The responses of division by 0 that we have been taught is undefined I actually find to be unintelligible and completely unacceptable. Take this into consideration, when we look at a fraction $\frac{n}{d}$ where $n$ is the numerator and $d$ is the denominator and $d$ happens to be $0$ let's apply this concept to a combination of linear equations and some trigonometric functions. Before I begin with that I will clearly state that some functions or equations that are normally used in different contexts mean exactly the same thing in all contexts. And I will show that the actual operation of division and fractions are nothing more than the slope of a line or the tangent of some angle with regards to the horizontal and that the slope of a vertical line or $\tan(90°)$ is completely defined!
Equivalent Equations
- slope of a line $m = \frac{\left( y_2 - y_1 \right)}{\left(x_2 - x_1\right)}$ is equivalent to the $\tan\theta$ where $\theta$ is the angle above the horizon when the angle is in standard form.
- The Pythagorean Theorem and the Equation of the Circle are the same exact thing: $A^2 + B^2 = C^2$ and $X^2 + Y^2 = C^2$
- The $\cos\theta$ between two vectors is equivalent to $\frac{\vec A \cdot \vec B}{|\vec A||\vec B|} $
Assertions
- Both the $\sin$ and the $\cos$ functions have the same domain and range:
Domains are the set of $\mathbb R$ and the ranges are $\left[-1,1\right]$
and they have a period of $2\pi$. They are continuous circular or rotational functions.
- $\tan\theta = \frac{\sin\theta}{\cos\theta}$
- When plotting on the Unit Circle where the radius has a value of 1: The $(x,y)$ pair is defined as $(\cos\theta,\sin\theta)$ where the radius is above the horizontal $x$ axis in standard position.
Assessments
- Let's consider that we have 3 points $a,b,c$ where $a = (0,0)$ and $b = (1,0)$ and these points are fixed and point $c$ starts at point $b$'s location $(1,0)$. These 3 points will form vectors or line segments between each other. Initially there are only 2 valid vectors $\vec A$ and $\vec B$ where $\vec A$ is $b - a$ and $\vec B$ is $c - a$. $\vec C$ doesn't exist yet or is the $\vec 0$ since $\vec C$ is defined as $c - b$ and both points coincide.
- We can technically rotate either direction but we will rotate $CCW$ and we will do a few observations in the process.
- 1 - We will observe point $c$ as it rotates around the unit circle.
- 2 - We will observe vector $\vec C$ as point $c$ rotates around the circle.
- 3 - We will observe the area of the triangle that is generated from vectors $\vec A, \vec B,$ and $\vec C$
- 4 - We will observe the slope of $\vec B$ as this is the line that rotates.
- 5 - We will observe the angles between $\vec A$ and $\vec B$
Intuitive Declarations
- When the point $c$ rotates to the position $(-1,0)$ the slope is $0$, the area of the triangle is $0$, but the length of $\vec C$ is at its longest which is $2$. Here you have nothing but horizontal translation with $0$ or no incline or change in height or elevation.
- When the angle is $45°$ or $\frac{pi}{4}$ radians the slope $\frac{rise}{run}$ and the $\tan\theta$ or $\frac{\sin\theta}{\cos\theta}$ are 1. You have an equal amount of vertical translation as you do horizontal translation.
- The $\cos\theta$ represents the $x$ value on the unit circle but also your horizontal displacement.
- The $\sin\theta$ represents the $y$ value on the unit circle but also your
vertical displacement.
- When the Angle is $0°,180°,360°$ or an multiple of them the Slope and the Tangent are also $0$. This means we have $0$ rise and the $\sin$ or the $y$ has an output of $0$ for its range at that angle.
- When the Angle is $45°$ both the $\cos$ and the $\sin$ have equivalent values since both legs of the triangle here are equal thus giving you both a slope and a tangent of $1$ and if you graph both the sine and cosine functions they will intersect at this point.
Considerations - Considering that both the sine and cosine are continuous circular functions neither of them at any point in their range nor their domain become undefined nor has any discontinuity in it.
Generalization - Imagine yourself walking down an alley between two skyscrapers and the alley starts off being level. You have $0$ slope or no incline but you are traveling either $N,E,W,S$ which doesn't matter because the ground you are walking on is a 2D plane. So you do have 2 degrees of dimension to travel upon. However since you are in a tight alley in this demonstration you are only heading in one arbitrary horizontal direction. Then the alley or the road has a hill that you have to walk up, now you have slope because you are rising in elevation. Then it levels off again and your slope is back to $0$ at the new elevation. This makes sense because two horizontal lines at different heights are parallel so both their slopes will be the same. Finally the alley comes to an end as their is another skyscraper in front of you and you can not go left nor right and there is no turning back, but the building in front of you has a ladder and you begin to climb it rung for rung. When you start to climb straight up your angle is $90°$ which is perpendicular and orthogonal to the ground. This means that you no longer have any horizontal displacement but you have incremental and continuous vertical translations. So in this case your elevation or height is forever increasing until you reach the roof and start to walk across a horizontal or sloped plane.
The argument - In the case where the $\sin$ component of the tangent or the $(y_2 - y_1)$ component is evaluated to $0$ you have no incline and the slope is $0$ because here $n = 0$ and $d = (x_2 - x_1)$ or $d = \cos\theta$ where this means you only have horizontal translation and this is valid because a numerator can be $0$. Let's reverse the case. This time the $\cos\theta$ component of the tangent or the $(x_2 - x_1)$ is $0$ which simply means just the opposite where we have no run yet we do have continuous rise, yet from the fallacies if what we were taught this is undefined because of division of $0$ because here the $d$, $(x_2 - x_1)$, or the $\cos\theta$ evaluates to $0$. I argue that these are valid outputs and acceptable domains of fractions or division. Division by $0$ is completely defined.
Conclusion - If $\frac{0}{d}$ means $0$ slope or no rise with infinite run then the opposite must be valid as well where $\frac{n}{0}$ means no run with infinite rise. The correct answer for division of $0$ would be $\infty$ since when the numerator is $0$ the slope is $0$ because there is no change in elevation. We can not evaluate $\frac{d}{0}$ as $0$ because here we have no run but we do only have rise and slope is defined as the change in the elevation and here the slope is ever increasing vertically up without any horizontal displacement and this does make complete sense. In trigonometry we do know that the tangent's graph has vertical asymptotes at periods of $\frac{pi}{2}$ or $90°$. We are taught that the tangent is undefined here. I think this is a wrong assessment because those vertical asymptotes are parallel vertical lines to the vertical $y$ axis and these are perpendicular and orthogonal to the horizontal or $x$ axis. When a line has a slope $\frac{a}{b}$ a line that is perpendicular to it is $-\frac{b}{a}$ Let's try this with a slope of $0$ then find its perpendicular.
$$\frac{0}{d} \implies 0$$ slope or a horizontal line therefor $$\frac{-d}{0} \implies \infty$$ vertical slope or a vertical line.
Let's apply the above with the trig functions again starting with $0$ slope.
$$\frac{\sin\theta}{\cos\theta} = \frac{0}{\cos\theta} \implies \tan\theta = 0$$
therefor it's perpendicular must be:
$$-\frac{\cos\theta}{\sin\theta} \implies -\cot\theta$$
since we have no run in vertical slope the $\cos\theta$ component must be $0$; then it suggests that:
$$-\cot\theta = \frac{0}{-\sin\theta}$$
which is also the same as:
$$\frac{0}{\sin\theta}$$
however, this does not evaluate to $0$ when regarding slope because we do have a change in height that is shown by the $\sin\theta$. The slope here is defined as being $\lim_\infty$ because the tangent has a vertical asymptote at $90°$ and cotangent is $0$. The cotangent has a vertical asymptote at $180°$ and the tangent is $0$ at $180°$.
It is these associations and relationships of division, fractions, slopes, trig functions and reciprocals with the use of the dot and even the cross products that define how two vectors are perpendicular and orthogonal to each other when they create a separation of $90°$ or $\frac{pi}{2}$ radians from each other. If you were to take just the $x$ and $y$ axis of a 2D Coordinate Cartesian Plane we know that the $x$ axis has $0$ slope because it is horizontal or level and that the $y$ axis has vertical slope. Vertical slope is NOT undefined! If we were to rotate these two axis together by $1°$ there is defined slope. Slope is always defined since one can always change their perspective to that system.
I think that people confuse what $0$ really is! $0$ is really not a number, it is a place holder, it also represents the empty or the null set, it has no value. So if you can divide any number by the empty set and it returns back to you the empty set. Is it not conceivable to divide anything into the empty set?
In this particular case and context division by $0$ here yields infinity because the problem pertains to slope or the change in height over distance.
In other contexts division by $0$ could mean, $0$ as the result is $0$ and is no different than when it is in the numerator.
It could also yield D.N.E. meaning that the function that it is being applied to just Does Not Exist at that location or context.
There is 1 special case and that is when both the numerator and denominator are $0$. This could yield $0$, $1$, and or $\infty$. This does satisfy that $0$ divided by any number equals $0$. It also satisfies the multiplicative identity that any number divided by itself is $1$. The infinity part also comes from the concept that if you have $0$ run and $0$ rise you are stationary and you are infinitely not moving in any direction which is no different than $0$. To try to make this understanding a little clearer then ask yourself this: Why does any number $n$ raised to the $0$ power always equal $1$? $n^0 = 1$.
For The Reader - It would be advisable to draw the unit circle and the points and vectors as described above in the assessments and to do several of them where the rotation around the unit circle are at different positions. Or you can visit this web page interactive graph of a graph that I made that shows all the relationships between these 3 points along with the tangent, area of the triangle, and even the volume if you increase the height factor, the coordinate pairs $(\cos\theta,\sin\theta)$ along the unit circle etc. And you will notice that when the angle is $0°$ the slope, area and volume are all $0$. They are also $0$ when the angle is $180°$ and $360°$. When the angle is 90° or $270°$ of course the "slope is labeled as undefined because that is what people have programmed it to be because of the fact that we were taught that division by $0$ is undefined!" However the Area and Volume of the triangle is at its maximum value. I have it set to where you can press the play button for the "t" value which stands for $\theta$ because this website at the time has not incorporated the use of the variable $\theta$ to be allowed into their functions or expressions to make graphs.