a surprising fact about numbers

alright thanks for watching and today i want to talk to you about a fact that it's guaranteed to surprise you and you have to understand if i'm excited about number theory it means it's a pretty cool fact okay so let's start with something that you probably thought of at some point in your lives namely if we pick two numbers at random let's say two and five what are the chances that they have no factors in common for instance here 2 and 5 have no factors in common but let's say 9 and 15 they have a factor in 3 in common and let's see how we can approach this well first of all what are the chances that the number is even so what are the chances that the number is divisible by two well one half because one half of numbers are even i know black pen return doesn't agree with it but i think probabilistically this is true and similarly what are the chances that a number is divisible by three well one out of three numbers is a multiple of three so one one-third and more generally what are the chances that the number is divisible by a prime number p well it's 1 over p and now look if we now pick 2 positive integers at random well it is a pretty independent choice so the chances that both of them have a factor of p in common is just one over p so for the first number times 1 over p for the second number which no it's not 1 over pi m but it's 1 over p squared in particular the chances of m and n not having factor of p in common is 1 minus p squared so now the question is what are the chances that m and n have no factors in common what does it mean it means they don't have 2 in common so 1 minus 1 over 2 squared they don't have 3 in common so times 1 minus 1 over 3 squared they don't have 5 in common 1 minus 1 over 5 squared etc etc so this is the probability that they have no factors in common and of course now the goal is to write this in closed form kind of have an explicit formula for this and for this let's try to cover a problem that seems to have nothing to do with this and surprisingly we'll see that they're related so part two consider the following weird sum one plus one over two squared plus 1 over 3 squared plus 1 over 4 squared dot dot and i have done a separate video on this and it turns out this sum is pi squared over 6. so using what's called parsevals so and this by the way if you're curious that's what's called zeta of two so the sum of reciprocals of squares and let's apply the following clever calculation what if we take zeta of 2 and we divide it by 2 squared so let's do 1 over 2 squared zeta of 2 then it turns out what we get all the denominators become even so it's 1 over 2 squared plus 1 over 4 squared plus 1 over 6 squared plus dot dot and now let's take the original sum and subtract it with this so let's try to do one minus one over two squared zeta two then it turns out if you subtract this from this all kind all the even terms disappear and what we're left with is simply also this and what we're left with is simply 1 plus 1 over 3 squared plus 1 over 5 squared so all the odd terms 7 squared et cetera et cetera and it turns out if you do the same spiel but with 1 minus 1 over 3 squared then all the multiples of 3 will disappear so in other words zeta 2 times 1 minus 1 over 2 squared times 1 minus 1 over 3 squared what this ends up being is simply let's say 1 plus 1 over 5 squared plus 1 over 7 squared plus 1 over 11 squared so not quite all the prime numbers but you can have 1 over 35 squared for instance etc etc and maybe now you notice the pattern here we eliminated all the multiples of two here we eliminated all the multiples of three and it turns out if you continue that process for every single prime number so zeta two times one minus one over two squared times one minus one over three squared times one minus one over five squared etc etc then everything disappears so bang bang bang bang bang bang except for this one and you must say okay this is a random calculation what does that have to do with the previous problem this is precisely the probability that we want it so in fact now we actually have a closed form solution for our probability namely again the chances of two integers being co-prime which is one minus one over two squared one minus one over three squared 1 minus 1 over 5 squared etc etc over all the prime numbers it's 1 over zeta 2 but remember zeta 2 was pi squared over 6 and this is 6 over pi squared did anyone guess that probably not again this is very very neat again it's not obvious that co prime has to do with six over pi squared but wait we are not done because it turns out there's an amazing geometric application of this which i'll cover now now let's try to cover another problem that seems to have nothing to do with co-prime namely consider the integer lattice like that so like this so just the lattice of positive integers and by the way so i got inspired by from this by a tweet by juliet bruce who talked about shadows and lattices and was like oh i should probably do that suppose you have the following lattice and let's say this is the origin 0 0 and let's consider the following fun game namely let's pick a random point and consider the ray going from the origin to that point so for instance this is one two three so one one two three four so this is the point one comma four but then notice here there's nothing blocking this ray so this ray goes directly from this point to this point but for instance if we pick the point uh let's say uh zero one two one two three four like let's say this point two comma four then oh no this middle point here is blocking the array so think like a kind of a shadow but if we pick let's say the point one two three one two three four five let's say here three comma five then it turns out there's also nothing blocking that point so and if you have this ray maybe it goes kind of like that and notice it goes directly from 0 0 to 3 5. and here's an interesting question namely um what is the probability that uh array goes directly from a point to another so among any of those lattice points what is the probability that there is no shadow there or in other words there is no blockage and it turns out surprisingly it has to do with numbers being co-prime because here for instance 3 5 3 and 5 have no factors in common so there is no blocking but if you pick let's say a 2 4 then precisely the point 1 comma 2 is blocking that ray so in other words the probability of having no shadow whatsoever is precisely six over pi squared who would have thought that and finally i mean this seems like a fun game but this is also very useful apparently um in cryptography and the other one is like machine learning machine design so apparently for designing machines this is useful and last but not least i do want to mention because this number remember is one over zeta two there's also some application i forgot exactly which one but if you have let's say three points and maybe there's no shadow or something then the probability would be one over zeta three and in general for a constellation of k points the probability would be one over zeta k how cool is that all right i hope you like this i know hope i promised my delivery okay if you like this and you want to see more math please make sure to subscribe to my channel thank you very much

Channel: Dr Peyam

Views: 48,037

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Id: FGdFxGBuIlI

Length: 10min 32sec (632 seconds)

Published: Tue May 18 2021