Find the "bug" - Two equals Four - a mathematical "paradox"(solipsys.co.uk)

67 pointsby RiderOfGiraffes15년 전69 comments

solipsys.co.uk

Find the "bug" - Two equals Four - a mathematical "paradox"

http://www.solipsys.co.uk/new/TwoEqualsFour.html?HN1

74 comments

It's not a paradox at all. It's simply not mathematical. The first question, which is completely ignored:

Is x^x^x^... well defined?

And does he mean x^(x^(x^...)) or ((x^x)^x)^x...

Because of his second statement, I guess he means x^(x^(x^...)). So he is looking at the limit of the sequence (x, x^x, x^(x^x), ...). So you have the function f

   x |-> limit of (x, x^x, x^(x^x), ...)

This function is not defined for all x (e.g. it diverges for x>=2) and (thus) its target set is clearly not R.

You can see that it does converge for x=sqrt(2), because 0 < sqrt(2)^a <= a for all a in 0..2) and thus it is correct that f(sqrt(2)) = 2 (by using his statements).

You can also see that 4 is not contained in the target set, because a^4 > 4 for all a>sqrt(2) and because f is monotone for x>0 and because f(sqrt(2))=2.

Thus, the set {x | f(x)=4} is empty.

I.e., his assumption that there exists such an x results in the wrong conclusion.

ced15년 전

You clearly know more than I do about these things, but:

0 < sqrt(2)^a <= a for all a in 0..2)

is not true. It fails for a=1. If you can make it right, could you explain the reasoning that gets you there? I never know how to tackle sequences like x, x^x, x^(x^x),...

albertzeyer15년 전

Whoops, yes you are right.

Fixed, more complete version of this part:

1 < sqrt(2)^a <= 2 for all a in [1,2]. I.e. the sequence (with x=sqrt(2)) is lower-bounded by 1 and upper-bounded by 2.

Also, because sqrt(2)^a >= a for all a>sqrt(2), the sequence is also monotone.

By that, it follows that it converges to some value y.

Now, given that it converges, you know that sqrt(2)^y = y and also y in [1,2]. And then you have y=2.

tassl15년 전

The function diverges for x>2, converges to 1 if x=1 and converges to 0 if x<1.

For the described problem, the incorrect assumption is that you can find a x st b = x^(x^(x^(x...

Where b>=4.

Edit: as RiderOfGiraffes pointed, the series does converge for x=sqrt(2)

RiderOfGiraffes15년 전

It's fairly easy to prove that the sequence x, x^x, x^(x^x), x^(x^(x^x)), ... converges for x=sqrt(2). Further, for x=sqrt(2) the sequence does converge to 2, so you're not entirely correct.

tmachinecharmer15년 전

Yes. It's simply not mathematical.

Suppose, x^(x^(x...) = y

At first the author says --> y = 2 and when he says "now consider the equation..." he equivalently says --> 4 = y.

Then obviously we can say 4 = 2. Means, you are assuming 4=2 and then concluding 4=2. Please point out if I am wrong. :)

Lozzer15년 전

You are wrong. When x is 1, y is 1. When x is sqrt(2), y is 2.

[deleted]15년 전

anigbrowl15년 전

Convergence is not an identity.

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

RiderOfGiraffes15년 전

I know that, and (clearly) you know that, but this is a nice demonstration as to why one must be careful.

I'll be interested to see how many times the page is accessed. I find that very few of my mathematical oddities get upvotes.

anigbrowl15년 전

Now I will look for other posts from you about this. BTW, was there any good discussion of the recent partition number proof by Ken Ono et al?

RiderOfGiraffes15년 전

It appears not. I don't remember seeing any.

[deleted]15년 전

pwagland15년 전

Along with convergence not being identity, any time you start playing with infinity, and then pretend that (∞-1 = ∞) you know that you are looking at a proof with problems…

RiderOfGiraffes15년 전

Provided it's done with care, the argument that removing a point from a countable set still leaves a countable set is perfectly valid. The trick is knowing that sometimes care is required.

There are many proofs that rely on removing finitely many points from an infinite set still leaving an infinite set. The definition of division on the equivalence classes of Cauchy sequences in the one of the definitions of the real numbers is just one example.

brown9-215년 전

Sorry, I'm not understanding a part of this statement and it sounds like you know math a whole lot better than I do: infinity is a "countable set"?

Could you explain this more?

RiderOfGiraffes15년 전

It stems from what you mean by "infinity". Most people use the term as if infinity is a number, and that gets you into trouble. You need to say what you mean.

He talked about \infty - 1, and really that means - "take an infinite-sized set and remove an element from it." Now you have the question: Which infinite-sized set?

In particular, there are different sizes of infinite sized sets. The set of positive whole numbers, the set of primes, and the set of rationals are all "the same size" (under a reasonable definition of the term). However, the set of real numbers, or the set of subsets of primes, are both "bigger" (in some very real sense).

So I was responding to the comment about "\infty - 1 being \infty" causing trouble. When you treat these things properly (whatever that means) then it doesn't cause trouble. At least, not any more, and not for people who understand the care required.

tl;dr : If you mention "infinity" you are probably talking about the size of a set.

lmkg15년 전

This gets into something that you and I discussed earlier, that "infinity" can refer to two different concepts. While your interpretation is well-defined, and reasonable, it's not what people usually mean when they think of that expression. Even knowing the math and notation behind both of them interpretations and the difference between them, I still tend to see \inf-1 as an ill-defined arithmetic (algebraic) expression rather than a well-defined set operation.

Note also, that you can treat \inf as an ordinal number, rather than a cardinal or a limit point, and subtracting one as the inverse of the ordinal successor operation. In this case, \inf-1 has a rigorous definition, which exists almost all of the time, and is somewhat closer to the naive, ill-defined algebraic interpretation. Under this definition, \inf-1 does not equal \inf.

snark15년 전

A "countable set" in that it has the same cardinality as the countable numbers (1, 2, 3, ...). For instance, the primes are countably infinite; you can map the set of primes onto the countable numbers (the first prime number is 2, the second is 3, the third is 5, ...). The term comes from Georg Cantor's work in the late 19th century.

Notably, while the set of rational numbers is countable, the reals (including transcendental numbers such as e and pi) are not; there's a fairly simply proof using diagonalization that you can Google, although that's not how Cantor originally demonstrated it.

lloeki15년 전

bug: 4=x^4 does not imply 4=∞(x), therefore you can't 'go back up' in the reasoning.

The psychological core of the "paradox" is that ∞(√2) actually equals 2, bringing credence to the second part. This does not actually makes the first part a valid proof of x=√2 either (as the bug still applies). All that it proves is that IF 2=∞(x) then it MUST be √2. Remains to be proven is that n(√2) converges when n->∞, and does indeed converge towards 2.

For the second part to be proven false, it is sufficient to show that convergent values are originating from [e^-e, e^1/e] and that for x in [e^-e, e^1/e] we have ∞(x) < 2.8. Therefore ∞(x) can't equal 4, ever.

amalcon15년 전

That's not quite what's going on here. The manipulations show that, if 4=x^(x^(...)), 4=x^4. The mistake is in beginning with the assumption that the former equation has a solution. If it doesn't, any manipulations that assume it does are invalid to begin with. Most people aren't exposed to the idea that an equation could have no solution all by itself, basically because polynomial equations always do.

lloeki15년 전

Well that's what I meant (or rather wanted to mean). I should have phrased the whole thing differently, but I'm still wrestling with English as a language, and rarely if ever used it in mathematics! I should definitely write more.

amalcon15년 전

Let's define x^x^(x...) as a closed form equation:

y=x^y

By simple substitution, y=x^(x^y), y=x^(x^(x^y)), and so on. But the closed form is easier to reason about. The equations given are equivalent to nonlinear systems of the form:

  y=x^y
  A=x^y

where a=2 or 4.

A little manipulation (logs are base x):

  y = x^y
  Assumption:  x>1 (log is not defined for all values if base is <1)
  log y = log x^y
  log y = y

Which is only true if y=1, * provided that x>1. But y is taken to be not 1 as a premise, and then x is derived to be greater than 1. The proof thus assumes a contradiction; naturally, it's not difficult to prove a contradiction if you assume one. The trick here was to hide the assumption in a strange equation.

* - Actually, on further review, this turns out to be false. Everything else was correct, though, and as it happens log(sqrt(2)) 4 != 4, so this still shows the flaw.

giberson15년 전

>But the part in brackets is the same as the whole, and hence is equal to 2. Thus we have 2=x^2

The bug is in the third line, the statement is invalid.

Obviously (x^x infinitum) is not the same as (x^x infinitum-1). For the same reason that x^x^x^x != x^x^x.

Sadly, I'm not math savvy. In fact the highest math level I passed in college was calc 3 and that was after taking calc 1 and calc 2 twice. So, my answer is based on logical reasoning rather than mathematical proof. Spock would likely indicate my logic is flawed, and only co-incidentally leads to the correct conclusion. Either way, this doesn't seem like a very interesting "math paradox".

RiderOfGiraffes15년 전

I'm really pleased that you've been willing to stick your head above the parapet and say this. What you've said seems to me to be a common line of reasoning, but it always confuses me.

In particular, you say:

    > Sadly, I'm not math savvy.

OK, that's cool, and I'm always willing to engage with people who want to learn.

    > Obviously (x^x infinitum) is not the same as (x^x infinitum-1).

Given that you're not math savvy, that's a pretty definite thing to say.

    > ... my answer is based on logical reasoning rather than mathematical proof.

Well, actually your answer seems to be based on intuition, and not on reasoning at all. You intuitively say that x^x^x^x != x^x^x, which is fair enough, and then leap to the conclusion that the infinite case must be like the finite case.

And there's a problem. The infinite case is not like the finite case. Similarly, the infinite sum 1/2+1/4+1/8+1/16+1/32+... insofar as it can be given any meaning at all, must be set equal to 1, even though all the finite partial sums are strictly less than 1.

    > this doesn't seem like a very interesting "math paradox".

But your "reasoning" is wrong. Perhaps you've dismissed it as "uninteresting" because you haven't really understood it? That's fair enough - I don't have a problem with that, but it would be nice to believe that you realised it.

Let me just say again that I'm really pleased you expressed your thoughts on this. I'm sure you're not the only one, and I find it useful to hear the way people think about these things. I give lots of presentations and masterclasses on math to all ranges of ages and experience. I need to know the intuitions people are using. So thanks.

giberson15년 전

Thanks for your response Rider, and let me re-iterate that I'm already aware my reasoning is wrong. Hence my allusion to how I think Spock would reply to me.

For clarification, I'm considering my observation that x^x infinitum is not the same as x^x infinitum-1 as based on reason rather than intuition because there is reasoning behind my statement. This may be an argument of semantic and thats certainly what I want to avoid since it's useless. But my answer isn't based on an instinctive knowing (intuition) it was based on the following considerations:

  3^1 = 3
  3^2 = 9
  3^3 = 27

For any n^x I could* think of, n^(x-1) was a different value.

If I think of infinity as some unbounded large number, then I think of x-1 as a smaller (by 1) unbounded large number. It may be fair to consider this an intuition, but since I'm basing it off of observations of smaller number examples I think its closer to an implication of the previous observed behavior.

With that argument I contest that my conclusion is in fact based on reason. However, I will and still do submit that my reasoning is that of a layman and may be (and probably is) incorrect. But, my conclusion seems to be inline with that of the other comments to this thread regarding the overall evaluation 2=4 through this number game. Now, I'm making an assumption here and by all rights perhaps making an ass of myself, but from my understanding of the majority responses is an agreement that 2=x^x(infinitum) does not lead to a conclusion of 2 != 4. Their arguments are more technical and perhaps outside my understanding level, but I think this is where my point lies about the math paradox being interesting. I would think that the paradox would only be interesting if the mathematical reasoning would be counter intuitive than a layman's intuition. That is to say, if the layman says it should be yes, mathematically it should be no in order for it to be interesting. (Again, I'm making the assumption original post argument doesn't hold water based on the refutation of the comment thread)

Thanks for taking the time to reply, level headed even without being insulting, to my [mathematically incompetent] comments. :)

  * while formulating this reply, it occurred to me that if n = 1, 1^x = 1^(x-1). 
  Here during this argument I've discovered a flaw in my reasoning. 
  Which I am conceding in my foot note as my comment above is meant to explain my original reasoning.

jerf15년 전

This isn't the only or necessarily the best way to think of infinity, but I think it's one of the better introductions to the idea: Think of infinity not as a concrete number (which it isn't, but without a replacement concept it's very easy to treat it as one because you know nothing else), but a statement that the answer to the question "Is there another?" is always "Yes". If you have 3, you can ask "Is there another?" and get three yeses, then you get no more. For 5, you get 5 yeses, then no more. For infinity, you never get told no.

There's no difference between infinity and infinity minus 1, because even after asking once "Is there another?" of your infinity, you've still got something that will forever answer "Yes" just as you did before.

This is also useful in things like proofs, because of that property. For any given finite instance of whatever thing you're trying to prove, you will never fully "use up" the infinity, but because of the way infinity never says "no", it allows you to write the proof without having to handle the cases where you run out of something, which can make things a lot trickier.

Similarly, "taking the limit as x goes to infinity" is said that way because x never reaches infinity, it's never "equal to" infinity, but for any number you choose, you can ask the question "Is x there yet?" and the answer is "No", always.

giberson15년 전

I contest that infinity != infinity-1. I base this on an assumption that there is more than one "infinity". Additionally, I will contest infinity != infinity necessarily. If we count on the number line from 1 to X, never stopping, that would be an infinite process. If we count on the number line from 1 to 2 by using fractional numbers that infinitely decrease, that is also an infinite process. However, both these infinite outcomes involve different components (different number values) and as such I consider them "different" infinities.

I admit this is a fairly week argument, and much closer to "intuition" than my original comment.

But, can you provide a mathematical proof that infinity = infinity-1?

jerf15년 전

One infinity is not necessarily equal to another. You are correct. Equality is not a very useful concept for infinities, really, you need other concepts like one-to-one mappings. Take one away from your infinity and you haven't done anything to its cardinality. A one-to-one mapping between the old infinity and the new one is still trivially available.

Words are useless. English is absolutely incapable of dealing with infinities in math, full stop. If you are arguing with English words, you're already wrong. That is why I said that I am presenting one useful way to think of them, and it is absolutely not the best or only way. What I showed was just a somewhat intuitive way to think of the infinity most commonly encountered in otherwise-conventional math and isn't totally correct there, either, it's just intuitively closer. Arguing about my intuitive shortcut misses the point entirely. You must go learn the math. I can't use the math terms to explain it to you here, either, because the correct understanding of the terminology would be tautological to understanding my point.

Also, I am not trying to be mean here. It took mathematicians decades to work out consistent definitions of infinity, and if you are going to insist on banging on the definitions and exposing the inconsistencies, an admirable enterprise, the only definitions that will stand up to that scrutiny are the ones that have been honed over the decades, and they do not fit well into HN posts. (Even if they are posted, they are thoroughly based on other mathematics that they would be meaningless until you understand those other things.) There's no Royal Road to infinity.

giberson15년 전

Very good points. Obviously, math is a language in itself and trying to make arguments from the trappings of an inexact translation to another language like English is folly.

>I am not trying to be mean here No worries, no maliciousness interpreted.

Thanks,

RiderOfGiraffes15년 전

This is not the place to give a complete introduction to transfinite arithmetic. However, let me try to give you a sense of what's happening.

You would agree that a collection of three cars is not the same as a collection of three cats, and yet you would agree that 3 equals 3. Just because we are counting different objects, that does not mean that the number of objects is necessarily different.

Specifically, we regard two collections as being the same size if we can put them in one-to-one correspondence with nothing left over. When we are talking about the size of a collection, it doesn't matter what is in the collection.

Similarly when we talk about the collection of positive even numbers and the collection of positive whole numbers. The collections are different, but it's possible to put each into one-to-one correspondence with the other. In this sense we feel that it's natural, right and appropriate to say that the two collections are the same size.

Using the same reasoning we can show that the collection of positive whole numbers (all of them) is the same size as the collection of positive even numbers. We pair off each number with its double, and that gives us a one-to-one correspondence. Given that such a correspondence exists it would be perverse to claim that the collections are of different sizes. The size of each is the same "infinity", even though the objects themselves are different.

And so we proceed. We can show that all sorts of collections are (in some cases surprisingly) the same size. Primes, even integers, rationals, algebraic numbers, all are collections that can be put into on-to-one correspondence.

Having become comfortable with that, it then comes as a shock to discover that there are collections that have infinitely many objects, but which cannot be put into one-to-one correspondence. If you try to pair up the real numbers with the positive whole numbers you always have some reals left over. Always. The generally accepted conclusion is that the collection of real numbers is genuinely bigger.

And so to your closing question.

Suppose we (quite reasonably) define X-1 as "take a collection of size X, remove one object, and X-1 is the number of objects left." If you start with an infinite set (any infinity, and any example) and remove one element, then the result can be put into one-to-one correspondence with the original. This can be proved. It's not hard, but it's a bit icky, and I won't do it here.

But under reasonable definitions of the terms, yes, I can provide a proof that infinity-1=infinity.

I hope that helps.

pbhjpbhj15년 전

>Using the same reasoning we can show that the collection of positive whole numbers (all of them) is the same size as the collection of positive even numbers. We pair off each number with its double, and that gives us a one-to-one correspondence. Given that such a correspondence exists it would be perverse to claim that the collections are of different sizes. The size of each is the same "infinity", even though the objects themselves are different.

This assertion has always troubled me. I think the trouble arises out of the reality² (in a metaphysical sense) of infinities and that the standard proof ("Cantor's backslash" is the name I learnt it as I think) is I think a case of petitio principii (proves the conclusion by assuming it to be true). The proof is beautiful but just always nags me as being a little too cunning.

I'm glad you wrote this as it's pretty much my mental understanding of transfinites (all the stuff about aleph-0 and beth numbers having been largely eroded from my memory) and I've been teaching my lad about infinities based on the cardinality of the set of natural numbers - he wrote this sum on the blackboard (at home) the other day:

  ∞+8 = ∞

and followed by crowning it with a picture of a number 8 turned in to a person wearing a hat.

tl;dr glad for confirmation as my young son is just grasping some of these results (though I don't imagine he's understanding the reason) and I hate to think I'm teaching him falsely¹.

---

² - like infinite coastlines in fractal geometry, the quantum limits of the real world hang there and tell me it's not consistent to maintain this mathematical model as useful given it's unreality.

¹ - yes I'm probably screwing up his learning basic arithmetic.

giberson15년 전

Thanks!

I'd like to say !Ding, I get it--but that wouldn't quite be accurate. Suffice to say, I understand how you can use a one-to-one correspondence method to proof two collection's equality--despite not really understanding how you are generating those values (as its clearly above my math level). Therefore, from now on, I will at least not use the argument infinity!=infinity-1 when attempting to reason some conclusion.

Thanks again, for taking the time to address my questions.

jules15년 전

This makes me wonder what are the fixpoints of f(a) = x^a, and whether they are attractive or not. For example if you take x=sqrt(2) then it appears that 2 is an attractive fixpoint while 4 is a repelling fixpoint.

Mahh15년 전

This reminds me of that 'troll math' post some time back. There was a square of edge length 1, but the corners were bent such that the total length(4) around the shape would stay the same. And then to infinity the corners were raggedly bent to keep the length around the shape the same while trying to say that as the bending went to infinity, the square actually became a circle, and that the circle's circumference was 4.

Logic when dealing with infinity is quite odd to deal with.

ianferrel15년 전

I don't see why this has anything to do with an infinite progression. The problem is that he has a system of two equations that are not consistent.

Replace the x^x^x^x... with "y". Now you have:

2 = y and 4 = y

OMG, 2 = 4!

Not every two equations can be solved with the same assignment of variables.

RiderOfGiraffes15년 전

You don't seem to have understood the point. Let's do it without re-using the same variable name.

Solving 2=y^y^y^... lets us find that y=sqrt(2). Therefore sqrt(2)^sqrt(2)^sqrt(2)^... = 2

Solving 4=z^z^z^... lets us find that z=sqrt(2). Therefore sqrt(2)^sqrt(2)^sqrt(2)^... = 4

Therefore 2=4.

Now, could you explain your resolution of the "paradox" more clearly?

vessenes15년 전

To restate; we can say:

1) Assume that an x exists such that 2=x^x^x^x^...

2) Then x = sqrt(2)

3) sqrt(2)^10 = 32

Therefore there is no such x.

In very brief, we can say that there is no number so infinitesimally close to 1 that infinite exponentiation of it does not 'blow up'.

Or to pull a Calc 1 term: x^x^x^x^... does not converge for all x >1. Simple.

RiderOfGiraffes15년 전

I don't understand why your step 3 is relevant. Observe that using r=sqrt(2) we have that the sequence r, r^r, r^r^r, r^r^r^r, ... does converge to 2. We are not performing arbitrarily large exponentiation of x.

Or, to pull a Calc 1 term, x^x^x^... does converge for x=sqrt(2).

Lozzer15년 전

It's the difference between x^(x^(x... and ((x^x)^x)...

ianferrel15년 전

Yes, I did misread the proof. Thanks for the explanation.

manmanic15년 전

2=x^x^x^x... cannot be solved, since x^x^x^x... will only converge to 0 (if x=0) or 1 (0<x<=1) or -1 (x=-1), ignoring complex numbers. So the original equation is false.

BoppreH15년 전

Wikipedia says it (roughly) converges if e^(−e) ≤ x ≤ e^(1/e)

http://en.wikipedia.org/wiki/Tetration#Extension_to_infinite...

[deleted]15년 전

RiderOfGiraffes15년 전

As a simple counter-example to your claim, the sequence:

  x, x^x, x^x^x, x^x^x^x, ...

when x = sqrt(2) is strictly increasing and bounded above, and therefore converges. It's not hard to show that it's bounded above by 2, because x^x^2 > x^x^x, and x^x^2 = x^2 = 2. Repeat for any length sequence of exponentiation.

fxj15년 전

maybe you meant 4=x^x^x^x^x^x... cannot be solved because there exists no x that solves the equation y=x^x^x^x^x... for a maximum value of y which is a positive real solution of the equation y = x^y. Thus, x = y^(1/y). The limit defining the infinite tetration of x fails to converge for x > e^(1/e) because the maximum of y^(1/y) is e^(1/e). The maximum value is smaller than 4 and thus no solution exists.

fxj15년 전

The equation 4=x^x^x^x^x^x^x... is not solvable. Read the wikipedia article. x^x^x^x^x... either converges to a value smaller than 4 or does not converge at all.

mfukar15년 전

It diverges for all values of x >= 2, iirc.

gersh15년 전

The first line assume 2=x^x^x^x^... which is impossible. We assume the author is asking: for what x \in R does 2=x^x^x^x^x^... Cognitive bias confuses us.

RiderOfGiraffes15년 전

I'd love to hear what you mean by this. In what sense is it impossible to have 2=x^x^x^x^... ??

gersh15년 전

There is x\in R such that 2=x^x^x^x^x^... Furthermore, I believe it should be possible to prove there is no field S such that R\in S, whereby 2=x^x^x^x^x^.... Complex numbers don't help.

sdizdar15년 전

In layman terms, infinity can't be a number, and neither can something times infinity. The problem is that infinity isn't really a number at all.

icandoitbetter15년 전

Don't want to sound condescending, but I don't think that seeing this in layman's terms has any merit. All of mathematics, although often built out of some intuitive, natural notions, is about definition: describe an object and then see what follows from your description. The description is not, in general, something that's "out there" and can be deduced by common-sense.

(Infinity, in fact, can be defined as a number and can be defined as a non-number in different contexts. But infinity is only something when it's part of a strictly defined mathematical universe with specific axioms.)

RiderOfGiraffes15년 전

Yes and no. Your point is true, but not entirely relevant, becuase infinite sums, infinite products, and infinite process can (sometimes) be shown to have definite and consistent meanings and interpretations.

The infinite tower of exponentiation is not (of itself) the problem.

arethuza15년 전

Infinitely small values (infinitesimals) can be quite useful:

http://en.wikipedia.org/wiki/Non-standard_analysis

VMG15년 전

The example doesn't claim that infinity is a number

dedward15년 전

This part doesn't ring true....

But the part in brackets is the same as the whole, and hence is equal to 2. Thus we have 2=x2

RiderOfGiraffes15년 전

No, that bit works. If 2=x^x^x^x^x^... then we can raise x to the power of both sides, giving

x^2=x^(x^x^x^x^...)

But the RHS is equal to x^x^x^x^... and hence equal to 2. Thus x^2=2.

dedward15년 전

Im' still thinking the original equation is not solvable = 2=x^x^x^x^x^x... is invalid.

RiderOfGiraffes15년 전

But (purely for brevity) using r=sqrt(2) we find that the sequence r, r^r, r^r^r, r^r^r^r, ... converges to 2. To my mind that makes it perfectly reasonable to say that 2=x^x^x^... has a solution with x=sqrt(2).

pbhjpbhj15년 전

I think implicit in the parent is that the equation is not solvable for ∀x, whilst you're saying ∃x where x converges toward a "solution" of 2 = x^^\inf.

cstross15년 전

Rolls Eyes

Someone here hasn't heard of Georg Cantor.

(More to the point, this is only a paradox if you assume that there is only one transfinite number. A brisk tour of "Infinity and the Mind" by Rudy Rucker sufficed to disabuse me of that idea some years ago. Time for a re-read, I think.)

Almaviva15년 전

That's either a nice troll or you don't know what you think you know!

cstross15년 전

Possibly. What do you think I don't know?

RiderOfGiraffes15년 전

Interesting question. The problem is that you've mentioned there being more than one transfinite number, when this question is actually about convergence. It appears that you've simply gone "Alleged paradox - infinity - must be ignorant about Cantor and the uncountables."

In fact, Cantor and the uncountables have nothing to do with it. If you think otherwise then I'd be interested in seeing a more complete explanation of your comment.

pbhjpbhj15년 전

It's 4am here and I need some sleep - choose 2 numbers from within the convergent range and do the same slight of hand with them. Clearly we can argue that x^^\inf = 4 has no convergence and so makes no sense but what if the equations are both in the converging region x € [e^−e, e^1/e] about [0.066, 1.445]:

  {1}    x^^\inf = 2

we've shown in the OP that x^2 = 2 is a solution.

Consider,

  {2}    x^^\inf = e^(2/e)

similarly to the OP we do

  {3}    x^ (x^^\inf) = x^ (e^(2/e))

but this is just

  {4}    x^^\inf =  x^ (e^(2/e))

and substituting on the left from {2}

  {5}    e^(2/e) = x^ (e^(2/e))

gives us (I think)

  {6}    x = e^ (2e^ (-1-(2/e))) ; ~~1.42267

Hence we've shown that

  {7}    2 ~~ 1.423

Which isn't as neat as 2==4 but equally troublesome.

There is no issue now with convergence is there (or there shouldn't be if I chose my numbers correctly)?

---

Notations

  ^^ : ascii version of Knuths up-arrow notation
  \inf : first countable infinity, aleph-0 
  € : in the range, it's really a Euro symbol, but close enough
  ~~ : approximately equal to

===

Solution

  y = x^2 
    = 9 give solution (3,9)
  y = x^2
    = 4 gives solution (2,4)

9=4?

They're just different points on the same curve ... doh! We're not showing that 9=4 we're showing that (3,9) and (2,4) are point solutions of y = x^2.

In the same way we have above 2 solution of y = x^^\inf.

cstross15년 전

Well, there's a transfinite implicit in the series of exponentiations -- we have aleph-0 exponentiations, don't we? And if we knock off that first one, we've got (aleph-0 minus 1) exponentiations.

(NB: I don't understand your use of the term "convergence" -- must be either something I've forgotten in the third of a century since I studied maths, or something above the level I reached. (Sub A-level.) (See also: Dunning-Kruger effect.))

gjm1115년 전

Convergence is what happens when you have a sequence of numbers, and they get closer and closer to some fixed number. Some sequences converge and some don't, and part of what's going on with this "paradox" is that one of the sequences it implicitly asks you to consider doesn't converge.

Yes, there's a transfinite number implicit in the sequence of exponentiations, namely aleph-0. (It might be better to say omega-0 -- an infinite ordinal rather than an infinite cardinal -- but that's a technicality that doesn't really matter here.)

But the problem with the "proof" doesn't have anything to do with the existence of infinite cardinals larger than aleph-0.

JoachimSchipper15년 전

Technically, f(x) converges to y as x -> infinity if for every epsilon there is an X such that |f(x) - y| < epsilon for all x > X ("the error becomes arbitrarily small"). For instance, 1/x converges to 0 as x -> infinity (pick X = 1/epsilon), but the sequence 0, 1, 0, 1, 0, ... does not converge to any y (pick epsilon < 1/2 to see this).

In particular, the sequence r, r^r, r^r^r, ..., where r^2 = 2, does converge (to 2; the first terms are approximately 1.41, 1.63, 1.76, ...); the sequence 2, 2^2, 2^2^2, ..., however, does not (the first terms are 2, 4, 16, ... - and it only grows faster from there.)

jules15년 전

That this doesn't have much at all to do with there being multiple transfinite numbers.

india15년 전

No infinity other than Aleph-naught appears to be involved in this. Why do you think the existence of other aleph numbers matters here?

[deleted]15년 전

waterhouse15년 전

I tried setting 9=x^x^x^x^..., and by similar logic, we can take x^[both sides] and get x^9 = x^x^... = 9. Which means x is a ninth root of 9; I think we may as well pick the real root. However...

  arc> (expt 9 1/9)
  1.2765180070092417
  arc> (expt (expt 9 1/9) that)
  1.3656685405664577
  arc> (expt (expt 9 1/9) that)
  1.3957179500635313
  arc> (expt (expt 9 1/9) that)
  1.405994788813235
  arc> (expt (expt 9 1/9) that)
  1.409526784112037
  arc> (expt (expt 9 1/9) that)
  1.4107427255115605
  arc> (expt (expt 9 1/9) that)
  1.4111615739170247

This doesn't look like it's converging to 9. This tells me that you can't just say "N = x^x^x^..." and expect there to be an x satisfying that condition, any more than there's an x satisfying "3 = 0*x". Actually, that one's arguably solvable with x = ∞. A better example is "|x| = -2", or perhaps "3 = f(x) where f(x) = 1 if x>0, 0 otherwise".

So I guess the takeaway is that the function "x^x^x^..." is a kind of "decision" function, if you know what I mean--one that requires making some sharp distinction somewhere, like the absolute value and "if x>0 then 1 else 0" functions. The x^x^... function is actually a limit, and evaluating it requires deciding that it does approach a limit in the first place. You can't expect these "decision" functions, as I call them, to always be as robust and invertible as usual mathematical functions. This result is probably surprising only because the limit is implicit.