The infinite monkey theorem states that a monkey hitting keys at random on a typewriter keyboard for an infinite amount of time will almost surely type a given text, such as the complete works of William Shakespeare.   Source:   My question is actually if pi is a "random" sequence of numbers. (I've read in other posts that it's likely to be such a sequence, and that e.g. irrationality isn't a sufficient condition, as can be seen in 0.01001000100001...) But is there an elegant mathematical way to proof if a number is such a random sequence? Or can it be proven with statistics/numerical methods with a kind of certainty that it's such a number?  EDIT: people stated that the exact mathematical term for "random" that I was searching for, should've been "normal". So my questions boils down to:  Pi is likely to be a normal number; if it is, it contains every sequence of numbers. But if it isn't, does it then contain every sequence of numbers (although in this case not with the same likelihood)? Or is this not sure (or even a contradiction)?
284
I came across the :    Which states:     $\pi$ Pi      Pi is an infinite, nonrepeating $($sic$)$ decimal - meaning that   every possible number combination exists somewhere in pi.  Converted   into ASCII text, somewhere in that infinite string of digits is the   name of every person you will ever love, the date, time and manner    of your death, and the answers to all the great questions of    the universe.   Is this true? Does it make any sense ?
41284
Define the sequence $P(n)=\sum_{i=1}^{n}p_i$, where $p_i$ is the $i$-th prime number.  I observed for some small $n$ that sometimes this sum evaluates to a prime number, for example $P(2)=2+3=5$ and $P(4)=2+3+5+7=17$ and $P(6)=2+3+5+7+11+13=41$.  So it is natural to ask:     Is it true that there is a sequence of natural numbers $\{n_i:i \in \mathbb N\}$ such that all numbers in the set $\{P(n_i):i \in \mathbb N\}$ are prime numbers?
56314
Q: Is every element of $\mathbb Z/p\mathbb Z$ represented by a prime number? More generally, let $m, n \in \mathbb Z$ be coprime. Is there a prime number congruent to $m$ modulo $n$?   the affirmative answer is given by .     A: Yes, in fact there are infinitely many such primes.   I don't want to prove Dirichlet's theorem. Is it obvious that there always exists at least one such prime?
75474
In Euclid's proof that there are infinitely many primes, the number  $p_1 p_2 ... p_n + 1$ is constructed and proved to be either a prime, or a product of primes greater than $p_n$.  Trivially, we could also use the number $R_n=p_1 p_2 ... p_n - 1$ to prove the theorem, for n>2.  Intuitively, as $n$ grows, the probability that $R_n$ is prime gets smaller. Is there a proof that $R_n$ is not prime for any $n$ greater than some integer $M$ ? Or conversely, that there is an infinite number of prime $R_n$ numbers?  A possibly equivalent question: Is there a prime number greater than $p_n$ and smaller than $R_n$ for any $n$ greater than some integer $M$ ?
6674
My understanding is that a random variable is actually a function $X: \Omega \to T$, where $\Omega$ is the sample space of some random experiment and $T$ is the set from which the possible values of the random variable are taken.  Regarding the set of values that the random variable can actually take, it is the image of the function $X$. If the image is finite, then $X$ must be a discrete random variable. However, if it is an infinite set, then $X$ may or may not be a continuous random variable. Whether it is depends on whether the image is countable or not. If it is countable, then $X$ is a discrete random variable; whereas if it is not, then $X$ is  continuous.  Assuming that my understanding is correct, why does the fact that the image is uncountable imply that $Pr(X = x) = 0$.  I would have thought that the fact that the image is infinite, regardless of whether it is countable or not, would already imply that $Pr(X = x) = 0$ since if it is infinite, then the domain $\Omega$ must also be infinite, and therefore  $$Pr(X = x) = \frac{\text{# favorable outcomes}}{\text{# possible outcomes}} = \frac{\text{# outcomes of the experiment where X = x}}{|\Omega|} = \frac{\text{# outcomes of the experiment where X = x}}{\infty} = 0$$  What is wrong with my argument?  Why does the probability that a continuous random variable takes on a specific value actually equal zero?
71721
Do you a digit efficient way to approximate $\pi$? I mean representing many digits of $\pi$ using only a few numeric digits and some sort of equation. Maybe mathematical operations also count as penalty.  For example the well known $\frac{355}{113}$ is an approximation, but it gives only 7 correct digits by using 6 digits (113355) in the approximation itself. Can you make a better digit ratio?  EDIT: to clarify the "game" let's assume that each mathematical operation (+, sqrt, power, ...) also counts as one digit. Otherwise one could of course make artifical infinitely nested structures of operations only. And preferably let's stick to basic arithmetics and powers/roots only.  EDIT: true. logarithm of imaginary numbers provides an easy way. let's not use complex numbers since that's what I had in mind. something you can present to non-mathematicians :)
95910
I am from a physics background and my mathematics is not very good, so pardon my insolence with the question.  Editing based on the comments : We know that $\pi$ in decimal (i.e. base 10) is transcendental. Is it possible to have a radix base in which $\pi$ can be rational in that base ?  PS : My question is in general for any transcendental number.
109545
The sum of a finite number of rational numbers is of course a rational number, but the sum of an infinite number of rational numbers might be an irrational number.  Can someone give me some intuition why this sum might be irrational?  I just "don't feel it."
97166
Problem description:  Suppose we use a set of digits $\{0,1,2\}$ to form a sequence, for example \begin{equation} 120210120102012102010210120212010\cdots  \end{equation} The length can be finite or infinite.  The sequence is invalid if there are two finite subsequences appears consecutively. For example, the following sequences are invalid  \begin{equation} 0{\bf 11}\cdots, \quad 010{\bf 201\, 201 }\cdots  \end{equation} where the consecutive subsequences are highlighted.   Question: Does there exist a valid sequence of infinite length? (a non-constructive proof is also acceptable; but a constructive one would be better.)  One way to think about it:  We can transform the sequence to a (base 3) real number,  \begin{equation} 120\cdots \rightarrow 0.120\cdots  \end{equation} Then the only possible infinite sequence should be a irrational number. Irrational number has no period, which means the pattern will not repeat infinite many times, but may repeat finite many times. A famous example is the ,      The Feynman point is a sequence of six 9s that begins at the 762nd   decimal place of the decimal representation of $\pi$.   So not all irrational number($\le 1$) are valid sequences.   I use a short program to do enumeration, but it quickly falls in exponential growth of candidates. I failed to come up with a contraction.   Generalization:  Can we take a larger set like $\{0,1,2,3, \cdots, n\}$? Can we relax the constraint to be three consecutive repetition or more?  Motivation  This problem origins from  in chess,     In chess and some other abstract strategy games, the threefold   repetition rule (also known as repetition of position) states that a   player can claim a draw if the same position occurs three times, or   will occur after their next move, with the same player to move. The   repeated positions do not need to occur in succession. The idea behind   the rule is that if the position occurs three times, no progress is   being made.   It reduces to the proposed problem if we modify the this rule to be "the same position occurs two times with the same sequences of moves".   Thanks for your help.  Edit: I find a related question:
39556
I think I have the basic framework for this proof, but I am having trouble putting everything together in a convincing way.    I plan on showing that by the definition, an unbounded sequence $(a_n)$ has infinitely many $n$ such that $a_n&gt;M$ where $M&gt;0$, for some natural number $n$. I'm not sure how it is implied that there exists some $a_{n_k}$ such that $a_{n_k}&gt;M$ for some natural number $k$.  Let me know if I'm on the right track, and please help me string these ideas together!
121480
I just had an (intellectual) panic attack.     A continuous random variable that follows a uniform in a closed interval $U(a,b)$: a comfortably familiar statistical concept.   A continuous uniform r.v. having support over the extended reals (half or whole): not an r.v. proper, but a basic Bayesian concept for an improper prior, useful and applicable. A discrete uniform taking a finite number of values: let's throw a geodesic dome, no big deal.     But what about a function that has as its domain all the rationals that are included in a closed interval with integer bounds (start with the $[0,1]$ if you wish)? And we want to use it in a probabilistic framework, requiring that each possible value has equal probability with all the others?   The number of possible values are countably infinite (which characterizes a lot of discrete distributions), but how to express the probability of a single value given that we want probabilities equal?  Can we say-show-prove that such an entity is (is not) a random variable?    If not, is this another incarnation (perhaps already well-known) of an "improper prior"?  Is it possible that this entity is in some well-defined sense, however special,  "equivalent" to a continuous uniform r.v.? Or I just committed a cardinal(ity) sin?  It appears that the fact that the domain is a closed interval does not let me let go. Bounded things are usually manageable.  The questions are many in order to be indicative of the internal maelstrom- I am not asking to get answers to each one of them.    At any time that I may come up with any insights, I will update.    UPDATE: the present question just acquired a constructivist sequel
92443
This question was inspired by an answer and some comments to .   Recall that the Champernowne constant is obtained by concatenating all natural numbers written in base 10 and then put $0.$ in front, that is, $$C_{10}=0.123456789101112131415161718192021222324\cdots$$  The question is does the decimal expansion of $\pi$ occur as a tail of this number? By which I mean is there some $n$ such that $$\pi=C_{10}\cdot10^{n+1}-10\cdot\lfloor C_{10}\cdot10^n\rfloor$$  Now it is obvious that any finite initial string of $\pi$ occurs in $C_{10}$. Not only that, it also occurs infinitely often (that's just because any finite string of numbers occurs infinitely often in $C_{10}$).  Hagen von Eitzen (who's answer inspired this question) also notes that proving that $\pi$ does occur as the tail of $C_{10}$ would imply that $\pi$ is base-$10$-normal which is an open question see e.g. , so it is very unlikely we can prove that. He also notes "that if such a position exists [one where $C_{10}$ starts giving the digits of $\pi$] (somewhere in the middle of an $n$-digit integer, say) then the first $10^{n/2}n$ or so digits of $\pi$ turn out nearly regular and this should give rise to an unusually good rational approximation."  I tried to consider if there might be some relationship with relative algebraicity, but even in that direction little seems to be known since it is unknown even whether $\pi$ is algebraic over $e$.   In conclusion it seems intuitively extremely unlikely that $\pi$ would occur in $C_{10}$, but I can't think of a proof that it doesn't. A good answer would also be a reduction of this problem to some known open problem (note that we need to reduce that $\pi$ is not the tail of $C_{10}$).  PS If someone can come up with better tags please have a go at it.
106254
$\newcommand{\P}{\mathbb{P}}$Is it valid to interpret $\P(B|A)$ as "infinite probability" when $\P(A) = 0$?
115723
A Professor prepares n letters and puts them in envelopes and closes them. Then realizes he has to write addresses, now he randomly writes addresses in the closed envelopes. What is the probability that he sends exactly k letters with correct addresses?  Attempt: Fix the k correct letters. We can choose them in $ n\choose k$ ways. Total number of combination is n!. Could I say answer is $\frac {n\choose k} {n!}$.  I am just getting started with probability, please post answers considering this.
113972
Are there any simple methods for calculating the digits of $\pi$? Computers are able to calculate billions of digits, so there must be an algorithm for computing them. Is there a simple algorithm that can be computed by hand in order to compute the first few digits?
12015
I came across a combinatorial proof of the Fermat's Little Theorem which states that    If $p$ is a prime number then the number ($a$$p$-$a$) is a multiple of $p$ for any natural number $a$.  Let me write down the proof.    PROOF  We have pearls of $a$ colours . From these we make necklaces of exactly $p$ pearls .   First,we make a string of pearls . There are $a$$p$ different strings.If we throw away  the $a$ one one -coloured pearls  ($a$$p$-$a$) strings will remain.We connect the ends of each string to get necklaces.We find that two strings that differ only by a cyclic permutation of its pearls result in indistinguishable necklaces.But there are $p$ cyclic permutations of $p$ pearls on a string . Hence the number of distinct necklaces is    [($a$$p$-$a$)/$p$].  Because of this interpretation this is an integer.Thus ($a$$p$-$a$) is a multiple of $p$ for any natural number $a$. [HENCE PROVED]     MY CONFUSIONS    What is the use of the fact that $p$ is a prime in this proof?  If this proof is true then why is this not true for every positve integer $p$?
68700
I came across this when learning about sequences and series. No one proved it to me tho! Is there a link anywhere? Or would it be beyond what I know? (First term of analysis)