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What a pity, you blunder on your very first point.Obviously, but no matter.
Consider:
If my point is that 0/0 = 1,
No it is crap with no evidence of any reasoning.No, that is circular reasoning:
"If 0/0 = 1 the 0/0 must be equal to 1".
No it allows you to conclude that it is meaningless.does not allow you to conclude that 0/0 = 1.
But _does_ allow me to conclude that any other number, no matter how
small, and no matter what its sign will, when divided by itself, give
a quotient of 1?
OK, look at it this way:
Chuckle....0/0 is meaningless and you are wrong and bordering onAnd if those x's were zeroes when they were cancelled, that still
results in a quotient of 1, so 0/0 = 1
0/n where n=any number except 0 = 0. You seem to be trying to makeThe point is that you can't cancel them when x may be 0, because
division by 0 is undefined. However, in the lim x->0 case x approaches
0, but never really equals zero, so the cancellation may be performed.
But, but Mati, nobody is arguing with you.Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
No it is meaningless gibberish.1/0 is a small number?
Then what would be the proper way to write it, please?
There isn't one. Division by 0 is meaningless.
John Fields wrote:
Then what would be the proper way to write it, please?
Write what? That 1/x gets big as x gets small?
1/x -> +infinity as x -> +0
1/x -> -infinity as x -> -0.
Both are meaningless. Just crap piled higher and deeper as in Ph D.
Wow! We started from the trollish "Re: Is zero even or odd?" and nowBut for me that is not the issue. I do not LIE on newsgroups - there is
no point - but I do make mistakes; everyone does. I suggest that your
Besides, there is also the definition from algebra, in which x^n is
defined whenever x is a member of a monoid M and n is a natural number.
In particular, x^0 = e, the identity in M.
x^0 = 1 except when x=0 which doesn't exist.
---John Fields schrieb:
On Mon, 27 Dec 2004 20:58:44 +0100, Michael Mendelsohn
invalid@msgid.michael.mendelsohn.de> wrote:
Below, you remove the short from my diagram.
However, you also remove the power supply, which achieves the same
thing.
---
I don't know what you mean, since the + and - terminals are there and
I refer to the voltage across the resistance as being 1V.
You eventually lower the voltage to 0V.
That's what I achieved with the short.
The proper circuit:
+---(V)---+
| |
(-)---o---[R]---o---(A)---o---(+)
Will yield the proper results if examined using Ohm's law.
Assuming that the voltage across the resistance is 1V and the current
through it is 1A, then the resistance will be:
E 1V
R = --- = ---- = 1 ohm (1)
I 1A
Assuming that the voltage across the resistance is 2V and the current
through it is 1A, then the resistance will be: 2 ohm.
---
Why would I want to do that? I'm specifically setting up a set of
conditions to illustrate _my_ point, not yours.
I am trying to illustrate that I can make a point that 0/0=2.
You cannot discard my point without adding extra information about your
set of conditions.
---This extra information is not present in the 0/0 term, but it _is_
explicitly written in lim x->0 x/x and lim x->0 2x/x , respectively.
Now, if we go to the more general case of:
x
y = ---
x
we can see that for any value of x, as x goes to zero, y will remain
constant, and exactly equal to 1. Therefore,
0
--- = 1
0
This is only true because you assumed a resistor of 1 ohm. If you assume
a resistor of 2 ohm, then 0/0 = 2.
---
Yes, of course. But I didn't "assume" a resistance of one ohm, I
selected the voltage and current to force the resistance to one ohm.
---
Again, you can only state with concidence that 0/0 = 1 in this case
because you already *know* that the resistance is 1; you have not
computed it from 0/0, because the 0/0 quotient doesn't help you to know
that the resistance is 1 ohm.
---
The game being to prove that 0/0 = 1, I'm not looking so much for a
resistance of 1 ohm as I am a set of values which when divided by
themselves will result in a quotient of 1.
You want 1 ohm, that's what you bring into the computation. You're
setting everything up so that 1 ohm results, which means it's circular
reasoning.
---If you hadn't set everything up that way, the 0V/0A measurement would
leave you stumped as to the value of the resistor, and 0/0=??? then.
---Again, the set of values that are divided by itself is x/x for all x in
R\{0}, and lim x->0 x/x = 1 as well.
Let's make them each equal to 1E-40:
1E-40
x = ------- = 1
1E-40
Damn! That x is still equal to 1!
It seems that no matter what we do, as long as the numerator and
denominator are equal, the quotient will always be 1. So, if the
smallest number we can come up with is 0, and if 0 = 0, then it seems
we can say:
0
x = --- = 1
0
You are doing the limit argument nicely.
Now you arrived at x=1 not by dividing by 0 outright, but by taking
larger values and going closer to 0. I would explicitly write what you
did, so your last formula would change to
r
x = lim --- = 1
r->0 r
This is mathematical shorthand for your reasoning that any nonzero value
plugged into r/r is 1, so it makes sense to take r/r=1 for r=0 as well.
---The value of x is unimportant. What does matter is that the numerator
and denominator be numerically equal.
No, it matters that they are algebraically equal.
---
Yeah, good point. they have to have the same sign in order for the
quotient to come out positive.
You misunderstood my point. My point is that you have to arrive at x/x
or r/r before you plug in the zero. Conmsider: if you had x=2r/r, put in
r=0 to get x=2*0/0 and use 2*0=0 to simplify to x=0/0, then the
numerator is algebraically 0=2r and the denominator is algebraically
0=r, and then
2r
x = lim ---- = 2
r->0 r
which would lead you to conclude that 0/0=2, in this case. You set your
case up so that x=r/r, i.e. numerator and denominator are algebraically
the same *before* you plug in the zero.
---This is not true for R=E/I, which is why you can't make a measurement in
the E=0, I=0 case (unless you have extra information).
If I give you a resistor to measure, but no power supply, you will
measure 0A and 0V, but must the resistor always be 1 ohm, then?
---
Since R = E/I, I won't be able to make a measurement, so the cat will
be both dead and alive.
You seem to grasp that so well. You can certainly measure 0V and 0A, but
you can't determine R. For what reason do not grasp the analogy that 0/0
represents the same condition in mathematics?
No 0/0 is undefined, it has _no_ value.snip> hence
0
--- = k
0
for all finite k.
Indeed lim x -> A f(x) can exist even when f(A) doesn't. Example: f(x)It is not always the case that
lim x -> A f(x) = f(A).
Your `arguments' are meaningless.No 0^0 is meaningless .
Your rants are meaningless, since you never include supporting arguments and
you never answer contrary arguments, even when documented with authoritative
references.
Nope, you are already there.Just a few more steps, then,
Meaningless comment.If you have no money, how large is the set of that which you can't
afford?
Fog never lifts on meaningless crap as posted by you.I see. In order not to have to deal with what you don't undestand you
label it 'crap' declare it 'meaningless' and wait for the fog to lift.
You bore with this nonspeak garbage. Go troll elsewhere Tonto.John Fields
'Undefined' is a human artefact
Not in mathematics, you idiot. Run along and troll elsewhere.