April 4th, 2019, 04:03 AM  #31  
Senior Member Joined: Jun 2015 From: England Posts: 915 Thanks: 271  Quote:
One of the issues you need to consider in assigning a meaning to 0/0 is the deeply philosophical question Why is $\displaystyle \frac{{dy}}{{dx}} \ne \frac{0}{0}$ For instance consider Bolzano's example $\displaystyle {y^3} = a{x^2} + {a^3}$ $\displaystyle {\left( {y + dy} \right)^3} = a{\left( {x + dx} \right)^2} + {a^3}$ So $\displaystyle \frac{{dy}}{{dx}} = \frac{{2ax + adx}}{{3{y^2} + 3ydy + d{y^2}}}$ set dy, dx = 0 $\displaystyle \frac{{dy}}{{dx}} = \frac{{2ax}}{{3{y^2}}}$  
April 4th, 2019, 05:32 AM  #32 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,161 Thanks: 734 Math Focus: Physics, mathematical modelling, numerical and computational solutions  I think this example is a better indication about the thornery of division by zero. I'll use this from now on instead of the sinc x example. 
April 4th, 2019, 06:34 AM  #33 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 Math Focus: Mainly analysis and algebra 
Surely the point here is that the OP proposed the idea that since $\displaystyle \lim_{x \to 0^+} \frac1x = +\infty$ and $\displaystyle \lim_{x \to 0^} \frac1x = \infty$, taking the "average" of these two "yields zero" for division by zero. Obviously, there are many problems with his reasoning, and so there are many ways to counter his argument, but one perfectly valid one is to demonstrate that there are other limits involving "division by zero" that do not result in the same value. The fact that you prefer a different rebuttal doesn't make everybody else wrong, and implying so simply exhibits an unfortunate arrogance on the part of the poster making that claim. There is space on this forum for many people to demonstrate their knowledge and understanding, each contributing to the knowledge and understanding of the OP. 
May 1st, 2019, 11:09 AM  #34 
Newbie Joined: Apr 2019 From: North Carolina Posts: 2 Thanks: 0 
8/2 = 4 is means that 4 is the unique number such that 4 times 2 =8. 8/0 is undefined because there does not exist any number n such that n times 0 equals 8. 0/0 is undefined because any number times zero equals zero, so there is no unique number n such that n times zero equals zero (emphasis on the word unique). 
May 1st, 2019, 04:04 PM  #35 
Senior Member Joined: Jun 2014 From: USA Posts: 564 Thanks: 43  
May 1st, 2019, 11:23 PM  #36 
Senior Member Joined: Oct 2009 Posts: 863 Thanks: 328  
May 2nd, 2019, 08:27 AM  #37 
Senior Member Joined: Jun 2014 From: USA Posts: 564 Thanks: 43  Their measure is undefined, not “0”. How many times do you need to give me 0 apples before I have 1 apple?: 1/0 is also undefined, not 0, as is the focus per the OP. So I said (somewhat jokingly), that you would need to give me 0 apples a countably infinite number of times as though equating the measure of a Vitali set to 1/0.

May 2nd, 2019, 10:24 AM  #38  
Senior Member Joined: Oct 2009 Posts: 863 Thanks: 328  Quote:
 
May 2nd, 2019, 02:57 PM  #39 
Senior Member Joined: Jun 2014 From: USA Posts: 564 Thanks: 43  It must be greater than 0, but also must be less than all real numbers greater than 0, so why not compare 1/0 ~ 1/[the measure of a Vitali set] ? It’s not a lot unlike considering the limit as x approaches 0 of 1/x, as was done earlier in the thread, even if the result of 1/0 is seemingly independent of anything involving limits.

May 2nd, 2019, 03:56 PM  #40  
Senior Member Joined: Jun 2014 From: USA Posts: 564 Thanks: 43  Quote:
Some Crank Talking About Vitali Sets  

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