Understanding the Controversy: Is 0^0 Really Equal to 0?

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SUMMARY

The discussion centers on the mathematical expression 0^0 and its validity. Participants debate whether 0^0 equals 0 or is an indeterminate form. Key points include the equations x^x = x and x^[1/x] = x, which suggest solutions of x = 0 and x = 1. Ultimately, the consensus leans towards 0^0 being undefined rather than definitively equal to 0.

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  • Understanding of exponential functions and their properties
  • Familiarity with limits in calculus
  • Knowledge of indeterminate forms in mathematics
  • Basic algebraic manipulation skills
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  • Research the concept of indeterminate forms in calculus
  • Study the properties of limits, particularly lim_{x → 0} x^x
  • Explore the definitions and implications of 0^0 in different mathematical contexts
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Mathematicians, students studying calculus, educators teaching algebra, and anyone interested in the nuances of mathematical definitions and their implications.

mathelord
a friend [no longer a user of the forum]showed me this and i felt i should as well show this to you all
let x^x=x
x^[1/x]=x
x^[1/x]-x=0
x[x^(1/x-1)-1]=x
so x is 0 and 1.
which evntually gives 0^0 as 0
in sloving with anither method,he also got -1 as x.
how true is the topic 0^0=0.
he explained that if you have nothing and you raise it to nothing you eventually get nothing.
 
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mathelord said:
...
x^[1/x]-x=0
x[x^(1/x-1)-1]=x
...
How did you get from the first step to the second ?
 
If you have x = 0, you get 1/0 which is invalid.
I also fail to see how you got from the first to the second equation.
 
I don't actually see how the first equation is valid , never mind the second one.

Could somebody please explain ?
 
It's valid because he wants to solve x^x = x, for x.
 
mathelord said:
a friend [no longer a user of the forum]showed me this and i felt i should as well show this to you all
let x^x=x
x^[1/x]=x
x^[1/x]-x=0
x[x^(1/x-1)-1]=x
so x is 0 and 1.
which evntually gives 0^0 as 0
in sloving with anither method,he also got -1 as x.
how true is the topic 0^0=0.
he explained that if you have nothing and you raise it to nothing you eventually get nothing.

mathelord, this is all pretty trivial. What is it that's confusing you? If you "have nothing and you raise it to nothing", then you don't have anything to exponentiate! You don't "eventually" get nothing, it was always nothing.
 
Last edited:
benjamincarson said:
mathelord, this is all pretty trivial...
Factoring x from 0 and getting x is some pretty nontrivial algebra. :smile:
 
hypermorphism said:
Factoring x from 0 and getting x is some pretty nontrivial algebra. :smile:
I don't see how he was "factoring x from 0 ". Anyway, It wouldn't be hard to construct a proof that shows that the equation x^{x}=x is only valid for 0 and 1. So...
0^{0}=0
1^{1}=1

Fiddle-dee-do
 
Well:

\lim_{x \rightarrow 0} x^x = 1

\lim_{x \rightarrow 0} 0^x = 0

Some times 00 = 1 is defined as for usefulness, but in general its not determined.
 
  • #10
benjamincarson said:
I don't see how he was "factoring x from 0 "...
Look at his fourth step.
benjamincarson said:
It wouldn't be hard to construct a proof that shows that the equation LaTeX graphic is being generated. Reload this page in a moment. is only valid for 0 and 1...
It would be impossible. 00 is an indeterminate form. See here and here.
 
  • #11
x^0=1 any number raised to zeor power equals 1. x*x*x*x...*x zeor amount of times equals 1 because 1 is the null value in multiplication. a number times himself zero times is equal to 1.
0^x=0 zero raised to any power equals zero. 0*0*0*0*0*...0=0 because of multiplication property. no matter how many zeors you have...you still have zero.

0^0=? Well...is it 1 becasue it is raised to zeor parts? or is it 0 becasue the 0 is raised to a power? Can't be both, but it can be none. answer: undefined.
 

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