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

## Welcome to BrainDen.com - Brain Teasers Forum

Welcome to BrainDen.com - Brain Teasers Forum. Like most online communities you must register to post in our community, but don't worry this is a simple free process. To be a part of BrainDen Forums you may create a new account or sign in if you already have an account. As a member you could start new topics, reply to others, subscribe to topics/forums to get automatic updates, get your own profile and make new friends. Of course, you can also enjoy our collection of amazing optical illusions and cool math games. If you like our site, you may support us by simply clicking Google "+1" or Facebook "Like" buttons at the top. If you have a website, we would appreciate a little link to BrainDen. Thanks and enjoy the Den :-) |

Guest Message by DevFuse

# n/0

Started by unreality, Sep 20 2007 12:29 AM

3 replies to this topic

### #1

Posted 20 September 2007 - 12:29 AM

### #2

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].

Read more about that here: http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule

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.

[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].

Read more about that here: http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule

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.

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #3

Posted 20 September 2007 - 05:44 AM

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

### #4

Posted 21 September 2007 - 10:03 AM

ROTFL

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

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

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

#### 0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users