Empty sum

The empty sum is defined to be equal to the additive identity, 0, as numbers are concerned. This is as straightforward a concept as the total cost of buying 0 items from a shop being 0.

Applications of the concept of empty sum

With the concept of empty sum we don't need to make many other conventions, e.g.

Partitions of 0

The set of partitions of 0 is a set containing the empty set (the sum of elements of the empty set being the empty sum, defined as the additive identity, i.e. 0)

$P(0)=\{\{\}\},\,p(0)=|P(0)|=1.\,$

Without the concept of empty sum, we would have to make the convention that $\scriptstyle p(0)\,$ is set to 1.

Prime factorization of 1

For the unit, 1, there are no primes with nonzero exponents (the set of prime factors of 1 is the empty set) and we get the empty product, defined as the multiplicative identity, i.e. 1. (Cf. Empty product#Prime factorization of 1.)

Since the set $\scriptstyle {\rm {DPF}}(1)\,$ of distinct prime factors of 1 is the empty set

$\scriptstyle {\rm {DPF}}(1)\,=\,\{\}\,$

and the number of distinct prime factors $\scriptstyle \omega (n)\,$ is the cardinality of the set of distinct prime factors of $\scriptstyle n\,$

$\scriptstyle \omega (n)\,=\,|{\rm {DPF}}(n)|\,$

we get the cardinality of the empty set, i.e. 0, for $\scriptstyle n\,$ = 1.

Number of distinct prime factors

The number of distinct prime factors of $\scriptstyle n\,$ is (tautologically) given by

$\omega (n)=\sum _{i=1}^{\omega (n)}{\alpha _{i}}^{0},\,$

where we get the empty sum, defined as the additive identity, i.e. 0 (the final value of the index being lower than the initial value) for $\scriptstyle n\,$ = 1.

Number of prime factors (with repetition)

The number of prime factors (with repetition) of $\scriptstyle n\,$ is given by

$\Omega (n)=\sum _{i=1}^{\omega (n)}{\alpha _{i}}^{1},\,$

where we get the empty sum, defined as the additive identity, i.e. 0 (the final value of the index being lower than the initial value) for $\scriptstyle n\,$ = 1.

See also

• Empty product