My Math Forum  

Go Back   My Math Forum > High School Math Forum > Elementary Math

Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion


Thanks Tree16Thanks
Reply
 
LinkBack Thread Tools Display Modes
April 3rd, 2019, 06:46 AM   #21
Senior Member
 
Joined: Oct 2009

Posts: 912
Thanks: 354

Quote:
Originally Posted by v8archie View Post
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.
Micrm@ss is offline  
 
April 3rd, 2019, 08:09 AM   #22
Math Team
 
topsquark's Avatar
 
Joined: May 2013
From: The Astral plane

Posts: 2,345
Thanks: 986

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
Originally Posted by Micrm@ss View Post
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
Thanks from v8archie
topsquark is offline  
April 3rd, 2019, 08:24 AM   #23
Global Moderator
 
greg1313's Avatar
 
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.
greg1313 is offline  
April 3rd, 2019, 08:51 AM   #24
Senior Member
 
Joined: Apr 2014
From: Glasgow

Posts: 2,166
Thanks: 738

Math Focus: Physics, mathematical modelling, numerical and computational solutions
Quote:
Originally Posted by Micrm@ss View Post
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.
Thanks from topsquark

Last edited by Benit13; April 3rd, 2019 at 09:11 AM. Reason: Note about the limiting value and grammar errors.
Benit13 is offline  
April 3rd, 2019, 05:58 PM   #25
SDK
Senior Member
 
Joined: Sep 2016
From: USA

Posts: 684
Thanks: 459

Math Focus: Dynamical systems, analytic function theory, numerics
Quote:
Originally Posted by Benit13 View Post
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.
Thanks from topsquark and Benit13
SDK is offline  
April 3rd, 2019, 11:44 PM   #26
Senior Member
 
Joined: Oct 2009

Posts: 912
Thanks: 354

Quote:
Originally Posted by topsquark View Post
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.
Micrm@ss is offline  
April 3rd, 2019, 11:47 PM   #27
Senior Member
 
Joined: Oct 2009

Posts: 912
Thanks: 354

Quote:
Originally Posted by SDK View Post
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.
Thanks from Benit13
Micrm@ss is offline  
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!!
Micrm@ss is offline  
April 4th, 2019, 03:17 AM   #29
Senior Member
 
Joined: Apr 2014
From: Glasgow

Posts: 2,166
Thanks: 738

Math Focus: Physics, mathematical modelling, numerical and computational solutions
Thanks for the clarification. I agree with what you say, but I'm still puzzled with this part:

Quote:
Originally Posted by SDK View Post
... 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
Benit13 is offline  
April 4th, 2019, 04:48 AM   #30
Senior Member
 
Joined: Oct 2009

Posts: 912
Thanks: 354

Quote:
Originally Posted by Benit13 View Post
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.
Thanks from Benit13
Micrm@ss is offline  
Reply

  My Math Forum > High School Math Forum > Elementary Math

Tags
dividing



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Dividing by 0 afk Algebra 3 December 26th, 2014 10:18 PM
Dividing by zero Chikis Elementary Math 4 March 12th, 2013 10:03 AM
Is this dividing by zero? maxgeo Algebra 1 September 12th, 2012 08:18 AM
Dividing by zero? Jo15 Calculus 7 January 1st, 2012 08:51 PM
Dividing by zero :) Sidways8 :) Number Theory 14 February 16th, 2011 05:44 PM





Copyright © 2019 My Math Forum. All rights reserved.