# Undefined Numbers!!!

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I have an almost perfect theory but......

Which is which; 0^0 = 0 or 0^0 = 1 or 0^0 = ∞

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There's so many different schools of argument that I won't attempt to list them. Usually in real analysis, choosing 0^0 = 1 is the most sensible option for a lot of reasons.

You can find a more complete list of reasons here.

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sometimes I just want to kill myself

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The readers Polymathmatics may have an unconventional take on the concept of humour, but it's quite worrying that you could be so dumb, Amod, not to spot a joke when it's so painfully obvious.

Not that I say worrying, not surprising, as you have already made quite a feat of demonstrating your stupidity.

Now I know that you have "issues" when it comes to directly awnsering questions, but why do you continue to post here?

CPL.Luke, please don't, the forces of sanity cannot risk further losses.

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This is like the question does 0/0=0 as 0/n=0; or 0/0=1 as n/n=1; or 0/0=infinite as n/0=infinite.

This questions hardly have a definite answer but I like to think that the answer is all three at the same time - its more fun that way!!!!

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I would have thought that it was undefined... 0/0=undef because you can't divide by zero... because I had always thought of x^(-1) as x/x/x... and x^(-2) as x/x/x/x and x^0 as x/x

Or is that just a shortcut to getting the answer or where did I pick this up?

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Hmm... I thought I saw a proof that 0/0 was undefined or something. I'm not sure, but it went something like this:

0/0=a/b.

Multiply both sides by 0. Since 0 times anything = 0, the equation is satisfied for all values of "a" and "b." Therefore, 0/0 can equal anything and everything.

I'm not sure if that made sense. I may be making a feat demonstrating my stupidity. Oh well.

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0/0 is undefined. One cannot say it is 0, a non-zero number or infinite. One can define limits of the form f(x)/g(x), with f and g going both to zero for a certain x, and then one can determine the value of such limits, but just a statement 0/0 equals XXX is nonsense, because nothing is specified over here.

0^0 can best be written as 1: take a limit of a^a, with a approaching zero from above, and you'll see that for non-zero, but very small a this goes to 1 arbitrarily close.

For the rest, I think this issue is settled here and does not need further discussion.

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Just as a quick thought that came into my brain 0/0 is like a random possibility. If you look at 0/0 simply the top 0 tells you that the number is zero whereas the bottom zero makes the number infinite so we at least know one thing, it's not zero and it's not infinite, according to that argument anyway. Also according to that argument the form 0/0 is actually a range of numbers from 0< x< infinite so with your choices, on the spur of the moment just like this thought, I would have to choose 1 as being the answer to 0/0

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Well, that isn't what happens. Plenty of thngs around here do tell you what is going on. Please read them.

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surely Any function of Zero is pointless anyway?

n^0 = n

n+0 = n

n/0 = n

etc...

and if n is Zero too, its even More pointless

some folk dont half Over-Complicate things!

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there are at least two factual errors in there, not to mention a matter of opinion that is very moot... ok, down right wrong. the value of a function at zero is incredibly useful, and is the whole basis of Taylor/Maclaurin series.

Some folk don't half not help things.

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I cant believe there is any argument here at all, and that matt grimes has crushed everyone. of course 0^0 is undefined. If you choose any of your solutions, 0, 1 or infinity, there are proof from those axioms that a=b where a and b are any number. so even if your perfect theory shows what 0/0 is, your going to have to persuade us 0.35424=4353452

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I have an almost perfect theory but......

Which is which; 0^0 = 0 or 0^0 = 1 or 0^0 = ∞

Dear 0^0 is not equal to all of three...but it is undefined I can proove it for you...

lets have a look over it...

for example we take 2^0

so 2^0 =1 i think u will agree over it...but how??

see here

2^0 can be written as 2^(1)*2^(-1)...........m I right??

If yes then 2^(-1)=1/(2^1)

so 2^1/2^1=1 its proved...

So in similar way...........

we can write 0^0= 0^(1)*0^(-1)

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

so in this way 0^0 =0^1/0^1 =0/0 =undefined....not 0 ,1 and

Regards:

Sanjay Kumar Kingrani

B.E (batch 08), Sukkur IBA

Edited by sanjay kingrani

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