15
$\begingroup$

How do we explain that dividing a positive number by $0$ yields positive infinity to a 2nd grader? The way I intuitively understand this is $\lim_{x \to 0}{a/x}$ but that's asking too much of a child. There's got to be an easier way.

In response to the comments about it being undefined, granted, it is undefined, but it's undefined because of flipping around $0$ in positive or negative values and is in any case either positive or negative infinity.

Yet, $|\frac{2}{0}|$ equals positive infinity in my book. How do you convey this idea?

$\endgroup$
  • 93
    $\begingroup$ Don't explain wrong things to a 2nd grader. Division by zero is not defined. Period. $\endgroup$ – Hagen von Eitzen Nov 21 '12 at 21:33
  • 68
    $\begingroup$ Perhaps it is OK to pretend to a little kid that there is a tooth fairy. But it is not OK to tell the kid that one can divide by $0$. $\endgroup$ – André Nicolas Nov 21 '12 at 21:38
  • 13
    $\begingroup$ Division by zero is not always defined - but I once explained to my daughter (10 years old at the time) how to wrap the real line around a circle and make the join using the point at infinity, and went onto the Riemann Sphere. Youngsters can understand these things before they have the mathematical sophistication to understand them precisely - which makes it really important that we guide their understanding accurately. And then there is projective geometry (including finite projective planes) rich territory to explore, just get it right. $\endgroup$ – Mark Bennet Nov 21 '12 at 21:54
  • 6
    $\begingroup$ Infinity is not a real number. That does not mean it is not a number. Many, including myself, would consider numbers the elements of the projective numbers and of the extended real numbers. On that note, the word "infinity" is also ambiguous. $\endgroup$ – Excluded and Offended Nov 21 '12 at 23:50
  • 6
    $\begingroup$ @Marcus My kid came home talking about his friends having a competition about "who can name the biggest number". One said "infinity!" Another said "infinity + 1!" My kid said "infinity to the infinity!" I gave him a time-out for that. Infinity is not a number. $\endgroup$ – Fixee Nov 22 '12 at 0:34

14 Answers 14

37
$\begingroup$

The first thing to point out is that division by zero is not defined! You cannot divide by zero. Consider the number $1/x$ where $x$ is a negative number. You will find that $1/x$ is negative for all negative $x$. As $x$ gets closer and closer to zero, $1/x$ gets bigger and bigger in the negative direction: $1/x \to -\infty$ as $x \to 0$ from the negative side. Next, consider the number $1/x$ where $x$ is a positive number. You will find that $1/x$ is positive for all positive $x$. As $x$ gets closer and closer to zero, $1/x$ gets bigger and bigger in the positive direction: $1/x \to +\infty$ as $x \to 0$ from the positive side.

$$\lim_{x \to 0^-} \frac{1}{x} \neq \lim_{x \to 0^+} \frac{1}{x}$$

Informally: what does $6 \div 3$ mean? It means, how many times do you add $3$ together to get $6$, and the answer is $2$. What does $7 \div 2$ mean? It means, how many times do you add $2$ together to get $7$, and the answer is $3\frac{1}{2}.$ What does $1 \div 0$ mean? It means, how many times do you add $0$ together to get $1$? Well: $0 = 0+0 = 0+0+0$, etc. You have to keep adding zeros for all of eternity. In reality, you never get to $1$ and so there is no answer. It is not infinity: you can't have "infinitly many" zeros. But some people might say "You add $0$ together infinitely many times".

$\endgroup$
  • 14
    $\begingroup$ Even if you could add $0$ infinite times, with any reasonable definition of infinite addition it still wouldn't be $1$. $\endgroup$ – Javier Nov 21 '12 at 22:24
  • 2
    $\begingroup$ Ehh, Javier, to me that makes it sound like you're saying $0\cdot \infty=0$. I think we'd have to leave an infinite addition of zero as undefined. (NOT to be confused with the limit of infinite addition of zeros. That's definitely zero). $\endgroup$ – user18862 Nov 21 '12 at 23:01
  • $\begingroup$ What's stopping us from simply defining $1/0 = \infty$ just as we once defined $\sqrt{-1} = i$? Would that lead to "bad things"? $\endgroup$ – JesperE Nov 22 '12 at 6:18
  • 3
    $\begingroup$ @JesperE Once we have defined $1/0$ as you've proposed, what will we do with it? Can we ask about adding it to itself? Multiplying or dividing it by itself? This is not to say there is no way of making sense of such an idea, but a new definition like that will bring with it more questions than it answers. $\endgroup$ – Benjamin Dickman Nov 22 '12 at 9:19
  • 5
    $\begingroup$ @JesperE yes, it does lead to problems. The problem with $\infty$ is that things like $0\infty$, $\infty/\infty$, $\infty - \infty$, and $0/0$ all can't be consistently defined. For example, if you want $\infty - \infty=0$ and $\infty + 1 = \infty$, then you get $\infty + 1 - \infty = (\infty + 1) - \infty= \infty - \infty= 0$ or $\infty + 1 - \infty = (\infty - \infty) + 1 = 1$. So you lose even the most basic rules, like associativity. $\endgroup$ – asmeurer Nov 30 '12 at 3:20
20
$\begingroup$

When one works in the set of real numbers, division by $0$ does not yield infinity. It is undefined. The reason is this: What would $\frac{1}{0}$ be? It would be the number which when multiplied by $0$ gives you $1$, but there is no such number.

Your book saying that $|\frac{2}{0}|=+\infty$ without further qualification is incorrect. We have $\lim_{x\to 0^+}\frac{2}{x}=+\infty$ and $\lim_{x\to 0^-}\frac{2}{x}=-\infty$ and $\lim_{x\to 0}|\frac{2}{x}|=+\infty$, that is all.

$\endgroup$
  • 13
    $\begingroup$ One does not simply divide by zero. $\endgroup$ – 000 Nov 22 '12 at 1:47
  • 2
    $\begingroup$ Let's take another tack. Let $\varepsilon$ be the "special" number that results when you divide anything by zero; that is, $x/0=\varepsilon$ for any $x$. Unfortunately, if we want to do this, we will have to dispose of the convenient notion that $0\times x=0$ for any $x$... $\endgroup$ – J. M. is a poor mathematician Nov 22 '12 at 5:50
16
$\begingroup$

Take a glass jar/glass/something, and a bunch of small objects (ping pong balls, bouncy balls, marbles, whatever is the best size for this).

Suppose your jar holds ten balls, and it's easy to see it holds exactly 10 of these. Demonstrate that if you're dividing by one, you can put one ball in, 10 times. You divided the jar into 10 sections. If you're dividing by two, show that you can put two balls in 5 times. If you're dividing by five, you can put five balls in 10 times. Associate "divided by" as equal to "how many in my hand each time I put something in the jar".

Now ask "What's 10 divided by zero? How many times can I put zero balls in at a time until it's full?" Take an empty hand, pantomime dropping it in the jar, and repeat. Keep going frantically/comically for bonus points. You can keep doing this forever and never fill the jar up. That's infinity.

(I realize this may not pass peer reviewed journals for accuracy, but for the target audience of 2nd graders, I think this is going to be close enough)

$\endgroup$
  • 6
    $\begingroup$ Or else how long will it take to get to school walking at $0$ miles per hour? Infinitely long? The commonsense answer is that we will not get there. $\endgroup$ – André Nicolas Nov 22 '12 at 4:39
5
$\begingroup$

I agree with the responses you received, but you might also want to read the following: http://www.merga.net.au/documents/MERJ_12_2_Tsamir%26Sheffer.pdf

$\endgroup$
  • $\begingroup$ Very nice link; I hadn't seen this one before. $+ \left(\lim_{n \to \infty} \sqrt[\Large n]{n}\right)$ $\endgroup$ – Namaste May 18 '13 at 0:55
  • $\begingroup$ @amWhy: That was a very nice write-up. I would love for someone to ask kids in kindergarten, 1st, 2nd, 3rd grade all the way through high school graduation and then compile the results of what they say. Wow, would that make for hours of laughter and maybe even some awesome nuggets! :-) $\endgroup$ – Amzoti May 18 '13 at 1:04
5
$\begingroup$

let us consider that any number divided by zero is undefined.

You can let the kid know in this way:

Division is actually splitting things, for example consider you have 4 chocolates and if u have to distribute those 4 chocolates among 2 of your friends, you would divide it(4) by 2(i.e : 4/2) = 2.

Now consider this, you have 4 chocolates and if u don't want to distribute among any of your friends, (that is like distributing to 0 friends) division does not even come into picture in such cases and also division(4/0) makes no sense. Hence in such cases its told UNDEFINED.

$\endgroup$
4
$\begingroup$

One way to explain that division of $x$ by $0$ is undefined is by contradiction. Suppose $x/0 = a$ and suppose $x$ is a non zero value. Then, by cross multiplication, we get $0\cdot a = x$. At this point ask the child what number times $0$ equals a non zero number. After a little thought the child will most likely say that any number times zero is $0$ so that $0\cdot a = x$, $x$ a non zero number is not possible. Next consider $x = 0$ so you have $0/0$. Let $0/0 = b$, where $b$ is a non zero number. Then you cross multiply to get $0\cdot b = 0$. Now ask the child to come up with a number that satisfies this equation. The child will most likely realize that any number will do and pick one, say $5$. $0\cdot 5 = 0$, true. Now say, what about $0\cdot 6$? The child will say that equals zero too. So, going back to $x/0$, there is no solution and in the case of $0/0$, in effect, any solution will do. Neither of these are allowed in mathematics. The above is not a proof of course but it might help a little. Note: the explanation doesn't really work for the case where $x/0 = 0$ or $0/0 = 0$. I imagine this observation would have to be modified a lot to be useful but perhaps it would be a good starting point for explaining that division by $0$ is undefined.

Also, a way I use to think of limit is to imagine what happens when you place smaller and smaller number under, say $1$. $1/(1/2)$, $1/(1/100)$, $1/(1/1000000)$ etc. I imagine that any child will know that you flip the denominator to simplify these equations so you get $1\cdot (2/1) = 2$; $1\cdot (100/1) = 100$; $1\cdot (1000000/1) = 1000000$ etc.

These two approaches combined might be used to explain one reason for limit. One thing that limit allows is to go as close as you would like to something that is not possible, ie. division by zero.

$\endgroup$
  • 2
    $\begingroup$ Dear Patrick, I think that this approach presupposes a fair amount of mathematical maturity for a second grader; cross-multplication is not something they will know at this age (unless they are quite unusually mathematically advanced); indeed, they will not be all that comfortable with fractions unless the denominators are quite small. Regards, $\endgroup$ – Matt E Nov 30 '12 at 4:02
3
$\begingroup$

Show It Visually

  • Draw a graph of 1/x.
  • Pick a point on the x axis (2, for example)
  • show that you can approach 2 from the left side and the right side
  • in both cases, you can hit x=2 exactly, and the y value is .5
  • i.e. approaching from the left or from the right, you get close to the same y value.
  • Now do the same thing for x=0.
  • Approach from the right, and you can see that the y value gets larger and larger the closer you are to zero.
  • Approaching from the left, the y values gets smaller and smaller the closer you are to zero.
  • Unlike all the other numbers on the number line, approaching from the left doesn't get you to the same value as approaching from the right.
  • So, we generally say division by zero is not allowed, because there's not one clear answer as to what that value might be.

This allows someone to get the idea without having to understand a lot of notation.

enter image description here

When she's in 3rd grade, you can teach her about the indeterminate forms 0/0 and inf/inf. :-)

$\endgroup$
  • 1
    $\begingroup$ I'm pretty sure a 2nd grader won't know what a graph is. $\endgroup$ – asmeurer Nov 30 '12 at 3:07
  • $\begingroup$ @asmeurer, most second graders don't know what division and multipication are, lol. I interpreted the question as being "how to explain to someone not sophisticated in math or math notation." Apologies to all second graders on math.se, of course! $\endgroup$ – Mark Harrison Nov 30 '12 at 7:44
  • $\begingroup$ I assumed from the question that they at least knew what division was (and thus they probably know what multiplication is as well). $\endgroup$ – asmeurer Nov 30 '12 at 15:31
2
$\begingroup$

Ok, I am assuming the 2nd grader knows the basic of division here i.e. when you divide 4 by 2.. you actually find out how many times you can add 2 to get to 4 in this case... answer is 2. But when you divide 4 by 0... you can keep on adding 0 as many times as you like but you will never reach 4. Hence its becomes infinity.....hope this helps.

$\endgroup$
2
$\begingroup$

This might be useful to you for inspiration http://www.khanacademy.org/math/arithmetic/number-properties/v/why-dividing-by-zero-is-undefined

$\endgroup$
2
$\begingroup$

Does the child understand what it means to divide by a number smaller than 1? If so, just explain it like the following $$\frac{1}{1/2} =2,$$ $$\frac{1}{1/3} =3,$$ $$\frac{1}{1/4} =4,$$ $$\vdots$$

So as you divide by numbers that get smaller and smaller, you get numbers that get bigger and bigger. In fact, if you divide by $1/n$, you just get $n$, so it's hopefully easy to see that this can be as big as you want. As the numbers in the denominator go to 0 (also something that hopefully won't be too hard to see), their reciprocal goes to $\infty$.

I've always found that the concept of $\infty$ is best understood as "as big as you want" (or "arbitrarily large" or "unbounded", to use more mathematical terms).

$\endgroup$
  • $\begingroup$ Dear asmeurer, This is the natural way to explain it, but unless the second grader is rather advanced, they may not be very comfortable with fractions such as $1/n$ when $n$ is large, and so $1/ (1/n)$ will be even trickier for them to compute with. (My experience is that what we consider trivial arithmetic is still in the non-trivial state for kids of this age. But for some reason, they love both infinity and zero, and seem to begin talking about them at a very early age!) Regards, $\endgroup$ – Matt E Nov 30 '12 at 3:59
  • $\begingroup$ I seem to remember being able to understand such things, but then again, I was much better at math than my peers in 2nd grade. $\endgroup$ – asmeurer Nov 30 '12 at 5:05
1
$\begingroup$

The notion of division of numbers is pretty well understood, at least intuitively, from early in our mathematical educations. To say that the number $a$ divides the number $b$ means there is a number $q$ such that $b=qa$. Of course, if $b \neq 0$ and $a=0$, then no such $q$ exists, and in this sense, division by zero is undefined. However, if both $a$ and $b$ are zero, we have a different problem, as $b=qa$ is true for any number $q$. I suppose we could say that in this case, division by $0$ is over-defined. Since we usually want a quotient to be unique, if it exists, we usually say simply that division by zero is undefined. However, the $0/0$ case comes back to haunt us as an indeterminate form in limits, and we see the vestiges of our over-defined case of division by $0$. It seems to befuddle students somewhat that a limit that has the form $0/0$ can have any real number value, provided that the limit does, indeed, exist.

This discussion can be pursued in any algebraic system in which we have a notion of division.

$\endgroup$
1
$\begingroup$

An Even Simpler Explanation: Why We Don't Allow Division by Zero

This will make sense to somebody (e.g. the hypothetical second grader) who understands division (fractions) but nothing more sophisticated.

  • Consider a fraction 0/x, where zero is divided by any number. The answer is always zero.

  • Consider a fraction x/x, where a number is divided by itself. The answer is always one.

  • Now consider 0/0. Is t answer zero by the first rule, or one by the second rule?

  • Since this causes us to have two conflicting answers, we disallow division by zero.

  • When you're older and have a bit more math, you'll understand one of the other explanations given here.

$\endgroup$
0
$\begingroup$

If the second-grader can understand a little algebra, then I think this would do the trick:

For any $x,y,a$

$x=a/y$,

$x*y=a$

right?

Now you can show that if given two nonzero numbers, you can find the missing number.

But what if $y=0$?

Then it would be:

$x=a/0$

$x*0=a$

But any number multiplied by zero is zero right? Yet $a$ is nonzero. So $x$ is not a number!

In fact, $x$ keeps running away (it keeps increasing). This way, zero can never "catch" $x$, and turn $a$ to zero.

Matematically, this makes little sense. But this is how I would explain this to a second-grader.

$\endgroup$
0
$\begingroup$

I say that division by $0$ is defined, that it can yield multiple results in different contexts. People need to stop treating $0$ as a number or an integer for it is not! It is a place holder, the empty or null set. First lets start with what the operation of division or a fraction is! It is the inverse or opposite operation of multiplication; simple enough!. What is multiplication? It is the repetitive operation of addition! What is addition? Addition is a 1 dimensional linear transformation specifically translation!

Consider this: we have an arbitrary point $p(n)$ where n is any value. A point itself is similar to but different than a vector. A point itself can have an infinite amount of dimensional coordinates but the point itself is $0$ dimensional where it has no association or awareness of any other point, direction (rotation), or dimensions. Once you introduce a second point and you associate the two, then you have a line segment that has distance, displacement or length. The addition or subtraction which is horizontal linear translation of two points that creates that line segment transforms those two points into a single 1 dimensional vector. A vector does not become $2D$ until it is introduce to another vector that has a separation by a angle of some degree. So at minimum you need 3 points at different locations to create at minimum two different vectors to where one of those points becomes a vertex or intersection between two line segments, lines or vectors.

So what does points, vectors, lines, rotations and different dimensional spaces have to do with division by $0$ and whether it is undefined or defined? Everything!

To explain this we need to understand some geometry, some algebra and some trigonometry and the relationships of vectors and angles.

Things need to know:

  • Slope of a line: $$m = \frac{rise}{run} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\sin\theta}{\cos\theta} = \tan\theta$$
  • If a slope has the form $\frac{n}{d}$ then its perpendicular or the line orthogonal to it is its negative reciprocal as in $-\frac{d}{n}$
  • If $\tan\theta$ represents the slope of the line according to $\theta$ where $\theta$ is the angle above the horizontal in standard position then we can say that the perpendicular must be: $$-\frac{\cos\theta}{\sin\theta} = -\cot\theta$$
  • Both the $\sin$ and $\cos$ have a domain of $\mathbb R$ and a range of $[-1,1]$, they are circular functions or are continuous and repeat forever with a period of $2\pi$ making them periodic wave functions.
  • Pythagorean Theorem & Equation of Circle Are The Same $$A^2 + B^2 = C^2$$ $$X^2 + Y^2 = r^2$$
  • The relationship between the cosine of the angle between two vectors: $$\cos\theta = \frac{\vec a \cdot \vec b}{|\vec a||\vec b|}$$

When we think of slope and its definition that we are referring to it is the rate of the increase of height or elevation over distance. So when the ground that we walk on is perfectly flat and we have $0$ elevation we have $0$. To demonstrate this use the Unit Circle and the coordinate pairs on the circle $(x,y) = (\cos\theta, \sin\theta)$ and the vector or line has a length of $1$. So when their is $0$ rise the $\sin\theta$ is $0$ so the equation of the slope of the line or the tangent is as follows: $\frac{0}{\cos\theta} = 0$. This simply means we have no elevation but we have any amount of horizontal translation.

Now when we rotate the line $CCW$ around the unit circle and we take a pit stop at $45°$ or $\frac{pi}{4}$ we end up with a slope of $1$.

This takes the form of: $$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{ \frac{\sqrt{2}}{2} } { \frac{\sqrt{2}}{2} } = 1$$

So here we are going up 1 in height for every one in length. So far so go.

Let's continue to rotate this vector in the same direction and make a pit stop at $90°$ or $\frac{pi}{2}$.

In conventional math teachings and what many of us have been taught was oh-no we have a problem; vertical slope is undefined due to division by $0$.

I say we don't have a problem and that it is not undefined. Why?

Well it was okay to say we had $0$ rise with infinite horizontal translation resulting in $0$ slope. And now that we are at its perpendicular that slope isn't defined?

In the first case we had $$\frac{\sin\theta}{\cos\theta} = \frac{0}{\cos\theta} = 0$$ but now that we have the opposite where we have no run but we do have rise we can not seem to defined this? $$\frac{\sin\theta}{\cos\theta} = \frac{\sin\theta}{0} =$$ $$? ... undefined$$

This is vertical slope and because of these assessments it also causes us to treat the tangent and cotangent to be undefined at specific values because of division by $0$.

However by definition of words and thought and sound intuition we know that our height is always increasing, but we are no longer moving horizontally. This is infinite slope or vertical slope. This is well defined if you ask me. Let's remove the $0$ for a moment and consider the following with some arbitrary variables.

If the slope = $\tan = \frac{\sin}{\cos}$ then its negative reciprocal must be $-\frac{\cos}{\sin} = -\cot$

So if the slope of a horizontal line is $0$ in the form of $\frac{0}{\cos}$ then to find the negative reciprocal to get its perpendicular we have to say that: Instead of using $0$ directly I will use the angle of the trig function that evaluates to $0$ since it has a valid range and domain to produce this value of $0$.

$$-1 = \frac{n}{d} * -\frac{d}{n}$$

Let $$\frac{n}{d} = \frac{\sin\left(0°\right)}{\cos}$$ and $$-\frac{d}{n} = -\frac{\cos}{\sin\left(0°\right)}$$

then

$$-1 = \frac{\sin\left(0°\right)}{\cos} * -\frac{\cos}{\sin\left(0°\right)}$$

Initially this would appear to produce $-1 = 0 * 0$ if you multiplied the fractions out which wouldn't make any sense. However when you look closely the two $\sin$ functions without any known input values intuitively cancel each other out leaving you with $-1 = -\frac{\cos\theta}{\cos\theta}$. Let's prove this by looking at the dot product of two vectors and applying trig substitutions. Let's consider this: $p1 = (0,0), p2 = (1,0), p3 = (0,1)$ which are the three points on the unit circle that would make a right triangle with the axis being the perpendicular. We need two vectors $\vec A \vec B$ and in this case we need the vectors to be $\vec A = p2 - p1$ and $\vec B = p3 - p1$ and this will give us the vectors as: $\vec A = (1,0)$ and $\vec B = (0,1)$.

If we take the dot product of these two vectors and divide them by their magnitudes it should look like this:

$$\cos\theta = \frac{\vec A(1,0) \cdot \vec B(0,1)}{|\vec A||\vec B|}$$

and this simplifies to: $$\cos\theta = \frac{0}{1} \implies \arccos(0) = 90°$$

Now that we have a relationship with $\arccos$ and we know that the these vectors do in fact intersect each other at a $90°$ angle making them perpendicular and orthogonal to each other, they are also both a normal to each other since their magnitudes or the length of these vectors both have an absolute value of $1$.

We know through trigonometry that there are relationships between the trig functions and since slope is defined as the $\tan$ of the angle $\theta$ above the horizontal in standard angle form where the vertical vector or leg that creates the right angle on the right side of the triangle with $\theta$ being the vertex on the left when the trig functions are used within the context of the unit circle with the unit circle being defined by the equation $X^2 + Y^2 = C^2 = X^2 + Y^2 = 1$. Since the radius vector of the unit circle has a magnitude of length $1$ we can apply some of these relationships between the $\sin$ and the $\cos$ to convert from one to the other based on the fact that the natural graphs of the two are a $\pm90°$ angle or a $\pm\frac{\pi}{2}$ radians horizontal translations of each other. We can use this to substitute back into the equation where we found the negative reciprocal of the horizontal normal vector with a magnitude of $1$.

NOTE: Before I do the substitution I will make the statement here that the sign of the vectors in this case only implies opposite direction or reflection of symmetry or the rotation of either $180°$ degrees or $\pi$ radians about a fixed point and in this case the fixed point is $(0,0)$. The reason I state this is because $\vec A$ that starts at the point $(0,0)$ and points to the point $(1,0)$ has the positive direction stemming out from $(0,0)$ while another vector $-\vec A$ starts at the point $(0,0)$ and points to the point $(-1,0)$ again stemming out from $(0,0)$. This relationship of vectors $\vec A$ and $-\vec A$ are the same exact vector when vector $\vec A$ is rotated $180°$ or $\pi$ radians the point at the end of the initial vector before the rotation located at $(1,0)$ is the same exact point that ends up at $(-1,0)$ along with all the points of the line or vector in between the fixed point $(0,0)$ and vector $\vec A$. This is evident by two supporting functions, the first was already used once but with different values and that is the relationship of $\cos\theta$ and the dot product between vectors $\vec A$ and $-\vec A$ which will yield $-1$; also if you take the $\arccos(-1)$ it will evaluate to $180°$ of rotation which is also the angle measure of any and all $1D$ Cartesian Linear Lines of the form $y = mx + b$. The second supporting function is the definition of the circumference of a circle $c = 2\pi r$ and since $r = 1$ this simplifies to $2\pi$ which also happens to be the period of the $sine$ and $cosine$ functions. Now that we have $2\pi$ which measures the arc length around the circle when the radius $r = 1$ when we divide that in half it gives us $\pi$ radians or $180°$.

Let's now apply the substitution of the $\sin$ and $\cos$ functions in order to remove the component of the fractional part of the $\tan$ that represents the axis directional component of vertical slope.

Given that when $-1$ is applied to the $\arccos()$ function and returns a value of $180°$ and that the first or left fraction represents the slope of a line whose slope is $0$ and that the second or the right negative fraction represents the slope of the line that is perpendicular to the first shown again here: $$-1 = \frac{\sin\left(0°\right)}{\cos} * -\frac{cos}{\sin\left(0°\right)}$$ and knowing the trigonometric Cofunction identities with $$\sin\left(\frac{\pi}{2} - \theta \right) = \cos\theta$$ and $$\cos\left(\frac{\pi}{2} - \theta \right) = \sin\theta$$ we can use these to make that substitution and since $\sin(0) = 0$ and it is found in the negative reciprocal we need to substitute all $\sin()$ functions with $\cos()$ functions

$$-1 = \frac{\cos(\frac{\pi}{2} - \theta)}{\cos\theta} * \frac{\cos\theta}{\cos(\frac{\pi}{2} - \theta)}$$

If we partially ignore the negative sign attached to the $-1$ on the left hand side of the equation since it only implies the facing direction of the vector as in right versus left we can see that $$|-1| = \frac{\cos(\frac{\pi}{2} - \theta)}{\cos\theta} * \frac{\cos\theta}{\cos(\frac{\pi}{2} - \theta)}$$ still holds true.

Conclusion To sum this up when a slope is $0$ it is the numerator or the rise component that is $0$ which means it has no slope for there is no rise in height and this is okay. When the slope is vertical slope we are normally taught that it is undefined because of division by $0$. However without using values and just using real life concepts since mathematics is supposed to model real life concepts when we do have vertical slope I say that it is very much defined. It is the opposite of $0$ slope from the rise being $0$ in the numerator and when slope is vertical we now have $0$ run or horizontal displacement but we are ever increasing in the up and down direction does imply that the slope is not $0$ but the slope is infinite or the $\lim$ approaches $\infty$ just like the horizontal component of $0$ slope approaches $\infty$. Imagine walking down an alley between two skyscrapers and that the alley represents the relationship of different translations between different inclines of slope. At first the ground is level so there is no slope then the alley has a hill and the slope is $1$ this means perfect rise over run as the increase in elevation is equivalent to the change in horizontal translation where the angle above the horizon is $45°$ and $\tan(45°) = 1$. Then the ground levels out again with $0$ slope and is flat and horizontal. Then all of a sudden you can go no farther because there is another skyscraper in front of you, there is no turning left or right, and there is no going back. The outside wall of the building has a ladder and you begin to climb up. Now you have $0$ horizontal translation and continuous vertical translation for each rung of the ladder you climb. There for vertical slope is defined as it is always 100% straight up, orthogonal, perpendicular and normal to $0$ or horizontal slope.

One thing to be aware of is if you have the unit circle with 2 fixed points one at $(0,0)$ and one at $(1,0)$ and your third point can move freely along the circumference of the unit circle we can make 3 vectors to form a triangle. We do this in the first quadrant so that the initial right triangle is in standard position. Two of the vectors will always equal 1 and initially the first vector will be composed of the first 2 fixed points. The 2nd vector will virtually start off as the same exact vector of the fixed vector but it will rotate about the origin around the unit circle. The 3rd vector initially doesn't exist or is the $\vec 0$ since this has a tail at $(1,0)$ that will remain fixed but will have its head tracing around the unit circle with the rotating vector radius. This vector will change its length at each interval of rotation. Now imagine the right triangle in standard position with a $45°$ angle and the rotating radius with a slope of 1 this will end up being the hypotenuse. This radius vector will remain the hypotenuse until it coincides with the $Y-axis$. Once it does and the angle becomes $90°$ and the once standard position of the right angle no longer exists on the right side of the triangle as the intersection of the axis now becomes your new right angle for only this measure of degree and when $\theta$ also equals $270°$. The radius vector that was the hypotenuse no longer is and the variable length vector now becomes the new hypotenuse with a length of $\sqrt 2$. When the two radius vectors are in this position or when the rotating vector is at $(0,-1)$ both of these right triangles yields the maximum area of any triangle that can be made by the unit circle while having two legs with a magnitude of its radius. Another peculiar thing to notice here is when we have this reversed right triangle such that the angle of rotation is at $90°$ degrees and we take the origin point $(0,0)$ and reflect it over this
new hypotenuse it will give us the point $(1,1)$ which is outside of the unit circle. This creates a perfect unit square with an area of $1$, this also shows the relationship of the Pythagorean Theorem that pertains to triangles and how it relates to the equation of the Unit Circle, and this is also one of its proofs.

So in this context of linear transformations, rotations, slope etc., division by $0$ is defined as it represents vertical or infinite slope. Now in other contexts division by $0$ may represent other things such as $0,1,\infty,D.N.E.(Does Not Exist)$ We have to remember that $0$ is not really a number nor an integer. It is a place holder and has no value as it represents the empty or null set, an approach to infinity or even infinity itself. $0$ in its purest form is only a character, digit or glyph.

How to explain this to a 2nd grader? Hmm, good luck as you seen what I had to do to prove all of the massed printed text books wrong!

Edit -Additional things to consider.

Since we already seen that $\tan(45°)$ or $\frac{\sin(45°)}{\cos(45°)} = 1$ let's consider the very first expression or equation that we are taught $1 + 1 = 2$. Since all integer values can be represented as fractions we can do this as well: Ant that is to substitute values of $1$ with the appropriate trig functions with the appropriate angles and this would give us...

$$1 + 1 = 2 \implies \tan(45°) + \tan(45°) = 2 * \tan(45°)$$ which can also mean: $$1 + 1 = 2 \implies \frac{\sin(45°)}{\cos(45°)} + \frac{\sin(45°)}{\cos(45°)} = 2 * \frac{\sin(45°)}{\cos(45°)}$$

This holds true because the definition of slope is $\frac{\delta y}{\delta x}$ which is a fraction. So one can see that all numbers are in fact a form of slope including the approach to $\pm\infty$. So I tend to believe in this particular context that slope in the form of $\frac{\sin\theta}{\cos\theta}$ where the $\sin$ does not evaluate to $0$ and the $\cos$ does, the yielding result would be $\pm\infty$.

Added Note - Also some might say well it yields both $\pm\infty$ so it still has more than one answer so it still has to be undefined. This is a wrong assumption or assessment. It depends on the direction of the angle or rotation. If you angle $\theta$ is either $90°$ or $-270°$ then the answer is obviously $+\infty$ or vertical slope that is straight up and if your angle $\theta$ is either $-90°$ or $270°$ then it is $-\infty$ or vertical slope and it is straight down, and don't forget to count they periodicity of higher valued angles which would be $\theta + 180° * n$ where $n$ is some integer $>=1$ or $0$. This also works as a function because even though there are two inputs that give you a single output it is the nature of the $\sin()$ and $\cos()$ functions that gives this behavior of the slope of the line. This duality is due to the equation of a circle $x^2 + y^2 = r^2$ or the Pythagorean Theorem $A^2 + B^2 = C^2$.

$\endgroup$

protected by robjohn Nov 22 '12 at 15:00

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.