anonymous
  • anonymous
Quick Question: Exponents I know this has been asked so many times but what is the proper explanation to why \(\sf n^0=1\) and \(\sf 0^0=0\) ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
oh i get the first one.. from the law of exponents \(\sf \Large \frac{a^m}{a^n}=a^{m-n}\) If \(\sf m=n\), it will be like: \(\sf \Large \frac{a^m}{a^m}=a^{m-m}=a^0\) for example, let's say \(\sf 2^3\) \(\sf \Large \frac{2^3}{2^3}=2^{3-3}=2^0\) this is the same as\(\sf \frac{8}{8}=1\) But how about the second one.. 0^0 is equal to zero and not 1?
anonymous
  • anonymous
Is it because any number you multiply by zero is still zero or something like that lol 0/0 is not usually "undefined" from what i know zero is an interesting number hmm
anonymous
  • anonymous
It's not true that \(0^0=0\). Consider this rewriting: \[0^0=e^{\ln0^0}=e^{0\ln0}\] Let's say for the moment that \(0\ln0\) makes sense. Then that would mean that there's some number \(n\) such that \(e^n=0\). But there is no such \(n\).

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anonymous
  • anonymous
yes, that's the real point there. so if 0^0 = 0 is not true, is it okay to say that 0^0 = 1?
anonymous
  • anonymous
That wouldn't work either. Consider \(f(x)=x^0\) and \(g(x)=0^x\). What happens as \(x\to0\) for each?
zzr0ck3r
  • zzr0ck3r
0^0 is undefined just as 1/0 is. It makes no sense. x^0 = 1 for x!=0 for notation purposes :)
zzr0ck3r
  • zzr0ck3r
or you could look at it the other way, properties of exponents are the reason...
anonymous
  • anonymous
x^0 will always be in y=1, while the other one will be at y=0. I think I got it, but I just don't know how to explain it. This probably means that 0^0 doesn't have a "right" value for it. I just asked this because I saw an article which says that they set 0^0=1 as a standard answer for 0^0 :P
zzr0ck3r
  • zzr0ck3r
just as you would explain 1/0. It actually makes no mathematical sense.
anonymous
  • anonymous
a proof using properties of exponents? can you show it to me?
anonymous
  • anonymous
I suppose it can be convenient at times to define \(0^0\) to take on the value of 1. It depends on the context, in the same way that defining \(\dbinom nk=0\) for \(n
zzr0ck3r
  • zzr0ck3r
\(0^0=1\\0*ln(0)=0\) yikes!
anonymous
  • anonymous
The point is that the limits of \(0^x\) and \(x^0\) as \(x\to0\) suggest that \(0^0\) could actually take on two values, but that would break some axiom (possibly the reflexive? I forget the names; the one that says \(x=x\) or something to that effect).
anonymous
  • anonymous
The key word here is "indeterminate". The so-called number can behave one way from one perspective, and behave in a completely different way from another. Another indeterminate form that might be easier to grasp is \(\dfrac{0}{0}\). If we suppose there is some number \(x\) for which \(x=\dfrac{0}{0}\), then that would suggest that \(0x=0\), which is true for all \(x\). But we start off by assuming \(x\) has a definitive value. Yet \(0x=0\) is true for all \(x\). We can't *determine* its value, hence the name "indeterminate".
anonymous
  • anonymous
assume 0^0 is well defined. then by properties of exponents 0^0= 0^(3-3) = 0^3/0^3 = 0^3* 1/0^3 = 0*1/0 but 1 / 0 is undefined contradiction
anonymous
  • anonymous
Yep, got it! I think I'm satisfied with all your explanations. I hope I can give you all a medal, but I can't so I'll just fan you. Thank you guys for all your useful inputs :)
anonymous
  • anonymous
I just thought of this. I'm not sure if it's an airtight proof. Seems logical to me :)

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