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To help illustrate my question I modify $tan_0$ so it returns a number divided by zero at $\frac \pi 2$ radians. I also reviewed another post, however still have questions. There is an example and then questions follow. The example uses $tan$ only to provide the idea of line direction.

Find the slope for $\frac \pi 2$ radians:

$tan_0(\theta) = \begin{cases} \frac 1 0,& \text{if } \theta = \frac \pi 2\\ \frac 1 0,& \text{if } \theta = -\frac \pi 2\\ \tan(\theta),& \text{otherwise}\\ \end{cases}$

For a linear equation use a ratio:

$tan_0(\frac \pi 2) = \frac y x$, $f(x) = tan_0(\frac \pi 2) x = \frac {opp} {adj} x = \frac 1 0 x$

where $x = 0$ then $f(0) = \frac 1 0 0 = \frac 1 1 \frac 1 0 \frac 0 1 = \frac 0 0$

Then $f(0) = \frac a b = \frac 0 0 = b q = 0 q$, where $q$ is any number. This seems to agree with the line at $\frac \pi 2$ being non functional and consiting of all the infinite points on the $y$ axis. It would seem that $f(0) =$ "all values" and where $p \gt 0 \implies f(p) =$ "undefined". The later being expected. In some posts folks have called the $\frac 0 0$ case as being "indeterminate". The "indeterminate" language makes sense as there are an infinite number of solutions, however to remain functional only a single value could be assigned, yet there are multiple values. Therefore indeterminate.

Questions

  1. Does the math here make sense? Appreciate guidance.
  2. Does the indeterminate language fit this observation?

Side note: George Boole uses division by $\frac 0 0$ and $\frac 1 0$ in his algebraic logic book "Laws of Thought". $\frac 0 0$ is an indeterminate.

Nick
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    One of the great things about dividing by zero is that you can make it equal to any number you prefer (popular with "proofs" of 2=1, 3=1 etc). For a simple example calculate $ka/a$ as $a$ tends to zero. – Joffan Feb 03 '21 at 00:57
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    I thought the solution $\frac a b = \frac a 0$ is the set of all numbers $q$, where $bq = 0q = a$. I.e. not a single number – Nick Feb 03 '21 at 01:22
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    To be honest it makes my head hurt just reading expressions with a zero denominator. – Joffan Feb 03 '21 at 01:25
  • For $k \frac a a$, $a \rightarrow 0$. Is this $1$ for $a \gt 0$ and $\frac 0 0$ for $a = 0$? What should I be observing here? – Nick Feb 03 '21 at 01:27
  • Something to note: In modern language, “indeterminate” means something different from “undefined.” $\frac 10$ is undefined. $\frac00$ is undefined and indeterminate. “Indeterminate” has to do with limits. – Thomas Andrews Feb 03 '21 at 01:43
  • @ThomasAndrews I read Wiki https://en.wikipedia.org/wiki/Indeterminate_form. Your definitions agree. The page has $\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac {a x} x = a = L$. This example corresponds with my post, but suggests $a$ is the limit vs all values. I'm confused because $|f(x) - L| \not \lt \epsilon$ where $\epsilon$ is finite and $f(0) = \frac 0 0 = $ "all values". Perhaps I'm interpreting something incorrectly. – Nick Feb 03 '21 at 02:19
  • I think I understand now, $\lim_{x \to 0} \frac {a x} x = 1$ and this limit $1 \ne \frac {a x} x, 1 \ne \frac 0 0$. Therefore like you say, this is indeterminate based on this. My assumption is $\frac 0 0$ is undefined because it is all values and it is not functional. – Nick Feb 03 '21 at 02:44
  • To be pedantic with terminology, all usages of the word 'indeterminate' I have seen in this context refer to limits only. $\frac00$ by itself is not indeterminate, only undefined. $\lim_{x\to0}\frac xx$ is a limit written in an indeterminate form. – Shuri2060 Aug 03 '23 at 21:40

1 Answers1

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Yes. Zero divided by zero is indeterminate as any number of solutions are valid.

Let us assume that there is only one solution to $0\over0$. Let's call it $x$. $$x={0\over0}$$ The first step would be to get rid of the fraction, by multiplying both sides by zero $$0(x)=0$$ $$0=0$$ As such, the equation $x={0\over0}$ is an identity, valid for all real (and complex) numbers, and the value of $0\over0$ is indeterminate and undefined.

A similar proof can be made for $a\over0$ where $a\neq0$. $$x={a\over0}$$ $$0(x)=a$$ $$0=a$$ However, we already established that $a\neq0$, so this is a contradiction and has no solution.

In both cases, any number divided by zero is undefined. That simply means it doesn't have a single valid solution. However, $0\over0$ is also indeterminate as it could be equal to any number, algebraically at least. Any other number divided by zero does not have this problem, though, as no solution would work for it, so it is only undefined.

Jme
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    Zero divided by zero is undefined. Calling it indeterminate is misleading, as it gives the expression a legitimacy that it has no right to. – TonyK Aug 03 '23 at 19:25
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    @TonyK Technically $0/0$ is both undefined and indeterminate – Jme Aug 03 '23 at 20:05
  • I've only seen 'indeterminate' used to refer to limits of certain forms after substituting the point in. – Shuri2060 Aug 03 '23 at 21:36
  • Technically $0/0$ is simply undefined. Which, technically, means that ascribing any other properties to it, such as being indeterminate, is (technically) meaningless. – TonyK Aug 03 '23 at 21:49
  • @TonyK There is definitely merit to that argument, but for the most part $0\over0$ is considered to be indeterminate (along with being undefined). See https://mathworld.wolfram.com/Indeterminate.html – Jme Aug 03 '23 at 23:25