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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.

R. Emery
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    "So 0*0 = 1 therefore 1/0 = 0". How? – Sil Nov 04 '17 at 15:28
  • See Why does 0 factorial equal 1? for $0 !$ and $x^0$. – Mauro ALLEGRANZA Nov 04 '17 at 15:31
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    $0 \times 0 \ne 1$ because there is no number $n$ such that $0 \times n=1$. – Mauro ALLEGRANZA Nov 04 '17 at 15:32
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    Product $0\times 0$ is not an empty product. – drhab Nov 04 '17 at 15:33
  • 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. – R. Emery Nov 04 '17 at 15:33
  • there is no number n such that 0×n=1? Thats circular reasoning – R. Emery Nov 04 '17 at 15:34
  • Also the function 1/x goes to infinitity from one direction. and to negative infinity from the other. Half way between infinity and negative infinity is zero. – R. Emery Nov 04 '17 at 15:38
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    @R.Emery Your main mistake is saying that $0\times0=1$. It is a conclusion you draw, but it is based on nothing. – drhab Nov 04 '17 at 15:39
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    Where is the question about limits here? – jdods Nov 04 '17 at 15:40
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    Is this a question or do you want us to look at this nonsense and tell you that we’ve been wrong all these years and how smart you are for figuring this all out and why didn’t we think of that. Normally a question has an answer but for every answer you seem to change the question and move the goalposts. – Dan Robertson Nov 04 '17 at 16:03
  • So when you run out of logical arguments then you just resort to insults? – R. Emery Nov 04 '17 at 16:21
  • When you break the rules, you can get anything you want, $1=0=\infty$ etc. The important thing in math is that you follow established principles and carefully make sure each step obeys accepted standards. You are not following the rules. – jdods Nov 04 '17 at 19:47
  • What rule were the mathematicians following when they decreed that sqrt(-1) = i? – R. Emery Nov 04 '17 at 20:05
  • The rules that are consistent and so you can talk about sensible concepts within it (the $\sqrt{-1}$ might not make sense in real numbers, but it makes sense in larger theory, and can be done consistently). Unfortunately defining things like division by zero etc . lead to inconsistent math, contradictions if you like. That is why these things are usually left undefined. – Sil Nov 04 '17 at 21:21
  • Yes that is the conclusion that I had already come to. But it was a possibility worth considering. You make it sound like it wasnt. – R. Emery Nov 05 '17 at 01:12
  • It was worth, and believe me others have tried, but it just does not lead to anything useful unfortunately. Here are for examples some answers that show what goes wrong with division by zero: https://math.stackexchange.com/questions/125186/why-not-to-extend-the-set-of-natural-numbers-to-make-it-closed-under-division-by. You can find other similar questions on this site. – Sil Nov 05 '17 at 09:16
  • $0\times 0$ represents the empty sum ($0$ many summands of size $0$), not the empty product. The empty sum is $0$, not $1$. The empty product could be written $0^0$, which is indeed often taken to have the value $1$. – ziggurism Apr 12 '18 at 17:23

4 Answers4

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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.

wgrenard
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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.

jdods
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$0\times0 =\prod_{i =1}^20$ 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 $\frac12(1-\infty+1+\infty)=1$.

Dan Robertson
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    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. – jdods Nov 04 '17 at 15:35
  • More circular reasoning. limit of 0^x would not be zero because 0^x would be 1 for all x – R. Emery Nov 04 '17 at 15:42
  • @jdods this is not natural for powers of real numbers because by that reasoning, $0^x=1$ for all $x$ as there is exactly one mapping from $\varnothing$ to any set. Indeed the reason we can’t just use $0^0=1$ as “powers correspond to powers in the group so the zeroth power of anything is 1” is that in $a^b$ we have $a$ being in the group and $b$ being an integer (or something else if we are allowed roots and limits). But 0 is specifically not in the multiplicative group of any ring. – Dan Robertson Nov 04 '17 at 15:42
  • There is a more natural way to think about it when $x$ isn't zero, than in terms of mappings. I would just argue that with $0^0$, the most natural way to think about it is in terms of the empty mapping. Of course, that is simply an opinion and not fact. – jdods Nov 04 '17 at 15:46
  • x/0 = x * 1/0 = x * 1 = x – R. Emery Nov 04 '17 at 15:54
  • A moment ago you defined $\frac 10$ to be 0 not 1. It can’t be both (although it can be nothing) – Dan Robertson Nov 04 '17 at 16:00
  • You are correct – R. Emery Nov 04 '17 at 18:17
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A null product is a result of multiplying no factors. $0$ itself is actually a factor, so the multiplication of $0(0)\neq1$

aleden
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