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April 1st, 2019, 11:20 PM   #1
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Question 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?
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April 1st, 2019, 11:26 PM   #2
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Quote:
Originally Posted by Apple30 View Post
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.
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April 1st, 2019, 11:58 PM   #3
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Originally Posted by Micrm@ss View Post
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.


Last edited by Apple30; April 2nd, 2019 at 12:05 AM.
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April 2nd, 2019, 01:07 AM   #4
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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.
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April 2nd, 2019, 01:26 AM   #5
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Originally Posted by Apple30 View Post
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.
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April 2nd, 2019, 01:35 AM   #6
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Originally Posted by Benit13 View Post
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.
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April 2nd, 2019, 01:56 AM   #7
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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.
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April 2nd, 2019, 05:06 AM   #8
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4 apples divided by 2: Jim and Joe get 2 each
4 apples divided by 1: Jim gets 4
4 apples divided by nobody: ?
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April 2nd, 2019, 07:55 AM   #9
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Originally Posted by Apple30 View Post
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}
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April 2nd, 2019, 12:45 PM   #10
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Hey guys, has anyone noticed the date of the first and last posts by the martian OP?
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