Ram Murty 1/8 Partial summation formula and applications

What people said

Cites user572425 · 2018-10 · Mathematics Stack Exchange · on “Question about the probability of a number”

This is in reference to our video [this video] which is Ram Murty's video on Analytic Number Theory. On the 17th minute the professor says what is the probability that a random number is divisible by $p$ ? and he himself answers that the probability is $\frac{1}{p}$ ? However I dont understand how he found that? Am I missing something ?Is the answer so much easy?

Cites Luke Collins · 2020-04 · Mathematics Stack Exchange · on “probability that two primes $p_1$ and $p_2$ divide a positive integer”

@joriki I edited my answer. In number theory it's very common to talk about probabilities this way. For instance, the fact that the probability of $x$ being divisible by $p$ is $1/p$ leads to the fact that $\phi(x) = x\prod_{p\mid x}(1-\frac1p)$. Here [this video] is Ram Murty giving this example.

Cites aram · 2024-07 · Mathematics Stack Exchange · on “Probabilistic proof of $\varphi(n) = n\prod_{p \mid n} \left(1 - \frac{1}{p}\right)$”

In these video lectures [this video] Ram Murty gives a, I think, very intuitive to remember probabilistic proof of the following equality of the Euler $\varphi$ function: $$ \varphi(n) = n\prod_{p \mid n} \left(1 - \frac{1}{p}\right) $$ Proof: $$\dfrac{\varphi(n)}{n}$$ equals to the probability that a number is coprime with $n$ (ocurrences over total cases). On the other hand a number is coprime with $n$ if it doesn't divide any of the prime divisors of $n$ , the probability of $k \nmid p$ with $p \mid n$ is $$\dfrac{p-1}{p}$$ so because these are independent and we want the number to not divide any of the primes that divide $n$ we can multiply them to get the desired product. My issue: Although