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Author Topic: m/n where n=0  (Read 7964 times)

Armok

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Re: m/n where n=0
« Reply #30 on: August 15, 2011, 06:33:41 pm »

the thing that made it click for me was to instead of asking "why can't I divide by zero?" asking "what actually happens if I try to?", and getting back the set of Everything instead of a number.
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Virex

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Re: m/n where n=0
« Reply #31 on: August 15, 2011, 07:04:42 pm »

You don't quite get the "set of everything" IIRC, but rather the set of extended real numbers or whatever it's name is (The union of R and the set containing infinity and -infinity). No mater how many fallacies it does allow, I don't think you can go from the real numbers to the set of all polynomials using a division by 0 fallacy, but feel free to prove me wrong.
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G-Flex

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Re: m/n where n=0
« Reply #32 on: August 15, 2011, 07:27:10 pm »

It's less "the answer is everything" and more "a meaningful answer could be anything", as far as I understand it. I'm mostly approaching this from a calculus/limits sort of angle, though.
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Lagslayer

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Re: m/n where n=0
« Reply #33 on: August 15, 2011, 07:42:05 pm »

No no no. X/0= infinity. For years, the joke was always that dividing by zero destroys matter, but really every time you do it, you actually create a whole new universe. So remember, when you are dividing by zero, you become GOD. Dividing a cookie into zero piles gives infinite piles of cookies! Unfortunately, you will never be able to attain them. Godhood is a cruel paradox.

kaijyuu

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Re: m/n where n=0
« Reply #34 on: August 15, 2011, 07:47:39 pm »

Alright math nuts, riddle me this. Does 0/0 = 1?


Take this graph: y = x/x

Now obviously that's a straight horizontal line at y = 1. But what about at x= 0? Is it a hole (which would be true if 0/0 is undefined), or is it 1 (as simplifying x/x to 1 requires it to be true for every value of x)?
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Vector

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Re: m/n where n=0
« Reply #35 on: August 15, 2011, 07:54:23 pm »

I believe it's just a hole, because 0/0 is itself an undefined value, and the limits on both sides converge to 1 (so there's none of the usual telltale asymptotic behavior).

I may be wrong on that, though.
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Starver

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Re: m/n where n=0
« Reply #36 on: August 15, 2011, 07:58:56 pm »

Or else as 0/n=0 it overrides the undefinedness and sets it as zero.

(I think it's undefined/undefinable[1], but I'm just giving you another possibility given that 0*n=0.)


[1]fakeedit: Yep, wot Vector sed.
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ed boy

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Re: m/n where n=0
« Reply #37 on: August 15, 2011, 08:00:05 pm »

Alright math nuts, riddle me this. Does 0/0 = 1?

However, note that the operation "multiply by 0" is not injective (1*0=0=2*0), and it is not surjective (0*X=0 for all X). If an inverse were to exist, it could be be defined for any non-zero X (as there would be no Y such that Y*0=X). Furthermore, an inverse cannot be defined for zero, since it is not injective as zero (There are multiple numbers X such that X*0=0, so how do we know which one to give for 0/0?), and therefore an inverse to dividing by zero does not exist anywhere.

Take this graph: y = x/x

Now obviously that's a straight horizontal line at y = 1. But what about at x= 0? Is it a hole (which would be true if 0/0 is undefined), or is it 1 (as simplifying x/x to 1 requires it to be true for every value of x)?
The function you described would be equivalent to 1, except where x=0.

When x=0, the instruction "y=x/x" doesn't make any sense, and so we cannot find a value of y at x=0.

In theory, the graph would be a horizontal line with a gap in it where x=0. However, in practice, drawing such a line would be impossible, but that is to do with the structure of the graph and not with the maths behind it.
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G-Flex

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Re: m/n where n=0
« Reply #38 on: August 15, 2011, 08:03:45 pm »

Alright math nuts, riddle me this. Does 0/0 = 1?


Take this graph: y = x/x

Now obviously that's a straight horizontal line at y = 1. But what about at x= 0? Is it a hole (which would be true if 0/0 is undefined), or is it 1 (as simplifying x/x to 1 requires it to be true for every value of x)?

You can't derive "0/0 = 1" from the fact that it seems to equal 1 in one specific case. Hell, the most obvious example here: y = 2x/x would make it seem like 0/0 = 2. And y = x2/x would make it seem like 0/0 = 0. This is without even really getting into limits, per se. Here's more: http://en.wikipedia.org/wiki/Indeterminate_form#Discussion

I believe it's just a hole, because 0/0 is itself an undefined value, and the limits on both sides converge to 1 (so there's none of the usual telltale asymptotic behavior).

I may be wrong on that, though.

Yes, you can tell it's a removable discontinuity (hole) because you can algebraically manipulate the original function to something equivalent that doesn't have the whole. y = 1 and y = x/x are equivalent aside from that one hole, in this case.
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Re: m/n where n=0
« Reply #39 on: August 15, 2011, 08:09:28 pm »

Fair enough.

Just something that popped into my head. I haven't had a math class in about 5 years now (sometimes I wish I could remember calculus... it completely left me after my final exam in that class).
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Quote from: Chesterton
For, in order that men should resist injustice, something more is necessary than that they should think injustice unpleasant. They must think injustice absurd; above all, they must think it startling. They must retain the violence of a virgin astonishment. When the pessimist looks at any infamy, it is to him, after all, only a repetition of the infamy of existence. But the optimist sees injustice as something discordant and unexpected, and it stings him into action.

Duke 2.0

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Re: m/n where n=0
« Reply #40 on: August 15, 2011, 08:40:24 pm »

Let me give you a more practical example. You have, say, 10 of your largest denomination coins, whatever they are. Dollars, probably. To divide by 2, you'd arrange them into two piles, giving 5 in each. 10/2=5. Now try to divide them into 0 piles.
This is the real magic mathamagicians should be doing, none of this "oooooohhh, this number is now over HERE!" silliness. A wave of the hands and a tap of a piece of chalk and your money is now in zero equal piles. He then produces the last digit of pi from his or her sleeve.
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Starver

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Re: m/n where n=0
« Reply #41 on: August 15, 2011, 09:29:58 pm »

He then produces the last digit of pi from his or her sleeve.

I'm going to be awkward and demand that he produces the penultimate digit. :)
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darkrider2

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Re: m/n where n=0
« Reply #42 on: August 15, 2011, 10:03:58 pm »

This is the real magic mathamagicians should be doing, none of this "oooooohhh, this number is now over HERE!" silliness. A wave of the hands and a tap of a piece of chalk and your money is now in zero equal piles. He then produces the last digit of pi from his or her sleeve.

hope you don't mind if I sig this.

Also note that HE can produce a digit from HIS OR HER sleeve, this suggests that the mathmagicians current gender is a wavefunction that has yet to be collapsed. Or he has an assistant.
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Duke 2.0

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Re: m/n where n=0
« Reply #43 on: August 15, 2011, 10:08:18 pm »

 Or that I caught the ambiguity of my statement only there and didn't want to backspace to my previous use of a gender term on a tablet.
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Kay12

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Re: m/n where n=0
« Reply #44 on: August 16, 2011, 04:37:37 am »

No no no. X/0= infinity.

Too bad infinity works pretty badly with finite numbers...

For example, for finite numbers x + 1 > x is true without exception. When dealing with infinities though, you'll get situations like ∞ + 1 = ∞, and ∞ + ∞ = ∞ and so on.
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