I know that in school, we were always taught that $0*0 = 0$, because anything times zero is zero, but wouldn't it be true that you are saying you have zero quantity of zero, meaning you cannot end up with zero (because you just said you do not have zero).
Therefore, wouldn't $0*0$ be equal to any number except zero?
If we assume that $0 \times 0$ equals anything but $0$, then $$ 0 = 0 \times (0 \times 0) = (0 \times 0) + (0 \times 0) = 2 \times (0 \times 0) \ne 0 \implies 0 \ne 0 $$
You first get in touch with this fact in dealing with natural numbers. Here, the distribution law holds:
Given three natural numbers $a$, $b$ and $c$ the multiplication is distributive over the sum, i.e. $$ a \times (b \times c) = a \times b + a \times c $$
Furthermore, it is well known and (somehow) easily provable that in $\mathbb{Z}$, $\mathbb{Q}$ and $\mathbb{R}$ the element $0$ is the only one satisfying $a \times 0 = 0$ for each element $a$ - they're integral domains, where $0$ is the only so-called $0$-divisor. Over $\mathbb{N}$, you define the multiplication as $a \times 0 = 0$ and $a \times S(b) = a \times b + a$, given two naturals $a$ and $b$. Then, being $1 = S(0)$, you get $$ 0 = 0 \times 1 = 0 \times 0 + 0 = 0 \times 0, $$ since $0$ is the identity element of the monoid $(\mathbb{N},+)$. More about the natural numbers at Wikipedia.
In my humble opinion, the trouble you got into is a linguistical one: you took multiplication $x \times y$ as "$x$ quantity of $y$", which is a non-formal view of that operation. That can be extremely dangerous.
For example:
Let $x$ be "the smallest number which cannot be expressed with more than twenty alphanumeric symbols".
There, you're already expressing $x$ with way more than twenty alphanumeric symbols. ;-)
Back to your question: if I told you that $0$ meant "nothing", you would have ended up having no amount of nothing, which is...nothing!
As you can see, it can be convenient to have an intuitive or non-formal visualisation of mathematical concepts, but this can lead to wrong conclusions.
I'm trying to empathize with your lexical hurdle. The best way I can think of to convince you is that having zero groups, containing zero elements each, doesn't imply you have any other amount of groups containing any other amount of elements, turning you into having something! You have even more so zero elements!!
Suppose $a*b$ indicates the amount of milk you have in your house, where $a$ is the number of trips you take to the store and $b$ is the amount of milk you buy per trip. Then if both $a$ and $b$ are $0$, you have no milk.
We have that $$\lim_{x\to 0}0\cdot x=0$$ So it's useful to define $0\cdot 0=0$ for continuity. Also imagine dividing $0$ candies to $0$ people it's kinda absurd that you will actually get some candies from none.
The number $0$ has a special role when we do arithmetic. Let's consider the reals together with addition $+$ and multiplication $\cdot$. Then there are some (more or less) plausible rules or algebraic laws from which \begin{align*} 0\cdot0=0 \tag{1} \end{align*} necessarily follows.
- Identity element
The number zero has a really distinguishable property: We can add zero to any number without changing anything. \begin{align*} 0+a=a=a+0\qquad\qquad\text{for all } a\text{ in } \mathbb{R}\tag{2} \end{align*}
An element with this property is called identity element of addition.
Additive inverse element
To each element $a$ in $\mathbb{R}$ we can find an element $-a$ in $\mathbb{R}$ with the property \begin{align*} a+(-a)=0=(-a)+a\tag{3} \end{align*}
An element with this property is called additive inverse of addition.
Distributive law
Together with the distributive law \begin{align*} a\cdot(b+c)=a\cdot b+ a\cdot c\qquad\qquad \text{for all } a,b,c \text{ in } \mathbb{R}\tag{4} \end{align*}
we conclude \begin{align*} 0\cdot 0&=0\cdot(0+0)&\text{rule }(2)\\ 0\cdot 0&=0\cdot 0+0\cdot 0&\text{rule }(4)\\ \end{align*} The element $0\cdot 0$ has according to rule (3) an additive inverse element $-(0\cdot 0)$ and adding this element to the last equation gives \begin{align*} 0=0\cdot 0 \end{align*}
Note: We might conclude the logic behind this identity is encoded in those algebraic laws.
Perhaps we can agree that $0 \cdot 1 =0$ and $0+1=1$? If so, then for arithmetic to work as we want it to, we must have: $$0=0\cdot 1=0\cdot (0+1)=$$ $$0\cdot 0 + 0\cdot 1 = 0 \cdot 0 + 0 = 0 \cdot 0$$
All of these operations are substitutions or basic algebra we want to use for all numbers, 0 included.
I noticed that you used the "Logic" tag on your question.
If you are doing math using standard boolean logic, the one thing that means 'game over' is a contradiction - the statement $P$ is both true and false.
So, go for it! Assume you are onto something and insist that
$0 \times 0 \ne 0$
I doubt you will have much fun. You will have to twist yourself into contortions to avoid a contradiction, and I doubt anything useful will come of it. Perhaps you need to develop a whole new system of logic, and then things will work out.
You might want to read
Now I will attempt to explain why $0 \times 0 = 0$.
Are you on-board that we have the natural numbers with an operation called addition and zero is the additive identity? If you have a formal system, then $0$ represents a 'blank page" or the starting no-length tick on a ruler when you are adding things to each other.
OK, if you do, then multiplication starts out as just an abbreviation:
$m \times n$ is shorthand for adding $m$ $n's$ together:
$n + n + ... + n$ where $n$ occurs $m$ times.
Now, $0 \times n$ gets you back to that 'blank page' or zero.
So if you accepts that, why balk over accepting that $0 \times 0$ is a 'blank page', contributing nothing when we are adding up marbles or stick figures or other things.
Take a 'blank page' 'blank page' times and what do you get? If our formal system insists that we get back to a number, well it better be zero.
Let me focus on the bolded part here:
I know that in school, we were always taught that $0*0 = 0$, because anything times zero is zero, but wouldn't it be true that you are saying you have zero quantity of zero, meaning you cannot end up with zero (because you just said you do not have zero).
If I have $0 * 0$, does that mean that I don't have $0$ (and so I must have some other number besides $0$)? That may seem like a natural way to continue the pattern:
As a matter of fact, $0 * 0$ is $0$, so the pattern must not hold here. But why doesn't it hold?
The problem is that the English word "nothing" is a special word which, unlike most nouns and pronouns, doesn't refer to something. Instead, causes the verb to mean the negation of what it usually means. (If I say that "I hear nothing", I'm not saying that I hear; I'm saying that I don't hear.)
So we write down the sentence "I don't have nothing", and then when we read that sentence, the word "nothing" changes the meaning from "I don't have" to "I do have". But we don't want that! We want to keep the original meaning, which is "I don't have".
We need an interpretation that doesn't change meaning when we put the number $0$ into it.
As you know, $0 = 1 - 1$ ($0$ is the same number as $1 - 1$). If I have one item, and then you take away one item, I am left with zero items.
So this means we can interpret "I have $0 * x$" as meaning "I had $x$, but then you took away $x$".
This interpretation shows why $0 * 0 = 0$:
Any real number $r$ such that $r^2 = r$ also satisfies $r(r-1) =0$. So the RHS being zero implies the LHS for your question.
Well, imagine you have zero candy. You have zero lots of zero candy. Therefore, you have nothing.
Imagine you have a zero length stick. You stuck a zero length stick perpendicular to the first one. There is no area in between!
Basically, I am trying to say that to do $0*0$, just use real life problems for $xy $, where $x=y=0$
P.S. What were you thinking of, if you didn't think that it was zero?
