

If x is a fraction, x^x and x^x^x are irrational. But can x^x^x^x be rational? - xSwag
http://math.stackexchange.com/questions/430797/can-xxxx-be-a-rational-number

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carstimon
Let me try to explain Schanuel's conjecture a little bit, if you're lost on
words like transcendence degree. This turned out longer than I hoped so I hope
it's not just a wall of text.

sqrt(2), π, π/2, π^2, and e are all irrational. Now think about π and π/2:
these are very closely related. What I mean is:

π - 2(π/2) = 0

So we can add some combination of the two numbers, and get 0.

Now what about π and e? These feel different. After working with them one gets
the feeling that a _π + b_ e will never be zero, if a and b are rational[1].
If that's true, then π and e are called "linearly independent over the
rationals". We do not know whether π and e are linearly independent over the
rationals. (In particular, we don't know whether e+pi is rational.) It turns
out that stuff like that is hard.

Now, there's an even stronger thing than being "linearly independent", and
that's being "algebraicly independent". When talking about linear indepence we
only thought about multiplying by rationals and adding (linear relations). For
example, π and π/2 are NOT algebraicly independent. What about π and π^2?
Well: (π)^2 - (π^2) = 0 which is rational. This shows that π and π^2 are also
not algebraicly independent. (The equation is obvious, but think about what it
means: on the left of the minus, I have π, and I'm raising it to a power of 2.
on the right of the minus, I have π^2, and I'm raising it to a power of 1). We
do not know whether π and e are mutually transcendental.

So you should think: linear independence of rationals is a weak way of saying
"these numbers aren't that related." Algebraic independence means there even
less related. You can think of both as ways of proving huge amounts of numbers
are irrational. If we can prove that e and pi are algebraicly independent then
for example we would know e + 5e^3 + e^7 + pi + 4pi^4 is irrational.

Now, you can understand what this conjecture says. Take z_1, z_2, ..., z_n
which are complex numbers. If you want, you can just imagine them to be
irrational numbers. Suppose they are linearly independent over the rationals.
The conjecture is that: at least n of z_1, z_2, ..., z_n, e^(z_1), e^(z_2),
..., e^(z_n) are algebraicly independent.

[1] And not both 0 ;)

Related: If you know some linear algebra, think about this. sqrt(2) is
irrational. Think of 1 and sqrt(2) has "vectors" which you can multiply by
scalars, but the only scalars you're allowed to multiply by are rational
numbers. In this way you get a "vector space over the rationals" which
includes things like 3+sqrt(2), 5, and 7sqrt(2). This vector space is also a
field: we can multiply things together, and we get another element of the
vector space: (3+sqrt(2))(7sqrt(2)) = 3x7sqrt(2) + 7x2 = 14 + 21sqrt(2).

Because sqrt(2) is algebraic over the rationals, this vector space-field thing
is finite. But now do it with pi instead of sqrt(2), and you'll get an
infinite dimensional vector space with the vectors 1, pi, pi^2, pi^3....
Elements of this field are things like 4+pi+7pi^3.

~~~
swordswinger12
In the first equation involving pi - how can the sum of two positive numbers
be zero?

~~~
XaspR8d
I believe he meant minus. For the purposes of linear independence, there is no
significance of positive vs negative coefficients.

EDIT: Wow, go team.

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ctdonath
The proof given is fascinating, but...translucent.

Can we get a simple, and correct, "yes" or "no" here?

~~~
seliopou
The answer is no, assuming Schanuel's conjecture.

Any set of algebraically independent numbers over a field are going to be
transcendental over that field. In this case, being transcendental means
they're not rational. And the number in question is in that set, so it's not
rational.

~~~
cookingrobot
But Schanuel's conjecture hasn't been proven. So would you say this reduces to
Schanuel's conjecture?

~~~
ot
The Schanuel's conjecture is widely believed to be true. A friend of mine
actually did his Ph.D. on it, and the tools used to attack the problem are
truly mind-blowing.

Basically, Zilber used model-theory to construct a field that "behaves like C"
(the field of complex numbers), and where the conjecture is forced to be true.
Not only this field exists, but it is unique. It would be no surprise then if
this field was exactly C, and this would prove the conjecture for complex
numbers.

Note that to prove that the two fields are isomorphic it would suffice to
prove that the "artificial" field has a continuous automorphism of order two
(that would basically be the conjugation in C); then its fixed set would be
the real line, and it would be easy to construct the rest of the isomorphism.

The existence of such an automorphism is even less surprising, yet, nobody has
managed to prove it.

~~~
seliopou
That sounds so baller. But one question.

You say that he constructed a field that "behaves like C." Granted, you put
that phrase in quotes for a reason, but it seems to me as though if you pinned
down the behavior of multiplication in this artificial field, then conjugation
wouldn't be too far away. So what does it mean to "behave like C"? Or better
yet, do you have a link to you friend's thesis or other papers?

~~~
ot
Of course the details are very tricky :) "behaves like C" means that you
construct a first-order theory that all the properties of C that you want,
plus the conjecture.

I don't know much about this, but I think that there is no conjugation in the
language of exponential fields, and if you try to add it everything becomes
intractable (for one, every proof would first require to prove Schanuel's
conjecture).

The Wikipedia has a few references:
[https://en.wikipedia.org/wiki/Schanuel%27s_conjecture#Zilber...](https://en.wikipedia.org/wiki/Schanuel%27s_conjecture#Zilber.27s_pseudo-
exponentiation)

------
aidos
I don't think I've seen this notation before:

    
    
        (a,b) = 1
    

I guess it means gcd(a, b) = 1

~~~
vog
Yes, this is a short notation of gcd, and it is quite common among
mathematicians.

~~~
Dewie
Is there any good reason for this short notation other than saving three
letters for the well-initiated? Does it make sense in the context of other
mathematical notation conventions (ie could you guess that it meant gdc(a,b)
without a lot of incriminating context pointing towards it)? Mathematics is
hard enough for laymen as it is.

~~~
PieSquared
As I learned recently, there is actually a good reason why gcd(a, b) is often
abbreviated as (a, b). Roughly speaking, in ring theory, an ideal _I_ is some
subset of a ring (which you can think of as a number system, sorta) for which
a _i_ is in _I_ (if _a_ is any element of the ring and _i_ is in _I_ ). Given
any element _x_ , we can generate an ideal from _x_ by just taking every other
_y_ in the ring and computing _yx_ \- if every _yx_ is in the ideal, then
multiplying by some other _z_ to get _zyx_ is the same as just having _(zy)x_
,which is just a different multiple of _x_. This ideal - the ideal generated
by _x_ \- is written as ( _x_ ); if it's generated by multiple elements, it's
written as (x1, x2, ...).

If you consider the ring of the integers, then the ideals are multiples of
some integer. For instance, the multiples of three are an ideal, because
multiplying a multiple of three by any integer yields a multiple of three (so
if you multiply the set of multiples of three by any integer, you just get
back something that's in the ideal).

The final point is this: the ideal generated by two integers is actually just
the set of multiples of their gcd. Therefore, if _a_ and _b_ are integers, (a,
b) is the set of multiples of their gcd - just like (3) is the set of
multiples of three. This is why the gcd is often written in this way.

The cool thing is that this works in rings in general, not just integers. You
can extend the concepts of gcd, primality, divisibility, etc to rings in
general, and operate on things besides just integers, such as matrices,
polynomials, or rotations of a cube.

For more info:

    
    
      http://en.wikipedia.org/wiki/Ring_theory
      http://en.wikipedia.org/wiki/Ideal_(ring_theory)

~~~
vog
_> You can extend the concepts of gcd, primality, divisibility, etc to rings
in general_

Thats true, but you have a "real" GCD only in special types of rings
(principal ideal domains). In other rings, you only have Ideals as rough
generalization of GCDs.

 _> Therefore, if a and b are integers, (a, b) is the set of multiples of
their gcd - just like (3) is the set of multiples of three._

To make this more clear, in the notation of Ideals, you can write this:

    
    
      (15,6) = (3)
    

That is, the Ideal generated by 15 and 6 is same as the Ideal generated by 3.
And for nonnegative integers, this essentially means the same as saying that 3
is the GCD of 15 and 6:

    
    
      gcd(15,6) = 3
    

There is still some "unclean" step involved here (that is, identifying numbers
by their principal ideal, i.e. treating 3 and (3) as if these were equal), but
I think this justifies the notation nevertheless.

~~~
PieSquared
You're is exactly right. To add to what you were saying and expand a little
bit:

A principal ideal domain is a ring in which every ideal is generated by only
one element, so whenever we see (a, b), we know there is some element c such
that (a, b) = (c). I think this is what vog meant by having a "real" GCD -
only in a principal ideal domain is your gcd _unique_. Without uniqueness, we
can still define a greatest common divisor such that if gcd(a, b) = g, we know
that there is nothing we can multiply by g to get a divisor of both a and b;
that is, there's no extra factor we can add to g in order to get another
factor of both a and b. That is enough to call g a GCD - but it's not
necessarily unique! It turns out that in rings which _aren 't_ principal ideal
domains, you can have more than one GCD! It's bizarre to think exactly what
"greatest" means in this context, but you can also just think of it as "can't
add any more factor while still dividing both a and b".

Ring theory is fun! (And practical, sometimes - you can describe some
algorithms very elegantly via embedding the things you're working with in
unusual rings.)

More info, with some examples and counterexamples:

    
    
      https://en.wikipedia.org/wiki/Principal_ideal_domain

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bmm6o
This is nit-picky, but why does the HN title use "fraction" when the linked
question uses the more-correct "rational"? It doesn't replace the other
instance of "rational".

~~~
GeneralMayhem
"Fraction" is shorthand (at least to a layman, which the vast majority of HN
is in mathematics) for "rational non-integer." x must not be an integer for
the stated property to hold, but the result might be.

~~~
codeflo
This is at the very least highly confusing, because "fraction" doesn't
actually imply "non-integer". But maybe that's because I also know what
rational numbers are, which according to you makes me an expert mathematician.

~~~
vehementi
I knew all of those things and it wasn't highly confusing to me and any
concerns were checked by RTFA

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snake_plissken
x<1>=x and by induction x<n+1>=x^x<n>

How does one induct (induce?) this?

~~~
qiemem
Gro-Tsen is just defining x<n> recursively. "by induction" there doesn't mean
"proof by induction".

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tomrod
That is a beautiful proof. Thanks for sharing this xSwag.

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happywolf
Mathematics is a fascinating and interesting subject, but I just wonder if
this is the right place to post? If this is, are we expecting to see similar
stuff from biology, physics, etc?

~~~
JonnieCache
Yes. eg.
[https://news.ycombinator.com/item?id=5878664](https://news.ycombinator.com/item?id=5878664)

