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-   -   Why isn't dividing by zero = zero? (http://mymathforum.com/elementary-math/346134-why-isnt-dividing-zero-zero.html)

Apple30 April 1st, 2019 11:20 PM

Why isn't dividing by zero = zero?
 
In layman terms, why is x divided by 0 equal to undefined, and not 0?

If anything slightly above 0 is +infinity, and anything slightly below 0 is -infinity, can't we argue that the number in between both infinities is 0?

Micrm@ss April 1st, 2019 11:26 PM

Quote:

Originally Posted by Apple30 (Post 607741)
In layman terms, why is x divided by 0 equal to undefined, and not 0?

If anything slightly above 0 is +infinity, and anything slightly below 0 is -infinity, can't we argue that the number in between both infinities is 0?

How many times do I need to give you 0 apples in order for you to have 1 apple.

Apple30 April 1st, 2019 11:58 PM

Quote:

Originally Posted by Micrm@ss (Post 607742)
How many times do I need to give you 0 apples in order for you to have 1 apple.

According to this math professor, 0 equals 1, so if you give me 0 apples, I will have 1 apple. :)


Micrm@ss April 2nd, 2019 01:07 AM

Quote:

Originally Posted by Apple30 (Post 607745)
According to this math professor, 0 equals 1, so if you give me 0 apples, I will have 1 apple. :)


Cool. I hope you handle money that way too. I guess you owe me 0 dollars.

Benit13 April 2nd, 2019 01:26 AM

1 Attachment(s)
Quote:

Originally Posted by Apple30 (Post 607741)
In layman terms, why is x divided by 0 equal to undefined, and not 0?

Because it depends on what you're dividing by zero.

More specifically, if you consider a function

$\displaystyle f(x) = \frac{g(x)}{x}$

and then try and find the limit of that function as x tends to zero, then the limit obtained is not always zero. It changes depending on what g(x) is.

A typical example is $\displaystyle g(x) = \sin x$. If you look at the attached table, which shows somes evaluations of the function with decreasing values of x, the function tends towards 1, not 0.

Apple30 April 2nd, 2019 01:35 AM

Quote:

Originally Posted by Benit13 (Post 607748)
Because it depends on what you're dividing by zero.

More specifically, if you consider a function

$\displaystyle f(x) = \frac{g(x)}{x}$

and then try and find the limit of that function as x tends to zero, then the limit obtained is not always zero. It changes depending on what g(x) is.

A typical example is $\displaystyle g(x) = \sin x$. If you look at the attached table, which shows somes evaluations of the function with decreasing values of x, the function tends towards 1, not 0.

Yes, but an opposite example to that would tend towards -1, and the average of -1 and 1 (and -infinity and infinity) is 0.

Benit13 April 2nd, 2019 01:56 AM

Quote:

Originally Posted by Apple30 (Post 607749)
Yes, but an opposite example to that would tend towards -1, and the average of -1 and 1 (and -infinity and infinity) is 0.

Right... but that's a different function. That's nothing to do with dividing by zero.

Denis April 2nd, 2019 05:06 AM

4 apples divided by 2: Jim and Joe get 2 each
4 apples divided by 1: Jim gets 4
4 apples divided by nobody: ?

v8archie April 2nd, 2019 07:55 AM

Quote:

Originally Posted by Apple30 (Post 607749)
Yes, but an opposite example to that would tend towards -1, and the average of -1 and 1 (and -infinity and infinity) is 0.

On the other hand, the same function $f(x)=\frac{\sin x}{x}$ evaluated for successively smaller but negative values of $x$ gets closer to 1 too. This is the real equivalent of your suggestion for $\frac1x$.
\begin{array}{c | c}
x & \frac{\sin x}{x} \\ \hline
-0.1\phantom{000} & 0.998334166468282 \\
-0.01\phantom{00} & 0.999983333416667 \\
-0.001\phantom{0} & 0.999999833333342 \\
-0.0001 & 0.999999998333333
\end{array}

studiot April 2nd, 2019 12:45 PM

Hey guys, has anyone noticed the date of the first and last posts by the martian OP?


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