By slope I mean derivative. Is something being infinity the same thing as being undefined?
Is the slope of a vertical line negative infinity or positive infinity?
[Math] Is the slope of a vertical line infinity or undefined
algebra-precalculuscalculusslope
Related Solutions
See the plot:
First we have to define the slope of a straight line. Then "slope of a curved line at a point" means the slope of the tangent to the curve at that point and this is equivalent to bringing two points on the curve so close to each other that there will be negligible difference between them and then finding the slope of the line passing through the two infinitesimally close points and this line can be regarded as a tangent line.
Okay so let's repeat a definition here
Slope is calculated by finding the ratio of the "vertical change" to the "horizontal change" between (any) two distinct points on a line.
So first of all, you cannot get the slope of a line by taking two equal points and finding the ratio between their distance (which would be $\frac 00$). With your reasoning, you find the gradient of $y=x$ at the point $(0,0)$ by taking $(0,0)$ as reference point, and then calculating $\frac{0-0}{0-0}$, after which you proceed to argue that that should be equal to $1$. However, the line $y=2x$ for example, has slope $2$ (yes, also in $(0,0)$), yet with your reasoning, it should be $\frac{0-0}{0-0}$, which you said should be $1$.
Second of all, vertical lines have no "horizontal change", and as such, the ratio between the "vertical change" and the "horizontal change" cannot exist, since we'd be dividing by $0$.
Why can we not divide by $0$?
In the real numbers (this works for complex numbers too, but we'll just go with $\mathbb{R}$), we have the following property (for all $b\neq 0$): $$\frac ab=c\iff a=bc$$ When you say that division by $0$ should be possible, you implicitly say that the above rule shouldn't only work for all $b\neq 0$, but also for $b=0$. This makes no sense however, since with that, we have: $$\frac a0=c\iff a=0\cdot b$$ But it is known, even without division by $0$, that $0\cdot c=0$ for all $c$, and so that is still true when division by zero is possible; but now we encounter a problem. The property with $b=0$ now states that for all $a,c$, we must have $$\frac a0=c\iff a=0$$ This is problematic, because, if division by zero was possible, that would mean that if you divide any number $n$ by $0$, that number must've been $0$; moreover, we have $\frac 00=c$ for all $c$. See how that produces problems?
To sum it up, if you allow division by $0$, then all numbers suddenly become equal, and we're dealing with $\{0\}$ instead of the usual, much more useful $\mathbb{R}$. This is why we don't do division by $0$.
As a sidenote, scepticism makes a good mathematician; however, don't underestimate the mathematicians that've existed and thought about all sorts of things over the ages. With such basic results as this, you can assume very talented and respected mathematicians have thought about this, and especially if they all agree, they're probably right.
Edit
It seems like you view real numbers more like quantifiers than a set of mathematical objects; this view is fine, but often doesn't work. You say $0$ is "just a placeholder", but it really doesn't work that way. The set of real numbers, as well as the set of complex numbers, the set of fractions, the integers, and many more, are sets with elements that follow specific rules; those rules work for all numbers in the set, and so $0$ is being treated, as should be treated, as just "one of the elements" of said set. $0$ does not mean "nothing" nor "NaN", even though real-life applications suggest the nothingness of $0$. You have to keep in mind mathematics is purely theoretical, and there's applications in the real world that mathematics wouldn't fully agree with (take for example physics, which does a lot of things like "rounding off", "discarding terms because they're insignificant" which are fine for the real world, and it works, but mathematically speaking, some things would be considered non-rigorous).
Best Answer
You can only compute derivatives of functions $f:\Bbb R\to\Bbb R$ (at least in this context here). A vertical line is no such function. So one can consider it as undefined. At least as long as you insist in defining "slope" as derivative.
But infinities are not the same as something being undefined. This depends on the context.