If x is a fraction, x^x and x^x^x are irrational. But can x^x^x^x be rational?

http://math.stackexchange.com/questions/430797/can-xxxx-be-a-rational-number

by xSwag • 13 years ago

68 42 13 years ago

$Preview of If x is a fraction, x^x and x^x^x are irrational. But can x^x^x^x be rational?$ math.stackexchange.com

ctdonath2• 13 years ago

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

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

seliopou3• 13 years ago

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• 13 years ago

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

thomasbk2• 13 years ago

The given answer is "no"[1], all powers are irrational. I'm not sure how this simple "no" would be helpful to you without understanding any part of the proof.

[1] assuming https://en.wikipedia.org/wiki/Schanuel%27s_conjecture

ctdonath2• 13 years ago

A simple "no" is helpful because it answers the title question without having to grok the proof. Some of us appreciate math geekery without being one.

aidos• 13 years ago

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

    (a,b) = 1

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

acchow• 13 years ago

Yes. This is standard notation in many textbooks and university classes.

vog4• 13 years ago

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

Dewie• 13 years ago

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.

coherentpony• 13 years ago

It's actually standard notation for the highest common factor.

tomrod2• 13 years ago

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

snake_plissken• 13 years ago

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

How does one induct (induce?) this?

tomrod2• 13 years ago

Prove it for the base case (x<1> = x), then assume it works for n-1, and show it works for the nth case.

qiemem• 13 years ago

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

happywolf• 13 years ago

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• 13 years ago

Yes. eg. https://news.ycombinator.com/item?id=5878664

carstimon2• 13 years ago

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 aπ + be 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.

arithma• 13 years ago

This is the kind of comments that I love. Sorry for the meta-comment, but I thought it deserved it. An easy read, and still informative and most importantly perks your imagination if you're into this kind of thing.

swordswinger12• 13 years ago

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

gphilip• 13 years ago

It cannot be. What the OP meant is π - 2(π/2) = 0

dllthomas• 13 years ago

By the magic of typos.

XaspR8d• 13 years ago

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

EDIT: Wow, go team.

carstimon2• 13 years ago

Thanks, edited.

Someone• 13 years ago

Why are you involving the rationals? If ax + by is zero for some non-zero rationals a and b, there also are non-zero integers i and j such that ix + jy is zero (multiply a and b by the product of their denumerators to get one such pair)

ggchappell• 13 years ago

Why are you involving the rationals?

Number systems with a well behaved division operation (fields) are heavily studied and have lots of wonderful properties. If we allow for division, thus turning the integers into the rationals (which form a field), then we have access to all this.

So, while you are correct, to a mathematician it might be simpler to start with a field and then be able to wave our hands at centuries of work in field theory and just say, "All that stuff applies here."

In particular, the notion of linear independence is generally defined only for subsets of a vector space. And being a vector space requires that the scalars (allowable coefficients) form a field.

bmm6o• 13 years ago

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• 13 years ago

"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• 13 years ago

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.

If x is a fraction, x^x and x^x^x are irrational. But can x^x^x^x be rational?

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