Question

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 \(\sf 0^0=0\) ?

Answer 1

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?

Answer 2

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

Answer 3

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 \(n\).

Answer 4

yes, that's the real point there.
so if 0^0 = 0 is not true, is it okay to say that 0^0 = 1?

Answer 5

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

Answer 6

0^0 is undefined just as 1/0 is. It makes no sense.

x^0 = 1 for x!=0 for notation purposes :)

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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.

Answer 12

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

Answer 13

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).

Answer 14

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".

Answer 15

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

Answer 16

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 :)

Answer 17

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