## anonymous one year ago 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 $$\sf 0^0=0$$ ?

1. 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?

2. 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

3. 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 $$n$$.

4. 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?

5. anonymous

That wouldn't work either. Consider $$f(x)=x^0$$ and $$g(x)=0^x$$. What happens as $$x\to0$$ for each?

6. zzr0ck3r

0^0 is undefined just as 1/0 is. It makes no sense. x^0 = 1 for x!=0 for notation purposes :)

7. zzr0ck3r

or you could look at it the other way, properties of exponents are the reason...

8. 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

9. zzr0ck3r

just as you would explain 1/0. It actually makes no mathematical sense.

10. anonymous

a proof using properties of exponents? can you show it to me?

11. 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 $$n<k$$ might help in smoothing out some calculation.

12. zzr0ck3r

$$0^0=1\\0*ln(0)=0$$ yikes!

13. 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).

14. 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".

15. 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

16. 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 :)

17. anonymous

I just thought of this. I'm not sure if it's an airtight proof. Seems logical to me :)