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DeWayfarer · 61-69, M
Zero or any defined number, yes!
Now ask: To what is the imaginary number raised to the zero power equal?
[i]i[/i]^⁰ ?
Since [i]i[/i] is undefined, the result can not be defined. 🤷🏻♂️
[i]i[/i] is only defined with an even number power.
[i]i[/i]^ⁿ = -1, where n= 2,4,6... ∞
0 by definition is a member of the empty set, so it's neither even nor odd.
Now ask: To what is the imaginary number raised to the zero power equal?
[i]i[/i]^⁰ ?
Since [i]i[/i] is undefined, the result can not be defined. 🤷🏻♂️
[i]i[/i] is only defined with an even number power.
[i]i[/i]^ⁿ = -1, where n= 2,4,6... ∞
0 by definition is a member of the empty set, so it's neither even nor odd.
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DeWayfarer · 61-69, M
@SomeMichGuy [center][big][b][i][c=1F5E00]𝓗𝓪𝓿𝓮 𝓪 𝓰𝓸𝓸𝓭 𝓭𝓪𝔂! 😊[/c][/i][/b][/big][/center]
SomeMichGuy · M
@DeWayfarer lmao I am. I already helped steer others away from your obvious errors, so that's a definite win, so thanks for that opportunity.
DeWayfarer · 61-69, M
@SomeMichGuy [center][big][b][i][c=1F5E00]𝓗𝓪𝓿𝓮 𝓪 𝓰𝓸𝓸𝓭 𝓭𝓪𝔂! 😊[/c][/i][/b][/big][/center]
SomeMichGuy · M
@CestManan OUCH!
Pfuzylogic · M
SomeMichGuy · M
@Pfuzylogic That quote illustrates some pretty poor reasoning.
How about this:
Given:
x^p, for p > 0
we know that
x^{-p} = 1/(x^p)
so
(x^p) * (x^{-p}) = (x^p) * [1/(x^p)] = 1
but
(x^p) * (x^{-p}) = x^{p + (-p)} = x^0
So
x^0 = 1, for any positive x
Thus
lim_{x ‐> 0+} x^0 = 1
(since δ^0 = 1 for arbitrarily small δ).
How about this:
Given:
x^p, for p > 0
we know that
x^{-p} = 1/(x^p)
so
(x^p) * (x^{-p}) = (x^p) * [1/(x^p)] = 1
but
(x^p) * (x^{-p}) = x^{p + (-p)} = x^0
So
x^0 = 1, for any positive x
Thus
lim_{x ‐> 0+} x^0 = 1
(since δ^0 = 1 for arbitrarily small δ).
Pfuzylogic · M
@SomeMichGuy
i think your proof would require a post secondary math background. Just keeping it very simple!
i think your proof would require a post secondary math background. Just keeping it very simple!
SomeMichGuy · M
@Pfuzylogic The last part--the limit stuff--is from calculus, but the first part is just high school.
Adding exponents (when the base is the same) comes from there, as well as being shown that a number raised to a neg. exponent is just
1/(that number raised to the pos. exponent)
Those two things lead immediately to the general result.
The "limit" thing is also conceptually simple. Since x^0 = 1 is true for even a billionth or a trillionth, etc., it's true for ANY positive number arbitrarily close to 0. So, in the limit as x goes to 0 through positive values, it makes sense that
0^0 = 1.
Adding exponents (when the base is the same) comes from there, as well as being shown that a number raised to a neg. exponent is just
1/(that number raised to the pos. exponent)
Those two things lead immediately to the general result.
The "limit" thing is also conceptually simple. Since x^0 = 1 is true for even a billionth or a trillionth, etc., it's true for ANY positive number arbitrarily close to 0. So, in the limit as x goes to 0 through positive values, it makes sense that
0^0 = 1.
fakable · T
the answer depends on the system in which the question is asked.
if the expression "zero to the power of zero" has a mathematical meaning, then the answer could be number 1.
if the expression has no mathematical meaning, then the answer can be 0, an undefined answer or no answer.
if the expression "zero to the power of zero" has a mathematical meaning, then the answer could be number 1.
if the expression has no mathematical meaning, then the answer can be 0, an undefined answer or no answer.
KiwiBird · 36-40, F
Simple. However you may need to define the context of your question otherwise the answer will be undefined.
SomeMichGuy · M
@KiwiBird Sorry, no.
Tastyfrzz · 61-69, M
Interestingly my calculator says that 0^0 is undefined or 1. I always assumed it was 1.
Guess that answers where the dark matter went.
Guess that answers where the dark matter went.
Roundandroundwego · 61-69
Yes.
SomeMichGuy · M
Yes.
auris · M
zero