Defining Success: What Does it Mean?

[pivoxa15]

no, I am only at secondary school but it makes more sense for the answer to be 0.
the answer is only 1 for other intergers to the power of 0 becuse say take 3
3to the 3 = 27 3 to the 2=9 3 to the 1= 3 you are dividing by three each time so it makes sense for the nxt to be 0 .
you do not get this with 0

Your argument with base 3 is good but If you take 0 as your base than it doesn't work because you can't divide by 0. Going from 0^5 to 0^4 doesn't work by dividing by 0.

But if you are trying to say that since 0^n where n is any number except 0 is 0 than 0^0 ought also be 0. That's not a bad one but Universe_Man's argument although wrong seems more convincing if I had to choose between the two.

 

[Universe_Man]

Wait it would be 0. I apologize if this is repetition, but I would like to give my own interpretation, just to see how I did.

1/.01=100 right?

So what if you were to put more and more zeros in the decimal, and always having it end in 1? You could make your decimal incredibly mind numbingly small and your solution would get closer and closer to infinity. So 1/0 could be defined as infinity couldn't it? If so:

given 0^-1=1/0= /infty and 0^1=0

and (0^1)(0^-1)=(/infty)(0)=0^0

Then 0^0=0

Because it doesen't matter what you multiply 0 by, you will still get zero.
plus if you put 0^0 into the calculator on windows, it gives you 0 which confirms it. And since 0 is a definable value, 0^0 is defined.

 

[masudr]

Infinity is defined, in fact, as a limit:

\lim_{x\rightarrow0}\frac{1}{x}\equiv\infty.

You have just carefully defined a process of getting to infinity. That's what infinity is: the result of some kind of process, it's not a number. And then you have gone on to disregard your careful definition of the concept of infinity and manipulated it as a real number. The reason infinity is not a proper number is because it leads to contradictions unless treated correctly as a limit.

Don't worry about what 0^0 is; it's not a problem. The value of 0^0 does not logically follow from the idea of exponentiation as being repeated multiplication. If we define it as 1 and we can still be consistent with the rest of the rules of the game, then that's what we do.

And surely you're joking (Mr. Man): some people who use this forum are much, much smarter than the people who made Windows Calculator. I'd trust them much more than I'd trust that software.

 

[d_leet]

plus if you put 0^0 into the calculator on windows, it gives you 0 which confirms it.

Uhh.. Not that the value that the windows calculator confirms anything at all, but when I try that I get 1 not 0, and why on Earth would what a calculator returns as the value of some expression validate whether or not 0⁰ is defined?

 

[shmoe]

plus if you put 0^0 into the calculator on windows, it gives you 0 which confirms it. And since 0 is a definable value, 0^0 is defined.

Here's a hint- don't use software for mathematical definitions, and certainly not a built in cheapo like windows calculator. It's sometimes easier to arbitrarily assign values to otherwise undefined things when it comes to programs, it gives a less likely chance of your program exploding, but the user has to be aware of potentially funky results.

Infinity is defined, in fact, as a limit:

\lim_{x\rightarrow0}\frac{1}{x}\equiv\infty.

First it should be:

\lim_{x\rightarrow0^{+}}\frac{1}{x}=\infty

The left hand limit will give -infinity. Secondly I wouldn't call this "defining infinity", rather we can define the use of the infinity symbol to mean a specific kind of divergence when it comes to limits (or a kind of convergence in some version of the extended reals if you prefer), but this is maybe just my preference for how to think about this notation.

 

[Universe_Man]

Okay, I'll remember that, but I also got the same thing on my Casio Scientific calculator as well