10 ms·

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

anigbrowl 16y ago
Convergence is not an identity. http://en.wikipedia.org/wiki/Tetration http://en.wikipedia.org/wiki/Tetration
- RiderOfGiraffes 16y ago
  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.
  - anigbrowl 16y ago
    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?
    - RiderOfGiraffes 16y ago
      It appears not. I don't remember seeing any.
- deleted 16y ago
  [deleted]
manmanic 16y ago
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.
- BoppreH 16y ago
  Wikipedia says it (roughly) converges if e^(−e) ≤ x ≤ e^(1/e) http://en.wikipedia.org/wiki/Tetration#Extension_to_infinite_heights http://en.wikipedia.org/wiki/Tetration#Extension_to_infinite...
  - deleted 16y ago
    [deleted]
- RiderOfGiraffes 16y ago
  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.
- fxj 16y ago
  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.
pwagland 16y ago
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…
- RiderOfGiraffes 16y ago
  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.
  - brown9-2 16y ago
    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?
    - snark 16y ago
      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.
    - RiderOfGiraffes 16y ago
      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.
cstross 16y ago
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.)
- Almaviva 16y ago
  That's either a nice troll or you don't know what you think you know!
  - cstross 16y ago
    Possibly. What do you think I don't know?
    - RiderOfGiraffes 16y ago
      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.
      - cstross 16y ago
        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.))
        JoachimSchipper 16y ago
jules 16y ago
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.
deleted 16y ago
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albertzeyer 16y ago
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.
- ced 16y ago
  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),...
  - albertzeyer 16y ago
    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.
- tassl 16y ago
  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)
dedward 16y ago
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
- RiderOfGiraffes 16y ago
  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.
  - dedward 16y ago
    Im' still thinking the original equation is not solvable = 2=x^x^x^x^x^x... is invalid.
    - RiderOfGiraffes 16y ago
      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).
      - pbhjpbhj 16y ago
        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.
giberson 16y ago
>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".
- RiderOfGiraffes 16y ago
  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.
  - giberson 16y ago
waterhouse 16y ago
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.
fxj 16y ago
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.
- mfukar 16y ago
  It diverges for all values of x >= 2, iirc.
amalcon 16y ago
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.
ianferrel 16y ago
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.
- RiderOfGiraffes 16y ago
  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?
  - ianferrel 16y ago
    Yes, I did misread the proof. Thanks for the explanation.
  - vessenes 16y ago
    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.
    - RiderOfGiraffes 16y ago
      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).
      - Lozzer 16y ago
Mahh 16y ago
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.
sdizdar 16y ago
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.
- arethuza 16y ago
  Infinitely small values (infinitesimals) can be quite useful: http://en.wikipedia.org/wiki/Non-standard_analysis http://en.wikipedia.org/wiki/Non-standard_analysis
- RiderOfGiraffes 16y ago
  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.
- icandoitbetter 16y ago
  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.)
- VMG 16y ago
  The example doesn't claim that infinity is a number
lloeki 16y ago
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.
- amalcon 16y ago
  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.
  - lloeki 16y ago
    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.
gersh 16y ago
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.
- RiderOfGiraffes 16y ago
  I'd love to hear what you mean by this. In what sense is it impossible to have 2=x^x^x^x^... ??
  - gersh 16y ago
    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.

Convergence is not an identity. http://en.wikipedia.org/wiki/Tetration http://en.wikipedia.org/wiki/Tetration

RiderOfGiraffes 16y ago
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.
- anigbrowl 16y ago
  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?
  - RiderOfGiraffes 16y ago
    It appears not. I don't remember seeing any.
deleted 16y ago
[deleted]

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.

BoppreH 16y ago
Wikipedia says it (roughly) converges if e^(−e) ≤ x ≤ e^(1/e) http://en.wikipedia.org/wiki/Tetration#Extension_to_infinite_heights http://en.wikipedia.org/wiki/Tetration#Extension_to_infinite...
- deleted 16y ago
  [deleted]
RiderOfGiraffes 16y ago
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.
fxj 16y ago
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.

pwagland 16y ago

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…

RiderOfGiraffes 16y ago
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.
- brown9-2 16y ago
  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?
  - snark 16y ago
    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.
  - RiderOfGiraffes 16y ago
    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.

cstross 16y ago

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.)

Almaviva 16y ago
That's either a nice troll or you don't know what you think you know!
- cstross 16y ago
  Possibly. What do you think I don't know?
  - RiderOfGiraffes 16y ago
    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.
    - cstross 16y ago
      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.))
      - JoachimSchipper 16y ago

jules 16y ago

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.

deleted 16y ago

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albertzeyer 16y ago

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.

ced 16y ago
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),...
- albertzeyer 16y ago
  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.
tassl 16y ago
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)

dedward 16y ago

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

RiderOfGiraffes 16y ago
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.
- dedward 16y ago
  Im' still thinking the original equation is not solvable = 2=x^x^x^x^x^x... is invalid.
  - RiderOfGiraffes 16y ago
    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).
    - pbhjpbhj 16y ago
      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.

giberson 16y ago

>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".

RiderOfGiraffes 16y ago
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.
- giberson 16y ago

waterhouse 16y ago

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.

fxj 16y ago

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.

mfukar 16y ago
It diverges for all values of x >= 2, iirc.

amalcon 16y ago

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.

ianferrel 16y ago

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.

RiderOfGiraffes 16y ago
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?
- ianferrel 16y ago
  Yes, I did misread the proof. Thanks for the explanation.
- vessenes 16y ago
  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.
  - RiderOfGiraffes 16y ago
    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).
    - Lozzer 16y ago

Mahh 16y ago

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.

sdizdar 16y ago

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.

arethuza 16y ago
Infinitely small values (infinitesimals) can be quite useful: http://en.wikipedia.org/wiki/Non-standard_analysis http://en.wikipedia.org/wiki/Non-standard_analysis
RiderOfGiraffes 16y ago
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.
icandoitbetter 16y ago
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.)
VMG 16y ago
The example doesn't claim that infinity is a number

lloeki 16y ago

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.

amalcon 16y ago
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.
- lloeki 16y ago
  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.

gersh 16y ago

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.

RiderOfGiraffes 16y ago
I'd love to hear what you mean by this. In what sense is it impossible to have 2=x^x^x^x^... ??
- gersh 16y ago
  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.