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-0 paradox


morgsboi

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Yes I saw that and I think the threads could profitably be combined.

 

However you have avoided my point that in certain circumstances the ratios zero/zero and infinity/infinity can be evaluated.

 

Further there was no stated requirement to remain within either the reals or the complex numbers.

 

Both the limits in the series ratios I displayed exist and are zero or infinity.

I never said that indeterminate forms cannot be evaluated. I simply demonstrated why the limit[math], \ \lim_{(x,y) \to (0,0)} \frac{y}{x} \ ,[/math] is undefined.

 

Indeterminate forms, [math]\ \frac{0}{0}, \ \frac{\infty}{\infty}, \ \infty - \infty, \ 0 \times \infty, \ 0^0, \ \infty^0, \ 1^\infty \ ,[/math] can be evaluated using limits.

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Okay, I'm only here to learn. Please point out where I went wrong in my mathematics.

 

In the specific post I quoted, you started with the claim that

 

[math]\frac{\infty}{1} = 0.000...001 = 0[/math],

 

which doesn't make much sense, firstly because "a decimal, then infinite zeros with a one at the end" isn't valid, and secondly because [math]\frac{\infty}{1}[/math] would equal [math]\infty[/math] anyway. I'll assume you meant to claim that

 

[math]\frac{1}{\infty} = 0[/math],

 

which still isn't valid, but let's assume it is. Then we can manipulate this expression to show that

 

[math]1 = 0(\infty)[/math].

 

That's all well and good, but let's take a real number [math]n[/math] and do the following:

 

[math]1(n) = 0(n)(\infty)[/math].

 

This means

 

[math]n = 0(\infty) = 1[/math],

 

but clearly not all real numbers are equal to 1.

 

Your mathematical error lies with treating infinity as just another number (when it isn't). You can't just toss infinity into an equation and pretend it's just some really big real number. Now, it is true that, approaching zero from the right,

 

[math]\lim_{x\to0^{+}}\frac{1}{x} = \infty[/math],

 

but that doesn't mean [math]\frac{1}{0}[/math] is infinity (especially since, approaching from the left, the limit is [math]-\infty[/math]). It means that as x approaches (but never reaches!) 0, the value of [math]\frac{1}{x}[/math] increases without bound. It may seem like a subtle difference, but it is very important.

 

Edit: To answer your more general question, division by zero is not defined because any definition we give it results in logical absurdity like the above.

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I never said that indeterminate forms cannot be evaluated.

 

yet in the post immediately following my note that it is sometimes defined you post this

 

This is why division by zero is undefined:

 

 

How is this to be taken by someone trying to learn?

 

division by zero is not defined because any definition we give it results in logical absurdity like the above.

 

I do wish folks would stop being so categoric (and wrong) because it is preventing the OP moving on to understanding what can be evaluated (without calculus at his level).

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I do wish folks would stop being so categoric (and wrong) because it is preventing the OP moving on to understanding what can be evaluated (without calculus at his level).

 

If you would like to show me how my statement is wrong, please feel free. The fact that there are certain contexts in which infinity and division by zero are treated differently has been covered elsewhere and referenced here. However, the OP is proposing simple mathematical statements that are absurd. Those are what I was addressing.

 

Your examples near the beginning of the thread do not apply here, and I wish you wouldn't troll while referencing them as some evidence that division by zero is defined.

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Your own words in two successive posts.

 

To answer your more general question, division by zero is not defined because any definition

 

 

A general categorical answer brooking no exceptions.

 

 

The fact that there are certain contexts in which infinity and division by zero are treated differently

 

An exception.

 

and I wish you wouldn't troll while referencing them as some evidence that division by zero is defined.

 

So how should I take this insulting response?

 

Surely the correct response to someone who has a misconception is to say something along the lines of

 

"Division by zero is not defined in the circumstance you are proposing, but can be achieved in certain specialised cases."

 

with or without further amplification.

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Your own words in two successive posts.

 

 

 

 

A general categorical answer brooking no exceptions.

 

 

 

 

An exception.

 

You are of course correct there, and I worried after I posted it that someone would respond as you did. Perhaps I should have erased the edit. However, surely the fact that I myself had previously referenced exceptions should have indicated there was an understood "with the exception of the cases mentioned previously." In the context of the OP's post, my claim stands. There are certain areas in which division by zero can be defined (see this for an example in complex analysis, and note that, as with any case in which division by zero is defined, there are caveats), but the OP is tossing infinity into simple arithmetic expressions where it isn't valid. A defined value for division by zero, in the way the OP intended to use it, leads to absurdity. It pains me that I have to spell out what should have been the obvious interpretation, but, Internet forum.

 

So how should I take this insulting response?

 

Surely the correct response to someone who has a misconception is to say something along the lines of

 

"Division by zero is not defined in the circumstance you are proposing, but can be achieved in certain specialised cases."

 

with or without further amplification.

 

This is exactly what my posts have claimed. As for how to take the "insulting" response, I would suggest in future posting "you're wrong, and here's why" rather than "you're wrong and you're misleading someone" without any explanation.

Edited by John
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