I understand that you can't divide anything by 0, but I can see arguments why it could be 0 (0 divided by anything is 0) or 1 (anything divided by itself is 1). Personally, before I plugged 0/0 in my calculator, I thought the answer would be 0. I'm just curious if there's a special reason why 0/0 is undefined, like how there's a special reason why 1 is not prime.
Videos
Like 10/2- imagine a 10 square foot box, saying 10 divided by 2 is like saying “how many 2 square foot boxes fit in this 10 square foot box?” So the answer is 5.
But if you take the same box and ask “how many boxes that are infinitely small, or zero feet squared, can fit in the same box the answer would be infinity not “undefined”. So 10/0=infinity.
I understand why 2/0 can’t be 0 not only because that doesn’t make and since but also because it could cause terrible contradictions like 1=2 and such.
Ah math is so cool. I love infinity so if anyone wants to talk about it drop a comment.
Edit: thanks everyone so much for the answers. Keep leaving comments though because I’m really enjoying seeing it explained in different ways. Also it doesn’t seem like anyone else has ever been confused by this judging by the comment but if anyone is I really liked this video https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:foundation-algebra/x2f8bb11595b61c86:division-zero/v/why-dividing-by-zero-is-undefined
The other comments are correct: 
is undefined. Similarly, the limit of 
as 
approaches 
is also undefined. However, if you take the limit of 
as 
approaches zero from the left or from the right, you get negative and positive infinity respectively.
does tend to 
as you approach zero from the left, and 
as you approach from the right:
That these limits are not equal is why 
is undefined.
Those expressions are about limits, not about numbers.
We say that $\frac00$ is an indeterminate form because a limit of that form can take any value:$$\lim_{y\to0}\frac{xy}y=x,$$for any real number $x$.
On the other hand, a limit of the type $\frac10$ cannot take any value. If it exists, it can only be $\infty$ or $-\infty$.
In the context of limits, $0/0$ is an indeterminate form (limit could be anything) while $1/0$ is not (limit either doesn't exist or is $\pm\infty$). This is a pretty reasonable way to think about why it is that $0/0$ is indeterminate and $1/0$ is not.
However, as algebraic expressions, neither is defined. Division requires multiplying by a multiplicative inverse, and $0$ doesn't have one.
If $0/0$ were equal to $1$, then $1=\frac{0}{0}=\frac{0+0}{0}=\frac{0}{0}+\frac{0}{0}=1+1=2$.
In lay terms, evaluating 0/0 is asking "what number, when multiplied by zero, gives zero". Since the answer to this is "any number", it cannot be defined as a specific value.
The problem is, when taking a limit, we're not actually saying $x=0$; we're just taking numbers really close to $0$. So when you say
One of the properties of $0$ is that no power modifies it, so wouldn't the two functions be equivalent at $0$?
The difference here is that when we use $x=\varepsilon >0$, then both $\tfrac 1x$ and $\tfrac 1{x^2}$ are positive, but when $x=-\varepsilon<0$, then $\tfrac 1x$ is negative but $\tfrac 1{x^2}$ is positive. So really, this doesn't have anything to do with $0$, it's just that negative numbers squared give a positive number.
To expand on the fact that you say $\lim_{x\to0}\frac1x$ does not have a limit and $\lim_{x\to 0}\frac 1{x^2}$ does; this is simply because, when approached from the left, the first limit should be $-\infty$, but when approached from the right, the first limit should be $+\infty$. This is why the first limit is indefinite.
The second however gives $+\infty$ no matter how you approach $0$, but this, as explained above, has to do with the fact that $(-x)^2=x^2$. There are no hole in algebra, nor are there in calculus; I suggest you read into the definition of limits again to refresh your view on this.
There are a couple of things here that are a bit confusing, so let's break them down. We say: $$\lim_{x\to 0} \frac{1}{x^2} = \infty$$
But we don't actually mean that it "equals infinity". The limit is still indefinite. The notation $\lim_{x\to a} \_\_\_ = \infty$ is actually shorthand to mean "as $x$ goes to $a$, the limit goes to an arbitrarily large positive."
The second part you have here is how zero functions:
One of the properties of $0$ is that no power modifies it, so wouldn't the two functions be equivalent at $0$?
You are correct! No power of zero modifies its value, and those two functions are "equivalent" at zero, in that they are both undefined.
The difference here is the limit. In a limit, you can approach the value (or indefinite value) from the "left" or "right" side of the variable. Whereas the limit in $\frac{1}{x^2}$ is positive no matter which direction it's approached from, the limit in $\frac{1}{x}$ is negative on the left side and positive on the right side. So, we can say that the limit for $\frac{1}{x^2}$ is arbitrarily positive, but $\frac{1}{x}$ can be positive or negative depending on which side you approach it.
If I lower down from positive numbers it can be positive infinite, if I lower down from negative ones, it can be negative infinite so one of the proves of it being undefined, why not both, why not simply infinite