[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)

[1] 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... (pdf)

[j2kun]

I think your argument is avoiding the real issues. One problem is that the simplification of setting y = x^y introduces new solutions that are not solutions to the original equation. While any solution of the original must also satisfy y = x^y, the reverse is not true.

Even worse, the OP's algebraic manipulations assume there exists a solution to the equation. It's not logically sound to assume something exists and use that fact to prove that it exists.

[voicereasonish]

Also the big red flag in the reasoning is:

> "But take the square root of each side".

Sorry, but that's not really something you can do without recognising the fact that there are two square roots.

[j2kun]

This is a red herring.