0^0 = 1?


by The Rev
Tags: None
arildno
arildno is offline
#73
May28-05, 07:00 AM
Sci Advisor
HW Helper
PF Gold
P: 12,016
BoTemp was right in critizing me, shmoe; I hadn't made the proper restriction on x.
shmoe
shmoe is offline
#74
May28-05, 07:41 AM
Sci Advisor
HW Helper
P: 1,996
How you stated it is fine to me, you were talking about a right hand limit. There's no reason for anyone to assume (or even care) if your function was defined at 0.
Sabine
Sabine is offline
#75
May28-05, 12:09 PM
P: 41
i agree that 0^0= 1

a^x= 1/(a^(-x))

if a=x=0 then there is no way that 0^0= 0 because we'll have 0=infinite(1/0)
Hurkyl
Hurkyl is offline
#76
May28-05, 12:52 PM
Emeritus
Sci Advisor
PF Gold
Hurkyl's Avatar
P: 16,101
if a=x=0 then there is no way that 0^0= 0 because we'll have 0=infinite(1/0)
No, there's no way that 0^0 = 0 because 0^0 is undefined.


If you like heuristic reasoning from identities, what about 0^x = 0?
Rahmuss
Rahmuss is offline
#77
May28-05, 01:17 PM
P: 223
I don't think of 0 as being in the same number system as anything else really. It's more of a concept like infinity.

So you really can't do all of the same math with 0 as you can with other numbers. [tex]0^0[/tex] makes no sense. Nor would [tex]log(0)[/tex].

LOL... that's so funny... where I listed ['tex'] 0^0 [/'tex'] it put "infinity".

EDIT: And now it's back to 0^0. Hmmm....
Hurkyl
Hurkyl is offline
#78
May28-05, 03:11 PM
Emeritus
Sci Advisor
PF Gold
Hurkyl's Avatar
P: 16,101
So you really can't do all of the same math with 0 as you can with other numbers.
You meant arithmetic. And that comes directly from the definitions -- division is defined for any nonzero denominator.

Incidentally, though, for each operation which is undefined at zero (such as 1/x), there's a corresponding operation which is undefined at one. (such as 1/(x-1)) So, in a very real sense, you can do exactly as much with zero as you can do with any other real number.
cronxeh
cronxeh is offline
#79
May28-05, 04:41 PM
PF Gold
cronxeh's Avatar
P: 1,236
Quote Quote by Hurkyl
No, there's no way that 0^0 = 0 because 0^0 is undefined.


If you like heuristic reasoning from identities, what about 0^x = 0?

0^x = 0

defined: for all x > 0
undefined: for all x < 0, x = 0
Rahmuss
Rahmuss is offline
#80
May28-05, 05:14 PM
P: 223
Wait! Did someone already do this?

[tex]0 \approx (1/\infty)[/tex] Not quite; but approximately.

[tex]1/(1/\infty) \longrightarrow (1/1)/(1/\infty) \longrightarrow (\infty/1) * (1/1) \longrightarrow \infty*1 = \infty[/tex]

Though [tex]0[/tex] is a little less than [tex]1/\infty[/tex]. So what value is it really?
jtox
jtox is offline
#81
May28-05, 05:35 PM
P: 4
As far as math teachers are concerned, you may safely assume [tex]0^0 = 1[/tex] (/End deliberate hand-waving mode). As a precaution, most documents or proofs that require its use (most that I've seen, anyway) will still explicitly state it as a useful interpretation before applying it.
Zurtex
Zurtex is offline
#82
May29-05, 05:01 AM
Sci Advisor
HW Helper
P: 1,123
Quote Quote by Rahmuss
Wait! Did someone already do this?

[tex]0 \approx (1/\infty)[/tex] Not quite; but approximately.

[tex]1/(1/\infty) \longrightarrow (1/1)/(1/\infty) \longrightarrow (\infty/1) * (1/1) \longrightarrow \infty*1 = \infty[/tex]

Though [tex]0[/tex] is a little less than [tex]1/\infty[/tex]. So what value is it really?
You are wrong [tex]\frac{1}{\infty}[/tex] does not make sense for real numbers as [tex]\infty[/tex] is not an element of the real number set. As for sets in which it does exist, you need to learn their axioms and not assume that they are the same as the real numbers (otherwise they would be the real numbers).
Sabine
Sabine is offline
#83
May29-05, 05:25 AM
P: 41
hurkyl 0^x=0 if x is different from 0

0^m*0^-m=0^0=1

or as i said before a^x=1\(a^(-x)) so if a=x=0 this means

0^0= 1\(0^(-0))
in this case 0^0 should b equal to one

x^0= 1 even for x=0
matt grime
matt grime is offline
#84
May29-05, 09:06 AM
Sci Advisor
HW Helper
P: 9,398
Quote Quote by Sabine
0^m*0^-m=0^0=1

If you look very carefully you've just divided by zero: one of m or -m is negative (i'm assuming integral exponent)


Register to reply