# Is zero even or odd?

Discussion in 'Electronic Design' started by Gactimus, Dec 20, 2004.

2. ### ISU WINS!Guest

Dont ever say "must". Thatw as inthe Novice exam!

3. ### Guest

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,
| chances are he is doing just the same"

4. ### Guest

In such case you may conclude that the specific limit evaluated is 1.
Not that 0/0 = 1.

Mati Meron | "When you argue with a fool,
| chances are he is doing just the same"

5. ### Michele DondiGuest

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

6. ### Michael MendelsohnGuest

[/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

7. ### Michael MendelsohnGuest

You eventually lower the voltage to 0V.
That's what I achieved with the short.
I am trying to illustrate that I can make a point that 0/0=2.
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.

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

8. ### vonroachGuest

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

9. ### vonroachGuest

No 0^0 is meaningless .

10. ### vonroachGuest

Ah, another subject where you are poorly informed.

11. ### vonroachGuest

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?

12. ### vonroachGuest

Parentheses and exponents are dealt with first.

13. ### vonroachGuest

Both are meaningless and undefined.

14. ### vonroachGuest

What a pity, you blunder on your very first point.

15. ### vonroachGuest

No it is crap with no evidence of any reasoning.

16. ### Dave SeamanGuest

Your rants are meaningless, since you never include supporting arguments and
you never answer contrary arguments, even when documented with authoritative
references.

17. ### vonroachGuest

No it allows you to conclude that it is meaningless.

18. ### vonroachGuest

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

19. ### vonroachGuest

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

20. ### vonroachGuest

But, but Mati, nobody is arguing with you.  