Welcome to another Mathologer video.
Liouville's number the monster up there
consists of infinitely many isolated
islands of 1s at the 1! th, 2! th,
3! th, etc. digits with
exploding gaps of zeros between them. As
I promised you at the end of the last
video, today's mission is to show you a
nice visual way of seeing that this
number is transcendental. I'm pretty
convinced that as far as transcendence proofs
for specific numbers goes, what I've put
together here is as simple and as
accessible as it will ever get. Still, if
you make it to the end and understand
all my arguments I think you can be
pretty proud of yourself since not even
that many professional mathematicians
know any transcendence proofs. And, if you
make it to the end, I've got a special
treat for you. I'll show you how you can
use Liouville's number as a template to make
a clone of the real numbers within the
real numbers. This clone is made up of
transcendental numbers like this strange
cousin of Pi, but perhaps the craziest
thing about this clone is that although
it is as large at the set of real
numbers itself, it is of measure 0. This
means that in a sense it takes up no
space within the real numbers. For those
of you in the know this set is closely
related to Cantor sets.
What's a transcendental number again?
Well for a detailed introduction just
watch the last video. So here just real
quick: the transcendental numbers are
defined in terms of what they are NOT.
They are NOT among the real numbers that
you encounter when you're doing
classical algebra, like the integers, the
rational numbers, any of the real numbers
that can be written as rooty expressions
like these, or any of the real numbers
that pop up as solutions to polynomial
equations with integer coefficients like
these. This last set of numbers is called the
algebraic numbers and includes all the
other types of numbers that I mentioned
before: integers, rationals and rooty
numbers. Anyway, we call a real number
transcendental if it is not algebraic,
if it is not a solution of a
polynomial equation with integer
coefficients. Liouville's number or Liouville's
constant was shown to be transcendental
in 1851 by the great French
mathematician Joseph Liouville. It was one of
the first numbers shown to be
transcendental and Liouville's proof of this fact
is probably the easiest transcendence
proof for any specific number. Liouville's
proof is accessible to anybody who has
had some exposure to proofs at
university level real analysis and I have
concluded links to the original French
paper and a modern version of the proof
in English in the description. But since
many of you would struggle with any of
the proofs in these papers or in
textbooks, I've tried to come up with an
alternative way of seeing the
transcendence of Liouville's number that can be
understood and hopefully also enjoyed by
anybody who watches these sorts of maths
videos, including all Mathologer fans who
are still in high school. I'm still
working on the primary school version
but and I am probably never going to get there.
Okay, Liouville's number, lots and lots of
0s and 1s and the 1s are the 1! th,
2! th, 3! th, etc.
digits. Just a quick reminder, 1!
is equal to 1, 2! is 1 times 2 is
equal to 2, 3! is 1 times 2
times 3 is equal to 6, 4! is 24.
Then 120, 720, 5040, etc. a sequence of
numbers that grows rapidly which means
that the stretches of zeros between
consecutive 1s get longer and longer
extremely quickly. Okay, here then is my
proof that Liouville's number is
transcendental. Let me know in the
comments whether this proof works for
you. When dealing with a complicated
number like pi
approximate value often suffice or are at
least a good place to start. To get such
an approximation we often chop off the
decimal expansion at some point.
Now let's approximate the number pi
squared with the squares of these
truncations.
Then these approximations for pi squared
will be spot-on
to a certain digit and may be good enough
for some application we have in mind.
However from some point on these
approximations or go off target. For
example, the last approximation down
there coincides with pi squared in only
the first four digits and the remaining
digits that you can see here are wrong.
Liouville's number L behaves quite
different in this respect. But to start
with the only distinguishable
truncations are those that cut off
after the 1s and these are the first
four of these truncations of L. Now
squaring gives these numbers here and in
contrast to what we had before all the
digits of the squared truncations seem to
be correct. And looks are not deceiving,
this is actually always true and the
reason for this is not too hard to pin
down. It's because all the nonzero digits
of Liouville's
number are spaced further and further
apart and those further down the line
simply don't play any role in producing
the earlier digits of the square of our
number. For the moment I'll gloss over the
technical details but I'll return to
them after I've finished
outlining the proof. Anyway, what happens
if you raise L to a different power, say
5? Well then, as you can see, while the
approximations are still definitely damn
good, things are no longer spot on
for all digits. Here, for example, one digit
is wrong and here three but if you look
closely you'll find that the next
approximation is spot on again, all
digits are correct. And actually things
will be spot on from this truncation on.
So, all digits of truncation L_5^5, L_6^5,
and so on will be correct. In fact,
something similar will be true for all
powers. S,o as we've seen, all squared
truncations are spot on. The same turns
out to be the case for the cubed ones,
fourth powers too, fifth powers we've
already seen that all digits are spot-on
from a certain truncation on. Same thing
for sixth powers, except this happens a
little bit later, and so on, for all
powers. Now,
if instead of just power we look at any polynomial with positive
integer coefficients: all pluses no minuses up there, right? So, if instead of just powers
of the number L we're approximating a polynomial like this, evaluated at the
number L, then we find that all the digits of our approximations will be
correct from some truncation onwards. In particular, in the case of this
polynomial all digits will be correct from L_2 onward. We're getting pretty
close to the punchline. Remember what we have to demonstrate is that L does not
solve any equation like this. To get the proof going we first move all the
negative terms to the right side. Now this results in two of those special
polynomials that only have positive terms, one on the left and one on the
right side of the equals sign. Now let's assume that L actually is a solution of
this polynomial equation. This means that the number on the left of the equals sign
is actually equal to the number on the right. I'll call this number the number
at the top. Let's see what numbers we get on the left and right when we replace L
by one of its truncations. Let's start with L_1 and work our way up. Obviously
the higher the truncation the better approximations for the number at the top
will get on the left and on the right but we also know that in addition to
this all digits of these approximations will be correct from some truncation on.
For the sake of argument, let's say that this is the case from L_5 onward. What
this means is that now the first digit of the approximation on the left is equal
to the first digit of the approximation on the right, the second digit on the
left is equal to the second digit on the right, and so on, for a while, until.... Well
you'd expect one of the two sides to run out of nonzero digits and the other side
to keep on going before it stops. Alright and maybe that's exactly what will
happen. However, surprisingly, and we'll see why
when we go all technical, at this point our assumption that L
solves the equation also implies that, starting with a possibly higher
truncation, let's say L_6, we will always run out of nonzero digits at the same
time in both approximations. So this means that our two approximations are in
fact equal from this truncation onward and this means that just like L itself L_6
and all the following infinitely many truncations are also solutions to our
polynomial equation. But infinitely many different solutions are impossible
because a polynomial equation can only have as many solutions as its degree. For
example, our degree 6 polynomial can have at most 6 solutions and so the
assumption that our number L solves a polynomial equation with integer
coefficients leads to the absolutely impossible conclusion that this
polynomial has infinitely many solutions. And this means that our assumption that
L solves an equation like this was wrong in the first place and that means that L
is a transcendental number. Very neat proof by contradiction, right? Of course
what I did not show you is that all the approximations will eventually be spot on
no matter what polynomial equation we're dealing with. Okay, let's have a closer
look. The different 1s in our number L
stand for different powers of 10. For example, the first one stands for 10
to the minus 1, the second for 10 to the minus 2, the third for 10 to the minus 6,
and so on. And the number L is just the sum of these powers of 10. Now we started by
pondering L squared so let's have a look at how we would calculate this number.
Actually before we do that, let's calculate the square of the second
truncation. Okay, so here we have to multiply every term at the top with
every one at the bottom and then add up. So 10 to the minus 1 times 10 to the
minus 1 is 10 to the minus 2, 10 to the minus 1 times 10 to the minus 2 is 10 to
the minus 3. This times that that, 10 to the minus 3 again. This times
that 10 to the minus 4. Add up everything to get the square we are after. For L squared
we have to multiply again every term at the top with every one at the bottom and
then add up. Hmm, might take a while, right? So 10 ^ -1
times 10 ^ -1 is 10 ^ -2, minus 1 minus 2 minus 3,
minus 1 minus 6 minus 7, and so on. Minus 2 minus 1 minus 3, minus 2 minus 2 minus
4, and so on. And so on. Now why are the digits of the square truncation correct?
Well, its rightmost digit 1 comes from multiplying the rightmost term of the
truncation with itself. For the other terms to interfere with the digits of
the square truncation, one of the terms outside the box at the top multiplied
with one of the terms at the bottom should be at least as big as
10 ^ -4. But the largest number we can make this way is 10 ^ -6 times
10 ^ -1 which is 10 ^ -7, a lot smaller than e
10 ^ -4. Okay so there's no danger here. What about these critical numbers for
the next truncation. Also no danger and to gap between the exponents minus 12
and minus 25 is larger than before, so even less danger. One more. The gap has
increased even further. Now it's a one-liner to show that this trend
continues and I actually leave it to you to sort out the details, and, well,
actually you only get partial credit if it takes you more than one line. Let's
have a look at all this for the fourth truncation of the fifth power of L.
That's the next thing we looked at, right? As I said before, all digits are correct
for this fourth truncation. Let's double check this: yes, the green 124 is larger
than the yellow 120. Tick! Again, I leave it to you to fill in the details
to show that this will also be the case for the higher truncations. Remember that
there were problems with the lower trunations. Let's see where these show up
here. Okay the order has flipped, the 30 is now
larger than the 28 and so there will be wrong digits in the approximation. One
step down there's also trouble with the green terms interfering with the digits
of the power of the second truncation. Okay so here are the first digits of L
squared. I just like to highlight again the gap between the yellow lowest terms
of the squared truncations together with the corresponding worst-case-scenario
green terms. As we've already observed this gap is increasing. In fact, it will get
arbitrarily large. And the same is true for all the other powers once the gap
starts appearing from some truncation onward. For those of you who did those
one-line calculations earlier on will already have convinced yourself of this
fact. What about polynomials with positive integer coefficients. Well,
they're built from powers. Take, for example, L squared plus L. Highlight the
second truncation. First note that the corresponding green digits are basically
aligned. That's because the top one essentially comes about from multiplying
the bottom one by 10 ^ -1 that's a shift of one position. Anyway
this relative alignment will always be the same as we move to the right. Also, as
we move towards the right, the gap of zeros in front of the green digits gets
as large as you wish. Given this gap of zeros, it is clear that when we add up L_3
and L_3 squared none of the green digits or beyond will
add to the sum within L squared, and so the digits of the approximation will all
be correct. But what happens if the individual powers in the polynomial are
multiplied by non-trivial positive constants. For example, let's multiply the linear term
by something huge, say 10. The overall effect of this changed term are it's
nonzero islands possibly growing and the greens shifting to
the left. This may lead to the overall gaps in front of the greens shrinking or even
vanishing for really large coefficients like googleplex however since the gaps
grow unlimited in size they will eventually absorb the shrinkage and gaps
will always be present from a certain point on. And this means that no matter
what polynomial we're dealing with, all our approximations will be spot-on from
a certain truncation onward. Almost there. In the final part of the proof we assumed
that L solves a polynomial equation like the one up there. I claim that this
implies that from a certain point onward all truncations also solve this equation
which then gives the contradiction. Well, as just now, we can see that from a
certain truncation, say L_5 onward, we have the usual common gap of zeros for all
the terms in this equation. But then for the identity on the left to be true,
which is our assumption, the sum of the two approximations on the top must be
equal to the sum of the two at the bottom, which is exactly what we needed
to convince ourselves of. Anyway that's basically it. Of course I could still
fill in all the nitty-gritty calculations, for example about the ever
growing gaps but if you've made it up to here I'm sure you'll be able to fill in
those details yourself. Here are a couple of super cool facts that can be shown in
the same way. First even when interpreted as base 2, 3, 4, etc., Liouville's
monster will always be a transcendental number. Second even when we replace the
ones by other digits we get a transcendental number as long as
infinitely many of these digits are nonzero. Now adding an integer to a
transcendental number gives a number that is still transcendental. Maybe
someone prove this in the comments. So adding 5 to the transcendental
number up there give a new transcendental number: five point one two zero,
zero, zero, and so on. Here's a fun idea then. Let's take a real
number like pi and use its digits to create a new number like this. Well
it is super-tough to prove that pi is transcendental whereas proving the
transcendence of this weird clone is not any harder then what I've showed you
earlier. Ok, next trick. Let's do the same for
every real number. What this does is create a clone of the set of real
numbers within the set of real numbers. It consists entirely of transcendental
numbers. Even the clones of the algebraic numbers are transcendental. We just have
to be a little bit careful here when we translate a number that features a
terminating decimal expansion like for example the number 1.23.
Using that terminating decimal expansion would also result in a clone with a
terminating decimal expansion and any number like this is rational and
therefore not transcendental. Of course that is easily fixed because every number
with a terminating decimal expansion has a second decimal expansion with an
infinite tail of nines for example 1.23 is equal to
1.2999..., and so on. Using this alternative decimal
expansion which features infinitely many nonzero digits will then give a
transcendental clone. Well one mini problem remains: the number 0 cannot
be massaged in this way to give a transcendental number. But, okay, as it
happens, quite often 0 is a little exception here, big deal. Anyway, we now
have got a clone of the real numbers within the real numbers consisting of
transcendental numbers. Of course, since it is a clone, there is a one-to-one
correspondence between it and the original set. In other words, just like
the real numbers the clone is an uncountably infinite set. But, of course,
in addition all its elements are easy transcendentals super cool, right? However,
this uncountably infinite set also has the paradoxical property of having
measure zero. So although it is as large as the whole set of real numbers, it is
so well hidden within that in a sense it's not even there. To really be able to
appreciate this paradoxical set you should watch the previous video where I
also give a little intro to how you can assign a length or measure to a subset
of the real numbers, or you could head over to Infinite
Series where Kelsey is also discussing these sorts of sets at the moment. I've
also prepared a proof of the fact that the clone has measured zero but I think
this video is already getting quite long and so I'll put this proof on Mathologer 2
sometime in the next couple of days. Okay, let's finish here. I hope you all
understood and enjoyed this video. But, as usual, please let me know what worked for
you and what didn't and if you are struggling with anything just ask. And
that's it for today. [Music]
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