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Möbius Inversion Formula

Description: This quiz will test your understanding of the Möbius Inversion Formula, a fundamental result in number theory that relates multiplicative and additive functions.
Number of Questions: 14
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Tags: möbius inversion formula number theory multiplicative functions additive functions
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What is the Möbius function?

  1. A function that takes a positive integer as input and returns 1 if the integer is square-free, -1 if the integer has an even number of prime factors, and 0 otherwise.

  2. A function that takes a positive integer as input and returns the number of prime factors of the integer.

  3. A function that takes a positive integer as input and returns the sum of the digits of the integer.

  4. A function that takes a positive integer as input and returns the greatest common divisor of the integer and 10.


Correct Option: A
Explanation:

The Möbius function is defined as follows: ( \mu(n) = \begin{cases} 1 & \text{if } n \text{ is square-free} \ -1 & \text{if } n \text{ has an even number of prime factors} \ 0 & \text{otherwise} \end{cases} )

What is the Möbius Inversion Formula?

  1. A formula that relates the sum of a multiplicative function over a set of integers to the sum of the Möbius function over the same set of integers.

  2. A formula that relates the sum of an additive function over a set of integers to the sum of the Möbius function over the same set of integers.

  3. A formula that relates the sum of a multiplicative function over a set of integers to the sum of the same function over a different set of integers.

  4. A formula that relates the sum of an additive function over a set of integers to the sum of the same function over a different set of integers.


Correct Option: A
Explanation:

The Möbius Inversion Formula states that if ( f ) is a multiplicative function and ( g ) is defined by ( g(n) = \sum_{d|n} f(d) ), then ( f(n) = \sum_{d|n} \mu(d) g(\frac{n}{d}) ).

What is an example of a multiplicative function?

  1. The greatest common divisor function.

  2. The Euler phi function.

  3. The sum-of-divisors function.

  4. The Möbius function.


Correct Option: B
Explanation:

The Euler phi function is a multiplicative function, meaning that ( \phi(mn) = \phi(m) \phi(n) ) whenever ( m ) and ( n ) are relatively prime.

What is an example of an additive function?

  1. The greatest common divisor function.

  2. The Euler phi function.

  3. The sum-of-divisors function.

  4. The Möbius function.


Correct Option: C
Explanation:

The sum-of-divisors function is an additive function, meaning that ( \sigma(mn) = \sigma(m) + \sigma(n) ) whenever ( m ) and ( n ) are relatively prime.

Use the Möbius Inversion Formula to find a formula for the sum of the Möbius function over the divisors of an integer ( n ).

  1. ( \sum_{d|n} \mu(d) = 1 )

  2. ( \sum_{d|n} \mu(d) = n )

  3. ( \sum_{d|n} \mu(d) = \phi(n) )

  4. ( \sum_{d|n} \mu(d) = \sigma(n) )


Correct Option: A
Explanation:

Using the Möbius Inversion Formula with ( f(n) = 1 ) and ( g(n) = \sum_{d|n} \mu(d) ), we get ( 1 = \sum_{d|n} \mu(d) g(\frac{n}{d}) ). Since ( g(\frac{n}{d}) = 1 ) for all ( d | n ), we have ( \sum_{d|n} \mu(d) = 1 ).

Use the Möbius Inversion Formula to find a formula for the sum of the divisors of an integer ( n ).

  1. ( \sum_{d|n} d = n )

  2. ( \sum_{d|n} d = \phi(n) )

  3. ( \sum_{d|n} d = \sigma(n) )

  4. ( \sum_{d|n} d = \mu(n) )


Correct Option: C
Explanation:

Using the Möbius Inversion Formula with ( f(n) = d ) and ( g(n) = \sum_{d|n} \mu(d) d ), we get ( d = \sum_{d|n} \mu(d) g(\frac{n}{d}) ). Since ( g(\frac{n}{d}) = \sigma(\frac{n}{d}) ) for all ( d | n ), we have ( \sum_{d|n} d = \sigma(n) ).

Use the Möbius Inversion Formula to find a formula for the sum of the Euler phi function over the divisors of an integer ( n ).

  1. ( \sum_{d|n} \phi(d) = n )

  2. ( \sum_{d|n} \phi(d) = \phi(n) )

  3. ( \sum_{d|n} \phi(d) = \sigma(n) )

  4. ( \sum_{d|n} \phi(d) = \mu(n) )


Correct Option: B
Explanation:

Using the Möbius Inversion Formula with ( f(n) = \phi(n) ) and ( g(n) = \sum_{d|n} \mu(d) \phi(d) ), we get ( \phi(n) = \sum_{d|n} \mu(d) g(\frac{n}{d}) ). Since ( g(\frac{n}{d}) = \phi(\frac{n}{d}) ) for all ( d | n ), we have ( \sum_{d|n} \phi(d) = \phi(n) ).

Use the Möbius Inversion Formula to find a formula for the sum of the Möbius function over the divisors of a square-free integer ( n ).

  1. ( \sum_{d|n} \mu(d) = 1 )

  2. ( \sum_{d|n} \mu(d) = n )

  3. ( \sum_{d|n} \mu(d) = \phi(n) )

  4. ( \sum_{d|n} \mu(d) = \sigma(n) )


Correct Option: A
Explanation:

Since ( n ) is square-free, all of its divisors are also square-free. Therefore, ( \mu(d) = 1 ) for all ( d | n ). Using the Möbius Inversion Formula with ( f(n) = 1 ) and ( g(n) = \sum_{d|n} \mu(d) ), we get ( 1 = \sum_{d|n} \mu(d) g(\frac{n}{d}) ). Since ( g(\frac{n}{d}) = 1 ) for all ( d | n ), we have ( \sum_{d|n} \mu(d) = 1 ).

Use the Möbius Inversion Formula to find a formula for the sum of the Möbius function over the divisors of an integer ( n ) that has exactly ( k ) prime factors.

  1. ( \sum_{d|n} \mu(d) = 1 )

  2. ( \sum_{d|n} \mu(d) = n )

  3. ( \sum_{d|n} \mu(d) = \phi(n) )

  4. ( \sum_{d|n} \mu(d) = \sigma(n) )


Correct Option:
Explanation:

If ( n ) has exactly ( k ) prime factors, then all of its divisors have between 0 and ( k ) prime factors. Therefore, ( \mu(d) = (-1)^k ) for all ( d | n ). Using the Möbius Inversion Formula with ( f(n) = 1 ) and ( g(n) = \sum_{d|n} \mu(d) ), we get ( 1 = \sum_{d|n} \mu(d) g(\frac{n}{d}) ). Since ( g(\frac{n}{d}) = (-1)^k ) for all ( d | n ), we have ( \sum_{d|n} \mu(d) = \begin{cases} 1 & \text{if } k \text{ is even} \ -1 & \text{if } k \text{ is odd} \end{cases} ).

Use the Möbius Inversion Formula to find a formula for the sum of the Möbius function over the divisors of an integer ( n ) that is divisible by ( m ).

  1. ( \sum_{d|n} \mu(d) = 1 )

  2. ( \sum_{d|n} \mu(d) = n )

  3. ( \sum_{d|n} \mu(d) = \phi(n) )

  4. ( \sum_{d|n} \mu(d) = \sigma(n) )


Correct Option:
Explanation:

If ( m ) is divisible by ( n ), then all of the divisors of ( n ) that are divisible by ( m ) are also divisible by ( \frac{n}{m} ). Therefore, ( \mu(d) = 0 ) for all ( d | n ) that are divisible by ( m ). Using the Möbius Inversion Formula with ( f(n) = 1 ) and ( g(n) = \sum_{d|n} \mu(d) ), we get ( 1 = \sum_{d|n} \mu(d) g(\frac{n}{d}) ). Since ( g(\frac{n}{d}) = 0 ) for all ( d | n ) that are divisible by ( m ), we have ( \sum_{d|n} \mu(d) = \begin{cases} 1 & \text{if } m = 1 \ 0 & \text{if } m > 1 \end{cases} ).

Use the Möbius Inversion Formula to find a formula for the sum of the Möbius function over the divisors of an integer ( n ) that is not divisible by ( m ).

  1. ( \sum_{d|n} \mu(d) = 1 )

  2. ( \sum_{d|n} \mu(d) = n )

  3. ( \sum_{d|n} \mu(d) = \phi(n) )

  4. ( \sum_{d|n} \mu(d) = \sigma(n) )


Correct Option:
Explanation:

If ( m ) is not divisible by ( n ), then there is at least one prime factor of ( n ) that is not a prime factor of ( m ). Therefore, there is at least one divisor of ( n ) that is not divisible by ( m ). Since ( \mu(d) = 0 ) for all ( d | n ) that are divisible by ( m ), we have ( \sum_{d|n} \mu(d) = -1 ). Using the Möbius Inversion Formula with ( f(n) = 1 ) and ( g(n) = \sum_{d|n} \mu(d) ), we get ( 1 = \sum_{d|n} \mu(d) g(\frac{n}{d}) ). Since ( g(\frac{n}{d}) = -1 ) for all ( d | n ) that are not divisible by ( m ), we have ( \sum_{d|n} \mu(d) = \begin{cases} 0 & \text{if } m = 1 \ -1 & \text{if } m > 1 \end{cases} ).

Use the Möbius Inversion Formula to find a formula for the sum of the Möbius function over the divisors of an integer ( n ) that are relatively prime to ( n ).

  1. ( \sum_{d|n} \mu(d) = 1 )

  2. ( \sum_{d|n} \mu(d) = n )

  3. ( \sum_{d|n} \mu(d) = \phi(n) )

  4. ( \sum_{d|n} \mu(d) = \sigma(n) )


Correct Option: C
Explanation:

If ( d ) is a divisor of ( n ) that is relatively prime to ( n ), then ( \frac{n}{d} ) is also a divisor of ( n ) that is relatively prime to ( n ). Therefore, ( \mu(d) = \mu(\frac{n}{d}) ). Using the Möbius Inversion Formula with ( f(n) = 1 ) and ( g(n) = \sum_{d|n} \mu(d) ), we get ( 1 = \sum_{d|n} \mu(d) g(\frac{n}{d}) ). Since ( g(\frac{n}{d}) = \phi(\frac{n}{d}) ) for all ( d | n ) that are relatively prime to ( n ), we have ( \sum_{d|n} \mu(d) = \phi(n) ).

Use the Möbius Inversion Formula to find a formula for the sum of the Möbius function over the divisors of an integer ( n ) that are not relatively prime to ( n ).

  1. ( \sum_{d|n} \mu(d) = 1 )

  2. ( \sum_{d|n} \mu(d) = n )

  3. ( \sum_{d|n} \mu(d) = \phi(n) )

  4. ( \sum_{d|n} \mu(d) = \sigma(n) )


Correct Option:
Explanation:

If ( d ) is a divisor of ( n ) that is not relatively prime to ( n ), then there is at least one prime factor of ( n ) that is also a prime factor of ( d ). Therefore, ( \mu(d) = 0 ). Using the Möbius Inversion Formula with ( f(n) = 1 ) and ( g(n) = \sum_{d|n} \mu(d) ), we get ( 1 = \sum_{d|n} \mu(d) g(\frac{n}{d}) ). Since ( g(\frac{n}{d}) = 0 ) for all ( d | n ) that are not relatively prime to ( n ), we have ( \sum_{d|n} \mu(d) = 0 ).

Use the Möbius Inversion Formula to find a formula for the sum of the Möbius function over the divisors of an integer ( n ) that are perfect squares.

  1. ( \sum_{d|n} \mu(d) = 1 )

  2. ( \sum_{d|n} \mu(d) = n )

  3. ( \sum_{d|n} \mu(d) = \phi(n) )

  4. ( \sum_{d|n} \mu(d) = \sigma(n) )


Correct Option:
Explanation:

If ( n ) is a perfect square, then all of its prime factors appear with even exponents. Therefore, ( \mu(d) = 1 ) for all ( d | n ) that are perfect squares. Using the Möbius Inversion Formula with ( f(n) = 1 ) and ( g(n) = \sum_{d|n} \mu(d) ), we get ( 1 = \sum_{d|n} \mu(d) g(\frac{n}{d}) ). Since ( g(\frac{n}{d}) = 1 ) for all ( d | n ) that are perfect squares, we have ( \sum_{d|n} \mu(d) = \begin{cases} 1 & \text{if } n \text{ is a perfect square} \ 0 & \text{otherwise} \end{cases} ).

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