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June 18

Useless mathematics

After reading Von Neumann universe and some related articles, all horribly abstract, I came to think of the practical use of mathematics. Personally, I find the main reason to study mathematics being that it is fun, but other than that, are there areas of mathematics that aren't expected to be of any practical use in a long time (mabye hundreds of years), if ever, or are there areas that have been around for a long time but no one so far have found any use for? —Bromskloss 08:13, 18 June 2007 (UTC)[reply]

A lot of mathematics has no known practical use. For mathematics for fun, see recreational mathematics and Category:Recreational mathematics. PrimeHunter 12:07, 18 June 2007 (UTC)[reply]

This does not answer your question, but Number theory must be mentioned in this discussion. If I am not mistaken, for millennia it was thought to be completely useless, right until the advent of modern cryptography (perhaps some time before). This goes to show you that it is hard to predict what will or will not be useful in the future. -- Meni Rosenfeld (talk) 14:10, 18 June 2007 (UTC)[reply]

From Pure mathematics#Quotes: "There is no branch of mathematics, however abstract, which may not someday be applied to the phenomena of the real world." Nikolai Lobachevsky (1792-1856). PrimeHunter 14:19, 18 June 2007 (UTC)[reply]

I think that was true in his days, but had he heard of hyper-inaccessible cardinals, he would surely have changed his mind! --Lambiam ^Talk 14:56, 18 June 2007 (UTC)[reply]

I don't even remotely know what the heck they are... but I would still suspect there is an isomorphism or at least a similarity between them or the theories that produce them and some real phenomena. Maybe you can't directly reason using the entities, like you can't reason using the infinitude of digits produced by calculating the decimal value of the square root of two, but you can reason about them. If one can claim a theorem about some entity and prove it, who is to say that the discovery or creation of the proof itself doesn't mirror some possible phenomena? However, this idea is probably impossible to prove, and also I'm probably taking liberties with the meaning of the words of the original quote. Root⁴(one) 22:09, 19 June 2007 (UTC)[reply]

Interesting that LIA should choose that particular example. In fact, the assertion that hyper-inaccessible cardinals exist has very concrete consequences for the physical world. It predicts that certain Turing machines do not halt, and you can actually build physical models of those TMs, and watch to see if they halt or not. If one of them does, it would falsify the assertion (though not in an absolute sense, of course -- your physical model could have been built wrong, or something could have gone wrong with it, and those possibilities would have to be checked before you could claim the assertion falsified). All large cardinal hypotheses are falsifiable in this sense. --Trovatore 07:34, 20 June 2007 (UTC)[reply]

Even if you could prove that the TM does not halt, you would (in my opinion) not have proved that the corresponding cardinal exists, but only that the assumption of its existence is not incompatible with one (in my view not particularly meaningful) specific axiomatization of set theory. --Lambiam ^Talk 09:34, 20 June 2007 (UTC)[reply]

That's true, leaving out the "not particularly meaningful" part. But that's always true. You can always make an ad hoc explanation of observations, one that boils down to "well, that's just the way it is", and doesn't isolate an underlying reason. Analogously, maybe electrons don't exist; maybe all experiments designed to test their existence, just by coincidence, come out as though electrons did exist. --Trovatore 18:20, 20 June 2007 (UTC)[reply]

from talking to the godlike minds of the theoretical physics grad students in my college days, i got the solid impression that even such lofty applied math fields as tensor calculus represent just a drop in the ocean of the universe of mathematics, most of which has no relevance to the physical world. Gzuckier 19:08, 20 June 2007 (UTC)[reply]

Make that: No known direct relevance to the physical world. -- Meni Rosenfeld (talk) 19:16, 20 June 2007 (UTC)[reply]

Factorisation

Hi,
Is it possible to further factorise (x+1)^2 + 3? I vaguely have the impression you can be completing the square, but the article didn't shed any light. Thanks, --124.179.222.152 10:25, 18 June 2007 (UTC)[reply]

I think that's the best answer you'll find. When written out, the quadratic is

x^{2}+2x+4

. This has no real linear factors because its discriminant is negative. (However, you can factor it further via the complex roots) nadav (talk) 10:35, 18 June 2007 (UTC)[reply]

Yeah I had

x^{2}+2x+4

as my original equation. I may as well tell you the full question. In the first part to the question they tell me to factorise

x^{2}+2x+4

, and then hence I have to anti-differentiate 1/(

x^{2}+2x+4

). I think what I need to do is to use partial fractions to solve it. And so I'm really looking for something like (x+a)(x+b) as a factor for

x^{2}+2x+4

--124.179.222.152 10:51, 18 June 2007 (UTC)[reply]

You're so close!! If you have to antidifferentiate, go back to your

(x+1)^{2}+3

bit. Think about what the derivative of the inverse tangent function is... –King Bee ^{(τ • γ)} 10:55, 18 June 2007 (UTC)[reply]

Oh, right! Feel so silly :-) - thanks King Bee and Nadav! --124.179.222.152 11:10, 18 June 2007 (UTC)[reply]

Hooray! Glad to help. (Don't feel too silly. I find that my students often have troubles with questions like this. You did well!) –King Bee ^{(τ • γ)} 11:12, 18 June 2007 (UTC)[reply]

It has always annoyed me when teachers say "factorize [or factor] x² + 2x + 4" and want you to come up with (x + 1)² + 3. Where are the factors of the original polynomial? Tesseran 12:04, 18 June 2007 (UTC)[reply]

Yeah, I agree; the hint that the OP received is basically nonsense. To answer your question, the original polynomial has no real zeros, as mentioned above by Navdav (the dsicriminant is < 0). –King Bee ^{(τ • γ)} 13:15, 18 June 2007 (UTC)[reply]

What's the next real number after 0?

Because you can't have 0.an infinite number of zeros followed by a 1. Vitriol 12:14, 18 June 2007 (UTC)[reply]

There simply is no "next" number, just like there is no largest real number. We'll have to live with that. —Bromskloss 12:33, 18 June 2007 (UTC)[reply]

Yes. See also well-order which says: "The standard ordering ≤ of the positive real numbers is not a well-ordering, since, for example, the open interval (0, 1) does not contain a least element." PrimeHunter 12:49, 18 June 2007 (UTC)[reply]

True. Although, Zorn's lemma implies that there is a well-ordering of the real numbers (it's clearly not ≤). Under this ordering, there is a "next" number after zero. =) –King Bee ^{(τ • γ)} 13:13, 18 June 2007 (UTC)[reply]

Is there, by any chance, a way to find such a well-ordering? —Bromskloss 13:50, 18 June 2007 (UTC)[reply]

No way. You would effectively have proved the axiom of choice (which is not my axiom of choice). I expect that in intuitionistic mathematics the claim that such a well-ordering exists is actually contradictory. --Lambiam ^Talk 14:49, 18 June 2007 (UTC)[reply]

What do we then mean when we say that a well-ordering exists? —Bromskloss 15:16, 18 June 2007 (UTC)[reply]

Well, it can be proven that the Axiom of Choice, Zorn's Lemma, and the Well-Ordering principle are all equivalent. Hence, if you take one, you have to take them all. (Which leads to a funny joke: The axiom of choice is obviously true, the well-ordering principle is obviously false, and we're still undecided on Zorn's lemma...or something.)

Lambiam - I also hate the axiom of choice (in that you can prove wacky counterintuitive things), but you can prove wacky things in ZF without choice as well. Do you really reject the axiom (a la Banach, Tarski, and a host of other prestigious skeptics)? –King Bee ^{(τ • γ)} 15:20, 18 June 2007 (UTC)[reply]

I'm a closet intuitionist. Oops, now the cat is out of the bag. The main problem I have with AC is that it is false. It implies that there exists a real-valued function f from [0, 1] to itself such that |f(x) − x| > 1/3 for all x. However, as L. E. J. Brouwer already proved, all functions defined on [0, 1] are continuous, and so in this case f(x) = x for some x in [0, 1]. I don't understand why some people think ZF is a meaningful basis for doing maths. --Lambiam ^Talk 20:07, 18 June 2007 (UTC)[reply]

Er, what? The Dirichlet function (restricted to that interval) is continuous? (Obviously it has a fixed point, but consider it divided by e.) --Tardis 15:29, 19 June 2007 (UTC)[reply]

Just to add to the above example, notice that the Dirichlet function is not a bijection. If you'd like an almost nowhere continuous bijection from [0,1] to [0,1], you can use a modified version such as f(x) = x if x is rational, or (1-x) if x is irrational. Notice that f(x) is a bijection from [0,1] but is discontinuous everywhere except x=1/2. So not all bijections from [0,1] to [0,1] are continuous. Dugwiki 21:11, 19 June 2007 (UTC)[reply]

There are much simpler examples for non-continuous bijections from [0, 1] onto itself than modified Dirichlet functions (who ever mentioned nowhere continuous?). Though I cannot say I am sure what Lambiam meant, it has to involve something deeper than giving classical counterexamples. -- Meni Rosenfeld (talk) 21:32, 19 June 2007 (UTC)[reply]

I certainly wasn't implying the function I described is the only such discontinous function. I mentioned it because it was (I think) an interesting expansion on the counterexample Tardis described. Finally, I have no idea what Lambien was trying to say, other than what he actually said. And what he said was "all functions defined on [0, 1] are continuous". So my counterexample is to disprove both that statement and the similar statement that "all bijections from [0,1] to [0,1] are continuous", or even have a countable number of discontinuities. Dugwiki 22:12, 19 June 2007 (UTC)[reply]

You can be sure that Lambiam knows how to disprove these statements, within the framework of classical logic and ZF. The question (which Trovatore has now answered) is only what alternative framework he has in mind. -- Meni Rosenfeld (talk) 22:32, 19 June 2007 (UTC)[reply]

In Brouwer's system, which is what Lambiam was talking about, the function you describe does not exist at all. Needless to say, Brouwer's system is kind of weird; you have to rethread your head quite a bit to understand what it's talking about. As you do with intuitionism in general. But it's an interesting exercise, and it does "work", to some extent, on its own terms -- although at a price I'm not interested in paying, and with benefits that don't strike me as all that beneficial. --Trovatore 22:21, 19 June 2007 (UTC)[reply]

I have to say, I'm not all that impressed by intuitionism's fundamental belief that "the truth of a statement is taken to be equivalent to the mathematician being able to intuit the statement." Certainly, for example, the function I described is intuitive and well defined. Are we to restrict our notion of truth only to those sets of numbers and statements which are finitely described by linguistic and mathematical symbolism? Just my opinion, of course. Dugwiki 23:11, 19 June 2007 (UTC)[reply]

I understand this applies to my function below as well? Any useful link for information regarding this system? -- Meni Rosenfeld (talk) 22:32, 19 June 2007 (UTC)[reply]

Are you saying that

\{(x,1)|0\leq x\leq {\tfrac {1}{2}}\}\cup \{(x,0)|{\tfrac {1}{2}}<x\leq 1\}

is not a function? -- Meni Rosenfeld (talk) 15:36, 19 June 2007 (UTC)[reply]

That is a simpler example to treat than an everywhere-discontinuous function. Let's call this function sd ("step down"). As anything else in mathematics, it all depends very much on your definitions – in this case in particular of function and of real number. To start with the latter, let's agree that a real number is an equivalence class of Cauchy sequences of rational numbers, as explained at Construction of real numbers#Construction from Cauchy sequences. Now take the real number x whose representative is the Cauchy sequence whose n-th term x_n = ¹/₂ + p(n)·2⁻ⁿ, where p(n) = 1 if there is a counterexample to Goldbach's conjecture below 2n, that is, some number 2k < 2n that cannot be written as the sum of two prime numbers. Otherwise, p(n) = −1. Now what is SD(x)? Give me a Cauchy sequence in this equivalence class. If you succeed in giving an effectively computable sequence, together with a proof that it belongs to SD(x), I can transform it into a proof or disproof of Goldbach's conjecture. If you can't give me an effectively computable sequence, then why should I believe that SD(x) is a real number, or for that matter that SD is a function, if it fails to assign a real number as the outcome for a possible input? The classical argument now invokes the Law of the Excluded Middle: either Goldbach's conjecture is true, and SD(x) = 1, or it is false, and then SD(x) = 0. However, this argument, like all nonconstructive proofs, is rejected in intuitionism. Whether it is weird really depends on how you have been brainwashed. --Lambiam ^Talk 09:08, 20 June 2007 (UTC)[reply]

Hahaha, are you suggesting more or less brainwash? :-) —Bromskloss 13:21, 20 June 2007 (UTC)[reply]

Um, no. In the classical argument, regardless of whether the Goldbach conjecture is true or false, the sequence you give is co-Cauchy with (1/2, 1/2, 1/2, ...), thus x is 1/2, and by the definition of SD we have SD (x) = 1. So I would say that SD succeeds in assigning the value 1 to the real number x. I do not understand the problem. -- Meni Rosenfeld (talk) 10:57, 20 June 2007 (UTC)[reply]

I made a mistake. Here is a corrected version:

x_n = ¹/₂ + 2^−k, if there is a counterexample to Goldbach's conjecture below 2n, where the first such counterexample is at 2k;

x_n = ¹/₂ otherwise.

--Lambiam ^Talk 13:57, 21 June 2007 (UTC)[reply]

I'll have to think about this example some more.

Anyway, I think this discussion is outside the scope of this thread, so we should stop here. Instead, I'll ask if you know any good source of information on the subject for someone like me, who until a few days ago thought of intuitionism as little more than a joke (I meant that as a compliment)? -- Meni Rosenfeld (talk) 10:41, 22 June 2007 (UTC)[reply]

Another thing: This discussion started as a criticism of the axiom of choice, but I still do not understand how it relates to it. Clearly, in classical ZF, there is no need for AC to show that SD exists, is non-continuous, and |SD(x) - x | > 1/3. Are you saying that, in Brouwer's system, the negation of AC is a theorem of ZF, or something like it? -- Meni Rosenfeld (talk) 12:31, 20 June 2007 (UTC)[reply]

I haven't studied possible interpretations of ZF in intuitionism. A problem in general is that a single notion in classical mathematics often unravels into several distinguishable notions in intuitionistic mathematics. For a start, in general P and ¬¬P have a different meanings, and so do P → Q and ¬P ∨ Q. So there is no straightforward unambiguous "translation" of a classical theory into intuitionese; anyone attempting that will have to make some crucial choices. Brouwer himself got somewhat bogged down in trying to classify possible different interpretations involved in such translations. I know others have made the attempt for ZF (see for example here), but I have not taken a serious look at it. Intuitionism by itself defies formalization, and I would be somewhat surprised if Brouwers' proof of the continuity of functions defined on the continuum can be captured in existing partial formalizations of intuitionism. --Lambiam ^Talk 14:18, 21 June 2007 (UTC)[reply]

Alas, I am still none the wiser about how AC gave rise to this discussion. -- Meni Rosenfeld (talk) 10:41, 22 June 2007 (UTC)[reply]

Yeah, I like that joke too. :-) Anyway, I don't understand why we can be so sure no one will ever find a well-ordering of the real numbers. —Bromskloss 15:52, 18 June 2007 (UTC)[reply]

Because, for any reasonable definition of 'find', it's provably impossible. For example, there is no formula of ZF which defines a well-ordering of the reals. Algebraist 16:40, 18 June 2007 (UTC)[reply]

I've seen an informal argument this before, but what about this statement in Well-order: "However it is consistent with ZFC that a definable well-ordering of the reals exists—for example, it is consistent with ZFC that V=L, and it follows from ZFC+V=L that a particular formula well-orders the reals, or indeed any set."? 70.21.2.80 03:42, 19 June 2007 (UTC)[reply]

Wow, I don't even know how I could define "find". —Bromskloss 07:15, 19 June 2007 (UTC)[reply]

Algebraist has already given one "reasonable definition".

As for the apparent discrepancy, I think what this means is: There is a formula, for which it is consistent that it defines a well-ordering of the reals. However, it does not follow from ZFC that this formula, indeed, defines a well-ordering of the reals, so even if we knew what the formula was (which I'm pretty sure is impossible), we would have no right to present this formula and claim we have found a well-ordering. -- Meni Rosenfeld (talk) 07:23, 19 June 2007 (UTC)[reply]

Yes, I meant that no formula of ZF provably well-orders the reals. And I think (read guess wildly) that in ZF+V=L we could (if we were sufficiently mad) actually write down the formula that well-orders the universe: you order by birthday and then by some kind of lexicographical ordering of definitions. Algebraist 07:42, 19 June 2007 (UTC)[reply]

You are saying, I gather, that existence of a formula is consistent with ZFC, but that the question of whether a particular formula well-orders the reals is undecidable? nadav (talk) 08:40, 19 June 2007 (UTC)[reply]

I think so. After all, proving that a given formula defines a well-ordering on the reals means proving that something holds for every set of real numbers, which is pretty heavy. -- Meni Rosenfeld (talk) 10:55, 19 June 2007 (UTC)[reply]

This diff might help us understand the meaning: [1]. I've added a {{fact}} tag to the statement in the meantime. nadav (talk) 11:17, 19 June 2007 (UTC)[reply]

Hehe, clever way to use a diff. :-) Anyway, regarding "consistent"… If a statement is outside the scope of the axiomatic system at hand, is it then said to be "consistent" with it? If so, it seems to me that "consistent" doesn't necessarily mean a lot. —Bromskloss 19:23, 19 June 2007 (UTC)[reply]

(Outdent, responding to Bromskloss) What do you mean by "outside the scope"? Are you talking about the fact that, say, "every frobnitz is a gizmo" is consistent with ZFC, because the ZFC axioms never mention frobnitzes or gizmoes and therefore can't hope to falsify it? It's true, that sort of "consistency" is not terribly interesting.

But that's not the situation at hand. Wellorderings are defined in terms of sets, and ZFC does talk about sets. If sets are real, then the formula in question either really does or really does not wellorder the reals, and ZFC, in spite of the fact that it discusses the right sort of thing, is not clever enough to say which. --Trovatore 20:42, 19 June 2007 (UTC)[reply]

(Mediawiki needs better discussion pages!) The frobnitz–gizmo statement is exactly of the kind I thought of. I think I see your point about defined or undefined things. So, let's see, the well-ordering theorem has some truth value (whatever that means), which is unknown to us and which ZFC will never lead us to? At first, this seems to show that ZFC is inadequate and should be replaced, but perhaps this kind of problem is what Gödel's first incompleteness theorem asserts will always come up, is it? Cruel world. —Bromskloss 12:05, 20 June 2007 (UTC)[reply]

(Technical point, responding to Meni) actually you don't have to quantify over sets of reals to express the claim that a given formula wellorders the reals. If the formula fails to wellorder the reals, then (either the formula fails to define a linear order at all or) there's a (countable) infinite sequence of distinct reals that's descending in the ordering given by the formula. That countable sequence can be coded by a single real. So ultimately you can put the claim in Π¹₁ form -- that is, you start by saying "for every real", and all your later quantifiers are over things simpler than reals. --Trovatore 20:57, 19 June 2007 (UTC)[reply]

Whoops -- that is, when you're using the formula itself as a black box. Of course if the formula itself is more complicated than Π¹₁ (which it will have to be) then the claim that the formula wellorders the reals will also be more complicated than Π¹₁, once you expand out the reference to the formula itself. --Trovatore 21:20, 19 June 2007 (UTC)[reply]

Noted. And I will take this opportunity to further respond to Bromskloss: Had we not known that the negation of the continuum hypothesis is consistent with ZFC, you would no doubt find countless mathematicians trying to prove CH as a theorem of ZFC. This has to mean something, right? -- Meni Rosenfeld (talk) 21:32, 19 June 2007 (UTC)[reply]

Noted the postscriptum as well - I guess this still means that the requirement is "lighter" than I had in mind. -- Meni Rosenfeld (talk) 21:36, 19 June 2007 (UTC)[reply]

What I meant was that the continuum hypothesis was meaningless if ZFC will neither confirm nor deny it, but mabye there is a world outside ZFC after all. I am learning a lot from this discussion! —Bromskloss 12:05, 20 June 2007 (UTC)[reply]

Mean of circular quantities

Hi! I'm using the algorithm provided in Mean_of_circular_quantities. It works just fine for some sets but I have discovered that some sets gives a small error. For example: a mean of 90 deg, 90 deg and 45 deg I would expect to be (90+90+45)/3 = 75.00 deg, but the algorithm gives 75.36119... deg (using atan2). I just can't understand how this can be? Is it an incorrectness in atan2? some cumulative error? can it get bigger? can it be keept smaller? Or am I just wrong in assuming that it should be 75.00 deg? 82.182.115.156 12:30, 18 June 2007 (UTC)[reply]

You are wrong to assume that the answer should be 75 degrees. The algorithm decribed in mean of circular quantities does not calculate the arithmetic mean of a set of angles. It calculates the direction of the vector sum of a set of unit vectors. There is no reason, in general, for these two "means" to have the same value. For an even more extreme example of the difference between the two concepts consider the "circular quantities" mean of 0 degrees and 180 degrees - it is

\arctan \left({\frac {0}{0}}\right)

, which is undefined. Yet the arithmetic mean of 0 and 180 is clearly 90. Gandalf61 12:56, 18 June 2007 (UTC)[reply]

Ok, thanks for that! 82.182.115.156 13:01, 18 June 2007 (UTC)[reply]

Just for the record, your computation is correct. I get 75.36119340482171369... degrees. PrimeHunter 13:03, 18 June 2007 (UTC)[reply]

Standard deviation

Why do you have to divide by n-1 instead of n?--71.185.137.94 15:41, 18 June 2007 (UTC)[reply]

Great question, this is explained (but not very clearly) at Standard deviation. The formula for standard deviation

{\frac {1}{n-1}}\sum _{i=1}^{n}(x_{i}-{\bar {x}})^{2}

is an unbiased estimator of

\sigma ^{2}

, meaning the result of the formula is as likely to be above the actual value of

\sigma ^{2}

as below. Another formula

{\frac {1}{n}}\sum _{i=1}^{n}(x_{i}-{\bar {x}})^{2}

is biased, but it has other properties which some people prefer (it is the maximum liklihood estimate).

Some people also use the argument that since we know

{\bar {x}}

, we don't really have n points to calculate the standard deviation; If you know n-1 points and the mean, you can calculate the last value. So the "real" number of points is n-1. --TeaDrinker 16:12, 18 June 2007 (UTC)[reply]

The formulas above are estimators of variance. To get estimators of the standard deviation, you have to take the square root. The property of being unbiased for the first formula only holds for the variance; its square root is still a biased estimator of the standard deviation. For example, when n = 2, its expectation is only 80% of the population value. If you want a less biased estimator for standard deviation, you're better off using n – ³/₂ in the denominator. --Lambiam ^Talk 21:27, 18 June 2007 (UTC)[reply]

Surely the fact that an estimator is unbiased means that it has the right expectation (aka mean), not that it is as likely to be above the true value as below? Algebraist 16:37, 18 June 2007 (UTC)[reply]

gasp* you're quite right, I stand corrected. --TeaDrinker 18:15, 18 June 2007 (UTC)[reply]

Our variance article actually explains this rather well. In particular, you should read the sections titled "Elementary description" and "Population variance and sample variance". In the end, though, the following quote from the latter section sums it up rather nicely:

"In practice, for large

n

, the distinction is often a minor one. In the course of statistical measurements, sample sizes so small as to warrant the use of the unbiased variance virtually never occur. In this context Press et al. commented that if the difference between n and n−1 ever matters to you, then you are probably up to no good anyway - e.g., trying to substantiate a questionable hypothesis with marginal data."

—Ilmari Karonen (talk) 11:43, 22 June 2007 (UTC)[reply]

PayPal

As an example I charge $95 for a product plus $5 for shipping. If customers choose to use PayPal, however, I have to charge for that just like I have to charge for shipping. PayPal charges approximately 4% or $4 added for a total of $104. The problem should be obvious: the customer now pays $104 and 4% of $104 is 16 cents greater so the new required total payment is $104.16. But again PayPal will charge 4% so now I need to calculate on the basis of $104.16, ad infintium. My questions is what formula can I use to stop this maddness and get the final answer in the first shot? 71.100.3.132 16:31, 18 June 2007 (UTC)[reply]

What you want to do is say that you want a 4% discount of

x

to be $100. That 4% discount can be thought of as

0.96x

so now it's just a matter of solving the simple linear equation

0.96x=100

. Donald Hosek 16:44, 18 June 2007 (UTC)[reply]

Okay give me an eample: I have a product that is $188 plus $55.5 for shipping for a total cash payment of $243.5. PayPal charges 4% of $243.50 or $9.25 rounded which is added to $243.50 for a total payment of $253.25. But wait now PayPal is going to charge 4% of $253.25 when payment is made via PayPal or $10.13 insted of $9.25. Play this out and show me the linear result curve. I do not think that it will be linear at all. 71.100.3.132 16:59, 18 June 2007 (UTC)[reply]

Read Donald's reply again. PayPal charges 4% of what the customer pays. So if the customer pays x, paypal takes

0.04x

and you get

0.96x

. If you are supposed to get 100$, then the customer should pay

{\frac {100}{0.96}}\approx 104.17

dollars. 4% of that is 4.17$, and you get your 100$.

An alternative (yet hopelessly more complicated) way to look at it is along the line you suggest: If you start with 100$, then add 4%, then add 4% of what you added before, ad infinitum, you get an infinite geometric series, which also sums up to

{\frac {100}{0.96}}

.

buyer/seller relations

On a completely unrelated note, I think your customers will appreciate it if you do not charge extra for using PayPal... -- Meni Rosenfeld (talk) 17:08, 18 June 2007 (UTC)[reply]

Okay, thanks. It is much clearer to me now. Item Price / (1-%fee} = Amount to Charge Customer

As for my PayPal policy... Many e-tailers do not charge directly for using PayPal but simply raise the price of their product to cover PayPal and then fail to give a discount when payment by check or money order is made. I think this is dishonest and deceptive and such business practices should be discouraged since the advantage of PayPal is speedy payment associated with securing a bid immediately in an auction situation such as eBay or the stock market.

Although I have long since raised my prices to cover PayPal at 4% I have also long since offered a discount for payment by check or money order. In this way my customer has the right to the option of saving by not using a service which he does not need. 71.100.3.132 18:35, 18 June 2007 (UTC)[reply]

Good point. I withdraw my comment, then. -- Meni Rosenfeld (talk) 18:55, 18 June 2007 (UTC)[reply]

who should pay credit card fees

A little off topic, but I believe credit card companies also charge a fee (and American Express and Diners Club are typically a bit higher than Visa and Mastercard), and different merchants handle this in different ways - some just raise the price across the board, some disallow credit cards - or just the more expensive ones, some don't allow them at all. I think that as long as the merchant is honest about how they charge their customers, it's not necessarily dishonest. Confusing Manifestation 22:53, 18 June 2007 (UTC)[reply]

Credit cards certainly charge fees to the merchant. The base rate for Mastercard and Visa is 3.5%, I believe, and it's higher if the card has any associated rewards program. Other card companies are higher also. Tesseran 04:20, 19 June 2007 (UTC)[reply]

It's been a while since I've had a merchant account, but 3.5% is a high rate (but not atypical for, say, a mail-order business). If you've got a long history of low fraud, you'll get a lower rate and rates vary across brands (Visa/MC/Discover/Amex), but issuing banks. The merchant agreement generally prohibits offering cash discounts or credit card surcharges (or even credit card minimums). That doesn't keep merchants from breaking the rules, but those are the rules that they're supposed to live under. Donald Hosek 16:29, 19 June 2007 (UTC)[reply]

Awhile back and now, every once in awhile, the sale of gasoline will drop so low that credit card companies are the only ones making money. In the past this spurred many gasoline stations to charge customers for using a credit card or to raise prices while offering a discount for cash. Naturally when credit card companies decided that this was hurting their business they simply banned gasoline stations from doing either one instead of adjusting their fees. Some gasoline stations still won't accept credit cards for this reason while the rest have long since added the credit card fee to their price in ways the credit card companies can not prevent without doing something illegal. In the end, after all, it is the customer who is the primary beneficiary of the convenience of the use of the credit card and the one who should rightfully pay. Denying that such logic is rightful is one of the reasons most likely behind such mishaps as airplanes intentionally veering off course and crashing into the Pentagon and tall buildings. 71.100.3.132 20:07, 19 June 2007 (UTC)[reply]