my friend said he'd pay me money if i manage to correctly answer an exercise, i've been trying to do it for the past 4 hours but i think i'm not gonna be able to
is there some sort of forums where you can ask people to help you with something like this?
thanks :)
forums where i can for ask help with math?
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my friend said he'd pay me money if i manage to correctly answer an exercise, i've been trying to do it for the past 4 hours but i think i'm not gonna be able to
is there some sort of forums where you can ask people to help you with something like this?
thanks :)
is there some sort of forums where you can ask people to help you with something like this?
thanks :)
some people might be able to answer it on this website, if not then try stackexchange:
http://math.stackexchange.com/
or overflow:
http://mathoverflow.net/
some people might be able to answer it on this website, if not then try stackexchange:
http://math.stackexchange.com/
or overflow:
http://mathoverflow.net/
http://math.stackexchange.com/
or overflow:
http://mathoverflow.net/
Might as well post the problem here for people to offer some help with it.
Might as well post the problem here for people to offer some help with it.
Falcon0408some people might be able to answer it on this website, if not then try stackexchange:
http://math.stackexchange.com/
or overflow:
http://mathoverflow.net/
Yo, idk how appropriate it is to use the stack exchange for something like this, especially not the overflow one, which is specifically for professional mathematicians. They're not really there to deal with much low-level stuff or homework questions. I mean, they say, "for people studying math at any level," but that's not really true, in my experience.
But yeah, post it in the thread and we'll take a crack. >:)
[quote=Falcon0408]some people might be able to answer it on this website, if not then try stackexchange:
http://math.stackexchange.com/
or overflow:
http://mathoverflow.net/[/quote]
Yo, idk how appropriate it is to use the stack exchange for something like this, especially not the overflow one, which is specifically for professional mathematicians. They're not really there to deal with much low-level stuff or homework questions. I mean, they say, "for people studying math at any level," but that's not really true, in my experience.
But yeah, post it in the thread and we'll take a crack. >:)
http://math.stackexchange.com/
or overflow:
http://mathoverflow.net/[/quote]
Yo, idk how appropriate it is to use the stack exchange for something like this, especially not the overflow one, which is specifically for professional mathematicians. They're not really there to deal with much low-level stuff or homework questions. I mean, they say, "for people studying math at any level," but that's not really true, in my experience.
But yeah, post it in the thread and we'll take a crack. >:)
zxppost it here?
I'd like to see what it is to a) solve it myself, b) have a smarter person here solve it, or c) be able to more properly assess where to send you.
[quote=zxp]post it here?[/quote]
I'd like to see what it is to a) solve it myself, b) have a smarter person here solve it, or c) be able to more properly assess where to send you.
I'd like to see what it is to a) solve it myself, b) have a smarter person here solve it, or c) be able to more properly assess where to send you.
zxppost it here?
∀n∈ℕ: A(n)=card { k∈[1;n]/ gcd(k;n)=1} **(k is also an integer)**
1) calculate A(1) and A(13) and A(20)
2) n∈ℕ-{0;1}, prove that A(n)=n-1 only if n is a prime number
3) p is a prime number, and (k;m)∈ℕ*², prove that gcd(m;p^k)≠1 only if p/m **(p is a divisor of m)**
4) conclude that A(p^n)=(p^n)-(p^(n-1)) for all n ∈ ℕ*
formatting is hard sorry for the mess :(
but srsly tho if anyone can solve question 2, 3 and 4 i will be forever grateful
[quote=zxp]post it here?[/quote]
∀n∈ℕ: A(n)=card { k∈[1;n]/ gcd(k;n)=1} **(k is also an integer)**
1) calculate A(1) and A(13) and A(20)
2) n∈ℕ-{0;1}, prove that A(n)=n-1 only if n is a prime number
3) p is a prime number, and (k;m)∈ℕ*², prove that gcd(m;p^k)≠1 only if p/m **(p is a divisor of m)**
4) conclude that A(p^n)=(p^n)-(p^(n-1)) for all n ∈ ℕ*
formatting is hard sorry for the mess :(
but srsly tho if anyone can solve question 2, 3 and 4 i will be forever grateful
∀n∈ℕ: A(n)=card { k∈[1;n]/ gcd(k;n)=1} **(k is also an integer)**
1) calculate A(1) and A(13) and A(20)
2) n∈ℕ-{0;1}, prove that A(n)=n-1 only if n is a prime number
3) p is a prime number, and (k;m)∈ℕ*², prove that gcd(m;p^k)≠1 only if p/m **(p is a divisor of m)**
4) conclude that A(p^n)=(p^n)-(p^(n-1)) for all n ∈ ℕ*
formatting is hard sorry for the mess :(
but srsly tho if anyone can solve question 2, 3 and 4 i will be forever grateful
that is the euler totient function...
i'll try to answer your questions later :)
that is the euler totient function...
i'll try to answer your questions later :)
i'll try to answer your questions later :)
#2:
If n is prime, then for all numbers 1, 2, 3, ..., n-1, gcd(x,n) = 1, which follows from the definition of a prime number, so A(n) = n-1.
If n isn't prime, then for some number in 2, 3, ..., n-1, that number divides n, so A(n) < n-1.
#3:
If p is not a divisor of m, then p and m share no common factors, i.e. gcd(p, m) = 1. Therefore gcd(p^k, m) = 1.
If p divides m, clearly gcd(m, p^k) > 1.
#4: We want the number of integers between 1 and p^n that have no common factors with p^n. From #3, we know that unless x is a multiple of p, gcd(x, p^n) = 1. So we calculate the number of multiples of p between 1 and p^n, which is just p^(n-1). And the total number of integers between 1 and p^n is p^n, so the answer is A(p^n) = p^n - p^(n-1).
#2:
If n is prime, then for all numbers 1, 2, 3, ..., n-1, gcd(x,n) = 1, which follows from the definition of a prime number, so A(n) = n-1.
If n isn't prime, then for some number in 2, 3, ..., n-1, that number divides n, so A(n) < n-1.
#3:
If p is not a divisor of m, then p and m share no common factors, i.e. gcd(p, m) = 1. Therefore gcd(p^k, m) = 1.
If p divides m, clearly gcd(m, p^k) > 1.
#4: We want the number of integers between 1 and p^n that have no common factors with p^n. From #3, we know that unless x is a multiple of p, gcd(x, p^n) = 1. So we calculate the number of multiples of p between 1 and p^n, which is just p^(n-1). And the total number of integers between 1 and p^n is p^n, so the answer is A(p^n) = p^n - p^(n-1).
If n is prime, then for all numbers 1, 2, 3, ..., n-1, gcd(x,n) = 1, which follows from the definition of a prime number, so A(n) = n-1.
If n isn't prime, then for some number in 2, 3, ..., n-1, that number divides n, so A(n) < n-1.
#3:
If p is not a divisor of m, then p and m share no common factors, i.e. gcd(p, m) = 1. Therefore gcd(p^k, m) = 1.
If p divides m, clearly gcd(m, p^k) > 1.
#4: We want the number of integers between 1 and p^n that have no common factors with p^n. From #3, we know that unless x is a multiple of p, gcd(x, p^n) = 1. So we calculate the number of multiples of p between 1 and p^n, which is just p^(n-1). And the total number of integers between 1 and p^n is p^n, so the answer is A(p^n) = p^n - p^(n-1).
rocketslay#2:
If n is prime, then for all numbers 1, 2, 3, ..., n-1, gcd(x,n) = 1, which follows from the definition of a prime number, so A(n) = n-1.
If n isn't prime, then for some number in 2, 3, ..., n-1, that number divides n, so A(n) < n-1.
#3:
If p is not a divisor of m, then p and m share no common factors, i.e. gcd(p, m) = 1. Therefore gcd(p^k, m) = 1.
If p divides m, clearly gcd(m, p^k) > 1.
#4: We want the number of integers between 1 and p^n that have no common factors with p^n. From #3, we know that unless x is a multiple of p, gcd(x, p^n) = 1. So we calculate the number of multiples of p between 1 and p^n, which is just p^(n-1). And the total number of integers between 1 and p^n is p^n, so the answer is A(p^n) = p^n - p^(n-1).
thanks a ton man :) i can give you some hats if you want to (sry its the only thing i have)
[quote=rocketslay]#2:
If n is prime, then for all numbers 1, 2, 3, ..., n-1, gcd(x,n) = 1, which follows from the definition of a prime number, so A(n) = n-1.
If n isn't prime, then for some number in 2, 3, ..., n-1, that number divides n, so A(n) < n-1.
#3:
If p is not a divisor of m, then p and m share no common factors, i.e. gcd(p, m) = 1. Therefore gcd(p^k, m) = 1.
If p divides m, clearly gcd(m, p^k) > 1.
#4: We want the number of integers between 1 and p^n that have no common factors with p^n. From #3, we know that unless x is a multiple of p, gcd(x, p^n) = 1. So we calculate the number of multiples of p between 1 and p^n, which is just p^(n-1). And the total number of integers between 1 and p^n is p^n, so the answer is A(p^n) = p^n - p^(n-1).[/quote]
thanks a ton man :) i can give you some hats if you want to (sry its the only thing i have)
If n is prime, then for all numbers 1, 2, 3, ..., n-1, gcd(x,n) = 1, which follows from the definition of a prime number, so A(n) = n-1.
If n isn't prime, then for some number in 2, 3, ..., n-1, that number divides n, so A(n) < n-1.
#3:
If p is not a divisor of m, then p and m share no common factors, i.e. gcd(p, m) = 1. Therefore gcd(p^k, m) = 1.
If p divides m, clearly gcd(m, p^k) > 1.
#4: We want the number of integers between 1 and p^n that have no common factors with p^n. From #3, we know that unless x is a multiple of p, gcd(x, p^n) = 1. So we calculate the number of multiples of p between 1 and p^n, which is just p^(n-1). And the total number of integers between 1 and p^n is p^n, so the answer is A(p^n) = p^n - p^(n-1).[/quote]
thanks a ton man :) i can give you some hats if you want to (sry its the only thing i have)
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