Why is empty product defined to be $1$?

Question

For example $\prod_{2 \le j < 1} 2^j= 1.$ How does that happen?

Answer 1

Note that if $\displaystyle\prod_{i=1}^n a_i=a$ then $\displaystyle\prod_{i=1}^{n-1}a_i = \frac a{a_n}$. Now note that $\displaystyle\prod_{i=1}^1 a=a$, so $\displaystyle\prod_{i=1}^0a=1$.

Answer 2

It's a definition that respects breaking down the product into pieces, just like saying $1=1+0$. For example, you have that $\prod_{j=1}^n a_j = (\prod_{j=1}^n a_j)(\prod_{j\in\{\phi\}}a_j)=\prod_{j=1}^n a_j$. Similarly, the empty sum $\sum_{2<j<1}a_j=0$ so that linaerity is preserved: $\sum_{j=1}^n a_j=\sum_{j=1}^na_j+\sum_{j\in\{\phi\}}a_j$. This is particularly useful since products and sums are indexed $by$ some variable $j$ taking values from an index set $A$, and the empty set is always an element of $A$. In particular, you can break down a product into $A=B\cup C$, so that $\prod_{j\in A}a_j=\prod_{j\in B}a_j\prod_{j\in c}a_j$. Thus the only answer is $\prod_{j\in\{\phi\}}a_j=1$

Answer 3

The empty sum is zero.

The empty product is one.

Note that we have $0+x = x$ for all $x$ and $1\cdot x=x$ for all $x$, so that these are the corresponding values that doesn't change the final sum/product.

Answer 4

Every product should start with $1$. For instance $$\prod_{i=1}^2a_i:=1\cdot a_1\cdot a_2$$ This mode of thinking at least makes the empty product naturally work out to $1$. But it's more than that. If you think of the above in the usual way as $a_1\cdot a_2$,then it's not really symmetric. It reads that you started with the object $a_1$, and then brought in the operation "$\cdot a_2$". If you start with $1$, then you are bringing in both factors as operations: "$\cdot a_1$" and "$\cdot a_2$". This is arguably more symmetric. (Similarly, define $\sum_{i=1}^na_i:=0\;\overbrace{{}+a_1}\;\overbrace{{}+a_2}\;+\cdots$.)

Answer 5

An empty sum is defined to be $0$; an empty product is defined to be $1$. It's the same thing.

For more concrete motivation, suppose $S$ is a finite set of integers, and $n$ is an integer not in $S$. Then we want the following to hold, in general:

$$\prod_{i\in S \cup \{n\}} i = n \cdot \prod_{i\in S}i$$

But if $S$ is the empty set, then this equation only holds if we define

$$\prod_{i\in\emptyset} i = 1$$

Answer 6

Alex R. and Asaf Karagila pointed out that it fits well with the usual identities. This might be a more structural refinement of the argument:

For any set of factors $A$ you have the identity $A ∪ ∅ = A$.

You would like to have $\prod_{a ∈ A} f_a · \prod_{b ∈ B} f_b = \prod_{x ∈ A ∪ B} f_x$ (for disjoint sets $A$ and $B$). You need this all the time for usual reasoning with splitting up products and such. Defining the empty product to yield the neutral element with respect to multiplication saves this.

Answer 7

When we add all elements of two sets, the result is the same as adding the sum of the values in the individual sets, if we add no elements to the mix we don't change the value and as such the empty sum shouldn't change the value.

With the empty product, we have two sets and the result of multiplying all elements of both sets together is the same if you took the product of each set and then multiply those products together.

For the sum, we need to find a number that doesn't change the value, so we chose $0$.
It's the exact same for the empty product, we simply need to find a value that doesn't change what it's multiplied with, so we chose $1$.

Answer 8

It's a convenient convention. Often times you could be manipulating an abstract product expression, which in some specific instances would have no terms in it (like your example). In almost every such case it turns out that the correct answer is obtained by treating such a product as 1. So rather than constantly need to handle this case separately it is more sensible to define it as 1.

Another, non-rigorous way of thinking about it is that an empty product is a multiplication by something which does nothing: it's not halving things, or tripling them, etc., it is doing nothing by virtue of being nothing. And that's the same as multiplying by 1.

Similarly, an empty sum is defined, by convention, to be 0: there's nothing to sum, and the only thing you can add to anything else without changing it is zero.

Answer 9

The logarithm of the empty product is the empty sum, which is zero. Assuming the product is real, it has to be one.

In your example: $$\prod_{j\in\emptyset}2^j=2^{\sum_{j\in\emptyset}j}=2^0=1.$$

If this doesn't convince you, then you should be asking, why is the empty sum defined to be zero?