Division by Zero and proofs that the slope is defined.

Question

Most people think that when they encounter a fraction number as such: $\frac{n}{0}$ as being an error due to division of $0$ and it is considered undefined. I have expressed this or a similar question with the use of points, vectors, and trig functions that vertical slope should be defined found here: https://math.stackexchange.com/questions/2177253/behaviour-of-perpendicular-lines-vertical-slope-and-divsion-by-0 and I want to re ask this question but in a different context using the same principles that I've applied before that involves the use of: slope $$m = \frac{y_2-y_1}{x_2-x_1} = \frac{\delta y}{\delta x} = \frac{\sin\theta}{\cos\theta} = \tan\theta$$

But this time I want to consider not just division by $0$, but also the infamous fraction of $\frac{0}{0}$ and what it should equal.

First let me state that $0$ does not have a value and that $0$ is actually a $NAN$ that means it is a place holder and represents either the null or the empty set.

Now when we have a set, the set it self is something tangible and when we divide anything tangible by itself the result is $1$ do to the identity property. For example consider this with the set without taking any of its elements into consideration: Set $a$ when divided by itself as in $\frac{a}{a} = 1$ should be a valid statement?

Let's consider the linear equation $y = x$ that has an understood slope of $1$ which makes the slope also $\frac{1}{1} = \frac{\sin\theta}{\cos\theta} = \tan\theta$ where $\theta = 45°$ and for every coordinate pair that belongs to this line has a slope of $1$ at that point on the line. This set of coordinate pairs is the set of values of the domain and the range to this function, equation or expression.

I will use a few of points belonging to $y = x$ where slope $m$ here is understood as $1$:

$$ \begin{array}{c|lcr} \text{x} &\text{y} \\ \hline -3 & -3 \\ -2 & -2 \\ -1 & -1 \\ -0 & +0 \text{ (-) is comging from and (+) is going to} \\ +1 & +1 \\ +2 & +2 \\ +3 & +3 \\ \end{array} $$

Since slope is defined as $\frac{rise}{run}$ or $\frac{\delta y}{\delta x}$ or $\frac{\sin\theta}{\cos\theta}$ or $\tan\theta$, and knowing that we can take the coordinate pairs and make them as slope since it is understood as $1$ have the form of $\frac{y}{x}$ when $y = x$? So does this not generate this sequence of values: $$\frac{-3}{-3} = \frac{-2}{-2} = \frac{-1}{-1} = \frac{0}{0} = \frac{1}{1} = \frac{2}{2} = \frac{3}{3} \implies 1$$

Does the point at $(0,0)$ on this line not have a slope of $1$? Yes! However the input and output are both $0$. The reason that $\frac{0}{0} = 1$ works is because of 2 main reasons: First is the Identity Property, anything divided by itself is of course itself! Second is the fact that the point at $(0,0)$ with a $(+)$ slope of $1$ is a point of reflection or point of symmetry, an origin of rotation, and a point in which the function has as a starting location that it expands out from making this the ROOT of the function. In the natural form of $y=x$, its graph in the first quadrant is approaching $+\infty$ while it is approaching $-\infty$ in the third quadrant. So here the coordinate point for slope $m$ that has a value of $1$ is represented by the coordinate point at the origin of $(0,0)$. Thus this does define $\frac{0}{0}$ to have a value of $1$. So if $\frac{0}{0}$ is defined by this then why is that $\frac{n}{0}$ should be or must be undefined? A line never looses it's slope! A lines slope has a range of $\pm\infty$ and for the domain we will use domains of the $\sin$ and $\cos$ independently of each other that has a domain of $\mathbb R$ instead of referring to the $\tan$ that is typically claimed to be undefined at an angle of $90°$ or $\frac{\pi}{2}$ due to the current assumption that division by $0$ is undefined. So how is it that "vertical slope" is Undefined? When the change in height = $0$ this is horizontal slope and it is defined. So when the opposite or the perpendicularity occurs where we do have change in $y$ but the change in $x$ stops we have a problem. This labeling of division by $0$ as being undefined is made from a long list of wrong assumptions. Does this not also support and back up the proofs I showed from before why this is valid although everyone else is hell-bent on claiming that it isn't?

Answer 1

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 $0$.

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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 $0$.

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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).

Answer 2

I just rewrite your argument word by word, using $y=2x$ instead of $y=x$. All typos are intensionally leaved unchanged.

I will use a few of points belonging to $y=2x$ where slope $m$ here is understood as $2$: \begin{array}{c|lcr} \text{x} &\text{y} \\ \hline -3 & -6 \\ -2 & -4 \\ -1 & -2 \\ -0 & +0 \text{ (-) is comging from and (+) is going to} \\ +1 & +2 \\ +2 & +4 \\ +3 & +6 \\ \end{array} ... So does this not generate this sequence of values: $$\frac{-6}{-3} = \frac{-4}{-2} = \frac{-2}{-1} = \frac{0}{0} = \frac{2}{1} = \frac{4}{2} = \frac{6}{3} \implies 2$$ Does the point at $(0,0)$ on this line not have a slope of $2$? Yes! However the input and output are both $0$...

You can indeed assume that $0/0=1$, but you cannot avoid others to assume that $0/0=m$ for any $m$. Mathematics avoids ambiguity. So normally we say that $0/0$ is undefined.

You said that $0$ does not have a value, it is a placeholder or the empty set. That's correct. More correctly, NOTHING in mathematics has a value. Everything in mathematics is nothing but sets.

For example, we use $0$ (or more commonly, $\varnothing$) to denotes the empty set, and $$1=\{0\}, 2=\{0,1\}, 3=\{0,1,2\},\ldots$$

By the way, $a/b$ is not defined by slopes, but slopes is defined by $a/b$.