Euler's Totient Function -- Number Theory 13

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14 окт 2024

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@goodplacetostop2973 3 года назад

34:54 End of example left for the viewer 35:00 Homework 35:25 Good Place To Stop

@AliKhanMaths 3 года назад

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@leif1075 3 года назад

Really? At what level of math are you ? I'm curious.

@youssefamen6872 3 года назад

Wow first time ?!!

@changjeffreysinto3872 3 года назад

Um please consider with caution before starting a YT channel

@AliKhanMaths 3 года назад

@@leif1075 I'm at GCSE level - but I enjoy learning beyond the scope and sharing my ability with others!

@AliKhanMaths 3 года назад

@@changjeffreysinto3872 I have considered it but it's something I really enjoy!

@elias_toivanen 3 года назад

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@charlottedarroch 3 года назад

φ(n) = 8 iff n is in {15,16,20,24,30}. The case of φ(n) = 10 is easier. If prime p|n and φ(n) = 10, then (p-1)|10, so p is in {2,3,11}. The formula for phi(n) can be expressed n = phi(n)*Product[p/(p-1),primes p]. But φ(n) = 10 is divisible by 5 and n = 2^a 3^b 11^c is not. So we must have that 5|(p-1) for a prime dividing n, so we must have that 11|n. Therefore we need only check the four cases for whether or not a and b are zero. Two of these give solutions and we obtain that φ(n) = 10 iff n is in {11,22}.

@iabervon 3 года назад

I don't think you need as much work in cases for the example. If 3^2 or 5^2 divides n, phi(n) is a multiple of 3 or 5, and 8 is not. So b and c are each 0 or 1, and you just have to pick the appropriate power of 2 to finish each case.

@rbnn 3 года назад

Let R be the pairs (x,y) where x is in Z_m and y in Z_n under coordinatewise addition and multiplication. The map f:Z_mn->R that sends z into (z mod m, z mod n) is injective since m and n are coprime. Hence f is an isomorphism by counting. The units in Z_m (the invertible elements) are the phi(m) numbers coprime to m. Similarly with Z_n. Since the units in R are the pairs (u,v) where u is a unit of Z_m and v of Z_n, the number of units in R is phi(m)phi(n). Because Z_mn and R are isomorphic, they have the same number of units, so phi(m)phi(n)=phi(mn).

@freecky1621 3 года назад

Great Video! However, I would like to point out a possible confusion. At around 10:00 , you're explain that the m on RHS implies LHS is divisible by m. That is only true because you're doing mod mn. Of course, for people who is familiar with number theory, we would notice this being the needed criterion. However, as this video is targeting people who aren't familiar with this topic, I believe that doing such deduction without saying that it is base on the fact that we're doing mod mn might cause confusion for audience. Aside from that, great video as always!

@eliapini1117 2 года назад

You could as well consider the isomorphism between Zn x Zm and Znm in order to demonstrate the multiplicative property. This tell us that they have the same number of invertible elements and (a,b) is invertible iff a and b are invertible, so Φ(nm)=#{(a,b)|gcd(a,n)=1 and gcd(a,m)=1}=Φ(n)Φ(m).

@aboodahmad8236 2 года назад

I have question, Why when prove S={a(j)n+b(s)m|0

@VictorHugo-xn9jz 7 месяцев назад

Then I will ask : where does it say that the size of S is equal to phi(m) phi(n) ? Elements constructed from 0 < j < phi(m) + 1 and 0 < s < phi(n) + 1 may be redundant. So, yes, you have proved that S = reduced residue system of mn. But now, you have to prove its size.

@darpmosh6601 2 месяца назад

@@VictorHugo-xn9jz But we just said that, by the definition of S, 1

@darpmosh6601 2 месяца назад

nevermind I finally understand: The reason why phi(mn) may not be phi(m)phi(n) by the proof alone is because there may be other ways to construct the reduced residue system and with the addition of those possible other configurations, phi(mn) may be greater than phi(m)phi(n)

@naringrass 8 месяцев назад

so what's the complex numer version of Euler's Totient function?

@weisanpang7173 8 месяцев назад

At 10:00, whoever wonders why the a's term on the LHS is multiple of m, and the b's term on RHS is multiple of n : a*n congruent b*m (mod m*n) -> a*n - b*m = (m*n) * k, k = some integer. This is the defn of congruence relation. -> a*n/(m*n) - b*m/(m*n) = k -> a/m - b/n = k -> a needs to be multiple of m and b needs to be multiple of n in order for the LHS to result in integer.

@Iovemath 6 месяцев назад

There is another easier explanation without expanding the congruence into definition form, that is, if x = y (mod mn), then they are congruent to m or n because gcd(m, n) = 1. So, in that case, we have an = bm (mod m), hence a = 0 (mod m), and symmetrically b = 0 (mod n )

@weisanpang7173 8 месяцев назад

18:20, for gcd(cy,m)=1, you said cy is congruent to a_j (mod m) where a_j is from the reduced residue set of m. For this statement, i cannot find any of your previous video explain that. All you video said was if gcd(a,n)=1, then ax=b(mod n) has exactly one solution. It never said b and n must be coprime as well. Please consider some rigor in your explanation, quite frustrating.

@JM-us3fr 2 года назад

It’s probably easier to prove multiplicativity directly; that is, representing phi as a sum of 1’s for each coprime number. Then just multiply it out, and analyse some facts about divisors, and viola!