# Find the value of x in this equation:

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I put this challenging puzzle in the “Brain Teaser and Puzzles” forum but nobody tried to solve it so far, so I decided to post it in this forum as well. I hope it does not amount to a “spam”.

…….._

….../……………......x

…..|………….......x

…..|…………......

<………........

…..|…...…….x

…..|……....x

…...\_....x

..........x = 2

In case it is not clear enough: x is raised to the power x, which is raised to the power x, which is raised to the power x and so on. The number of exponents to the power of x is infinite.

Have fun !

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Although I'm still pondering whether there is a solution, it would be a lot easier for you to learn to use the TeX features of this forum.

$\underbrace{x^{\cdots^x}}_{\infty \, \mbox{times}}=2$

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Although I'm still pondering whether there is a solution, it would be a lot easier for you to learn to use the TeX features of this forum.

$\underbrace{x^{\cdots^x}}_{\infty \, \mbox{times}}=2$

Hum... interesting feature.

Do you want a hint about the problem ?

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My solution

$x=\sqrt{2}$ I'm too lazy to check if $x=-\sqrt{2}$ works

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would you mind explaining how it is the square root of 2?

i'm just saying this because if i take root 2 to the power of root2 to .... 8 times... i get an answer much bigger than 2 and it'll keep on increasing.

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would you mind explaining how it is the square root of 2?

Starting again with the equation, x^x^x^…x = 2.

If you isolate the first x then the remaining sequence of exponents remains infinite:

(x) ^ ( x^x^x^…x) = 2

Therefore the second part, between parenteses, would still be equal 2 because it is still a infinite sequence ( ∞ - 1 = ∞ ). So by replacing the second part by 2 you get :

(x) ^ 2 = 2

Therefore:

x = sqrt(2).

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would you mind explaining how it is the square root of 2?

i'm just saying this because if i take root 2 to the power of root2 to .... 8 times... i get an answer much bigger than 2 and it'll keep on increasing.

You need to make sure you're calculating it properly. To calculate $y= x^{x^{x^x}}$, you calculate $x_1 = x^x$, then $x_2 = x^{x_1}$, then $y = x^{x_2}$ (i.e. you start from the top and work down).

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dave, i can do simple exponents.

i get

1.41

1.61

2

2.67

4

7.1

16

50.45

256

2545.46

65536

6479347.02

4294967296

41981937869758.1

1.84E+019

1.76E+027

3.40E+038

3.11E+054

1.16E+077

9.65E+108

1.34E+154

9.31E+217

#NUM!

#NUM!

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But the 2 as your 3rd term is a strong indication that you did $\left( \sqrt{2} ^{\sqrt{2}} \right)^{\sqrt{2}} =2$ instead of ${\sqrt{2}^{\sqrt{2}}}^{\sqrt{2}} = \sqrt{2}^{1.6\dots} < \sqrt{2}^2 = 2$ (where I took the 1.6 from your numbers without checking it).

@Blue_crystal: Very nice question. Two comments: (1) Perhaps don't give out the solution too early (I'm still not completely sure if it really works, but your explanation looks ok, so far). (2) Don't write stuff like "$\infty - 1 = \infty$", even if it's just meant as a handwaving argument (unless you can properly define an addition on R $\cup \{ \infty , - \infty \}$, which I spontaneously doubt) - it confuses others into thinking this type of addition did exist.

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atheist, an addition does exist on $\left[-\infty, \infty\right]$ - just treat the added points as zeros under addition. Its what the extended real system is all about

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atheist, an addition does exist on $\left[-\infty, \infty\right]$ - just treat the added points as zeros under addition. Its what the extended real system is all about

When talking about addition, I talk about addition as a group operation. So depending on which way round I understand you statement, I get either:

(1)$\infty + 1 = 1 +\infty = 1$. That violates the uniqueness of the neutral element.

(2) $\pm \infty + x = \pm \infty \, \forall \ x \in$ R. What's the inverse of infinity, then? It can only be $\pm \infty$ as far as I see. But then, the addition you just defined violates associativity: Assuming the inverse of $\infty$ was $-\infty$: $(\infty + \, -\infty) + 1 = 1 \neq \infty + (-\infty + 1 ) = \infty + \, -\infty = 0$.

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But why must the extended reals act as a group? So we declare $\left[-\infty, \infty\right]$ to be a semigroup with $\infty$ seen as a zero (so it has no inverse, so what?).

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(1)$\infty + 1 = 1 +\infty = 1$. That violates the uniqueness of the neutral element.

Atheist, who said that $\infty + 1$ OR $1 +\infty$ EQUALS 1 ?

Both, $\infty + 1$ OR $1 +\infty$ EQUALS $\infty$.

This is derived from a more general rule:

If x is a Real number such that $-\infty < x < \infty$ then:

$\infty + x = \infty$

$x + \infty = \infty$

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Atheist, who said that $\infty + 1$ OR $1 +\infty$ EQUALS 1 ?

River_rat's post was slightly disambiguous considering that matter, so I covered this possibility in case (1). The case you meant is covered in (2), as you certainly saw.

This is derived from a more general rule:

If x is a Real number such that $-\infty < x < \infty$ then:

$\infty + x = \infty$

$x + \infty = \infty$

Of course you can define almost arbitrary non-invertible, non-closing operations on any sets - it's just not what I understand under "properly defined addition" and not what I expect when reading a plus sign. I assume that I'm not the only one thinking of at least a group operation when reading a plus sign, hence my comment that you shouldn't write such stuff to avoid confusion.

@River_rat: Not only do not all elements have an inverse, but your "addition" additionally () also doesn't close. As soon as you define $\infty + (-\infty)$ to equal something, you run into exactly the problem with associativity I mentioned.

To me, the "addition" you defined looks like little more than an implementation of "but I want this operation to exist" (don't get me wrong; that's an acceptable motivation) that quickly runs into its limits. With the limited range the operations might serve as some shortcut-operations like when checking the limits of composite functions where "undefined" then means "analyse with proper methods".

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Of course you can define almost arbitrary non-invertible, non-closing operations on any sets - it's just not what I understand under "properly defined addition" and not what I expect when reading a plus sign. I assume that I'm not the only one thinking of at least a group operation when reading a plus sign, hence my comment that you shouldn't write such stuff to avoid confusion.

Atheist, it is immediately obvious to anyone that if you add or remove a small quantity from a infinity number of the same objects the total of objects would continue to be infinite.

I think that the excess of mathematical semantic formalism is blurring your mind.

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Atheist, it is immediately obvious to anyone that if you add or remove a small quantity from a infinity number of the same objects the total of objects would continue to be infinite.

I wasn't complaining about the part in parantheses in the term x^(x^x^...) to be infinite, I telling you that you shouldn't write "because $\infty -1 = \infty$" as a justification because that does seem to need more justification than the original statement.

I think that the excess of mathematical semantic formalism is blurring your mind.

Telling someone that an excess of mathematical formalism (not exactly sure what you meant by semantics) is blurring the mind seems counter-productive to the purpose of a mathematics section of a science forum to me.

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Telling someone that an excess of mathematical formalism (not exactly sure what you meant by semantics) is blurring the mind seems counter-productive to the purpose of a mathematics section of a science forum to me.

No offence intended, but what else could explain why someone would conclude this absurdity "$\infty +1 = 1 + \infty = 1$." ?

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I'm afraid I have side with Atheist on this one, statements that don't quite work or that require a twisting of common functions have no place in a mathematical process. And Blue's proof holds without it anyway.

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No offence intended, but what else could explain why someone would conclude this absurdity "$\infty +1 = 1 + \infty = 1$." ?

He was reading river_rats post:

atheist, an addition does exist on $\left[-\infty, \infty\right]$ - just treat the added points as zeros under addition.
He didn't "conclude" it, he just demonstrated why it couldn't be correct.

(so it has no inverse, so what?).
Addition is a GROUP operation.
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Addition is a GROUP operation.

Personally, I'd accept addition as a semi-group operation, too. But I do not really know what the common stance is, there. Problem is, that the given rules (and I'm not talking about the $\infty = 0$ version here ) either violate closure or -if you define $\infty + \, -\infty$ to equal something- associativity. Hence, the operation defined can't be the operation on a semi-group, either.

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(2) Don't write stuff like "$\infty - 1 = \infty$", even if it's just meant as a handwaving argument (unless you can properly define an addition on R $\cup \{ \infty , - \infty \}$, which I spontaneously doubt) - it confuses others into thinking this type of addition did exist.

I do not know a lot about mathematical formal language ( though I doubt that the lack of it significantly prevents someone to think logically ). I am not a professional or dedicated mathematician. I just play with it sometimes as a hobby.

But since the number 1 is obviously a Real number, what is the point of defining an addition on R $\cup \{ \infty , - \infty \}$ ?

Are you saying that I only can formally make operations between real numbers and $\infty$ and $-\infty$ if I formally extend its "population" to include $\infty$ and $-\infty$ ?

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No offence intended, but what else could explain why someone would conclude this absurdity “ $\infty + 1 = 1 + \infty = 1$." ?

He was reading river_rats post:

He didn't "conclude" it, he just demonstrated why it couldn't be correct.

Oops.. I apologise for the mistake, Atheist.

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But since the number 1 is obviously a Real number, what is the point of defining an addition on R $\cup \{ \infty , - \infty \}$ ?

Are you saying that I only can formally make operations between real numbers and $\infty$ and $-\infty$ if I formally extend its "population" to include $\infty$ and $-\infty$ ?

I don't understand what you're asking.

When you give a mathematical expression involving an operator like x+y=z, then I assume the operation to be defined for the x, y and z being members of some common set. It is very unlikely that someone adds a tree (or the tree ) to seven Volt. I assumed the "1" in your equation stood for an element of the real numbers or a subset thereof (the rationals or the integers). Infinity is not a member of those sets; it's usually not even considered a number at all.

Therefore, for the statement $\infty + 1 = \infty$ to make any sense to me, I expect a set that consists of at least the two elements 1 and $\infty$ (the minus case makes it a bit more complicated but let's forget that for now). R $\cup \{ \infty, -\infty \}$ simply seemed like the most straightforward one of the possible sets. The notation simply means "all the real numbers and two additional elements called plus and minus infinity".

Well, the rest is written already: Your calculation rules on this extended set didn't convince me to be obvious enough that they wouldn't need further explanation on (a) what exactly they are and (b) on where the big limits and differences to the calculation rules on the reals are (lack of some inverse elements under addition or no associativity, no closure at least under multiplication).

Mathematical definitions possibly worthwhile looking up (for those who don't understand the terms "inverse", "closure", "associativity" and their relevance here): Group, Body.

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Atheist, perhaps if I describe briefly my mathematics background you would understand me better.

First of all, english is not my first langauge ( it is the 3rd one ).

The last time that I studied mathematics seriously was on year 1973 ( 34 years ago ) in Brazil. It was maths of secondary school level. Furthermore, at that time, I was taught that infinity was considered a “normal” number that could be operated with ordinary numbers ( 1, 2, 3, etc ) following a given set of rules.

So I was not quite aware of this more modern mathematics and the way it formalizes operations and classifies numbers in different sets.

I sporadically learned little bit about it but I assumed that the old way of operating with infinity is still valid.

Now I realize that it may not be the case.

OK, then let me start fitting myself in this new context. I apologize if I still don’t deal properly with maths current formalities. So, assuming that I reasonably understand the new math notations I would say that :

If all the operands belong to the set R $\cup \{ \infty, -\infty \}$ then $\infty - 1 = \infty$.

Do you agree on that or do you still have some formal / logical objections ?

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Personally, I'd accept addition as a semi-group operation, too. But I do not really know what the common stance is, there. Problem is, that the given rules (and I'm not talking about the $\infty = 0$ version here ) either violate closure or -if you define $\infty + \, -\infty$ to equal something- associativity. Hence, the operation defined can't be the operation on a semi-group, either.

Take $\mathbb{R}$ and adjoin two points to it, namely $+\infty$ and $-\infty$, denote this new set by $\left[-\infty, +\infty\right]$.

Define addition as follows :

$a + b = \begin{cases} a + b & \text{ if } a, b \in \mathbb{R} \\ +\infty & \text{ if } a = +\infty \text{ or } \left( a \in \mathbb{R} \text{ and } b = +\infty \right) \\ -\infty & \text{ if } a = -\infty\ \text{or } \left( a \in \mathbb{R} \text{ and } b = -\infty \right) \end{cases}$

This operation is closed and associative so we have a semigroup with both $+\infty$ and $-\infty$ acting as zeros on $\mathbb{R}$ under addition and is an extension of the normal addition on $\mathbb{R}$. In fact both added points are left-zeros of the entire space

The trick here is to not demand that addition be commutative for all possible values, only for real values (else define a local semigroup and work in the two point compactification of $\mathbb{R}$ if you want to keep the commutativity.) For our original problem though we only actually need to add one point to $\mathbb{R}$ so the issue is actually a moot point.

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