0/0

0/0

Posers and Puzzles

Cookies help us deliver our Services. By using our Services or clicking I agree, you agree to our use of cookies. Learn More.

Joined
26 Apr 03
Moves
26771
01 Nov 04
2 edits

Originally posted by royalchicken
True. The WHOLE POINT is that ''0/0'' doesn't mean anything. The point of that post was to show that we can find functions which would give 0/0 at some point but tend to any limit we like as x approaches that point. In that sense, 0/0 is anything we like.
i.e. 0/0 is indeterminate. If I am trying to find the bearing to travel from one point to another, I can use atan(x/y), if x and y happen to both be zero this indicates not that there is no way of getting between the two points, but that I can travel at any bearing I like as I am travelling zero distance. It is important to note that infinity is a potential answer, otherwise I couldn't opt to travel zero metres at a bearing of 90 degrees. In this sense 0/0 means quite a bit more than asin(2).

r
CHAOS GHOST!!!

Elsewhere

Joined
29 Nov 02
Moves
17317
01 Nov 04

Originally posted by iamatiger
i.e. 0/0 is indeterminate. If I am trying to find the bearing to travel from one point to another, I can use atan(x/y), if x and y happen to both be zero this indicates not that there is no way of getting between the two points, but that I can travel at any bearing I like as I am travelling zero distance. It is important to note that infinity is a potentia ...[text shortened]... l zero metres at a bearing of 90 degrees. In this sense 0/0 means quite a bit more than asin(2).
Exactly, and a good example too.

T
Kupikupopo!

Out of my mind

Joined
25 Oct 02
Moves
20443
04 Nov 04

Originally posted by fearlessleader
i dont see how f(x,y)=0/0 w/out both x and y being 0.


i have to ask knoptfel about gallus nobelus's post, i'll get back to it later.
That's the idea. How do you define x*y/(x^2 +y^2) in the point (0, 0)?

Whatever you choose for that point (0/0), the function will not be continuous in that point.

G

Joined
15 Oct 04
Moves
1995
10 Nov 04

Originally posted by TheMaster37
That's the idea. How do you define x*y/(x^2 +y^2) in the point (0, 0)?

Whatever you choose for that point (0/0), the function will not be continuous in that point.
x*y/(x^2 +y^2) is an expression, not a function or even a relation...

that expression would have to equal something in order to place it in the cartesian plane...

if you wish to say y=x*y/(x^2+y^2)

then y/y= x/(x^2+y^2)

1= x/(x^2+y^2)

x^2+y^2=x

y^2=x-x^2

y= sqrt(x-x^2)
&
y= sqrt(x^2-x)

0,0 then is a real and defined point.

f

my head

Joined
03 Oct 03
Moves
671
10 Nov 04

Originally posted by TheMaster37
That's the idea. How do you define x*y/(x^2 +y^2) in the point (0, 0)?

Whatever you choose for that point (0/0), the function will not be continuous in that point.
lets set it equal to z, to get a three quardent equation. as x or y approches 0, z approches infinity because the denominator is shrinking faster. this is with one variable changing, and the other remaining constant. if the other variable is set as 0, then z=0 regardless of the other vaiable. so, if x or y=0, then z=0, but the limit as either approches 0 is infinet. (note, i'm sort of ignoring negativs.)
however, if both apporch 0 simeltaniously, then z will be approching 1, because the the denominator and numerator will be shrinking in unison. so there are a number of ways of defining the limit of this expression as x and y both approch 0.

someone with 3d graphing capabilities could probably draw more spicific conclusions.

sorry about the rambeling nature of this post, i was sort of making it up as i went along.

T
Kupikupopo!

Out of my mind

Joined
25 Oct 02
Moves
20443
16 Nov 04

More exact:

let (x,y) -> (0,0) as (1/n,1/n) with n -> inf.

then the function above gives (1/n * 1/n) / (1/n*1/n + 1/n*1/n) = 1/2

let (x,y) -> (0,0) as (0,y) with y->0 then the function above gives 0

let (x,y) -> (0,0) as (1/n, a/n) with n -> inf.

then the function gives a / (a*a + 1)

now as (b/n, a/n) wich gives ab / (a*a + b*b)

there are so many possible discontinuations there 🙂

f

my head

Joined
03 Oct 03
Moves
671
18 Nov 04

Originally posted by TheMaster37
More exact:

let (x,y) -> (0,0) as (1/n,1/n) with n -> inf.

then the function above gives (1/n * 1/n) / (1/n*1/n + 1/n*1/n) = 1/2

let (x,y) -> (0,0) as (0,y) with y->0 then the function above gives 0

let (x,y) -> (0,0) as (1/n, a/n) with n -> inf.

then the function gives a / (a*a + 1)

now as (b/n, a/n) wich gives ab / (a*a + b*b)

there are so many possible discontinuations there 🙂
i'm seeing that the implied value of 0/0 is singular for the direction from which it is approched.
just a hunch, but what happens if we express this same thing in polar cordenets. the rules of the polar plane i think have special rules regarding the nature of the origin that minght be of some help.





in regards to the overall argument of this thread: i'm seeing that 0/0 can be defined as a single real number, but that number can be any number.
a simmiler solution was presented regarding the nature of the following set of statements:
Statment A: Statment B is faulse.
Statment B: Statment A is true.

Statment A, this argument claims, can be true or faulse, but never both.

r
CHAOS GHOST!!!

Elsewhere

Joined
29 Nov 02
Moves
17317
18 Nov 04
1 edit

Originally posted by fearlessleader
i'm seeing that the implied value of 0/0 is singular for the direction from which it is approched.
Very good 😀!

In general, the limit of a function of several variables as some point is approached depends on the curve along which the point is approached. Functions for which the curve does not matter are particularly special and nice.

T
Kupikupopo!

Out of my mind

Joined
25 Oct 02
Moves
20443
18 Nov 04

Originally posted by fearlessleader
a simmiler solution was presented regarding the nature of the following set of statements:
Statment A: Statment B is faulse.
Statment B: Statment A is true.

Statment A, this argument claims, can be true or faulse, but never both.
Sorry, but i fail to see how a statement can be true AND false at the same time in any situation.

And apologies again for not seeing how statement A can be true OR false. Assuming either one leads to a contradiction in the argument, so statement A is neither true or false.

Our argument is; no matter what you choose for 0/0 to be, it will always lead to discussions, since no single number can be chosen to that 0/0 has a sensible meaning.

Joined
26 Apr 03
Moves
26771
19 Nov 04

Originally posted by TheMaster37
Sorry, but i fail to see how a statement can be true AND false at the same time in any situation.

And apologies again for not seeing how statement A can be true OR false. Assuming either one leads to a contradiction in the argument, so statement A is neither true or false.

Our argument is; no matter what you choose for 0/0 to be, it will always lead to discussions, since no single number can be chosen to that 0/0 has a sensible meaning.
The only solution that works is:

A: Undecidable
B: Undecidable

f

my head

Joined
03 Oct 03
Moves
671
19 Nov 04

Originally posted by iamatiger
The only solution that works is:

A: Undecidable
B: Undecidable
well yes of course we can't say that either statment is really true or really faulse, just as we cant really say that 0/0=an actual number, but in trying to find a solution, we will assign diffrent validities to the statments, but A will always be either true or faulse, same for B.

this isn't too great an anaolgy for 0/0, but it is similar. we can say that 0/0=a number, and we can show why this is the case, but this is true of any number, but never more than one at a time

Joined
26 Apr 03
Moves
26771
19 Nov 04

Originally posted by fearlessleader
well yes of course we can't say that either statment is really true or really faulse, just as we cant really say that 0/0=an actual number, but in trying to find a solution, we will assign diffrent validities to the statments, but A will always be either true or faulse, same for B.

this isn't too great an anaolgy for 0/0, but it is similar. we c ...[text shortened]... w why this is the case, but this is true of any number, but never more than one at a time
Well, if I say A is definitely true or false - but I don't know which (because I haven't looked at statement B yet), then A is currently undecidable - likewise if I can't decide A then I can't decide B - and if I can' decide B then I am still right in saying I can't decide A! So the state is self consistent and perpetuating. It's a bit different from 0/0 in that way.

O

Joined
22 Jun 01
Moves
89084
20 Nov 04

i would recommend that you guyrls stop eating the cheese late at night.
0/0 has to be indeterminate, because we can never truly appreciate the exact nature of either 0 or infinity. does anyone here know of any applications where solving or defining this thing would be useful?

T
Kupikupopo!

Out of my mind

Joined
25 Oct 02
Moves
20443
20 Nov 04

I have to object a little here, seems as if things are going against intuition (but with logics, that's often the case)

What i'm trying to say, is that statement A is NOT either true or false, wich you (adressing fearlessleader) keep saying 🙂

If it is either true or false, it's exactly one of them. A cannot be true, A cannot be false. So it's NOT either true or false.

I wouldn't call A undecidable, i would call it a paradox.

Undecidable is the following satement;

C: The decimal expansion of Pi contains a block of 99 nines in a row.

Joined
26 Apr 03
Moves
26771
20 Nov 04

Originally posted by TheMaster37
I have to object a little here, seems as if things are going against intuition (but with logics, that's often the case)

What i'm trying to say, is that statement A is NOT either true or false, wich you (adressing fearlessleader) keep saying 🙂

If it is either true or false, it's exactly one of them. A cannot be true, A cannot be false. So it's ...[text shortened]... the following satement;

C: The decimal expansion of Pi contains a block of 99 nines in a row.
I beg to disagree.
It's not a paradox until you decide whether A (or B) is true or false. Then it's a paradox. If I don't decide whether they're true or false I'm unparadoxed (see explanation above), that's why I said that they are then undecidable. If you don't allow me to use the value undecidable then you have a paradox - but you didn't forbid that value.