Liouville’s number, the easiest transcendental and its clones (corrected reupload)

by birtanpublished on August 24, 2020

welcome to another mythology video labels number the monster up there consists of infinitely many isolated islands of ones at the one factorials 2 factorial 3 factorial 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 as fast 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 pros and if you make it to the end I've got a special treat for you I'll show you how you can use levels 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 as 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 counter sets what's the 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 root expressions like this 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 algebraic numbers and includes all the

Other types of numbers that I mentioned before integers rationals and routine numbers anyway we call a real number transcendental if it is not algebra if it is not a solution of any polynomial equation with integer

Coefficients labels number or levels constant was shown to be transcendental in 1851 but the great French mathematician Rose F Louisville it was one of the first numbers shown to be transcendental and Lil's proof of this

Fact is probably easiest transcendence proof for any specific number liberals proof is accessible to anybody who has had some exposure to proofs at university level real analysis and I've included 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 Louisville's number that can be understood and hopefully also enjoyed by anybody who watches these sorts of math videos including all my fellow Jew fans were still in high school I'm still working on the primary

School version but I'm pro not never going to get there okay levels number lots and lots of zeros and ones and the ones are the 1 factorial 2 factorial 3 factorial etc digits just a quick reminder one factorials equal to 1 2

Factorial is 1 times 2 is equal to 2 3 factorial is 1 times 2 times 3 is equal to 6 4 factorial is 24 then 120 725,000 40 etc a sequence of numbers that grows rapidly which means that the stretches of zeros between consecutive ones get

Longer and longer extremely quickly ok here then is my proof that Louisville'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 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 maybe 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 coincide was PI squared and only the first four digits and the remaining digits that you can see here are wrong Lee Will's number L behaves quite different in this respect but to start

With the only distinguishable truncations are those that cut off after the ones 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 square 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 Li ovals 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 a 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 we raise L to a different power say five well then as you can see while the approximations are still definitely damn good things are no longer spot-on four digits here for example one digit is

Wrong and here three but if you look closely you find that the next approximation is spot-on again all digits are correct and actually things will be spot-on from this truncation also all digits of truncation l v ^ v l

6 and so on will be correct in fact something similar will be true for all powers so as we've seen all squared truncations are spur on the same turns out to be the case for acute ones fourth powers to fifth powers we've

Already seen that all digits are spot-on from a certain truncation on same thing for six powers except this happens a little bit later and so on for all powers now we find that the same is true if instead of just power

We look at any polynomial with positive integer coefficients or pluses no – this they all right so if instead of just powers of the number L we are 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 punch line 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 equal 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 equal sign is actually equal to the number on the

Right I call this number the number at the top now let's see what numbers we get on the left and right when we replace L but one of its truncations let's start with L 1 and work our way up obviously the higher the truncation the

Better approximations to 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 a sake of argument

Let's say that this is the case from l5 onward what this means is that now the first digit of the approximation that 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 awhile 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 – all right 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 the 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

Approximation will eventually be spot-on no matter what polynomial equation we're dealing with ok so let's have a closer look so the different ones in our number L just 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 a 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 here 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 everyone 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 10 to the minus 3 again this times that 10 to the minus 4 add up

Everything to get this square we're after 4l squared we have to multiply again every term at the top with everyone at the bottom and then it up yeah might take a while right so 10 to the minus 1 times 10 to

The minus 1 is 10 to the minus 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 was one of the terms at

The bottom should be at least as big as 10 to the minus 4 but the largest number we can make this way is 10 to the minus 6 times 10 to the minus 1 which is 10 to the minus 7 a lot smaller than 10 to minus 4 okay so there's no danger here

What about these critical numbers for an X truncation also no danger and to get 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 forced truncation of the fifth

Power of L it'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'll 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 truncations 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 the first digits of l squared I just like to highlight again the gap between the yellow lowest terms of the square truncations together with the corresponding worst-case-scenario green terms as if already observed this

Gap is increasing in fact it will get arbitrarily large and the same is true for all the other powers wants to gap starts appearing from some truncation onward those of you who did those one-line calculations earlier on will

Already have convinced yourself of this fact okay now 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 ten to the minus one 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 get 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 Square 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 constants for example it's multiplied a linear term by something huge say 10 the overall effect of this change 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 proof we assumed

That L solves a polynomial equation like the one up there I claimed 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 a couple of super cool facts that can be shown in the same way first even when interpreted as base 2 3 4 etc numbers Liris 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 somewhere approved is in the comments so adding 5 to this transcendental number up there gives 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's super tough to prove that pious transcendental whereas proving the transcendence of this weird clone is not

Any harder than what I've showed you earlier okay 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 one point two three 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's easily fixed because every number was a terminating decimal expansion has a second decimal expansion with an infinite tail of nines for example one point two three is equal to one point two two nine nine nine nine nine 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 zero cannot be massaged in this way to give a

Transcendental number but okay as it happens but often zero is a little exception here big deal anyway we now got a clone of the real numbers within the real numbers consisting of transcendental numbers of course since

It is a clone there's a one-to-one correspondence between it and their 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 in 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 mythology to 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're struggling with anything just ask and that's it for today you

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