4

Edit:

My question has been requested to close due to its apparent lack of clarity. My question is below under "Problem". If the information above it is redundant, please let me know in a comment. I will try my best to make my question as clear as possible.

I was looking at the equation $x^y=y$. If I want to isolate $x$, we would have $$x=y^{1/y}\stackrel{\small\rm{or}}{=}\sqrt[y]{y}.$$

Now let's consider $x^{x^y}=y$. I did not know how to do this, but I noticed something at first. If we let $x^y=y$, then this means $x^{x^y}=y\implies x^y=y$ by order of substitution. Notice that the same equation we substituted is in fact yielded; and, as mentioned previously, this equation would arrive at the fact that $x=y^{1/y}$. If we substitute this equation in, we obtain:

$$({y^{1/y}})^{(y^{1/y})^{y}}=y$$ which is clearly true.

But that means...

$$x^y=y\implies x^{x^y}=y\implies x^{x^{x^y}}=y\implies x^{x^{x^{x^y}}}=y\implies \cdots\tag1$$

It is now clear that we can look at this from a different perspective: if $x^y=y$ then similarly, $y=x^y$. We can substitute this on the LHS of the former equation, therefore resulting in that $x^{x^y}=y$. Clearly, this can be done ad infinitum, and this angle of looking at it is much easier to mentally grasp.

However.... if $x^y=y$ and $x^{x^y}=y$ then this means $x^y=x^{x^y}$ which doesn't make sense. Perhaps I am missing the part that $$x^y=x^{x^y}\color{red}{\iff x^y=y}$$ for otherwise we could let $x$ and $y$ be equal to anything, even each other, in the equation $x^y=x^{x^y}$ and quickly arrive at some problems.

Problem:

Now I have a problem when it comes to a certain substitution. If we can carry out the implications as shown in $(1)$ forever and ever, we will get the following equation:

$$x^{x^{x^{x^{x^{\,\,\style{display: inline-block; transform: rotate(60deg)}{\vdots}}}}}} =y.\tag2$$

Considering that $\sqrt[4]{4}=\sqrt{\sqrt{4}}=\sqrt{2}$ then one can quickly deduce from Eq. $(2)$ that $2=4$ which is clearly wrong. Let $y=2$ then $x=\sqrt{2}$. Let $y=4$ then $x=\sqrt [4]{4}=\sqrt{2}$. Thus $2=4$.

But since we just discussed the entire concept of this equation and there doesn't seem to be any flaws with it, why am I getting the absurd flaw now that $2=4$? Clearly I did something wrong with the substitution, but that's the easy part, so I believe I am overlooking something, something likely obvious $-$ but I don't understand!

May someone please correct me on this? Any help would be much appreciated (particularly hints).

Thank you in advance.

Mr Pie
  • 9,726
  • 4
    You have $y=2\to x=\sqrt{2},$ and then you have $y=4\to x=\sqrt{2}.$ But if $A\to C$ and $B\to C,$ why is $A$ equivalent to $B?$ That would be like saying, "If you're a dog, then you're a mammal. If you're a cat, you're also a mammal. Therefore, if you're a dog, you're a cat." In logic, we call this the fallacy of the undistributed middle. – Adrian Keister Jul 01 '19 at 18:19
  • @PeterForeman $a=-1$ and $b=2$, $c=0$ – Kat Jul 01 '19 at 18:20
  • You are getting an absurd thing because any x and y won't fit in the equation . You can do that only on the solutions . – Chinmaya mishra Jul 01 '19 at 18:21
  • 1
    @AdrianKeister that is a very good point. Why did I not think of this earlier? Maybe I should go to bed... it is 4:23am for me right now xD – Mr Pie Jul 01 '19 at 18:22
  • @PeterForeman Still fails if $a = 1$. It should work whenever $|a| \neq 1$. – CyborgOctopus Jul 01 '19 at 18:24
  • @AdrianKeister I will delete my question now after a minute. Thanks for your input. I will work on this tomorrow so I can get some rest. In the meantime, you can have some reputation. Thanks for the help, intuitive as it may be! I also appreciate everyone else's efforts here :) – Mr Pie Jul 01 '19 at 18:25
  • 1
    @MrPie: Thanks, have a good one! – Adrian Keister Jul 01 '19 at 18:30
  • 1
    Yesterday YouTube suggested me this video. O haven't watched yet but I hope it helps you. – Dog_69 Jul 01 '19 at 18:58
  • 1
    @Dog_69 thank you!! I did some research myself and came across this video. But the one you recommended to me is far better suited for this question, so again, thank you very much! :) – Mr Pie Jul 01 '19 at 21:22
  • 1
    @Mr.Pie Fantastic. It's a pleasure for me to be able to help you. – Dog_69 Jul 01 '19 at 23:22

1 Answers1

3

Your error mainly occurs in assuming that $y$ can be equal to $4$. It is known (see here) that the infinite 'power-tower' only converges to values within the range $(e^{-1},e)$ for real $x$ (where $x$ is chosen such that the 'power-tower' converges) which does not include $4$.

Peter Foreman
  • 20,095
  • Ha, I didn't know that! I was just about to delete this question in the thought that this was a ridiculous question merely wasting everyone's time; but now that there is an official answer, namely this one, I will refrain from doing that. I am also glad I learnt something, before I head off to bed. Unless there is another potentially better answer, you will earn the checkmark, sir! Thank you for your explanation! :) – Mr Pie Jul 01 '19 at 18:30
  • I think the tower part isn't necessary to show the contradiction. Simply take $y=x^y$, and observe that $y=4, x=\sqrt{2}$ satisfies this. It's natural to assume that you can substitute to obtain $y=x^{x^y}$, but the aforementioned values of $x$ and $y$ no longer work. What exactly forbids this self-substitution? – Vedvart1 Jul 01 '19 at 18:31
  • 1
    @Vedvart1 The aforementioned values of $x,y$ still work. $4=(\sqrt{2})^{(\sqrt{2})^4}$ – Peter Foreman Jul 01 '19 at 18:34
  • @PeterForeman No, they don't: $\sqrt{2}^{(\sqrt{2}^4)}=\sqrt{2}^2=2\not=4$. – Vedvart1 Jul 01 '19 at 18:38
  • 1
    @Vedvart1 Um, I am sorry to say that you are mistaken: $\sqrt{2}^4=2^{4/2}=2^2=4$. – Mr Pie Jul 01 '19 at 18:39
  • 1
    @MrPie Wow, not sure what I was thinking here... nevermind. Thank you for spelling it out lol – Vedvart1 Jul 01 '19 at 18:42
  • @Vedvart1 Haha, it's called being human. Everyone makes mistakes, don't stress :) – Mr Pie Jul 01 '19 at 21:24