're in new york don't. simple set of rules just like mathematics is. apart from the simple bit. don't confuse mathematics with its application in modelling something. 0*5 is not what you've got if you've got no lots of 5 apples. 1*x=x for all non zero x in R, 0+x=x for all x in R, using this we can consistently define 1*0=0 since 1*0=1*(0+0) =1*0+1*0, subtract 1*0 and we get 1*0 =0 as for callipygous, those three squirrels, are they the same squirrel or 3 different squirrels? what's even a squirrel really? red or grey? why not another rodent very much like a squirrel?

Exactly, AL. I can define the symbol 0*inf to be anything I choose. The problems arise when I try and make this consistent with the properties of other multiplications in that ring: there is no useful way of doing this so we don't bother as it is unnecessary. Let me give a silly analogy (there are no good analogies as we all know) Why isn't there a single word in the English Language that means 'yellow stone eating monster that lives at the centre of the Earth' i could possibly invent one etymologically, say, ge'centric'etio'pedi'phag'osaur but that doesn't mean I've done anything good or useful.

Once more the question is ill posed as all these things are. Multiplication is dependent on the set (RING) we're talking about. Infinity isn't in the RING of integers, or rationals, or reals or complexes.... and so asking what infinity times zero is is exactly as meaningful as asking what 3 times squirrel equals, it just looks more plausible, unless *you* specify what *you* mean by multiply.

That all depends on what you are trying to do by claiming a definition of 0/0. NB thinking about 0 fitting into 0 some number of times is not mathematics, and has no place in the discussion at all. Simply there is no way to define 0/0 in side any field, or ring, that is consistent with the axioms of a field, where consistent has the obvious meaning.

Countability is taught in the first term at both Cambridge and Bristol Universities to mathematicians. Surely the most elegant proofs are the simplest: send a/b to 3^a.5^b if it is a positive rational 2.3^a.5^b if it is negative. Clearly this shows the rationals are countable, and is a very powerful technique that works on many such 'show something is countable' questions.

You could try proving that if S is a complete set of eigenfunctions for some function space that the sum over all elements of S is a delta fucntion. I don't think that that is a necessary condition, but I've not proved it. It also depends on how liberal you are with convergence issues, obviously, but thinking about L^2(T^1) and fourier expansions I don't think that you can meaningfully even do that sum.