
Two Equals Four (2011) - ColinWright
http://www.solipsys.co.uk/new/TwoEqualsFour.html?HN0
======
throwaway13qf85
To explain the flaw in the reasoning, it's easiest to generalize -

    
    
        y = x ^ (x ^ x ^ x ^ ...)
          = x ^ y
    

and hence

    
    
        x = y ^ (1/y)
          = exp( 1/y * log(y) )
    

which is defined for y > 0.

Graphing this using Google plot [0] or Wolfram Alpha [1], or just
differentiating, reveals that it has a maximum at x = e, and that x takes the
value sqrt(2) at both y = 2 and y = 4.

Therefore the inversion, which is the original equation

    
    
        y = x ^ (x ^ x ^ x ^ ...)
    

is dual-valued, i.e. it is not a mathematical function. For it to be a
function, you need to specify whether you are on the upper branch (so that y =
4) or the lower branch (so that y = 2). Deliberately obfuscating the
difference between the two branches leads to the conclusion that 2 = 4.

The square root function can lead to a similar confusion, since there are two
solutions to the equation

    
    
        y = x^2
    

and hence the "function"

    
    
        x = sqrt(y)
    

is not really a function unless we specify whether we are on the upper (x > 0)
or lower (x < 0) branch. By convention we interpret sqrt(y) to be the upper
branch, and write -sqrt(y) for the lower branch, but there's nothing that
forces that choice.

Obfuscating the difference between the two branches could lead one to conclude
that 1 = -1, although the fallacy is more obvious in this case, since everyone
is familiar with the fact that the square root function is dual-valued.

[0]
[https://www.google.co.uk/search?q=y%5E(1%2Fy)](https://www.google.co.uk/search?q=y%5E\(1%2Fy\))

[1]
[http://www.wolframalpha.com/input/?i=y%5E(1%2Fy)](http://www.wolframalpha.com/input/?i=y%5E\(1%2Fy\))

~~~
jds375
It's certainly not a mathematical function. That being said, to even write 4 =
x ^ (x ^ x ^ x ^ ...) is a bit misleading. The function diverges for y not in
the range of e^-1 < y < e. I think that's a bit simpler and elegant way to
look at it. A proof can be found on page 240:
[http://www.maa.org/sites/default/files/pdf/upload_library/22...](http://www.maa.org/sites/default/files/pdf/upload_library/22/Chauvenet/Knoebelchv.pdf)
(pdf)

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belovedeagle
I don't believe all the comments pointing out that x^(x^...) converges only in
a certain range are the best way to understand what's going on here. The way I
see it, the "solve for x" puzzle really just serves to obscure the true
reasoning used:

    
    
        if y=sqrt(2)^y and z=sqrt(2)^z, then y=z.
    

When put like this, we can see that this is clearly false: solutions to the
equation need not be equal any more than all zeros of a polynomial need be
equal.

~~~
j2kun
Well if you observe the tower power is an injective function then that just
fixes everything :)

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jds375
This does not work because it does not converge for the number 4. The equation
x^x^x^x^...=y is only valid for y in the range e^-1 < y < e, although proving
this is a bit complicated. More info on hyperpowers can be found here:
([http://www.faculty.fairfield.edu/jmac/ther/tower.htm](http://www.faculty.fairfield.edu/jmac/ther/tower.htm)).
The proof can be found here:
([http://www.maa.org/sites/default/files/pdf/upload_library/22...](http://www.maa.org/sites/default/files/pdf/upload_library/22/Chauvenet/Knoebelchv.pdf))

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bartkappenburg
Just as wrong as saying:

1) Inf - 2 = Inf

2) Inf - 4 = Inf

Hence: 2 = 4

Alway be careful when doing operations on infinite numbers, sequences etc.
These are the kind of things that roll out of that.

The author just found a way to write something similar in a more fancier way.

~~~
jds375
The difference is that the equation is actually correct for y=2. It doesn't
apply for y=4 because it only converges in the range e^-1 < y < e.

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Kiro
> But the part in brackets is the same as the whole

Can someone explain what this means?

~~~
jds375
So the logic goes, consider x^x^x^...=2. We have an infinite tower of x's.
Now, suppose we group all but the first x as so: x^(x^x^x^..)=2. The number of
x's in the parenthesis is still infinite (infinity minus 1 is still infinity).
We already know by hypothesis that x^x^x^...=2. So we can substitute this in
for the term in parenthesis since they are both an infinite tower of x's and
get: x^2=2.

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cousin_it
So to recap:

1) If we define f(x)=x^x^x^... as the limit of the sequence x, x^x, x^x^x, ...
then f(sqrt(2))=2, not 4, and in fact there's no such x that f(x)=4.

2) If we instead "solve" the equation y=x^x^x... by dubious manipulations with
infinite expressions, we get x=y^(1/y), so y=2 and y=4 both lead to x=sqrt(2).

When playing with infinite expressions, you must always keep in mind how
they're defined in terms of limits. If you learn to do that, many puzzles with
infinity will stop bothering you forever, like Zeno's paradox, 0.999...=1,
etc.

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vezzy-fnord
For more pseudomathematics like this:
[http://www.crank.net/maths.html](http://www.crank.net/maths.html)

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Houshalter
EDIT: Ignore this.

I tried doing this for x = sqrt(2) and it quickly approaches infinity. What is
the actual solution for x?

~~~
ColinWright
I don't see what you're doing to make it approach infinity. Let r0=sqrt(2) and
r(i+1)=sqrt(2)^r(i). Then

    
    
      r0 = 1.4142135624
      r1 = 1.6325269194
      r2 = 1.7608395559
      r3 = 1.8409108693
      r4 = 1.8927126968
      r5 = 1.9269997018
      r6 = 1.9500347738
      r7 = 1.9656648865
    
      ...
    

This is clearly approaching 2. Not sure what you were doing.

 _Edit: Ah! I see - you 've got the association the wrong way round. The
convention is that a^b^c is a^(b^c). This is because (a^b)^c can just as
easily be expressed as a^(bc), so you get more expressive power._

~~~
Houshalter
You're right. I just completely abused math and did previous^x rather than
x^previous. Fixed the code and it does become 2. Thanks.

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aschwa32
why did you write this?

~~~
ColinWright
It's a puzzle, and a good one. There are puzzles where the challenge is to
work out what the insight is needed to see the solution, and there are puzzles
where the challenge is to work out exactly what calculation is needed (and
then to do it).

In this case the puzzle is to find the error in the reasoning. Other examples
of this include the two-envelope problem, the unexpected hanging, and the case
of the missing dollar. Sometimes puzzles like this reveal deep insights into
the subject, sometimes they point out that "obvious" things aren't obvious,
and sometimes they reveal weaknesses in our understanding or reasoning.

This is - for some students - an excellent introduction to why we need to be
careful when dealing with certain types of mathematical manipulation.

