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

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

April 4th, 2019, 04:03 AM   #31
Senior Member

Joined: Jun 2015
From: England

Posts: 905
Thanks: 271

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.

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: 2,150
Thanks: 730

Math Focus: Physics, mathematical modelling, numerical and computational solutions
Quote:
 Originally Posted by Micrm@ss 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.
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: 7,640 Thanks: 2624 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.

 Tags dividing

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post afk Algebra 3 December 26th, 2014 09:18 PM Chikis Elementary Math 4 March 12th, 2013 09:03 AM maxgeo Algebra 1 September 12th, 2012 07:18 AM Jo15 Calculus 7 January 1st, 2012 07:51 PM :) Sidways8 :) Number Theory 14 February 16th, 2011 04:44 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top