April 3rd, 2019, 05:46 AM  #21  
Senior Member Joined: Oct 2009 Posts: 796 Thanks: 295  Quote:
Limits to $0^0$ are also indeterminate. But $0^0=1$ is still true and widely used.  
April 3rd, 2019, 07:09 AM  #22 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,194 Thanks: 897 Math Focus: Wibbly wobbly timeywimey stuff.  
April 3rd, 2019, 07:24 AM  #23 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,943 Thanks: 1132 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, 07:51 AM  #24  
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,156 Thanks: 731 Math Focus: Physics, mathematical modelling, numerical and computational solutions  Quote:
$\displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$ 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 08:11 AM. Reason: Note about the limiting value and grammar errors.  
April 3rd, 2019, 04:58 PM  #25  
Senior Member Joined: Sep 2016 From: USA Posts: 619 Thanks: 391 Math Focus: Dynamical systems, analytic function theory, numerics  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.  
April 3rd, 2019, 10:44 PM  #26  
Senior Member Joined: Oct 2009 Posts: 796 Thanks: 295  Quote:
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, 10:47 PM  #27  
Senior Member Joined: Oct 2009 Posts: 796 Thanks: 295  Quote:
 
April 3rd, 2019, 10:51 PM  #28 
Senior Member Joined: Oct 2009 Posts: 796 Thanks: 295 
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, 02:17 AM  #29  
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,156 Thanks: 731 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:
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:
Last edited by Benit13; April 4th, 2019 at 02:19 AM. Reason: Typos  
April 4th, 2019, 03:48 AM  #30  
Senior Member Joined: Oct 2009 Posts: 796 Thanks: 295  Quote:
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:
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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