Two Equals Four (2011)

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)

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

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.

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). The proof can be found here: (http://www.maa.org/sites/default/files/pdf/upload_library/22...)

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.

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

Can someone explain what this means?

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.

For more pseudomathematics like this: http://www.crank.net/maths.html

EDIT: Ignore this.

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

why did you write this?