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Someone told me that the proportion of integers that contain the digit "3" (anywhere in the number) increases as you get higher.
Indeed, as you approach infinity, the proportion of integers containing "3" approaches 100%.
Therefore, all numbers contain the digit "3". :rolleyes:
What do you guys think about that?
Um, I think this is probably where this tremendous feat of logic falls over ;)Quote:
Originally posted by simonm
Therefore, all numbers contain the digit "3". :rolleyes:
But if 100% of integers contain a "3", then mustn't they therefore all contain "3"?
Discussions relating to infinity and transfinite numbers bother me. I vaguely understand the logic, but find the results hard to believe.
I am not sure about the percentage of threes approaching 100%, but it could be so without all integers containing a three.
The problem with this sort of subject matter is that common notions which are valid for ordinary numbers are not valid for transfinite numbers, which results in seemingly absurd conclusions. In dealing with this subject, it is important to remember the following.In the late 19th century, a man named Cantor developed the concept of sets which could not be counted, and coined the term transfinte. His concepts are based on various definitions and assumptions, including the following.
- Everybody understands simple arithmetic with ordinary integers and considers all the conclusions of simple arithmetic as self-evident and true.
- If every property of transfinite numbers was also a property of ordinary numbers, then transfinite numbers would be ordinary numbers.
- Therefore there are some properties of transfinite numbers which are different from ordinary integers. Said properties will seem weird and untrue.
Now consider the set containing all the even integers and the set containing all the integers. Try to put them into a one to one correspondence.
- Two sets have the same number of members if the members of the sets can be put into one to one correspondence, with no member of either set ignored and no member of either set unpaired. This certainly seems self evident and true, until you start investigating the conclusions based on it.
- If an attempt to put two sets into one to one correspondence exhausts the members of one set and there are unpaired members of the other set, the set with unpaired members has more members. This seems obvious, but apparently, it needs be said.
- If a subset of a set can be put into one to one correspondence with the entire set, the set has a transfinite number of members. This one is tuff to grok.
1 2
2 4
3 6
4 8
. . .
n 2*n
Gee, the members match up!!! You provide an integer, I can provide the corresponding even integer and vice versa. No member of either set ignored & no member of either set left over.
The set of all integers has the same number of members as the set of all even integers. It does?? Yes, it seems to fit the definition of sets with same number of members (see above)!!!
This seems strange, but not stranger than the percentage of integers containing a three approaching 100% as you consider more and more integers.
Note that the above does not imply that all integers are even. I am pretty sure that the percentage of threes approaching 100 does not imply that all integers contain a three.
Can u explain how this first stage in ur irreverent hypothesis is true? Is it the amount of numbers divisible by three or actually containing 3? I would think that any integer ending (i know u said anywhere) in a 3 when 1 was added would provide an integer ending in 4.Quote:
Originally posted by simonm
Someone told me that the proportion of integers that contain the digit "3" (anywhere in the number) increases as you get higher.
I am probably ( :D ) a simpleton but I would appreciate some indication as to how this claim can even be devised... is it also applicable to any other number?
Regards
Stuart
BeachBum
Firstly, I did say numbers that contained (anywhere in the string) the digit "3".
I could equally well have said any other number but three was picked on arbitarilly.
And it can be shown that, statisticly, the percentage of integers containing "3" does (generally) increase without bound towards 100%.
Crazy, I know.
Hi Simon
yeah i know u said anywhere.. i just used the final digit as an example as it really makes no difference. I see that the 'theory' is essentially the same as Guv's example of all numbers being even. I guess when using those 'transfinite' (love the name :D ) numbers you can sorta show any result as theoretically true. Maybe you could work for the govt on their warped statistics!! :D
"Statistically, 99% of Irish people are gits!" lol that is there just for mr Plender!!
Regards
Stuart
Beachbum,
The proof of Simon's original statement is as follows.
consider a string of n digits, what proportion of them contain the number 3?
we assign F(n) to be the proportion of strings of n digits which don't contain a 3
we can split these into 2 catagories, the ones that contain a 3 as the first digit and the ones which don't contain a 3 as their first digit but contain a 3 in one of their subsequent digits.
clearly 1/10 of them have a 3 as their first digit, and of the remaining 9/10 (which don't have a 3 as the first digit) F(n-1) of them don't contain a 3 in their last n-1 digits.
So
F(n+1) = (1/10) + (9/10)F(n)
to prove this converges to 1 as n grows without bound we look at the sequencewhich clearly tends to 1Code:F'(n) = 1 - F(n)
= 1 - (1/10) - (9/10)F(n)
= (9/10)(1 - F(n))
= (9/10)F'(n)
= (9/10)^n
=> F(n) = 1 - (9/10)^n
I think what Guv was saying was that the Idea of a percentage of a transfinite number is meaningless, rather than saying that everything is true. Transfinite Number Theory is considered valid and has been used to prove results in a few areas of maths.
Guv,
This statement is falseQuote:
If an attempt to put two sets into one to one correspondence exhausts the members of one set and there are unpaired members of the other set, the set with unpaired members has more members. This seems obvious, but apparently, it needs be said.
what you ment to say was.
For any 2 sets A and B: A has more members than B if and only if it can be shown that the members of A cannot be paired off 1 to 1 with B, or any subset of B.
I heard that if two infinite sets cannot have their members mapped in one-to-one correspondance with each other, one has a higher cardinality than the other.Quote:
For any 2 sets A and B: B has more members than A if and only if it can be shown that the members of A cannot be paired off 1 to 2 with B, or any subset of B.
I got my statement a bit mixed up, I've corrected it now.
Um, this seems a bit silly, really.
Think of it as flipping a coin N times and recording the sequence of flips. What you're saying is analogous to me saying that as N goes to infinity, the probability of a heads being in the sequence of coin flips goes to 1.
The only thing different with numbers is that you have 10 digits... the probability of the number '3' being in an infinitely long number approaches 1 as you proceed towards the "infinite digits."
What would seem mad is if ALL of the numbers where 3 as you approached an infinitely long number. However, I guess you could write infinity as 333333333333333333......... if you truely had an infinitely long sequence of 3's.. But you get into a circular argument, and then everything goes out the window. :P
Destined
Sam: Your proof seems confusing.The bolding was done by me. Is the function the percent with threes or the percent without threes? Later you come up with the following formula, which suggests that the function is the percent of integers with threes.Quote:
we assign F(n) to be the proportion of strings of n digits which don't contain a 3
we can split these into 2 catagories, the ones that contain a 3 as the first digit and the ones which don't contain a 3 as their first digit but contain a 3 in one of their subsequent digits.
The above formula is not obvious. You remind me of the professor who writes a few lines of a proof on the blackboard and prior to the next line says.Quote:
F(n+1) = (1/10) + (9/10)F(n)
He interrupts himself and spends ten minutes scribbling on several pieces of paper, after which he says.Quote:
It is obvious that . . .
After which he writes the next line of the proof.Quote:
I was correct, it is obvious that . . .
BTW: Isn’t your proof a complex version of the following?
- For an n-digit number, the probability of a given digit not being a three is 9/10.
- The probability of no digit being a three is (9/10)^n
- Limit[ (9/10)^n ] is zero as n grows without bound.
- Proability( of a digit 3) = 1 - Probability( of no digit being a 3)
Wow with your math it seems like all the end results are 99%, 100%, 0%, or 1 (the actual number)
You can also split all people into 3 categories... those that can count and those that can't..... badaboom... ok so it was old and lame! :pQuote:
Originally posted by Guv
we can split these into 2 catagories, the ones that contain a 3 as the first digit and the ones which don't contain a 3 as their first digit but contain a 3 in one of their subsequent digits.
good one
:rolleyes:
VB Code:
So, You are saying that, F'(n) = (9/10)*F'(n).
Hmm, So this means 1 = (9/10).
I think we need to revise something here.
-Lou
Hmm, Seems I didn't write the proof out very well at all. Here's a revised version.
Quote:
Consider a string of n digits, what proportion of them contain the number 3?
We assign F(n) to be the proportion of strings of n digits which don't contain a 3
We can divide the numbers into 2 catagories.
Catagory a contains the strings of length n of which 3 is the first digit.
Catagory b contains all the strings of length n which are not in catagory a
And if we define
A(n) = the proportion of strings of length n in catagory a
B(n) = the proportion of strings in catagory b which contain a 3
Clearly A(n) = 1/10
and as all strings in catagory b don't have a 3 as their first digit then the proportion of them which have a 3 in the remaining n-1 digits is B(n) = F(n-1)
So The proportion of strings of length n which contain a 3 is
(The proportion of strings which are in catagory a) + (the proportion of strings which are in catagory b and contain a 3)
which is
(The proportion of strings which are in catagory a) + (the proportion of strings which are in catagory b * the proportion of strings in catagory b which contain a 3)
which is
A(n) + (1 - A(n))(B(n)) = (1/10) + (9/10)F(n-1)
F(n+1) = (1/10) + (9/10)F(n)
to prove this converges to 1 as n grows without bound we look at the sequence
which clearly tends to 1Code:F'(n) = 1 - F(n)
= 1 - (1/10) - (9/10)F(n-1)
= (9/10)(1 - F(n-1))
= (9/10)F'(n-1)
= (9/10)^n
=> F(n) = 1 - (9/10)^n
And let that be a lesson to all of you who post without reading what you've written.
Guv and destined soul are right, the reason is exactly as desined soul pointed out, and my proof, or indeed any proof, is obviously a contortion of his argument. But that's the way I did it, so there.
One thing that really p*ssed me off at school was that mathematicians can't bring themselves to use more descriptive names for their functions...
F(x) or A(g) means bugger all to anyone.
I realise that in algebra a string of letters means the product of the values that all those letters represent. Fine.
But it would be loads easier if mathematicians were to drop this silly custom and actually write in the multiplication * operator where ever it should be and then they can get on with writing nice descriptive function names like GetReciprocal(x) instead of G(x).
God I hate single letter notation. Its flawed in so many ways, like only having a finite number of them for a start, this is not a problem with variable names longer than 1 character.
Oh yes, Newton, Einstein and Pythagoras, you guys think you so clever dont you! Smart arses in my opinion. (Grumble moan)
i suppose that's how it's been for a long time and that's how it'll continue to be:
Student.Add
Do
Call Teach(Student, SingleLetterNotation)
Call StudentBecomesTeacher(Student)
Student.Add
Loop
If Student.Name = "Wossname" then Student.Flunk = True
I have to both agree and disagree with this statement. True, it could be better defined as StdDev(x) instead of S(x), but when you get into upper-level math, using more than one character to represent a function or variable would become extremely time-consuming. Seeing as 50 minutes of notes already takes an average of 6 to 7 pages to copy down, I'm thankful that the professor only uses g(x)...Quote:
Originally posted by wossname
One thing that really p*ssed me off at school was that mathematicians can't bring themselves to use more descriptive names for their functions...
F(x) or A(g) means bugger all to anyone.
(It does, however, make it a lot more readable to programmers if functions have more than one character in their name.)
Anyhow, that's my 2 cents.
Destined
Quote:
"I think it is obvious that as the number of included values approaches infinity, the proportion of those values that meet any arbitrary criteria will also approach infinity.
How can it be otherwise? In an infinite universe everything is possible, and with infinite frequency." - Adam N. Ward
NotLKH: (Number of integers containing digit 3) / (Number not containing digit 3) grows without bound as the number of digits increases without bound. Consider the following.Limit[ (10 / 9)^n - 1 ] grows without bound as n increases without bound.
- 10^n is total number of n-digit integers.
- 9^n is total number of n-digit integers without using digit 3.
- (10^n - 9^n) is total number of n-digit integers containing at least one instance of the digit 3.
- x = (10^n - 9^n) / 9^n, where x is (Number with a 3) / (Number without a 3)
- x = 10^n / 9^n - 9^n / 9^n, which is [ (10 / 9)^n - 1 ]
Note that y = (10^n - 9^n) / 10^n, where y is (Number with a 3) / (Total number)
y = 10^n / 10^n - 9^n / 10^n, which is [ 1 - (9 / 10)^n ]
Limit[ 1 - (9/10)^n ] = 1 as n increases without bound, which is what was proven in previous posts.
Conclusion: As n grows without bound, almost every n-digit integer contains at least one instance of each of the ten digits.
A similar conclusion can be made for every radix notation. In binary the conclusion is obvious: There is only one n-bit integer with no zero bits; Only one n-bit integer (zero) with no one bits; And (2^n - 2) n-bit integers with at least 1 one bit and at least 1 zero bit.
Your reasoning is almost unflawed, but is.
The percentage will get close to 100%, but never actually be 100%:)
Think of it like this:
Start with the number 1
Halve it (0.5)
Halve that (0.25)
Halve that (0.125)
etc..keep halving.
You will NEVER get to 0. You will get very close, each step closer to 0 than the last, but you will never get to 0.
Similarly, the higher you go, the closer to 100% you will get but you will never get to 100%:cool:
which is why everyone is talking about the limit, or 'in the limit'.
Err, actually, you will get there after an infinite number of steps, just not after a finite number of steps.Quote:
Think of it like this:
Start with the number 1
Halve it (0.5)
Halve that (0.25)
Halve that (0.125)
etc..keep halving.
You will NEVER get to 0. You will get very close, each step closer to 0 than the last, but you will never get to 0.
This reminds me of a theoretical illustration once made by Rudy Rucker.
In this story there is a mountain that is infinitely high. Instead of a gradual incline, it is staggered into an infinite number of levels (or cliffs). The distance between each cliff is finite.
The character in the story manages to climb the mountain in two hours. He does this by climbing the first cliff in one hour, the seccond in half an hour, the third in quarter of an hour etc.
Basically, by climbing the first cliff in one hour and repeatedly halving the amount of time it took to climb each subsequant cliff, he climbed the lot in a finite time.
He used one infinity to cancel out the other.
Sql_Lall: You are absolutely correct. When a mathematician says Limit(Something) = LimitValue, he is not claiming that Something ever equals LimitValue. He is only saying that it gets closer than you can imagine to LimitValue. In some cases, he might claim that using the limit instead of the function will result in neither illogical conclusion nor noticeably erroneous computations.
To the best of my knowledge, formal mathematicians avoid making statements about a computational value or a geometric location called infinity, although they might occasionally refer to it as a shorthand jargon for more formal statements.
There are all sorts of common notions about infinity which are not really valid. I have heard it said that parallel lines meet at infinity. I can remember reading somewhere that it is convenient to define parallel lines as meeting at infinity so it can be claimed that any two lines define a point, which is their intersection.
A problem with this notion is that is it difficult to claim that the X-Axis and the line y = 1 meet at positive infinity without claiming that they also meet at negative infinity. Now you have the parallel lines meeting in two places, which seems more absurd than other claims about infinity and transfinite number.
I have even seem trigonometric proofs that parallel lines meet. They always start with the assumption that the lines meet, so that the proof can work with a triangle. Once you assume they meet, it is easy to come up with an argument consistent with their meeting. Hell, if I assume the moon is made of green cheese, I can prove that it is made of green cheese.
In complex analysis, there are references to a circle at infinity. Once again, this is a limit concept not a circle one can work with.
Guv & Sql_Lall
It seems to me that you two are both suffereing from infinity denial. You can't quite bring yourselves to say that infinities exist.
I know you have always maintained this but I have seen mathematical texts saying that (for example) an infinite sum does does equal it's limit.Quote:
When a mathematician says Limit(Something) = LimitValue, he is not claiming that Something ever equals LimitValue. He is only saying that it gets closer than you can imagine to LimitValue.
In the example I provided, Rudy Rucker is clearly saying that the limit is reached (not just approached).
It's like saying Pi does not have an infinite number of decimal places. That the number of decimals seem to be infinite but they aren't.
Well, They have found more than a quadrillion of pi's decimal places...
How do they know it is right?
Why do they feel the need to waste CPU time to this meaningless feat?
Wasn't 100 decimal places enough?
Thankyou for your suggestions, and i'll leave you with these thoughts:
:p 1:
In a theoretical world, one where you can do infinite calculations, we say limit(something) = limit value.
But, in the physical world, you can't say that, as there isn't a way to do infinte calculations.
take this example
1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 +.....
= 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) +.....
> 1/2 + 2/4 + 4/8 +.....
> 1/2 + 1/2 + 1/2 +.....
> 1/2 + 1/3 + 1/4 +.....
i.e is is greater then itself.
This works out because we are using a theoretical infinity combined with the physical operations of +, > etc.
:D 2: there will always be a power of ten, no matter how high you go. Powers of ten do not contain the number 3, not do
10^x + 1, +2, +4 etc. So it never can reach 100%.
Sql_Lall...
Are you saying that something like
1/2 + 1/4 + 1/8 + 1/16 + 1/32 .....infinity
cannot be solved?
I wasn't quite sure what you meant in your post.
It equals 1 if my A-Level maths textbooks tell me right!
This is probably because that particular sequence converges on y=1 rather than diverging away from it.
And its pretty easy to show its = 1.Quote:
Originally posted by wossname
Sql_Lall...
Are you saying that something like
1/2 + 1/4 + 1/8 + 1/16 + 1/32 .....infinity
cannot be solved?
I wasn't quite sure what you meant in your post.
It equals 1 if my A-Level maths textbooks tell me right!
This is probably because that particular sequence converges on y=1 rather than diverging away from it.
N = 1/2 + 1/4 + 1/8 + 1/16 + ....
So,
2*n = 1 + 1/2 + 1/4 + 1/8 + 1/16 + ....
2*n - 1*n = 1 + (1/2 - 1/2) + (1/4 - 1/4) + ....
so,
n*(2-1) = 1 => n = 1.
Nice when you don't need to remember limit theory.
-Lou
SimonM: You are correct in saying that I have infinity denial. There is a good reason for my denial.
The formal mathematics of algebra and analysis do not accept infinity as a number. Set Theory accepts it in the contexts of the transfinite cardinal numbers associated with the number of elements in various sets. For example: the set of all integers (Aleph-0, or Aleph Null), the set of all subsets of Aleph-0, the set of all real numbers, et cetera. BTW: At one time there was a group of mathematicians who did not accept the concepts of transfinite set theory.
Yes, there are textbooks which use the term infinity and the figure 8 symbol rotated 90 degrees. There was a time when formal mathematical literature used infinity as the name of some usable number. Near the start of the 20th century, formal mathematicians became very pedantic (picky) about the terms infinity and infinite. I will paraphrase from the 3rd Edition of Fundamentals of Mathematics (Moses Richardson, The MacMillan Company, published in 1969). As far as I know, later books take a similar position.
The book uses the term infinite sequences when introducing the subject of limits, referring to it as meaning an unending sequence. It gives formal definitions of the limit of a sequence and the limit of a function using terms like increases indefinitely and grows without bound, but avoids using the term infinity or infinite in these formal definitions.
After discussing some functions & sequences which grow without bound and sequences with an indefinite number of terms, he makes some statements similar to the following, which is not a precise quote.In a footnote to the above paragraph he says:Quote:
Intuitively, this means that the values of Function(x) get indefinitely large as x gets near to the critical value.
This is often written as Limit[ Function(x) ] = infinity (or approaches infinity) as x approaches the critical value.
Such statements sometimes mislead the unwary student into thinking that there is a peculiar number (infinity) which is the limit of Function(x) . . . . It would be even more misleading to write 1/0 = infinity, although some books write it. Of course, they use this notation to mean that 1/x grows without bound as x approaches zero. They do not mean that infinity is a number which you obtain by dividing one by zero.
As we have seen, 1/0 is a meaningless symbol, and furthermore the symbol (reclining 8) is not a number at all. Infinity is merely a way of describing the manner in which certain functions vary.
To show how really strange, picky, & compulsive a formal mathematician is, consider the following.Quote:
Historically, the statement 1/0 = infinity was taken literally, but we are now more familiar with the logical foundations of numbers than were the pioneers in the field.
Richardson points out that the function (x^2 - 9) / (x - 3) is defined everywhere except for x = 3. He further points out that it is equivalent to the function (x+3) everywhere but at x = 3, because it is undefined there. He points out that the limit of the function is 6 as x approaches 3. He then accuses the function of being discontinuous (at least failing the formal definition of continuous) because it is undefined at x = 3. This is too picky for me. I would forget I ever heard of the function and work with (x + 3) as the function.
In Complex Analysis Theory, there are references to a circle at infinity, but the formal definitions avoid use of the terms infinity and infinite. The circle at infinity is derived by mapping a sphere to a plane and vice versa. The south pole of the sphere rests on the origin of an XY-coordinate system. A line from the north pole to the plane defines the mapping. The points on the sphere and in the plane intersected by the line correspond to each other. Circles of constant latitude correspond to circles in the plane. These circles in the plane grow without bound as the latitude approaches the north pole.
If you accept the concept of an actual circle at infinity, then a plane parallel to the XY-Plane at the north pole has a circular intersection with the XY-Plane. This seems even worse than the common notion that parallel lines intersect at infinity, which implies that parallel lines have two points of intersection.
Thankyou for that nice responce
Infact, i was just using that series to show this:
answer this:
a + 1 = a
Solution: a = the aleph-0 infinity
(yes, there is more than one infinity) :cool:
aleph-0 + 1 = aleph-0 ;)
also: aleph-0 x 2 = aleph-0 ;) (I think)
That's what happens with infinity :p
Just reminding what this thread is about:
Question:
The percent of numbers that contain the digit "3" gets higher as the numbers get higher. Does that mean that eventuall all numbers contain the digit "3"?
My answer:
Yes theoretically . It can be said to equal it's limit.
No physically, as you can't actually say the number that after it, all numbers contain the digit "3"
I think some people are thinking theoretical, som physical. So you are both right.
The percentage of n-digit numbers containing the digit 3 approaches 100% as n grows without bound but all indefinitely long numbers do not contain a digit 3.
This concept of the percentage of numbers containing the digit 3 is a wonderful example of problems with limiting values.
As the number of digits grows without bound, the percent of integers containing the digit 3 approaches 100%. This is quite easy to prove, as has been shown in previous posts.
Yet there are examples of indefinitely long integers which do not contain the digit 3. In spite of 100% being the limit, it is obviously incorrect to say the percentage ever equals 100%.
It is due to examples like this that mathematicians avoid saying that a function equals its limit. Even in theory, this is incorrect.
As I said earlier, there are subtleties in the concepts of limits and unbounded functions which are contrary to common sense notions.
Guv -
I'll bet you knew this but anyway:
'figure 8 laying on it's side' = analemma
Jim McNamara: If you knew what analemma means, you not only know more than I do, you know more than an old dictionary I consulted, which did not include the word.
A newer dictionary mentions a figure 8 shape as one definition, but does not relate it to the infinity symbol.
I assume that you are correct and will try to remember that analemma is the infinity symbol
Thanx.
Very interesting! Excuse me now while my head explodes! :D
www.analemma.com
analemma being the shape but not the quantity/reference/value/absence of value/number/notion that is infinity
I thought it was a porno title ;) ;) ;) ;) ;)