Empty product

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In mathematics, the empty product is the special case of a product with zero factors . It is usually assigned the value one .

In combinatorial , counting considerations, the empty product should normally be included, since there is exactly one possibility to multiply nothing , which is why it is also justified to speak of the empty product. It is to be distinguished from the product or a product with only a single factor (which is then equal to this factor). ${\displaystyle 0\cdot 0}$

In other areas such as group , ring or body theory , in which multiplication is viewed as a fundamental, internal connection , any definition with fewer than two factors initially makes no sense. Nevertheless, the empty product appears implicitly in several contexts, e.g. B. with potencies and the faculty and is there occasionally the reason for comprehension problems. Even the common value assignment to one is not always intuitively clear.

Relationship to powers and the empty sum

Similarly, the addition of 0 summands is called the empty sum and gives it the value zero . This can be clearly justified: When nothing is added, nothing is obtained ( nothing = zero is the neutral element of the addition).

For every finite product with factors and the logarithm to an arbitrary basis we now have: ${\displaystyle N}$ ${\displaystyle b\in \mathbb {R} ,b>0}$

${\displaystyle \prod _{i=1}^{N}x_{i}=b^{\sum _{i=1}^{N}\log _{b}x_{i}},}$ there ${\displaystyle \log xy=\log x+\log y.}$

If you set, you get the empty product on the left and the empty sum in the exponent on the right : ${\displaystyle N=0}$

${\displaystyle \prod _{i\in \emptyset }x_{i}=b^{\sum _{i\in \emptyset }\log _{b}x_{i}}=b^{0}=1.}$

Since the value assignment of the empty sum to 0 is very plausible, the empty product according to the needs consistency of the value of obtained at least for all be constant need. ${\displaystyle b^{0}}$ ${\displaystyle b>0}$

Problems of value assignment

It is common practice to define for real things . So that the real value is exponential steadily and analytically in point continued . In the complex numbers it is a bit more complicated, since there is a branch point there , for real numbers it remains correct there too. So nothing speaks against it ${\displaystyle b^{0}=1}$ ${\displaystyle b\neq 0}$ ${\displaystyle 0}$ ${\displaystyle 0}$ ${\displaystyle b}$ ${\displaystyle \prod _{i\in \emptyset }x_{i}=1.}$

A blemish becomes apparent when one tries to generalize this to. The power to set, is still compatible with the most common definitions, but as for all true: this provides for the function with a discontinuity at . See also “ zero to the power of zero ”. ${\displaystyle b=0}$ ${\displaystyle 0^{0}=1}$ ${\displaystyle a>0}$ ${\displaystyle 0^{a}=0}$ ${\displaystyle f\colon \mathbb {R} _{\geq 0}\rightarrow \mathbb {R} }$ ${\displaystyle f(a)=0^{a}}$ ${\displaystyle a=0}$

Empty Cartesian Product

The Cartesian product of two sets is defined as the set of all ordered pairs : . More generally, this can be defined for any index set as follows: ${\displaystyle A\times B}$ ${\displaystyle \left\{(a,b)\mid a\in A,b\in B\right\}}$ ${\displaystyle I}$

${\displaystyle \prod _{i\in I}A_{i}=\{f\colon \,I\to \textstyle \bigcup \limits _{i\in I}A_{i}\mid \forall i\colon \,f(i)\in A_{i}\}.}$

Applies now

${\displaystyle A_{i}=A}$ for all ${\displaystyle i\in I,}$

then the -th power of each set is (also ) given by ${\displaystyle I}$ ${\displaystyle A}$ ${\displaystyle A=\emptyset }$

${\displaystyle A^{I}=\prod _{i\in I}A=\{f\colon \,I\to A\}.}$

This results for the empty Cartesian product :

${\displaystyle \prod _{i\in \emptyset }A_{i}=A^{\emptyset }=\{f\colon \,\emptyset \to A\}=\{\emptyset \},}$

because as a special relation ${\displaystyle f\subseteq \emptyset \times A=\{(b,a)\mid b\in \emptyset ,a\in A\}=\emptyset .}$

Since the numbers can theoretically be defined as and , it follows: ${\displaystyle 0,1}$ ${\displaystyle 0:=\emptyset }$ ${\displaystyle 1:=\{\emptyset \}}$

${\displaystyle A^{0}=1}$and especially too . ${\displaystyle 0^{0}=1}$

Other connections

• Looking at the one that has no prime factors , it is consistent to assign the empty prime factorization to it, i.e. the empty product.
• Just as the empty sum is equal to the neutral element of addition, the empty product is equal to the neutral element of multiplication .
• From the definitions of empty product and faculty it follows: ${\displaystyle \textstyle 0!=\prod _{i=1}^{0}i=\prod _{i\in \emptyset }i=1.}$
• There is exactly one possibility not to select anything from pieces - the same applies to the binomial coefficients , in particular . They can be traced back directly to the factorial of zero. ${\displaystyle n}$ ${\displaystyle {\tbinom {n}{0}}=1}$ ${\displaystyle {\tbinom {0}{0}}=1}$

Web links

Wikibooks: Math for Non-Freaks: Empty Product  - Learning and Teaching Materials