My Math Forum Why isn't dividing by zero = zero?

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April 3rd, 2019, 06:46 AM   #21
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Quote:
 Originally Posted by v8archie Perhaps you should pay more attention to people's posts then, in particular the OP, so that you can get an insight into how they are thinking.
I read the OP. The only right answer is that limits are irrelevant to this issue.

Limits to $0^0$ are also indeterminate. But $0^0=1$ is still true and widely used.

April 3rd, 2019, 08:09 AM   #22
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Quote:
 Originally Posted by Micrm@ss But $0^0=1$ is still true and widely used.
Someone may adopt a convention that 0^0 = 1 but it is not at all correct too say that 0^0 = 1.

-Dan

 April 3rd, 2019, 08:24 AM #23 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,981 Thanks: 1164 Math Focus: Elementary mathematics and beyond Well, what if the concept of limits were introduced? Using L'Hospital's rule it's easily seen that $\displaystyle \lim_{x\to0}x^x=1$ I think the same holds for division by 0. If the numerator of a fraction that has a zero denominator is any limit other than 0 then the limit does not exist.
April 3rd, 2019, 08:51 AM   #24
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Quote:
 Originally Posted by Micrm@ss I read the OP. The only right answer is that limits are irrelevant to this issue.
I concede that I'm not exactly rigorous or specific, but, to be clear, what you're saying is that evaluations of the limits of certain functions at x=0, e.g.

$\displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$

is irrelevant to discussions about division by zero?

Note: it's routine in physics to define

$\displaystyle \text{sinc} \; x = \frac{\sin x}{x}$

and then $\displaystyle \text{sinc} \; (0)$ evaluates to 1 and $\displaystyle x \in \mathbb{R}$. The value of the function at x=0 is defined to be the limiting value.

Last edited by Benit13; April 3rd, 2019 at 09:11 AM. Reason: Note about the limiting value and grammar errors.

April 3rd, 2019, 05:58 PM   #25
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Quote:
 Originally Posted by Benit13 I concede that I'm not exactly rigorous or specific, but, to be clear, what you're saying is that evaluations of the limits of certain functions at x=0, e.g. $\displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$ is irrelevant to discussions about division by zero? Note: it's routine in physics to define $\displaystyle \text{sinc} \; x = \frac{\sin x}{x}$ and then $\displaystyle \text{sinc} \; (0)$ evaluates to 1 and $\displaystyle x \in \mathbb{R}$. The value of the function at x=0 is defined to be the limiting value.
This is fine and its done all the time for many reasons. This is called a continuous extension or "extending by continuity". However, the point is that nobody would claim that this is somehow defining a value for $\frac{0}{0}$. It has absolutely nothing to do with division by zero. Its just completely unrelated in any way whatsoever which is what micromass is trying to point out.

Moreover, in the typical contexts where one does allow division by $0$ such as, to make many theorems in measure theory easier to state and prove, one defines the extended reals to be $\mathbb{R} \cup \{\infty\}$ and declares that if $r \neq 0$, then $\frac{r}{0} = \infty$. In this instance, we aren't typically worried about losing the multiplicative group structure on our scalars. However, the expression $\frac{0}{0}$ is STILL undefined in this context. I don't know of any place in mathematics where $\frac{0}{0}$ has any meaning whatsoever. If there is such an application, it certainly doesn't have anything to do with computing limits of fractions or the $\operatorname{sinc}$ function.

April 3rd, 2019, 11:44 PM   #26
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Quote:
 Originally Posted by topsquark Someone may adopt a convention that 0^0 = 1 but it is not at all correct too say that 0^0 = 1. -Dan
It's not a convention. At least, it's not more of a convention than what $0\cdot 2$ is. It can be rigorously proven just like $0\cdot 2$.

Indeed, in set theory, we make the following definition of cardinal numbers (which include the natural numbers):
$n\cdot m$ is the cardinality of $n\times m$, the cartesian product.
$n^m$ is the cardinality of the set of functions from $m$ to $n$.

If you adopt these definitions, as is standard, it can be proven quite easily that $0^0 = 1$, at least in the system of natural numbers.

In much of the same way, a proof in category theory holds for exponential objects. So $0^0 = 1$ is more than a convention but is rather a fact that can be rigorously proven.

April 3rd, 2019, 11:47 PM   #27
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Quote:
 Originally Posted by SDK This is fine and its done all the time for many reasons. This is called a continuous extension or "extending by continuity". However, the point is that nobody would claim that this is somehow defining a value for $\frac{0}{0}$. It has absolutely nothing to do with division by zero. Its just completely unrelated in any way whatsoever which is what micromass is trying to point out. Moreover, in the typical contexts where one does allow division by $0$ such as, to make many theorems in measure theory easier to state and prove, one defines the extended reals to be $\mathbb{R} \cup \{\infty\}$ and declares that if $r \neq 0$, then $\frac{r}{0} = \infty$. In this instance, we aren't typically worried about losing the multiplicative group structure on our scalars. However, the expression $\frac{0}{0}$ is STILL undefined in this context. I don't know of any place in mathematics where $\frac{0}{0}$ has any meaning whatsoever. If there is such an application, it certainly doesn't have anything to do with computing limits of fractions or the $\operatorname{sinc}$ function.
A structure where $0/0$ is always defined is called a wheel. https://en.wikipedia.org/wiki/Wheel_theory So it kind of does exist, but I've never seen this theory used in practice. So it's very likely a completely useless theory.

 April 3rd, 2019, 11:51 PM #28 Senior Member   Joined: Oct 2009 Posts: 912 Thanks: 354 As for $0^0 = 1$. This can be seen to hold in any bicartesian closed category: https://en.wikipedia.org/wiki/Cartesian_closed_category A nice example of this, besides natural numbers with a certain class of morphisms, is propositional logic, where $x^y$ is the same as $y\Rightarrow x$. The fact that $0^0 = 1$ is then the same as $F~\Rightarrow~F$ being True!!
April 4th, 2019, 03:17 AM   #29
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Thanks for the clarification. I agree with what you say, but I'm still puzzled with this part:

Quote:
 Originally Posted by SDK ... the point is that nobody would claim that this is somehow defining a value for $\frac{0}{0}$. It has absolutely nothing to do with division by zero. Its just completely unrelated in any way whatsoever which is what micromass is trying to point out.
I'm not claiming that my sinc x example defines a value for $\displaystyle \frac{0}{0}$, but let's say that somebody new to mathematics tries to evaluate the function $\displaystyle f(x) = sin(x)/x$ at $\displaystyle x=0$. They substitute the value into the numerator and the denominator and obtain $\displaystyle \frac{0}{0}$. Now what?

Someone with no knowledge of mathematics might make an error, such as "Oh, I have 0/0 = 0, so the answer is 0" and then get confused when they look at a plot of the function and see a value of 1 at x=0.

However, someone with some knowledge of mathematics can remember that 0/0 is undefined and say "okay, it is undefined, but let's see what the limit is", then make some progress with their problem.

So I'm genuinely puzzled why you and Micromass assert that division by zero is irrelevant to this situation, when errors can be made specifically on this problem without knowing that 0/0 is undefined and without applying knowledge of limits, which allows for a resolution to the problem (at least in a way that is satisfactory to physicists working with the function).

Quote:
 Moreover, in the typical contexts where one does allow division by $0$ such as, to make many theorems in measure theory easier to state and prove, one defines the extended reals to be $\mathbb{R} \cup \{\infty\}$ and declares that if $r \neq 0$, then $\frac{r}{0} = \infty$. In this instance, we aren't typically worried about losing the multiplicative group structure on our scalars. However, the expression $\frac{0}{0}$ is STILL undefined in this context. I don't know of any place in mathematics where $\frac{0}{0}$ has any meaning whatsoever. If there is such an application, it certainly doesn't have anything to do with computing limits of fractions or the $\operatorname{sinc}$ function.
Okay, but how do you respond to somebody who claims that division by zero should have meaning, if not by looking at examples where imposing such a meaning gives erroneous results? Is there a better application or use case where errors would clearly be obtained? The apples example provided earlier clearly indicates an issue with division by zero, but doesn't seem to indicate a definitive error if someone forcefully overrides existing knowledge with their own definition of 0/0. After all, the OP responded to the apples example in a way that seems to suggest that they don't care that 0/0 is undefined.

Last edited by Benit13; April 4th, 2019 at 03:19 AM. Reason: Typos

April 4th, 2019, 04:48 AM   #30
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Quote:
 Originally Posted by Benit13 Thanks for the clarification. I agree with what you say, but I'm still puzzled with this part: I'm not claiming that my sinc x example defines a value for $\displaystyle \frac{0}{0}$, but let's say that somebody new to mathematics tries to evaluate the function $\displaystyle f(x) = sin(x)/x$ at $\displaystyle x=0$. They substitute the value into the numerator and the denominator and obtain $\displaystyle \frac{0}{0}$. Now what? Someone with no knowledge of mathematics might make an error, such as "Oh, I have 0/0 = 0, so the answer is 0" and then get confused when they look at a plot of the function and see a value of 1 at x=0. However, someone with some knowledge of mathematics can remember that 0/0 is undefined and say "okay, it is undefined, but let's see what the limit is", then make some progress with their problem. So I'm genuinely puzzled why you and Micromass assert that division by zero is irrelevant to this situation, when errors can be made specifically on this problem without knowing that 0/0 is undefined and without applying knowledge of limits, which allows for a resolution to the problem (at least in a way that is satisfactory to physicists working with the function).
It's the other way around. Knowing $\lim_{x\rightarrow 0}\frac{\sin(x)}{x} = 1$ is irrelevant in how to define division by 0.
But knowing that division by 0 is not allowed and could get us into troubles is very relevant and important when evaluating limits. The fact that $\lim_{x\rightarrow 0}\frac{\sin(x)}{x}$ gives you a $0/0$ is important information on how to proceed in the limit.

Quote:
 Okay, but how do you respond to somebody who claims that division by zero should have meaning, if not by looking at examples where imposing such a meaning gives erroneous results? Is there a better application or use case where errors would clearly be obtained? The apples example provided earlier clearly indicates an issue with division by zero, but doesn't seem to indicate a definitive error if someone forcefully overrides existing knowledge with their own definition of 0/0. After all, the OP responded to the apples example in a way that seems to suggest that they don't care that 0/0 is undefined.
That's a good point. For me personally, I think the apple question settles that we don't define 0/0. But let's say we have somebody who insists in defining division by 0. I still think looking at limits is not a good way to proceed, but it might give a hint. I think to proceed one has to recognize that if we define division by 0, then certain arithmetical rules, which we take for granted, will stop holding. For example:

We have that $n/m = k$ if and only if $mk = n$.
But we have $0\cdot 1 = 0$ and $0\cdot 2=0$. This means that $0/0 = 1$ and $0/0 = 2$. So $0/0$ is not a unique answer. This is problematic.

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