Derivation of Euler's formula for phi(m)

In summary, Euler's formula for phi(m) is a mathematical expression that calculates the totient function of a positive integer m. It states that phi(m) is equal to m multiplied by the product of 1 minus the reciprocals of all primes that divide m. This formula was derived by Swiss mathematician Leonhard Euler and has many applications in number theory, particularly in the study of prime numbers and their properties. It provides a simple and elegant way to calculate the totient function, which is used in various mathematical problems and algorithms.
  • #1
Eus
94
0
I read on http://www.cut-the-knot.org/blue/Euler.shtml that the derivation of the Euler's formula for [itex]\varphi(m)[/itex] requires that the following multiplicative property of [itex]\varphi[/itex] be established:

[tex]
\varphi(m_{1}m_{2})=\varphi(m_{1})\varphi(m_{2})\mbox{ for coprime } m_{1} \mbox{ and } m_{2}
[/tex]

The article proves that the multiplicative property holds in the following way:
Let [itex]0 \leq n < m[/itex] be coprime to [itex]m[/itex].
Find remainders [itex]n_{1}[/itex] and [itex]n_{2}[/itex] of division of [itex]n[/itex] by [itex]m_{1}[/itex] and [itex]m_{2}[/itex], respectively:
[tex]
n\equiv n_{1} \mbox{ (mod }m_{1}\mbox{) } \mbox{and } n\equiv n_{2} \mbox{ (mod } m_{2} \mbox{)}
[/tex]
Obviously, [itex]n_{1}[/itex] is coprime to [itex]m_{1}[/itex] and [itex]n_{2}[/itex] is coprime to [itex]m_{2}[/itex].

Although it says "obviously", I don't find that the relationship is obvious enough. That is:
[tex]
n \mbox{ is coprime to } m\mbox{ and }m=m_{1}m_{2},\ m_{1} \mbox{ is coprime to } m_{2} }\mbox{ and }n\equiv n_{1} \mbox{ (mod }m_{1}\mbox{) } \rightarrow n_{1} \mbox{ is coprime to } m_{1}
[/tex]
Is there any theorem that will guarantee that relationship?

Besides that, why is the value of [itex]\varphi(m_{1}m_{2})[/itex] found by multiplying [itex]\varphi(m_{1})[/itex] and [itex]\varphi(m_{2})[/itex] instead of by adding [itex]\varphi(m_{1})[/itex] and [itex]\varphi(m_{2})[/itex]?

Thank you.
 
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  • #2
Eus said:
I read on http://www.cut-the-knot.org/blue/Euler.shtml that the derivation of the Euler's formula for [itex]\varphi(m)[/itex] requires that the following multiplicative property of [itex]\varphi[/itex] be established:

[tex]
\varphi(m_{1}m_{2})=\varphi(m_{1})\varphi(m_{2})\mbox{ for coprime } m_{1} \mbox{ and } m_{2}
[/tex]

The article proves that the multiplicative property holds in the following way:
Let [itex]0 \leq n < m[/itex] be coprime to [itex]m[/itex].
Find remainders [itex]n_{1}[/itex] and [itex]n_{2}[/itex] of division of [itex]n[/itex] by [itex]m_{1}[/itex] and [itex]m_{2}[/itex], respectively:
[tex]
n\equiv n_{1} \mbox{ (mod }m_{1}\mbox{) } \mbox{and } n\equiv n_{2} \mbox{ (mod } m_{2} \mbox{)}
[/tex]
Obviously, [itex]n_{1}[/itex] is coprime to [itex]m_{1}[/itex] and [itex]n_{2}[/itex] is coprime to [itex]m_{2}[/itex].

Although it says "obviously", I don't find that the relationship is obvious enough.

the remainder [itex]n_{1}[/itex] is coprime to [itex]m_{1}[/itex] and the remainder [itex]n_{2}[/itex] is coprime to [itex]m_{2}[/itex] because otherwise the division of [itex]n_{1}[/itex] by [itex]m_{1}[/itex] and [itex]n_{2}[/itex] by [itex]m_{2}[/itex] should give another remainders, and in these cases we cannot call [itex]n_{1}[/itex] and [itex]n_{2}[/itex] remainders. For instance if a = qb + r, r being the remainder, then by definition r < b (otherwise the division of r by b should let another remainder, and then r shouldn't be a the remainder). So, if a = qb + r, r being the remainder, b and r are always coprimes.
 
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  • #3
So, if a = qb + r, r being the remainder, b and r are always coprimes.

Ah, thank you for the explanation.

But, why is the value of [itex]\varphi(m_{1}m_{2})[/itex] found by multiplying [itex]\varphi(m_{1})[/itex] and [itex]\varphi(m_{2})[/itex] instead of by adding [itex]\varphi(m_{1})[/itex] and [itex]\varphi(m_{2})[/itex]?
 
  • #4
this is a little bit long to prove

for a p prime, [itex]\varphi(p^a) = p^{a-1}(p-1)[/itex]

proof:

for every prime number, the amount of number from 1 to p such that all of them are coprimes with p is = p-1. Ex: p = 7 ==> 6,5,4,3,2,1 are coprime with 7

there are [itex]p^{a-1}[/itex] integers between 1 and [itex]p^a[/itex] which are divisible by p: p, 2p, 3p, 4p, 5p, ..., [itex]p^{a-1}p[/itex]

thus the set {1,2,3,4,...,[itex]p^a[/itex]} contain exactly [itex]p^a - p^{a-1} = p^{a-1}(p-1)[/itex] integers which are relatively prime to [itex]p^a[/itex]

from here one can use this result to prove that relation

to prove the relation [tex]\varphi(m_{1}m_{2})=\varphi(m_{1})\varphi(m_{2})\ box{ for coprime } m_{1} \mbox{ and } m_{2}[/tex] I think you can search the web and find probably the same proof that I have here from a book
 
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  • #5
Ah, I see.

Thank you very much for your help.
 

FAQ: Derivation of Euler's formula for phi(m)

What is Euler's formula for phi(m)?

Euler's formula for phi(m) is a mathematical relationship that connects the values of Euler's totient function (phi) to the prime factorization of a positive integer m. It states that phi(m) = m * (1-1/p1) * (1-1/p2) * ... * (1-1/pn), where p1, p2, ..., pn are the distinct prime factors of m.

How is Euler's formula for phi(m) derived?

The derivation of Euler's formula for phi(m) involves using the fundamental theorem of arithmetic, which states that every positive integer can be uniquely expressed as a product of primes. By applying this theorem and manipulating the expression for phi(m), the formula can be derived.

What is the significance of Euler's formula for phi(m)?

Euler's formula for phi(m) is significant because it provides a way to calculate the totient function for any positive integer, which has important applications in number theory and cryptography. It also helps in solving problems related to modular arithmetic.

Can Euler's formula for phi(m) be extended to other number systems?

Yes, Euler's formula for phi(m) can be extended to other number systems such as complex numbers. In this case, the formula becomes phi(z) = z * (1-1/p1) * (1-1/p2) * ... * (1-1/pn), where z is a complex number and p1, p2, ..., pn are the distinct prime factors of z.

Are there any other important mathematical concepts related to Euler's formula for phi(m)?

Yes, there are several other important mathematical concepts related to Euler's formula for phi(m). These include Euler's theorem, which states that a^phi(n) ≡ 1 (mod n) for coprime integers a and n, and the Euler product formula, which expresses the Riemann zeta function as an infinite product involving primes. Additionally, Euler's formula for phi(m) has connections to the Chinese remainder theorem and the notion of primitive roots.

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