Maker Pro
Maker Pro

Is zero even or odd?

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.
 
I

ISU WINS!

Jan 1, 1970
0
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

Dont ever say "must". Thatw as inthe Novice exam!
 
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.
Sigh. Forget machines. Is the definition of division clear to you?
the result of the division a/b is such a number c that c*b = a.
That's *all* there is to it. So, the result of 0/0 is a number c such
that c*0 = 0. Can you tell me what c is?

OK, here is a little exercise for you. At x = 0, sin(x) is also zero.
And, so is any positive power of either x or sin(x). So, using the
Taylor expansion for sin(x), evaluate the following limits:

1) lim_x->0 sin(x)/x
2) lim_x->0 sin(x)/x^2
3) lim_x->0 (sin(x))^2/x

Mati Meron | "When you argue with a fool,
[email protected] | chances are he is doing just the same"
 
M

Michele Dondi

Jan 1, 1970
0
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

Since this was (cross)posted -for reasons that I can not imagine- also
to clpmisc, to bring it _slightly_ more on topic:

$ perl -le 'print 0**0'
1

BTW: my good 'ol faithful HP-28s tells me the same.


PS: the abondance of crankery this thread is revealing is tending to
make me feel homesick of my sci.math days... (has JSH pop out with an
FLT-based proof about 0^0 being... <whatever>?!?)


Michele
 
M

Michael Mendelsohn

Jan 1, 1970
0
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.
[/QUOTE]

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.

It is desirable for mathematics not to allow two contradictory
statements such as 0/0=1 and 0/0=2 to be true at the same time; and your
resoning allows these statements to be true in some cases (we did this
in the Ohm's law episode, where you never said my reasoning was wrong,
it was merely not the thing you wanted to prove).

0/0=1 and 0/0=2 leads to 1=2, which is undesirable.

lim x->0 x/x = 1 and lim x->0 2x/x = 2 doesn't have this problem.
It is downright fine and dandy, because
2 lim x->0 x/x = lim x->0 2x/x

It is more inconvenient to write, so maybe you could agree to rephrasing
that 0/0=1 where 0/0 is shorthand for lim x->0 x/x ?
(You would decidedly be in a minority if you used that shorthand).

Cheers
Michael
 
M

Michael Mendelsohn

Jan 1, 1970
0
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.

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.


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.

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).

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?

Cheers
Michael
 
V

vonroach

Jan 1, 1970
0
The task at hand is to show whether the rule
is valid or otherwise. Can you do that?

Otherwise? Accept your opinion? Crazy George is really crazy!
 
V

vonroach

Jan 1, 1970
0
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.

No 0^0 is meaningless .
 
V

vonroach

Jan 1, 1970
0
Not really, simply cooling a resistive material
won't usually reduce the resistance to zero. The
superconducting state is fundamentally different.

Ah, another subject where you are poorly informed.
 
V

vonroach

Jan 1, 1970
0
0 = k * 0
What a profound observation. How many years have you ben working on
this concept? I suppose k*k= k^2 will be next? Have you carefully
assured yourself that 1+1=2?
 
V

vonroach

Jan 1, 1970
0
Ah, but :) the normal order of precedence dictates that the
multiplication be performed first,

Parentheses and exponents are dealt with first.
 
D

Dave Seaman

Jan 1, 1970
0
On Tue, 28 Dec 2004 23:21:43 +0000 (UTC), Dave Seaman
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.
 
V

vonroach

Jan 1, 1970
0
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:

No it allows you to conclude that it is meaningless.
 
V

vonroach

Jan 1, 1970
0
And if those x's were zeroes when they were cancelled, that still
results in a quotient of 1, so 0/0 = 1

Chuckle....0/0 is meaningless and you are wrong and bordering on
lunacy.
 
V

vonroach

Jan 1, 1970
0
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.

0/n where n=any number except 0 = 0. You seem to be trying to make
something hard out of a triviality.
 
Top