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

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

by RiderOfGiraffes • 15 years ago

67 69 15 years ago

solipsys.co.uk

anigbrowl2• 15 years ago

Convergence is not an identity.

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

RiderOfGiraffes15• 15 years 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.

anigbrowl2• 15 years 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?

manmanic• 15 years 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• 15 years ago

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

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

RiderOfGiraffes15• 15 years 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.

fxj2• 15 years 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• 15 years 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…

RiderOfGiraffes15• 15 years 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• 15 years 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?

cstross3• 15 years 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• 15 years ago

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

cstross3• 15 years ago

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

jules2• 15 years 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.

albertzeyer2• 15 years 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• 15 years 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),...

albertzeyer2• 15 years 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• 15 years 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)

RiderOfGiraffes15• 15 years ago

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.

tmachinecharmer• 15 years ago

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. :)

Lozzer2• 15 years ago

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

dedward2• 15 years 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

RiderOfGiraffes15• 15 years 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.

dedward2• 15 years ago

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

giberson5• 15 years 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".

RiderOfGiraffes15• 15 years 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.

giberson5• 15 years ago

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.

jerf2• 15 years ago

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.

giberson5• 15 years ago

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?

waterhouse• 15 years 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.

fxj2• 15 years 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• 15 years ago

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

amalcon2• 15 years 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.

ianferrel2• 15 years 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.

RiderOfGiraffes15• 15 years 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?

ianferrel2• 15 years ago

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

vessenes• 15 years 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.

Mahh• 15 years 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• 15 years 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• 15 years ago

Infinitely small values (infinitesimals) can be quite useful:

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

RiderOfGiraffes15• 15 years 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• 15 years 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• 15 years ago

The example doesn't claim that infinity is a number

lloeki2• 15 years 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.

amalcon2• 15 years 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.

lloeki2• 15 years 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.

gersh2• 15 years 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.

RiderOfGiraffes15• 15 years ago

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

gersh2• 15 years 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.

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

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