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Paradoxes with infinity
By Brad Rodgers (P1930) on Monday, February 21, 2000 - 10:59 pm:

I have two simple paradoxes, Both of which I don't understand.

The first of them involves a library of books that each contain 500 pages of 1000 spaces for characters. In this library, every possible combination of characters is given in a book. Thus at least one, or a series of, book(s) in here can describe every single possible event to take place. Despite this there are only a finite number of books. But wouldn't being able to describe every possible event mean that a finite number of books can describe an infinite amount of time? And wouldn't this mathematically prove that history repeats itself or that the universe can never be infinite in time?

The second paradox is slightly more normal, but still rather odd: In the equation
S=1+a+aa+aaa+aaaa+......,
S=1/(1-a)

Given that 2=a, does 1+2+4+8.....=-1?

Is there an answer to either of these questions?

Brad

By Dan Goodman (Dfmg2) on Tuesday, February 22, 2000 - 02:11 am:

The first one is not a paradox. If you only allow yourself to read a book picked up at random, you will limit yourself to books of length 500 pages. In fact, the Bible probably won't be in there for instance, let alone many modern books on the Big Bang theory etc. However, you also allow series of books. Well, how about 30 or so books say, each one containing only one letter or punctuation mark. Suitable combinations of these books will produce every concept expressible in the English language, and by extension, probably every thought that a human can convey to another human. However, these books can be reduced to just two books, binary 1 and binary 0. In fact, why use books, lets use rocks and cabbages say, where a rock = 0 and a cabbage = 1.

The problem is twofold, firstly, not every event can be described in language. For example, there are numbers which cannot be expressed in language, so there are physical events you cannot express in language defeating your paradox. The more fundamental problem is that the books do not contain any information themselves, they only contain the building blocks of information. You need to know which combinations of books to look at to actually get information. To put it another way, I can write a computer program to spew out (in sequence) every combination of letters there is, however, there is no way of getting it to distinguish which sequences (a) make sense, this could be solved, (b) assert true statements, this cannot be solved as was proved by Alan Turing. The building blocks of information do not contain information in themselves. Does this solve the first paradox for you?

The second one is much less philosophical, and more concrete. The formula you gave is not quite true, it should read 1+a+a2+a3+...=1/(1-a) if 0<a<1. Therefore 2 is not a problem. The reason for this is as follows, let Sn=1+a+a2+...+an-1, you can show that Sn=(1-an)/(1-a). So, if 0<a<1, and n big, an=0, Sn=1/(1-a) as you stated. However, for a>1 and n big, an is big, so Sn is very big, in fact, it gets bigger and bigger the bigger n gets, going off to infinity. Solving this sort of puzzle is the sort of thing you do much more precisely at university, although depending on your age / level you might be interested in studying them now.

There are some other interesting paradoxes along these lines, another way of getting your result above is this:

x=1+2+4+8+...
2x=2+4+8+...
2x+1=1+2+4+8+... (but the right hand side is just x)
2x+1=x
x=-1
-1=1+2+4+8+...

Here's an example where (amazingly) it does work. The problem, find a solution x to the equation:
xxxx... = x^(x^(x^...))) = 2

(where a^b means a to the power b)

i.e. x to the power of itself infinity times. Firstly, if a=b then xa=xb, so we can write
x( x^(x^(x^...))) = x2

but the left hand side of this is just 2 (from the first equation), so

2=x2
x=sqrt(2)

You can try this on your calculator, and you will find that it actually works! There's a more general result, x^x^x^x^...=m has a solution x=m1/m (see if you can show how). There are other examples of this too, for instance "Nested Radicals":

x=sqrt(2+sqrt(2+sqrt(2+...)))

Square both sides to get

x2=2+sqrt(2+sqrt(2+...))

Subtract 2 to get:

x2-2=sqrt(2+sqrt(2+...))

But the right hand side is just x, so:

x2-2 = x
x2-x-2=0

Which we can solve using the quadratic formula:

x=(1+sqrt(1+8))/2=2

Which you can check works on your calculator.

If you like, you could try finding the general solution to x=sqrt(m+sqrt(m+sqrt(m+sqrt(m+...)))) or (more tricky, involves squaring, subtracting, squaring again, and subtracting again), x=sqrt(1+sqrt(2+sqrt(1+sqrt(2+sqrt(1+...))))). If you can't do these and are interested to know how, ask me and I'll show you.

To finish off with, just to show that this method doesn't always work, here is one which cannot be solved:

x=sqrt(1+sqrt(2+sqrt(3+sqrt(4+sqrt(5+...))))

Hope you found that little digression interesting, personally, I love this sort of thing, and I particularly like showing people than infinity = -1 using your argument above, hardly anyone knows what to do about it. Try it on your maths teacher and see what they say :)

By Dan Goodman (Dfmg2) on Tuesday, February 22, 2000 - 02:13 am:

One more comment, solving x=sqrt(1+sqrt(2+sqrt(1+sqrt(2+...)))) you probably won't be able to find an exact answer, but just finding a simple equation for x is probably good enough, you could get your computer or a calculator to find an approximate solution (or an exact solution if you have some good maths software).

By Brad Rodgers (P1930) on Tuesday, February 22, 2000 - 11:06 pm:

Are you sure that x=sqrt(1sqrt(2sqrt(3sqrt(4...)))) can't be solved? By Ramanujuan's work, sqrt(3sqrt(4sqrt(5sqrt(6sqrt(...))))=3. So wouldn't the answer be 61/4.

How did you obtain the result of S(n)=(1-an)/(1-a)?

Also, when finding the answer to m=1 in x=sqrt(msqrt(msqrt(...))), i found that x2-x-m=0, thus
x=(-1±sqrt(1-4m))/2. I don't know if this can be further simplified. I haven't worked on the other yet.

Thanks

Brad- age 14

P.S. you did resolve my paradox about the books in case you were wondering

By Dan Goodman (Dfmg2) on Wednesday, February 23, 2000 - 12:16 am:

I think you are misquoting ramanujan, I checked out sqrt(3sqrt(4sqrt(5sqrt(6sqrt(...)))) and it was bigger by the 10th level than 3.5 and was getting bigger. Also, that wasn't the one I put on, I put sqrt(1+sqrt(2+sqrt(3+...))) (additions inside the radicals).

For the general case, you are nearly right, you got the quadratic formula slightly wrong, it should be x=(1+sqrt(1+4m))/2.

Finally, to answer the question about S(n). Do you know what mathematical induction is? If not, here is a quick overview. If you are trying to prove something is true for every number n, here is a method for doing so. First, prove it is true for n=1. Second, prove that if it were true for some number n, then it would also be true for n+1. Now you know it must be true for all n>=1, because of this. True for n=1, therefore true for n=2. True for n=2, therefore true for n=3, etc.

For the S(n) case, first let's prove it for n=1:

S(1)=1=(1-a1)/(1-a)

So it is true for n=1.

If it were true for n, then we have

S(n)=1+a+a2+...+an-1 = (1-an)/(1-a)
Then
S(n+1)=(1+a+a2+...+an-1)+an = (1-an)/(1-a)+an = (1-an+(1-a)an)/(1-a) = (1-an+1)/(1-a)

So it is true for n+1 if it is true for n, therefore it must be true for all n>=1. Let me know if you can't quite follow that.

By Brad Rodgers (P1930) on Wednesday, February 23, 2000 - 09:30 pm:

Thanks, I think I understand the proof by induction. I do have some experience in that method of proof. I also think that sqrt(1sqrt(2sqrt(3sqrt(4...))) does equal 61/4. The fourth equation in chapter 14, page 105, of Ramanujan's first notebook reads:

x+n+a=sqrt.(ax+(n+a)2+x sqrt.(a(x+n)+(n+a)2+(x=n) sqrt.(etc.))

Plugging x=2, n=1, and a=0 into the equation gives sqrt.(3 sqrt.(4 sqrt.(5 sqrt.(6 sqrt.(...)))))=3. In my calculations I recieved a value lower than 3 when worked out to 10 places. But, in the equation you originally gave me, I certainly can't think of a way to solve it- so I must trust you that it can't be solved.

By Dan Goodman (Dfmg2) on Wednesday, February 23, 2000 - 09:52 pm:

Aha, I think I see the problem. Ramanujan's equation with x=2, n=1 and a=0 expands to 3=sqrt(1+2sqrt(1+3sqrt(1+4sqrt(1+...)))) which seems to work on my calculator. However, this is a slightly different problem from sqrt(1sqrt(2sqrt(3sqrt(4...)))), I'm finding your notation slightly confusing here. I think that your mistake is this, 1+2sqrt(blah) you are writing as 3sqrt(blah) and so forth? Forgive me if I'm misunderstanding you here...

By Brad Rodgers (P1930) on Wednesday, February 23, 2000 - 10:24 pm:

Yep, that was my error. Thanks for pointing that out. when I originally wrote it out I had it as 1+2sqrt.(x), but when copying the problem down to this page my brain suddenly stopped working and I mistook the statement at hand for (1+2)sqrt(x).
Thanks,

Brad

By the way, I did try that on my math teacher and he had no idea what to think of it.

By Dan Goodman (Dfmg2) on Wednesday, February 23, 2000 - 10:36 pm:

Would you like me to tell you a bit about how we resolve difficulties like the infinity = -1 when doing higher maths? Basically, the problem is that when doing infinite sums, we require that the sum "converges" or gets closer and closer to some particular number as we add more and more terms to the sum. With 1+2+4+8+... it doesn't get closer and closer to any number. (If it was getting closer to x say, then there has to be some number of terms where 1+2+4+8+...+2N is within x±1. But then, 1+2+4+8+...+2N+2N+1 is withing x+2N+1±1, so it isn't getting closer and closer to x after all.)

By Brad Rodgers (P1930) on Friday, February 25, 2000 - 05:34 pm:

Just wondering, does the "paradox" about the books prove that not everything can be expressed in the american language or any language composed of a finite number of characters? I'll write what the answer to the second question is tonight. I would write it now, but I'm in school and don't have much time left before this class is over.

By Dan Goodman (Dfmg2) on Friday, February 25, 2000 - 06:42 pm:

Yes, there are things that cannot be described in english, or in any finite (or even countably infinite) language. I can explain this to you very simply if you know the difference between countably infinite and uncountably infinite, do you know about this? If not I'll explain, but it might take a while.

By Dan Goodman (Dfmg2) on Friday, February 25, 2000 - 07:36 pm:

Actually, I may as well tell you about it rather than wait for you to say yes. The "natural numbers" are the numbers 1,2,3,4,5,6,7,... and there are an infinite number of them. However, it turns out that the number of "real numbers" (i.e. any infinite decimal expansion, e.g. 3.14159... 1.00000... etc.) is also infinity, but in this case it is a bigger infinity. If it was the same infinity, then we could find a way of pairing them off with one another, i.e. for every natural number there would be a real number corresponding to it. Imagine we write just the real numbers between 0 and 1 out in a list, which might look like this:

N R
1 0.102438957083...
2 0.598479857230...
3 0.357984359845...
4 0.587439857439...
5 0.857983759473...
6 0.985723987544...
etc.

Now, we can find a number which is not on this list anywhere as follows. Create a number x such that in the nth decimal place it is different from the nth decimal place of the nth number in the list. For instance, the 1st place of the first number is 1, so choose 2 as the first place of x. The second place of the second number is 9, so choose 4 as the second place of x, etc. Now, x cannot be the first number, as it is different in the first place. x cannot be the second number as it is different in the 2nd place. But, x cannot be the nth number as it is different in the nth place, so x is not on the list. So, the number of real numbers is bigger than the number of natural numbers. We say that the natural numbers are "countably infinite" and the real numbers are "uncountable".

Relating this to the book problem, the number of books is countably infinite. To see this, list all books of length 1 letter, then list all books of 2 letters, then list all books of 3 letters, etc. Clearly for any natural number n there is a book, so the number of books is countable. However, if everything was expressible by a book, then every real number would also be expressible, but the number of real numbers is more than the number of books, so there are real numbers not expressible by any book.

I had to write this in a bit of a hurry, but did you follow that?

By Brad Rodgers (P1930) on Sunday, February 27, 2000 - 12:52 am:

Yes, I get it. I have read much on cantor's set theory and very much of hilbert's follow up on it. I believe that the uncountable infinity is alef-one.

Sorry about my latency of this, I had to work on a project, but x -(x2-1)2 -2=0. I did this rather hastily though so it may be wrong. By this equation x is somewhere between 1 and 1.5 I think. How could I solve this further? I don't really see a way to do so, but I don't have too good of math software.

By Dan Goodman (Dfmg2) on Monday, February 28, 2000 - 04:56 am:

I get (x2-1)2-2-x=0, maybe you made a slight sign error? Don't worry about not being able to solve this, although you can do it. The exact answer, according to "Derive" a symbolic maths program is this:

x=SQRT(SQRT((133*SQRT(2193)/1990656+18713/5971968)^(1/3)+(18713/5971968-133*SQRT(2193)/1990656)^(1/3)+(133*SQRT(2193)/73728+18713/221184)^(1/3)+(18713/221184-133*SQRT(2193)/73728)^(1/3)+(SQRT(2193)/486+133/1458)^(1/3)-(SQRT(2193)/486-133/1458)^(1/3)+2/3)+(SQRT(2193)/1152+133/3456)^(1/3)-(SQRT(2193)/1152-133/3456)^(1/3)+2/3)+SQRT(-(SQRT(2193)/1152+133/3456)^(1/3)+(SQRT(2193)/1152-133/3456)^(1/3)+1/3)

An approximate answer is:

x=1.710644095045032935990634163335859456331

In fact, your equation has no solution as x-(x2-1)2-2<0 for all x.

By Dan Goodman (Dfmg2) on Monday, February 28, 2000 - 05:07 am:

Purely for my own amusement I just got MathCAD to display that previous equation, and here is the result...

Biiiiig equation

Amazing how such a simple equation can have such a complicated answer...

By Dan Goodman (Dfmg2) on Monday, February 28, 2000 - 05:11 am:

Also, just noticed a slight problem above. You write that the uncountable infinity is aleph one. Aleph one is uncountable, but it is not necessarily the size of the real numbers. The "Continuum Hypothesis" is that the real numbers have size aleph one. It has been shown that the standard axioms of set theory with the axiom "CH is true" are consistent. But, the standard axioms of set theory with the axiom "CH is false" are also consistent. So CH is independent of the standard axioms of set theory (Zermelo Fraenkel axioms).