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My simple proof of x^0 = 1 
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#1
May2907, 02:02 PM

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I always wondered, how can any number raised to the power of 0 be 1.
So, I came up with this! ( * = multiplication sign) 1 * 4 * 4 * 4 = 4^3 1* 4 * 4 = 4^2 1 * 4 = 4^1 Therefore, 1 = 4^0 


#2
May2907, 02:10 PM

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One can define exponentials in terms of the family of standard functions and it falls naturally that anything to the power zero is one. Exponentials are defined as follows:
[tex] n^x = e^{x\ln{n}} [/tex] Now put x = 0 and see what happens. 


#3
May2907, 02:14 PM

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#4
May2907, 02:16 PM

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My simple proof of x^0 = 1
[tex]1 = \frac{a^n}{a^n} = a^{nn} = a^0[/tex]
What could be more simpler than that? 


#5
May2907, 02:20 PM

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#6
May2907, 03:22 PM

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#7
May2907, 03:25 PM

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#8
May2907, 03:35 PM

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[tex] n^0 = e^{0\ln{n}} = e^0 = 1[/tex] 


#9
May2907, 03:44 PM

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Shing,
[tex]\forall a \ne 0,\;a^n = a^{n  1} a \Leftrightarrow a^n = \frac{{a^{n + 1} }}{a}[/tex] Given that a^{1}=a, [tex]a^0 = a^{  1} a \Leftrightarrow a^0 = \frac{{a^1 }}{a} = 1[/tex] 


#10
May2907, 04:10 PM

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4^3=4*4*4+0 4^2=4*4+0 4^1=4+0 4^0=0 Can you explain why your proof is better than mine? (it is, but you haven't shown why) 


#11
May2907, 06:00 PM

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4^3=1*4*4*4 4^2=1*4*4 4^1=1*4 4^0=1 


#12
May2907, 10:16 PM

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This is not a proof, it's a pattern. x^0 = 1 is defined.



#13
May2907, 11:19 PM

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Neutrino's proof is the basic elementary method. However, with Kurdt's method; I accept it. However one must be careful when taking logarithms, as this restricts the limit to which a function may take, as I found the hard way...
For example: x^x = y. Techniqually the domain has integers below 0. But when converting it to y=e^(xln(x)), it has immediately taken those integers away... Anyway, I think that here it is valid. But how can this be proven for sure? Or is it completely unneccessary to consider here... 


#14
May3007, 03:00 AM

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#15
May3007, 05:13 AM

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#16
May3007, 06:30 AM

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#17
May3007, 07:05 AM

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0^0 is either undefined or arbitrarily defined to be some value, typically one. It is clear when examining a plot of z=x^y that the value of z at (x,y)=(0,0) depends on the direction via which one approaches this (x,y) pair. All of the previous posts have approached this point in a direction that makes the value appear to be one. I could just as easily make it appear to be zero by looking at [tex]\lim_{x\to0}0^x[/tex] which is obviously zero. I could just as easily make it any complex number!
Given any complex number [itex]a[/itex], it is easy to come up with a form that reaches [itex]0^0[/tex] as some limit and evaluates to [itex]a[/itex]. Suppose [itex]0<a<1[/itex]. Using [itex]1=n/n[/tex], [tex]a=(a^{n/n})=(a^n)^{(1/n)}\equiv x^y[/tex] where [itex]x\equiv a^n, y\equiv 1/n[/itex] Note that both [itex]x[/itex] and [itex]y[/itex] approach zero as [itex]n\to\infty[/itex]. Thus [tex]0^0=\lim_{n\to\infty}(a^n)^{(1/n)} = a[/tex] ================== 0^0 is typically defined to be one for the sake of simplicity. This definition eliminates zero as a special case for power series, binomial expansions, etc. It is just a convention however, not a proof. 


#18
May3007, 07:15 AM

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DH I think you made the mistake I was about to, for this thread. This thread isnt about the commonly talked about 0^0 which one expects to find, its just about the exponent of zero in general. His original post said he proved that 4^0 = 1, which we know on solid ground =)



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