Controversial Facts In Mathematics

http://www.businessinsider.com/the-most-controversial-math-problems-2013-3?op=1

by edtechdev • 13 years ago

15 23 13 years ago

All good fun problems - and some less well-known as well as the old favourites. That said, calling these the "12 most controversial facts in mathematics" is like calling the truth of the moon landings the most controversial fact in astronautics. But I suppose "The Most Unintuitive Mathematical Results That Laymen Can Be Made To Understand" is not quite so catchy.

gordaco2• 13 years ago

Agree, this is mostly controversial for non-math people. Although the Monty Hall problem supposedly caused a lot of mail from people, some even math teachers, who got it wrong but didn't realize and/or admit it.

DerekL• 13 years ago

People often get the Monty Hall problem wrong because it is often stated ambiguously, and the listener fills in the details in his own way.

Here's the question as posed in Marilyn Vos Savant's column [http://marilynvossavant.com/game-show-problem/]: "Suppose you’re on a game show, and you’re given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat. He says to you, 'Do you want to pick door #2?' Is it to your advantage to switch your choice of doors?"

But what exactly is the procedure followed by the host? I can think of three:

A. If you pick a goat, then the host shows you the other goat. If you pick the car, then the host picks one of the goats at random to show it to you.

B. The host picks one of the other doors at random and shows it to you. So you might pick a goat, then he shows you the car, and gives you a chance to switch, but you already know that the remaining doors have goats, so there's no point.

C. If you pick the car, then the host shows you one of the goats picked at random. If you pick the goat, then the host says "You picked the goat!", and you don't get a chance to switch.

People assume that the procedure is A, but I don't see how the other procedures are excluded by the description.

thebear• 13 years ago

That was my first thought as well. Puzzling, vexing, yes, but there is nothing "controversial" here. Moreover, I'm not sure if the expression "controversial fact" even makes sense, linguistically speaking. Isn't it true that if something is a fact, it may be "disputed," but it can, by definition, not be "controversial?"

B-Con• 13 years ago

To be fair, in the introduction they do clearly point out that these are not technically "controversial" and just have a tendency to evoke controversial/confused reactions from laymen.

gordaco2• 13 years ago

I just knew that the first two were going to be the Monty Hall problem and 0.99999...==1. I'm actually surprised that so many people don't accept that last one; I understood it immediately when I learned it (I was 12).

The third blows your mind until you realize that, simply put, infinity is not a number and common arithmetic doesn't apply.

The sixth is not really math, although it perplexes a lot of people. I remember that a few years ago my sister didn't get that, if searching for "termX" wouldn't yield any results, searching for "termX and termY" also wouldn't give anything, no matter how much "termY" would.

I'd never seen the 8th, but it's beautiful, and very simple too! Does it really confuse people? Maybe I'm too mathy, but it's obvious to me.

The 9th one is a classic, but I feel it's less "controversial" because if you know (or have been told) that the series diverge, chances are that you have a relatively solid base of mathematical knowledge.

I encountered the 11th in a list of geometry problems recently; I had to prove it. The first time I saw the problem, just before starting a geometry course, I didn't have a clue. I shitted my pants thinking that I was going to fail the course if my exam was like that! Then, after a few weeks of study, I came back and immediately saw the proof, without thinking much (you just have to use the intercept theorem: each side of the parallelogram is parallel to one original polygon's diagonal).

The 7th and the last one don't seem too different than the Monty Hall... our brains are not very well wired for probability estimation and you can see that in many situations.

The rest are nice pieces of pop math (if you feel that "pop math" is not much of an oxymoron, anyway).

alok-g2• 13 years ago

>> 0.99999...==1

I am one of the people who don't accept the above as truth. Whether it is truth or not depends on the formal meaning of the ellipsis ... which no one provides to me (I have asked several people about this). Without a formal definition, the statement is meaningless to me.

Here is a way to present it formally.

0.9999... is sum of a geometric series a + a * d + a * d^2 ... a * d^n ... with a = 0.9 and d = 0.1.

In the limit n -> infinity, the above sum is 1. This is however only true in the limit.

It is not clear/obvious to me that ... really means "in the limit".

lutusp2• 13 years ago

> Whether it is truth or not depends on the formal meaning of the ellipsis ... which no one provides to me (I have asked several people about this).

Mathematical definitions aren't like dictionary definitions. A closed ellipsis:

    1...100

Is accepted as meaning a list of all the intermediate values.

An open-ended ellipsis is accepted as meaning an infinite repetition of what precedes it. So this ellipsis:

    0.9999...

Means an infinite decimal sequence of 9's. And .999... really is equal to 1.

http://en.wikipedia.org/wiki/0.999...

A quote: "In mathematics, the repeating decimal 0.999... (sometimes written with more or fewer 9s before the final ellipsis, or as 0.9, , 0.(9)) denotes a real number that can be shown to be the number one. In other words, the symbols "0.999..." and "1" represent the same number. Proofs of this equality have been formulated with varying degrees of mathematical rigor, taking into account preferred development of the real numbers, background assumptions, historical context, and target audience."

Remember that mathematics, like science, avoids ambiguity where possible. And mathematical notation strives for less ambiguity than mathematics itself.

strangestchild2• 13 years ago

I agree with you that the ... notation is ill-defined. It's common in mathematics to use conventions that are a little woolly, but only where everyone understands how to express the idea more correctly.

Here's one way of presenting this formally. Define a 'decimal' to be an ordered sequence of integers (called 'digits') a_1, a_2, a_3, and so on. (By 'and so on', I formally mean that for each positive integer k we have a digit a_k at the kth position in the sequence). Let's say each a_k has to be between 0 and 9 inclusive.

For each positive integer k, define the 'kth partial sum' of the sequence to be the sum from j=1 to j=k of (1/10^j) x a_j.

I'll skip over what it means if we say that the partial sums converge as k->Infinity, because it sounds like you understand what limits are and how they work. If not, I'd be happy to explain.

Now, if the partial sums converge to some value 'd', we say that the decimal has value equal to d. It can be shown that any decimal has at most one such d (which is good, because a decimal shouldn't have two values).

Now, I think you'd be happy to say that when we write 0.9999999..., what we mean is the decimal where a_k=9 for all k. Given this definition, it follows that the value of the decimal is precisely 1, using properties of geometric sequences.

It is up to you how you define '...', but all mathematicians would agree that 0.99999... should be interpreted as above if it is to have any meaning at all.

If you really want a more precise eplanation of '...' at the end of a truncated decimal, I would provide the following: "Writing 0.abcdef... asserts that the digits abcdef of the truncation provided have a pattern which should be obvious to the reader. Assign the first few digits a_1, a_2 etc as per the part of the decimal that is explicitly given; and then assign all subsequent digits values according to said pattern." - it's not a formal notation, as I say, but rather a convenient shorthand that is understood by working mathematicians. It is always possible to be more precise if one has to be.

rskar5• 13 years ago

Ok, would someone explain #6, "The Broken Water Heater Problem" to me? As expressed, there's nothing to say that the person (who repaired the heater) could not be just a plumber, or perhaps just a handyman and neither an accountant or a plumber. Yet, as demonstrated, the person is given to be an accountant.

Event A is "accountant and plumber". Event B is "accountant and not a plumber". Hence A unioned with B means "accountant"; or, if that's too general, "is an accountant and may or may not be a plumber too". In any case, both events A and B qualify the person to be an accountant.

So what's the controversy? When it comes to water heaters, we're all accountants?

scarmig2• 13 years ago

If I'm understanding it correctly, it's (ab)using the word "accountant" to mean "someone who can receive payments."

So the question is really "Is this person someone who accepts money, or is this person someone who accepts money and is a licensed plumber?"

If you interpret it literally, it uses the law of the excluded middle inappropriately and would be just as true if you replaced "accountant" with "President" or "extraterrestrial angel."

rskar5• 13 years ago

Thanks for the reply, but I doubt that the issue surrounds (ab)use of the word "accountant". If you go to the attributed source (at http://www.reddit.com/r/math/comments/12n04d/counterintuitiv...), instead of water-heater/accountant/plumber it's car/teacher/mechanic.

I'm more inclined to believe this is a (poor) illustration of the idea that a sub-set is likely to be smaller than the set it came from: the set of all accountants vs. the sub-set of all accountants who are also plumbers. In which case - duh! - P(accountant) > P(accountant and plumber). As long as one can indulge in the naivety of this pedagogical tale, that the narrator (the "I" who said "I went to a person") wasn't too particular on whom he hired for the job, then sure, why not?, 'twas likely the accountant that fixed it.

In the real world, which has teachers and mechanics in addition to accountants and plumbers and other occupations, and where people may tend to be pickier on who they'll let near their water heater - leave alone pay someone to mess with it - would this conclusion of "accountant" still prevail necessarily?

hackinthebochs2• 13 years ago

Your breakdown isn't quite right. Event A is "account and plumber" while Event B is "accountant" (leaving plumber status unconstrained). Thought of this way its easy to see that event B is completely contained in event A, therefore P(A) >= P(B).

rskar5• 13 years ago

Nope, take a closer look, Event B is pictured like a crescent moon, labelled "People who are accountants and not plumbers", and described thus: "There's also the probability 'B' that the guy who fixed my water heater is only an accountant and not a plumber." Plumber status within B is clearly "Not a plumber".

NhanH• 13 years ago

The way it was explained in the infographic wasn't clear. But what he meant is that between P(X = accountant) and P(X = accountant and plumber), the former is bigger. The fact that P(X = plumber) > P(X = accountant) is irrelevant, as we wasn't concerned with P(X= plumber) at all.

The implication of this "problem" is an interesting human biased: there appears to be cases where we judge the more detailed explanation as more probably, while statistically speaking, it isn't. If a person is described as "being concerned with discrimination and social justice", we tend to judge P(X = activist and Y) as more probable than P( X = Y). This is detailed in "Thinking, fast and slow" by Kahneman - a wonderful book if anyone haven't read it.

rskar5• 13 years ago

I think you've shown me the way here. My issue, I suppose, is the problem with word problems in math. Generally that problem is two-fold. Does a word problem sufficiently describe the situation? And do you sufficiently understand the situation it describes? I believe this to be at the core of all the supposed "controversial math problems". BTW, thank you for suggesting Kahneman's book, it is now on my reading list.

I did take notice that a sub-set is likely to be smaller than the set it came from: the set of all accountants vs. the sub-set of all accountants who are also plumbers. So therefore P(accountant) > P(accountant and plumber). Unless P(accountant) be zero, or P(plumber|accountant) be one, in which case neither is greater.

Reflecting on my own human biases, I think I ended up reading it as if this were a multiple-choice question, and I had to pick the "best" choice. In other words, it isn't enough that this person is an accountant, but that this person is also a plumber. Or put another way, if I had to wager on whether anyone would pay just any old accountant to go mess with their water heater, I would find it more surprising that the particular accountant in fact hired had no demonstrable plumbing skills. I'd imagine Kahneman could teach me more on this.

My favorite word problem joke goes something like this. If there are three cans on a fence, and you hit one with a thrown rock, how many cans remain on the fence? Answer: Two. Now, if there are three birds on a fence, and you hit one with a thrown rock, how many birds remain on the fence? Answer: None - they'd all start flying off when there's a rock hurtling towards them.

scarmig2• 13 years ago

They leave out Banach-Tarski?

I guess in all fairness it isn't so much a fact as a reason to reject the Axiom of Choice...

gizmo686• 13 years ago

The harmonic series was included as diverging, but there was no mention of conditionally convergents series (which, if you allow terms to be rearrange, can be made to converge to anything)

Controversial Facts In Mathematics

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