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I've got a student in a senior year maths class, who is adamant that it is incorrect to depict it as a hole, even though this is the standard way of depicting it.

I couldn't find any problems with his reasoning. So I've asked him to put his reasoning down in a paper, so someone who knows more than me about this topic can get to the bottom of it. Link below:

http://www.slideshare.net/MrIndererminate/indeterminate-is-not-synonymous-with-undefined

H Borrow
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    The slides end with a conclusion that is nonsense, so I would rather stop worrying about this "adamant senior year student". – Pedro Apr 14 '17 at 05:13
  • I wish it were that easy. He's pulling other students aside one by one in lunch hours and convincing them of his assertions. I'm going to have the whole class chanting "vertical line" in classes soon if I don't put this to bed with solid reasoning. – H Borrow Apr 14 '17 at 23:10

3 Answers3

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Note that just because all numbers "are equally valid" for the answer of $\frac00$ does not many any of them valid. In fact, an implication of his argument is that all answers are equally invalid.

Additionally, we define operations as functions (this will be important in a minute). Give your student an example from number theory:

Let the successor function be defined as thus:

$$s(n)=n+1$$

And so we define addition $n+k$ as $n$ plugged into the successor function $k$ times.

But there is a definition of a function that becomes very important here. A function, for each unique input, must have a single output (this output can be undefined).

So we can define:

$$\frac xy=d(x,y)= \begin{cases} z\text{ s.t. } zy=x, & \text{if } x,y \neq 0\\ \text{undefined}, & \text{if } x,y = 0\\ \end{cases}$$

Because otherwise, without that second case, we can't define division as a function, and we can't define it as an operation, and I think you can see the algebraic implications of that.

Another example is the square root function - $\sqrt{x}\geq0$ for all $x>0$, but doesn't it make sense that the square root should give us two answers, one for the positive, and one for the negative? No. Because we want operations to be functions. And so, we define the output to be the principal square root, the one that, if you plot it on the complex plane with polar coordinates, has the smallest $\theta$ value. When dealing with real numbers, this is always the positive number.

And that, in a nutshell, is why $\frac00$ is undefined. Tell your student that following this path of reasoning - i.e. holding to a point without understanding its implications - will make it impossible for him to get anywhere in the field of math - abstract algebra, real analysis, they'd all fall apart with his defintion. We wouldn't even have the rational numbers well defined. Mathematicians aren't idiots - we have our reasons.

In one sense, I'm being arrogant by calling myself a mathematician. But in another sense, anyone who pursues math voluntarily is a mathematician, and that questioning is exactly what makes math awesome.

But he does probably need to stop spouting off "revolutionary" theories about basic math facts.

Robin Aldabanx
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  • My initial counter-reasoning was right in line with yours!! I told him that functions need to produce single unique answers. In response he just grabbed a text book and flicked to a chapter on circle equations (i.e (x – h)^2 + (y – k)^2 = r^2). His counter to my counter was that just because it is not a function doesn't mean it it can't be graphed. "If imputing an x value can produce two y values, why not infinite y values?" he asked. how do I rebut that? – H Borrow Apr 14 '17 at 06:18
  • A circle is a geometric shape, not an operation or a function. The equation to describe it is not a function, and, $y$ is not a dependent of $x$ in that case. An operation, however, is an algebraic function (or manipulation, however you want to put it) that can only produce one answer. This is necessary for the construction of the rational numbers from the integers, and, therefore, the construction of the reals by the Axiom of Completeness and the complex numbers. – Robin Aldabanx Apr 14 '17 at 06:23
  • Additionally, tell him that graphing a function is the most basic and possibly the most useless ways to deal with a function. In fact, tell him that until he understands abstract algebra and the ramifications of making $\mathbb{Q}$ no longer well-ordered and a group under division (because this would destroy the idea of an identity element in $\mathbb{Q}$), he should accept the ideas given by those who have a broader scope than he does. What class is this, Precalculus, Calculus, or a college class? – Robin Aldabanx Apr 14 '17 at 06:25
  • For a real world implication, one can also consider computers. Computer can't really handle true real numbers, since they have finite memory, however IEEE floating-point numbers used in most computers were designed to specifically be able to encode 0. If a CPU were give two zeroes and asked to compute their division, what shoud it do ? 1/ Report an error because the value is undefined 2/ Give all possible results, i.e. every possible IEEE floating point value ? No, a processor computes a mathematical function and it's expected it gives at most one answer. – Lærne Apr 14 '17 at 12:21
  • Hi Robin, this is year 12 advanced mathematics class and unfortunately this has been going on for weeks.... You say that a " A circle is a geometric shape, not an operation or a function." but there are circle equations, which aren't functions (they don't pass the vertical line test) but still they can be plotted.... – H Borrow Apr 14 '17 at 22:58
  • A circle is a specific set of points in Euclidean Space. Something doesn't have to be a function to be plotted. Being plotted is irrelevant. The real question here is how we define division. And, for reasons that a kid in year 12 mathematics could not hope to understand, we have to have division as a function. If he cannot accept this, he cannot – Robin Aldabanx Apr 15 '17 at 01:47
  • hope to move forward in a world where you aren't right just because what you say makes sense to you. – Robin Aldabanx Apr 15 '17 at 02:27
  • I guess if you want to continue this discussion, my gmail is [email protected]. Maybe I could gmail him - I mean, he's not stupid, he believes what he is doing is right, but I don't know. To be honest, he almost had me half-convinced there until the end. – Robin Aldabanx Apr 15 '17 at 02:28
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There's nothing special about $\frac{0}{0}$ compared to $\frac{1}{0}$ or any other "divided by zero" number in this case. Division by zero is simply undefined: you can't do it. The reason you can't divide by zero is basically what the student explains, and is to do with the fact you can't get a unique result. But this doesn't mean any answer will do, this means that dividing by zero is an illegal operation. I can't make sense of $\frac{1}{0}$ any more than I can make sense of $\frac{1}{\text{cats and dogs}}$.

This means the function $f(x) = \frac{x^2 - 1}{x - 1}$ is not defined for all values of $x$, it's actually undefined when $x = 1$, since the equation does not make sense. A more explicit way of writing it would be $$f(x) = \begin{cases} \frac{x^2 - 1}{x-1} & \text{when } x \neq 1 \end{cases}$$ $f$ does not specify what to do when $x = 1$, and this is why we plot it with a "hole".

This is the same as, for example, the equation of a semicircle $g(x) = \sqrt{1 - x^2}$: the formula only makes sense for $-1 \leq x \leq 1$. When you plot the function $g(x)$, you plot only the semicircle, with no extra horizontal or vertical lines. Questions such as "what is the value of $g(x)$ at $x = 2$?" don't make sense, since $g(x)$ does not give a rule that works for that value of $x$.

The "removable" singularity is a red herring. You can "patch" the function $f(x) = \frac{x^2 - 1}{x - 1}$ up by specifying what to do when $x = 1$, and clearly the reasonable thing to do here is to patch the hole with the value $2$. But this is now a different function:

$$h(x) = \begin{cases} \frac{x^2 - 1}{x-1} & \text{when } x \neq 1 \\ 2 & \text{when } x = 1\end{cases}$$

You may observe that $h(x)$ is also given by the simpler formula $h(x) = x + 1$. But $h$ is not the same function as $f$, since $h$ knows what to do when given $x = 1$, while $f$ does not.

Joppy
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Imagine we want to calculate $\dfrac{0}{0}$.

From the sequence below, we can conclude that $\dfrac{0}{0} = 0 $

$\dfrac{0}{0.1} = 0, \dfrac{0}{0.01} = 0, \dfrac{0}{0.001} = 0, \dfrac{0}{0.0001} = 0,$

But from the sequence below, we can conclude that $\dfrac{0}{0}=1$

$\dfrac{0.1}{0.1} = 1, \dfrac{0.01}{0.01} = 1, \dfrac{0.001}{0.001} = 1, \dfrac{0.0001}{0.0001} = 1$

So it is not possible to understand what is $\dfrac{0}{0}$. even though the limit of the function $f(x) = \dfrac{x}{x}$ at $x\to 0$ can be calculated, but we cannot say anything about $\dfrac{0}{0}$, which means it's undefined.

Soroush khoubyarian
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