CauTi0N's Site

ID:111588

Apr 4 2011, 2:07 pm

For those of you that were recently discussing the debate over 0^0, I decided to do some research on the subject and this is the following I have come up with based on several mathematician's arguments.

The argument is posed because it questions the contradiction that 0 multiplied by anything must be 0, and x^0 must always be 1. So does 0^0 = 0, indeterminate, or 1?

Technically, it is impossible to say on either sides. It has been defined several times as indeterminate, and has also been defined several times as 1. Many mathematicians say pending your use, you can allow it to go either way - should your proof require it to be 1, that is acceptable. If it must be indeterminate, then that is acceptable as well. (These statements lead me to believe there may therefore be a flaw in mathematics, but that's a separate issue.)

However, considering the nature of the law of x^0, one can estimate that the limit as x->0 would still remain at 1, and therefore the general consensus of the debate is that 0^0=1.

#1 Apr 4 2011, 2:27 pm
Yut Put	I am studying this currently in algebra. What a coincidence, since when I was studying slope in algebra we both came out with 3D demos(And mine used slope for calculations)

#2 Apr 4 2011, 2:30 pm
CauTi0N	My demo was created with slope in mind, but I initially used angles to attempt it. Angles work, but they provide a funky graphic, so I switched over to a modified slope formula.

#3 Apr 4 2011, 3:07 pm
Masterdarwin88	Interesting, but kinda old news. 0 to the power of 0 cannout be 0, for 0 is a number but represents the absence of nothing. If something is to the power of 0, it is automatically one. In some cases though, as said by Caution, it is indeterminate depending on the siutation :c

#4 Apr 4 2011, 3:17 pm
ExPixel	Off course mathematics would be flawed, it evolved using the knowledge of people around the world living in different time periods. I try not to think about the power of 0 and such, but when I do it usally ends up like this in this odd little brain of mine: 2^5 or 1x1x2x2x2x2x2 So 0^0 is 1 until my teacher says otherwise :D

#5 Apr 4 2011, 3:25 pm
Fugsnarf	Looking at it simply, if I have nothing and multiply it by something then I still have nothing. I see no logical reason why raising 0 by 0 would change anything except to preserve your petty all-encompassing rules. Of course, if you want to say 0^0 is 1, I really have no problem with that. It's not like it's anything useful anyways. Might be better just to say it's undefined.

#6 Apr 4 2011, 3:40 pm

CauTi0N

Fugsnarf wrote:

Looking at it simply, if I have nothing and multiply it by something then I still have nothing. I see no logical reason why raising 0 by 0 would change anything except to preserve your petty all-encompassing rules.

I had your reasoning too, but considering taking the limit as x->0 of x^0, and applying a modified squeeze theorem, one could argue that the case of it equaling 1 is justified. However, indeterminate is the best solution IMO.

#7 Apr 4 2011, 4:36 pm (Edited on Apr 4 2011, 4:48 pm)
Toadfish	So the correct answer is: for your purposes, it's 1.

#8 Apr 4 2011, 4:40 pm
CauTi0N	No, either indeterminate or 1.

#9 Apr 4 2011, 4:48 pm (Edited on Apr 4 2011, 4:53 pm)
Toadfish	Er, meant to type 1 there. ;P To expand: it depends on how you want to define it. Defining it as 1 gives you many good properties which you can then use. Defining it as 0 can be convenient in different contexts, but those aren't contexts you would encounter in a HS calculus class. It is not indeterminate by default, but you can define it as such.