J
John Fields
- Jan 1, 1970
- 0
This is true, but it is true because
x
y = lim x->0 --- = lim x->0 1 = 1.
x
The "1" on the left of the equals sign is there because the two x's
cancelled.
This is true, but it is true because
x
y = lim x->0 --- = lim x->0 1 = 1.
x
The "1" on the left of the equals sign is there because the two x's
cancelled.
Vince Fiscus said:An even number plus an even number equals an even number.
An odd number plus an even number equals an odd number.
An odd number plus an odd number equals an even number.
0 + 1 = odd number
0 + 2 = even number, 2 is not odd, so zero must be even.
KB7ADL
Sigh. Forget machines. Is the definition of division clear to you?OK, look at it this way:
These two number lines are behind a screen, so the numbers can't be
seen, but arranged in such a way that when a number is picked from
one, the same number will be picked from the other and those two
numbers will be divided and the quotient presented for viewing.
-1 0 1
-|-------------------------|-------------------------|-
-1 0 1
-|-------------------------|-------------------------|-
I fail to see why, at one infinitesimal spot on the line(s), the
division becomes intractable and the machine refuses to output a 1.
using about half a dozen or so different C compilers on various platforms, and
every single one of them printed 1.000000. Similarly for most other
programming languages that I have tried. However, the mathematical definition
[/QUOTE]John said:This is true, but it is true because
x
y = lim x->0 --- = lim x->0 1 = 1.
x
The "1" on the left of the equals sign is there because the two x's
cancelled.
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.
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
The task at hand is to show whether the rule
is valid or otherwise. Can you do that?
0 factorial is an empty product, and therefore equal to 1.
0^0 is an empty product, and therefore equal to 1.
The same argument also shows that empty sums are equal to 0.
Not really, simply cooling a resistive material
won't usually reduce the resistance to zero. The
superconducting state is fundamentally different.
What a profound observation. How many years have you ben working on0 = k * 0
Ah, but the normal order of precedence dictates that the
multiplication be performed first,
So (k*0)/0 is not equal to k*(0/0), then?
What a pity!
Obviously, but no matter.
Consider:
If my point is that 0/0 = 1,
No, that is circular reasoning:
"If 0/0 = 1 the 0/0 must be equal to 1".
On Tue, 28 Dec 2004 23:21:43 +0000 (UTC), Dave Seaman
No 0^0 is meaningless .
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:
And if those x's were zeroes when they were cancelled, that still
results in a quotient of 1, so 0/0 = 1
The 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.
Mati Meron | "When you argue with a fool,
[email protected] | chances are he is doing just the same"