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# n/0

3 replies to this topic

### #1 unreality

unreality

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Posted 20 September 2007 - 12:29 AM

first i have to clear up my useage of n:

n = any real number and ALL real numbers

it could be pi, or 5, or -2 or .333repeating. It doesnt work with at least one type of nonreal number (see below)

One Exception: n may not be 0

******
RULES OF ZERO (by unreality) (lol)
******

EQUATION: n/x = y

FACT: as x gets smaller, y will get bigger. As x gets reduced to 0, y goes to infinity (?)
EXAMPLE:
5/1 = 5
5/.5 = 10
5/.0005 = 10,000
5/.0000000001 = 50,000,000,000
etc

FACT: n/0 = ?

FACT: n/? = 0

FACT: z/z = 1
FACT: ?/? = 1 (not 0)
THUS: since ? is nonreal, this only works with Real numbers or at least non-infinite

ANYWAY:
BACK TO FACTS:
MOST RECENT FACT: n/? = 0 since n/0 = ?

FACT: n*0 = 0
(divide by zero on both sides)
FACT: n = 0/0

The question is:
Is 0n over 0 (0n/0) equal to 0/0 or n? (I say BOTH, see sentence after next) Since you are dividing by 0, it might be 0/0 = 0/0 which casts no light on 0/0. But the nature of what I just said, before that, does mean that n = 0/0, right?

So my final facts:

FINAL FACT: n/0 = ? (n cant be zero here, remember)
FINAL FACT: 0/0 = n (here n could be zero)

***
i dunno... just ramblings of the 0
• 0

### #2 bonanova

bonanova

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Posted 20 September 2007 - 04:52 AM

For any non-zero, finite number, say 13, for sake of discussion, then usually:

[1] 0/13 = 0
[2] 13/0 = infinity
[3] 0/0 = undefined [that is, could be 242, for example]
[4] infinity/infinity = undefined [could be 0.327, for example]
[5] 0 x infinity = undefined [could be 69, for example]

But it turns out that both zero and infinity have different flavors, called orders.
There are 1st-order zeros, and 1st-order infinities, as well as 2nd, 3rd-order, etc. of both.

In these cases, the higher order predominates:
[3a] 0[2]/0[1] = 0; 0[1]/0[2] = infinity
[4a] infinity[2]/infinity[1] = infinity; infinity[1]/infinity[2] = 0.
[5a] 0[2] x infinity[1] = 0; 0[1] x infinity[2] = infinity.

where 0[n] means n-th order 0 and infinity[p] means p-th order infinity.

You can see what that means by changing one of these zeros a wee bit, then inching your way back.

Let x be a small number - say 1/100.

[6] x/x=1. no problem, since x isn't zero. Let x inch its way to zero, the ratio stays at 1.
[7] 2x/x = 2. no problem, since x isn't zero [yet]. Let x -> 0, the ratio stays 2.

That's why 0/0 is usually thought of as undefined: you could have used any number instead of 2.

But look at

[8] x^2/x = x. now the ratio is x. Let x ->0, and the ratio [always = x] ->0 as well.

What happened? both numerator and denominator -> 0, so the ratio goes to 0/0.
How come it's not undefined? How come in this case we know it's zero?

The reason is x^2 becomes a 2nd-order zero: 0^2 so to speak; it's stronger than just 0.

Same thing about the infinity you get by dividing by x^2 instead of by x as x->0.
There you get a stronger infinity - a 2nd-order infinity - infinity^2 so to speak.

There is a rule called L'Hopital's rule that gives you these answers for [6] - [8].

Basically you step away from the divide-by-zero situation and evaluate the division using a legal small number, then see what happens when it approaches zero, and that gives you a defined answer, for that particular case.

L'Hopital's rule can make some paradoxical problems intuitive.

Absent some expression like [2x/x] ->2, or [x^2/x] ->0, or [x/x^2] -> infinity, you're just left with undefined and you don't try to prove too much with expressions that contain infinities in the numerator or zeros in the denominator.
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Vidi vici veni.

### #3 Writersblock

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Posted 20 September 2007 - 05:44 AM

Why do I feel like the bastard child of Karl Childers and Forest Gump?
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### #4 bonanova

bonanova

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Posted 21 September 2007 - 10:03 AM

Why do I feel like the bastard child of Karl Childers and Forest Gump?

ROTFL

I'm with ya ... I don't know who Karl Childers is...
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Vidi vici veni.

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