The null product $0*0,\, 0^0$, and $0^x$ [closed]

Question

An empty product, or nullary product, is the result of multiplying no numbers. Its numerical value is 1, the multiplicative identity. Common examples are $0!$ and $x^0$.

So $0 \cdot 0 = 1$ therefore $1/0 = 0$.

But that means that $0^2 = 1$ which implies that $0^x = 1$

Look at it this way. $1/$positive number = positive. $1/$negtive number $= $ negative. $1/0$ can be neither positive nor negative. That only leaves one possibility

Also the function $1/x$ goes to infinity from one direction and to negative infinity from the other. Half way between infinity and negative infinity is zero

Edit: doesnt work because if $0^x = 1$ for all $x$ then $0^{-1}$ cant be $0$

I am probably just wrong but continuing my train of thought:

$1/0 = 0$

therefore $0^2 = 1$ (this is impossible because $0 \cdot (0+0) = 0 \cdot 0 + 0 \cdot 0$)

So we have:

$0^{-4} = \infty$

$0^{-3} = 0$

$0^{-2} = \infty$

$0^{-1} = 0$

$0^0 = 1$

$0^1 = 0$

$0^2 = 1$

Couldnt be periodic could it? I dont see how or why it would be.

Answer 1

An empty product, or nullary product, is the result of multiplying no numbers. Its numerical value is $1$ [. . .] So $0\cdot 0 = 1$.

Firstly, $0\cdot0$ is not an empty product. It is the product of two elements, not the product of zero elements. So, it is a mistake to use this reasoning to arrive at the conclusion $0\cdot0 = 1$. This is the main reason why your attempt at showing $1/0 = 1$ is incorrect.

But okay, let's say we just want to $\textit{define}$ $0 \cdot 0 = 1$. Certainly if we are going to do this, we'd better make sure that it is consistent with the rules we already have for arithmetic on the integers. We quickly run into issues. For one thing:

$$ 1 = 0\cdot 0 = 0\cdot(0 + 0) = 0 \cdot 0 + 0\cdot 0 = 2 $$

So something is wrong here. It seems like we (at least) either have to throw away the fact that $0 + 0 = 0$ or we have to throw away the distributive property. Either option doesn't seem very good. So, we see without much difficulty that your definition creates inconsistencies, which is why it is not a valid definition.

Answer 2

For real numbers with standard multiplication, $0\cdot 0 =0$ always. For real number $x> 0$, $0^x=0$ always, and us undefined for $x<0$, and $0^0$ is generally left undefined and is called an "indeterminate form".

The most natural choice is $0^0=1$, since $|B|^{|A|}$ = the number of mappings from $A$ to $B$. When they are both empty sets, $|A|=|B|=0$, there is only 1 mapping, the empty one (see here). However, this is not necessarily universal.

Here's how I like to think about it:

The identity of addition is $0$, so when you are adding, you always have a $0$ to start with. Hence and empty sum is still equivalent to zero. I.e. "nothing added" equals $0$, $\sum_{x\in\emptyset}=0$.

The identity of multiplication is $1$, so when you are multiplying, you always start with a $1$. Hence an empty product is $1$. I.e. "nothing multiplied" equals $1$, $\Pi_{x\in\emptyset}=1$.

Let me also add that this is not necessarily a universal convention regarding empty sums and products.

Answer 3

A null product is a result of multiplying no factors. $0$ itself is actually a factor, so the multiplication of $0(0)\neq1$

Answer 4

$0\times0 =\prod_{i =1}^20$ is a product of two values, not of zero values. $0\times0=0.$

Also note that $0^0$ is not defined because there is no good choice: $\lim_{x\to0}x^0 = 1$ but $\lim_{x\to0}0^x=0.$

Edit: you write that there is only one possibility for $\frac10.$ I contend that a second possibility is that it is undefined. Indeed if you want the laws of algebra to make sense then it must be undefined so I would say that there is in fact only one possibility: that $\frac10$ is undefined. Otherwise $\frac x0=\frac1x\frac10=0$ but then $x=\frac x0\times 0= 0\times0=1$ therefore all numbers are 0 or 1. Maybe you would like to live in a world where everything is black and white, 0 or 1 but I’m quite happy having more than 2 (err.. I mean 0) numbers to play with.

Edit2: note that $\frac1{re^{ix}}$ goes to $e^{-ix}\infty$ as $ r\to0$. Normally in mathematics one does not consider a signed $\infty$ but rather a projective $\infty$ which you can get to by going infinitely far in any direction on the complex plane. I would also suggest that even if you insist on a signed infinity, that zero needn’t be halfway. $-\infty=1-\infty$ and $\infty=1+\infty$ so surely halfway is $\frac12(1-\infty+1+\infty)=1$.