
Trapping a Transcendental: There are reals that are not roots of polynomials  - ColinWright
http://www.penzba.co.uk/Writings/TrappingATranscendental.html
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gjm11
This is a nice proof showcasing an important idea, but for this particular
theorem I prefer the beautiful _constructive_ proof found by Liouville. It can
be sketched in a few paragraphs, so let me do that.

0\. The basic idea is to show that an algebraic number -- i.e., one that's a
root of a polynomial with integer coefficients -- can't be very well
approximated by rational numbers ... and then to construct a number that _can_
be very well approximated by rational numbers.

1\. Suppose you've got a polynomial _p_ , of degree _d_ , with integer
coefficients. Let _x_ be one of its roots, and let _a_ / _b_ be any rational
number. How close can it be to _x_? Well, let's ask a slightly different
question: how close can _p_ ( _a_ / _b_ ) be to 0? Answer: it's always (some
integer)/ _b_ ^ _d_ , and unless _x_ actually equals _a_ / _b_ that integer
isn't zero, so is >=1 or <=-1; so abs(poly(rational)) >= 1/denominator^degree.

2\. OK, so how close can _a_ / _b_ be to _x_? Well, roughly speaking the ratio
between p( _a_ / _b_ )-0 (which we know isn't too small) and _a_ / _b_ - _x_
(which is what we care about) is close to _p_ '( _x_ ), provided _a_ / _b_ is
close to _x_. So (after filling in the details) we find that: if _x_ is a root
of an integer polynomial of degree _d_ then there's a constant _C_ such that
every rational number has abs(rational-x) >= C denominator^ _d_.

3\. Now let's construct a number that is better-approximated by rationals than
that allows. Specifically, take a decimal that looks like 0.110001000... where
the spacing between the 1s grows very fast. The traditional thing is to take
10^-1! + 10^-2! + 10^-3! + ... so let's do that. Then truncating after 10^-
_k_! leaves an error that's approximately 10^-( _k_ +1)!; in other words we
have a rational number with denominator _N_ =10^ _k_! and an error on the
order of _N_ ^( _k_ +1), so for any integer polynomial of degree _d_ , once
_k_ gets somewhat larger than _d_ this approximation is "too good", so this
number can't be a root of that polynomial. So it can't be the root of _any_
integer polynomial; in other words, it's transcendental.

~~~
mjw
This is a really nice proof, and I love this kind of "you could have
discovered X if you'd just had this core idea!" kind of exposition.

Perhaps a hard ask, but do you know of any similarly explicable proofs for
transcendental-ness of "well known" numbers?

~~~
gjm11
As Colin says, transcendence proofs tend to be hard except for numbers
carefully constructed to make transcendence proofs easy :-).

His mention of the (beautiful but difficult) Gelfond-Schneider theorem reminds
me of the following much easier, much less deep but rather pretty observation.

Question: Is it possible to take an irrational number, raise it to an
irrational power, and get a rational number?

Answer: Yes. Write s=sqrt(2) -- which is of course irrational -- and consider
s^s. If it's rational then we're done. Otherwise, (s^s)^s is an irrational
number to an irrational power; but it equals s^(s^2) = s^2 = 2, which is
rational!

Thanks to Gelfond and Schneider, we know that actually s^s is not only
irrational but transcendental.

(In 1919, the great mathematician David Hilbert mentioned three big unsolved
problems in number theory: the Riemann Hypothesis, Fermat's "last theorem",
and the transcendence of 2^sqrt(2). He said RH would probably be proved within
a few years, FLT maybe in his lifetime, and the transcendence of 2^sqrt(2)
probably not in the lifetime of anyone in his audience. In fact he got the
order exactly wrong: the transcendence of 2^sqrt(2) was proved in 1930, FLT in
1995, and RH _still_ isn't done despite its great importance.)

~~~
tzs
Another answer I've seen to the irrational^irrational question is
sqrt(10)^log(4)=2, where log is the common logarithm. This avoids upsetting
the constructivists.

~~~
gjm11
But upsetting the constructivists is _the whole point_! (That is: what's so
cute about that proof is that it tells you that there's a solution to
irrational^irrational=rational, and that it's one of two things, but it
doesn't tell you which. Also, I think it's a little shorter and more elegant
than using sqrt(10)^log10(4). Don't you?)

------
ColinWright
A quick note for those upvoting this ...

Thank you. I'm casting about for more things to write about like this. There
are lots of things hackers will have read, but not seen the details for, and
sometimes the details are very accessible.

If you liked this one, what else would you like? For example:

* Between every two rationals there's an irrational

* Between every two rationals there's a rational

* The rationals can be listed, in full, exactly once each

That kind of stuff. What assertions or claims have you seen made that you'd
like explained in more detail?

Feel free to reply here, or email - address in my profile.

 _Edited for layout._

~~~
tzs
I don't know how accessible it can be made, but the Khinchin-Lévy constant is
astonishing, both in its existance and its value [1].

[1] <http://mathworld.wolfram.com/Khinchin-LevyConstant.html>

~~~
gjm11
I can give a handwavy explanation for this. I'm afraid it's a bit heavy going
in places, but there's nothing too difficult and the more technical bits can
largely be skipped.

First, a reminder of what we're going to kinda-sorta-prove: Take any number
and compute its continued fraction, giving approximations p1/q1, p2/q2, etc.
Then for "almost all" numbers it happens that qn^(1/n) tends to the magical
number exp(pi^2 / (12 log 2)).

And a brief summary of the idea: the process of computing continued fractions
is basically one of applying a certain function again and again, and the
number qn^(1/n) can be related to a certain average involving the numbers you
get; a very general and powerful thing called the "ergodic theorem" tells us
that when you do this, the values you get look a lot like random numbers
chosen with a particular distribution; the corresponding average (over numbers
chosen with that probability distribution) is a fairly simple integral, equal
to an infinite series whose value is the thing we're looking for.

CONTINUED FRACTIONS

OK. So, how do you find the continued fraction for a number x? You repeatedly
subtract off the integer part and then take the reciprocal; the integer parts
you subtract off are the c.f. coefficients.

Forget about the integer parts for a moment and just consider the following
function: x -> fractional_part(1/x). Call this F. We'll apply it only to
numbers between 0 and 1. If you start with any x, compute fractional_part(x)
and then repeatedly apply F, the c.f. coefficients are exactly the bits you
throw away when computing the fractional parts.

Now suppose you pick a random x and compute F(x), F(F(x)), etc. These points
will be scattered over the interval (0,1) somehow. In this sort of setting it
often turns out that for almost all x, the density with which these points
turn up is independent of x. Let me be more specific.

ERGODIC THEORY

Suppose you've got a probability distribution on (0,1) with the following
magical property: _picking a number according to this probability distribution
and applying F to it gives you a number with the same probability
distribution_. For some F you can do this, and for others you can't. If you
happen to have done anything with Markov chains and run across the notion of
an "invariant measure" or "stationary distribution", this is the same idea but
in a continuous rather than a discrete setting. "Measure", "distribution", and
"probability distribution" all mean more or less the same thing here.

Well, anyway, if you have your function F and an invariant measure, and if
another technical condition called "ergodicity" applies (it says that F mixes
things up enough that you can't find nontrivial subsets that F maps to
themselves), then something wonderful follows: for almost all x, the sequence
x, F(x), F(F(x)), ... "samples the space fairly" according to your probability
distribution. There are several different "ergodic theorems" that make this
rigorous with different meanings of "sample fairly". Here's one: _let f be any
well-behaved function; then for almost all x, the averages
[f(x)+f(F(x))+...+f(F^(n-1)(x))]/n tend to the same value, which is the
average value of f(x) when you pick x from that magic probability
distribution_.

THE INVARIANT MEASURE

If you're still awake after all that, it will not surprise you that the
particular F we're talking about -- mapping x to frac(1/x) -- does indeed have
an invariant measure. Its probability density function is nice and simple:
(1/log2) 1/(1+x). Proving that this is indeed invariant under F is pretty
straightforward.

So all we need to do is to set things up so that the thing we're interested in
is of the form [f(x)+f(F(x))+...+f(F^(n-1)(x))]/n, and then compute the
average of f(x) with respect to our invariant measure. How do we do that?

CONTINUED FRACTIONS AGAIN

(There's a bit of magic here, but it's fairly easy magic.) Suppose the n'th
convergent to x is pk(x)/qk(x). Then it's not hard to show that pk(x) =
q{k-1}(Fx). So the following identity is trivial because it's just multiplying
by a lot of 1s:

1/qn(x) = 1/qn(x) pn(x)/q{n-1}(Fx) p{n-1}(Fx)/q{n-2}(FFx) ...
p2(F^{n-2}x)/q1(F^{n-1}x).

Now "shift the numerators left by one space":

1/qn(x) = pn(x)/qn(x) p{n-1}(Fx)/q{n-1}(Fx) ... p1(F^{n-1}x)/q1(F^{n-1}x).

Now remember that pk(y)/qk(y) is a pretty good approximation to y (at this
point in an actual proof there'd need to be some computation of what the
errors are, but I'll spare you), so

1/qn(x) ~= x F(x) F(F(x)) ... F^n(x)

which is beginning to look a lot like what we need. Take n'th roots and logs:

log qn(x)^(1/n) ~= -[log x + log F(x) + ... + log F^n(x)]/n

NOW IT'S JUST INTEGRATION

Aha. So now (thanks to the ergodic theorem) we just need to know the average
of log x when we pick x according to our probability distribution. That's

integral {0..1} of (1/log2) . log x . 1/(1+x)

which, given the form of the answer we're looking out, is crying out to be
done by expanding log x as a series. More specifically, first an integration
by parts shows that -(that integral) equals

integral {0..1} of log(1+x)/x

which equals

integral {0..1} of (x-x^2/2+x^3/3-x^4/4+...)/x

which, integrating term by term, equals 1 - 1/2^2 + 1/3^2 - 1/4^2 + ..., and
the fact that this equals pi^2/12 is basically the same as the slightly more
famous fact that 1 + 1/2^2 + 1/3^2 + ... equals pi^2/6. And now we're done.

------
calhoun137
Stuff like this is amazing. I am so happy I switched out of comp-sci freshman
year of college, and took a 10 year break from programming to study math and
physics. I think if I had stuck with comp-sci there is a big chance I would
hate computers, and always wish I had gone into science. Now that I am back to
programming all the time it feels like I am a kid again, and I don't have a
regret in the world. My career might be slightly fucked, but the joy of
knowing this kind of stuff, and still loving computers makes it worth it.

------
e3pi
Thank you. Got it, the ratio of algebraics to transcendentals is nothing. This
has been bugging me lately... can you tell me why the transcendental e, signs
it's name in the first significant digits past the decimal point...

...where s3 is root(3), and gg is Euler's gamma/Masceroni's 0.5772...
constant:

e^(s3 times gg)/(e times s3 times gg) = 1.00000002718?

Flirting? Fear of patent trolls, proprietary pride, just dumb luck, what?

Note: HN edit coughed when I first used '*' for 'times'

~~~
jerf
Somewhat more generally, the space of interesting numbers you can write
compactly with traditional notation ("4", "sqrt(2)", "log(7)"), combined with
the ways you can compactly write a relation between two numbers (+, /,
Ackermann's function), combined with allowing two or three or maybe four
numbers (even before they start showing up multiple times as you show there),
combined with the number of ways in which a result may be "interesting",
results in an inevitable population of combinations of numbers and operations
that have an "interesting" result that aren't "really" interesting, or perhaps
rather, are only interesting because of the selection function being used
("looks interesting to a human").

You may enjoy <http://en.wikipedia.org/wiki/Mathematical_coincidence> .

~~~
e3pi
"........that have an "interesting" result that aren't "really" interesting,
or perhaps rather, are only interesting because of the selection function
being used ("looks interesting to a human").

No sir, you are mistaken. My linux netbook's perl regex's determined this is
interesting. I myself, a human, merely scan a move-to-front list that suit's
this week's curious fancy.

