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Euler's totient $phi$ function as a product of primes

Infinite Prime Proof Using Euler's TotientEuler's totient function and complex numbersEuler's totient function for large numbersCalculating Euler's totient function values.Connection between Euler's totient function and Fibonacci numbersHow fast does Euler's totient function drop?Euler's Phi function works for non dividing primes as well, why?Let $phi$ be Euler's totient function, find all $n$ such that $phi(n) = frac13 n$.Euler's totient function applied to higher power triplesExpression for the inverse of Euler's totient function $phi^-1$

2

1

$begingroup$

Many web pages say that Euler's totient function $phi(n)$ can be given as

$$phi(n)=n prod_p biggl(1- frac1p biggr)$$

But $phi(1)=1$, and no primes divide $1$. Surely this gives

$$phi(1)=prod_p biggl(1- frac1p biggr)=0n=0$$

Is $phi(1)$ a special case, or am I missing something?

prime-numbers totient-function

share|cite|improve this question

asked yesterday

Richard Burke-WardRichard Burke-Ward

35318

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Empty product is considered $1$
  $endgroup$
  – J. W. Tanner
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That'll be the thing I was missing, then! Should have checked the definition for product series... Thanks.
  $endgroup$
  – Richard Burke-Ward
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It makes a bit of sense, really. If you have a sum of numbers, and you subtract them all away, one by one, you're down to zero when the last one is gone. If you do the same with a product, dividing the product until everything is divided out, then you've finally landed at 1.
  $endgroup$
  – G Tony Jacobs
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Actually, there is no harm to assume $n>1$ in the formula and consider $phi(1)=1$ separately.
  $endgroup$
  – Dietrich Burde
  yesterday
  
  

  
  
  
  

add a comment | 

2

1

$begingroup$

Many web pages say that Euler's totient function $phi(n)$ can be given as

$$phi(n)=n prod_p biggl(1- frac1p biggr)$$

But $phi(1)=1$, and no primes divide $1$. Surely this gives

$$phi(1)=prod_p biggl(1- frac1p biggr)=0n=0$$

Is $phi(1)$ a special case, or am I missing something?

prime-numbers totient-function

share|cite|improve this question

asked yesterday

Richard Burke-WardRichard Burke-Ward

35318

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Empty product is considered $1$
  $endgroup$
  – J. W. Tanner
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That'll be the thing I was missing, then! Should have checked the definition for product series... Thanks.
  $endgroup$
  – Richard Burke-Ward
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It makes a bit of sense, really. If you have a sum of numbers, and you subtract them all away, one by one, you're down to zero when the last one is gone. If you do the same with a product, dividing the product until everything is divided out, then you've finally landed at 1.
  $endgroup$
  – G Tony Jacobs
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Actually, there is no harm to assume $n>1$ in the formula and consider $phi(1)=1$ separately.
  $endgroup$
  – Dietrich Burde
  yesterday
  
  

  
  
  
  

add a comment | 

$begingroup$

Many web pages say that Euler's totient function $phi(n)$ can be given as

$$phi(n)=n prod_p biggl(1- frac1p biggr)$$

But $phi(1)=1$, and no primes divide $1$. Surely this gives

$$phi(1)=prod_p biggl(1- frac1p biggr)=0n=0$$

Is $phi(1)$ a special case, or am I missing something?

prime-numbers totient-function

share|cite|improve this question

asked yesterday

Richard Burke-WardRichard Burke-Ward

35318

$endgroup$

share|cite|improve this question

asked yesterday

Richard Burke-WardRichard Burke-Ward

35318

share|cite|improve this question

asked yesterday

Richard Burke-WardRichard Burke-Ward

35318

asked yesterday

Richard Burke-WardRichard Burke-Ward

35318

asked yesterday

Richard Burke-WardRichard Burke-Ward

35318

2 Answers
2

active

oldest

votes

3

$begingroup$

There's no problem here.

$phi(1)=1$ even according to the product definition;

there are no primes dividing $1$, so it's an empty product, which is $1$.

share|cite|improve this answer

answered yesterday

J. W. TannerJ. W. Tanner

2,9441217

$endgroup$

add a comment | 

1

$begingroup$

What is the sum of zero numbers? 0, of course. Most people have no trouble conceptually with this. Obviously $x + 0 = x$. So 0 is the identity element of addition.

Now, what is the product of zero numbers? This one I need to go through the logic to convince myself of the right answer, I think most people do. The identity element is not 0 because $x times 0 = 0$, not $x$. But if we try 1 instead, then bingo: $x times 1 = x$.

Therefore, since no primes divide 1, and if $phi(n)$ really is a product of numbers of the form $$fracp - 1p$$ with $p$ prime, it follows that $phi(1) = 1$.

---

There are good reasons why 1 is not a prime number, and there are also stupid reasons. With Euler's totient function, we get a hint of one of the good reasons why 1 is not a prime number: because in many fundamental ways, it behaves fundamentally differently from the prime numbers. Clearly $$1 - frac11 = 1 - 1 = 0,$$ so then $phi(n)$ would be 0 for all $n$. Then 1 being prime would require a special case for Euler's totient function to be of any use.

Lastly, I wanted to mention $phi(0) = 0$. Since 0 is divisible by all primes, even limiting to the positive primes, we'd have an infinite product on our hands. But then the whole thing is multiplied by 0, so there is no need to even begin computing an infinite product...

share|cite|improve this answer

answered 9 hours ago

Robert SoupeRobert Soupe

11.4k21950

$endgroup$

add a comment | 

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3

$begingroup$

There's no problem here.

$phi(1)=1$ even according to the product definition;

there are no primes dividing $1$, so it's an empty product, which is $1$.

share|cite|improve this answer

answered yesterday

J. W. TannerJ. W. Tanner

2,9441217

$endgroup$

add a comment | 

3

$begingroup$

There's no problem here.

$phi(1)=1$ even according to the product definition;

there are no primes dividing $1$, so it's an empty product, which is $1$.

share|cite|improve this answer

answered yesterday

J. W. TannerJ. W. Tanner

2,9441217

$endgroup$

add a comment | 

$begingroup$

There's no problem here.

$phi(1)=1$ even according to the product definition;

there are no primes dividing $1$, so it's an empty product, which is $1$.

share|cite|improve this answer

answered yesterday

J. W. TannerJ. W. Tanner

2,9441217

$endgroup$

share|cite|improve this answer

answered yesterday

J. W. TannerJ. W. Tanner

2,9441217

answered yesterday

J. W. TannerJ. W. Tanner

2,9441217

answered yesterday

J. W. TannerJ. W. Tanner

2,9441217

1

$begingroup$

What is the sum of zero numbers? 0, of course. Most people have no trouble conceptually with this. Obviously $x + 0 = x$. So 0 is the identity element of addition.

Now, what is the product of zero numbers? This one I need to go through the logic to convince myself of the right answer, I think most people do. The identity element is not 0 because $x times 0 = 0$, not $x$. But if we try 1 instead, then bingo: $x times 1 = x$.

Therefore, since no primes divide 1, and if $phi(n)$ really is a product of numbers of the form $$fracp - 1p$$ with $p$ prime, it follows that $phi(1) = 1$.

---

There are good reasons why 1 is not a prime number, and there are also stupid reasons. With Euler's totient function, we get a hint of one of the good reasons why 1 is not a prime number: because in many fundamental ways, it behaves fundamentally differently from the prime numbers. Clearly $$1 - frac11 = 1 - 1 = 0,$$ so then $phi(n)$ would be 0 for all $n$. Then 1 being prime would require a special case for Euler's totient function to be of any use.

Lastly, I wanted to mention $phi(0) = 0$. Since 0 is divisible by all primes, even limiting to the positive primes, we'd have an infinite product on our hands. But then the whole thing is multiplied by 0, so there is no need to even begin computing an infinite product...

share|cite|improve this answer

answered 9 hours ago

Robert SoupeRobert Soupe

11.4k21950

$endgroup$

add a comment | 

1

$begingroup$

What is the sum of zero numbers? 0, of course. Most people have no trouble conceptually with this. Obviously $x + 0 = x$. So 0 is the identity element of addition.

Now, what is the product of zero numbers? This one I need to go through the logic to convince myself of the right answer, I think most people do. The identity element is not 0 because $x times 0 = 0$, not $x$. But if we try 1 instead, then bingo: $x times 1 = x$.

Therefore, since no primes divide 1, and if $phi(n)$ really is a product of numbers of the form $$fracp - 1p$$ with $p$ prime, it follows that $phi(1) = 1$.

---

There are good reasons why 1 is not a prime number, and there are also stupid reasons. With Euler's totient function, we get a hint of one of the good reasons why 1 is not a prime number: because in many fundamental ways, it behaves fundamentally differently from the prime numbers. Clearly $$1 - frac11 = 1 - 1 = 0,$$ so then $phi(n)$ would be 0 for all $n$. Then 1 being prime would require a special case for Euler's totient function to be of any use.

Lastly, I wanted to mention $phi(0) = 0$. Since 0 is divisible by all primes, even limiting to the positive primes, we'd have an infinite product on our hands. But then the whole thing is multiplied by 0, so there is no need to even begin computing an infinite product...

share|cite|improve this answer

answered 9 hours ago

Robert SoupeRobert Soupe

11.4k21950

$endgroup$

add a comment | 

$begingroup$

What is the sum of zero numbers? 0, of course. Most people have no trouble conceptually with this. Obviously $x + 0 = x$. So 0 is the identity element of addition.

Now, what is the product of zero numbers? This one I need to go through the logic to convince myself of the right answer, I think most people do. The identity element is not 0 because $x times 0 = 0$, not $x$. But if we try 1 instead, then bingo: $x times 1 = x$.

Therefore, since no primes divide 1, and if $phi(n)$ really is a product of numbers of the form $$fracp - 1p$$ with $p$ prime, it follows that $phi(1) = 1$.

---

There are good reasons why 1 is not a prime number, and there are also stupid reasons. With Euler's totient function, we get a hint of one of the good reasons why 1 is not a prime number: because in many fundamental ways, it behaves fundamentally differently from the prime numbers. Clearly $$1 - frac11 = 1 - 1 = 0,$$ so then $phi(n)$ would be 0 for all $n$. Then 1 being prime would require a special case for Euler's totient function to be of any use.

Lastly, I wanted to mention $phi(0) = 0$. Since 0 is divisible by all primes, even limiting to the positive primes, we'd have an infinite product on our hands. But then the whole thing is multiplied by 0, so there is no need to even begin computing an infinite product...

share|cite|improve this answer

answered 9 hours ago

Robert SoupeRobert Soupe

11.4k21950

$endgroup$

share|cite|improve this answer

answered 9 hours ago

Robert SoupeRobert Soupe

11.4k21950

answered 9 hours ago

Robert SoupeRobert Soupe

11.4k21950

answered 9 hours ago

Robert SoupeRobert Soupe

11.4k21950