

Find the "bug" - Two equals Four - a mathematical "paradox" - RiderOfGiraffes
http://www.solipsys.co.uk/new/TwoEqualsFour.html?HN1

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albertzeyer
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.

~~~
tmachinecharmer
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. :)

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

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anigbrowl
Convergence is not an identity.

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

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

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pwagland
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
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
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?

~~~
RiderOfGiraffes
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.

~~~
lmkg
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_.

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lloeki
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
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
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.

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amalcon
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.

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giberson
>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
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
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.

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jules
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.

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Mahh
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.

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ianferrel
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
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?

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

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

[http://en.wikipedia.org/wiki/Tetration#Extension_to_infinite...](http://en.wikipedia.org/wiki/Tetration#Extension_to_infinite_heights)

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

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

~~~
gersh
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.

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sdizdar
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.

~~~
icandoitbetter
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.)

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

~~~
RiderOfGiraffes
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
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.

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

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

~~~
RiderOfGiraffes
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
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.))

~~~
gjm11
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.

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waterhouse
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.

