3 is everywhere - Numberphile

What people said

Recommends QBrute · 2015-07 · Mathematics Stack Exchange · on “Frazzle game question”

There is a really nice Numberphile video on [this video] , where he talks about occurrences of digits in numbers, which is directly related :)

Cites v010dya · 2014-08 · Mathematics Stack Exchange · on “Comparing density of countable infinite sets by examining the association”

The two questions that i am asking are in bold. To be clear, i am talking about whole number here. Having seen 3 is everywhere by Numberphile [this video] that shows that almost 100% of the whole the numbers have the digit three in them, i have been thinking of approaching the same problem from the different perspective: We have the following method of generating numbers with 3: Take any whole number without digit 3, add 3 to the left of it, and then keep adding whatever digits to the left of that 3 for as long as you want. What this tells me is that i can generate infinite amount of numbers with 3 for each one number without 3. I can then trivially show that each infinite set of numbers with 3 w

Cites Crostul · 2016-03 · Mathematics Stack Exchange · on “On the theorem "$3$ is everywhere"”

In this Numberphile video [this video] it is stated that "almost all natural numbers have the digit $3$ in their decimal representation", and a proof of this fact is proposed. A sketch of the proof follows: Denote by $D_3$ the set of natural numbers having a digit $3$ in their decimal representation. For all $n \ge 1$, denote by $$f(n) = | D_3 \cap \{ 1, \dots , n\} |$$ it is proved that for all $n$ $$f(10^n) = 10^n- 9^n $$ holds (and this is quite clear), hence $$\lim_{n \to + \infty} \frac{f(10^n)}{10^n} = 1$$ and this concludes the proof in the video. Now, this proof is clear and evident to me, but I think that it is incomplete, since we should prove that $$\lim_{n \to + \infty} \frac{f(n)}{n}

Cites ruby duby · 2016-08 · Mathematics Stack Exchange · on “How can the probability that a number contains the digit 3 be 1?”

Based on this Numberphile video [this video] which claims almost all integers contain a $3$, I have a few questions on the reasoning behind recurring decimal numbers like $0.9999\ldots =1$ What they have shown is that $$\lim_{n \to + \infty} \frac{10^n-9^n}{10^n} = 1$$ this basically means that probability of eg. a $3$ occurring in a set of numbers like for $1-10, 1-100,$ increases as the upper bound gets large. So you are more likely to see a $3$ when you take $1-100000$, than $1-10$ as the probability gets higher. So what I would like to know is as '$n$' approaches $∞$ does probability of seeing a '$3$' equals $0.99999....$? But since $0.9999... = 1$ wouldn't this not make sense, since there ar

Cites Truth-seek · 2018-02 · Mathematics Stack Exchange · on “Prove $10^{n+1} - 9^{n+1} = 9^n + 9^{n-1}10 + 9^{n-2}10^2 + ... + 10^n$”

In this video [this video] , James Grime shows that the number of $\mathbb{N}$ less than $10^{n+1}$ that have at least one $3$ among their digits is given by this recurrence relation: $$T_{n+1} = 9T_{n} + 10^{n}\; where\; T_0=1$$ But later in the video, he says it can also be written as: $$T_n = 10^{n+1} - 9^{n+1}$$ So, I set out to prove they are the same algebraically: Solving the recurrence relation, I got: $$T_n = 9^n + 9^{n-1}10 + 9^{n-2}10^2 + ... + 10^n$$ So, can somebody show me a proof to show that: $$10^{n+1} - 9^{n+1} = 9^n + 9^{n-1}10 + 9^{n-2}10^2 + ... + 10^n$$