# Is $$0^0=1$$ and can be applied everywhere in mathematics?

Note by Aditya Parson
5 years, 3 months ago

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exact zero raise to the power exact zero is NOT DEFINED.

- 5 years, 3 months ago

Read this funny yet informative article http://www.askamathematician.com/2010/12/q-what-does-00-zero-raised-to-the-zeroth-power-equal-why-do-mathematicians-and-high-school-teachers-disagree/

- 5 years, 3 months ago

- 5 years, 3 months ago

http://www.wolframalpha.com/input/?i=lim+x+is+approaching+to+0+x%5Ex

- 5 years, 3 months ago

In calculus and other branches of mathematical analysis, an indeterminate form is an algebraic expression obtained in the context of limits. Limits involving algebraic operations are often performed by replacing subexpressions by their limits; if the expression obtained after this substitution does not give enough information to determine the original limit, it is known as an indeterminate form. The indeterminate forms include 0^0, 0/0, 1^∞, ∞−∞,∞/∞, 0 × ∞, and ∞/0.........so it's also a indeterminate form you have given....

- 5 years, 3 months ago

0^0 = 0^(m-m) = 0^(m) . 0^(-m) .... now as, 0^(-m) = 1 / 0 which is undefined... so 0^0 = undefined .....well first I tried to prove it like this but then I found that it wasn't a good proof as it had contradictions . :(

- 5 years, 3 months ago

it is indeterminate form in chapter limits....

- 5 years, 3 months ago

but mathematicians prove through limits that it it comes to 1?

- 5 years, 3 months ago

ok......can u xplain me how?

- 5 years, 3 months ago

read the article posted by Saurabh.

- 5 years, 3 months ago

ok thanks....i'll do that...

- 5 years, 3 months ago

limits tell us that n/0 = inf but we know that n/0 is undefined

- 5 years, 3 months ago

well someone can give me the clear proof of any number raised to 0 is 1.

- 5 years, 3 months ago

yes we can prove it by log theorem? take log and calculate.

- 5 years, 3 months ago

0^0=0/0-------------not defined

- 5 years, 3 months ago

look at this.

4^2 = 16

4^1 = 4

4^0 = 1

why 4^0 = 1?

easy to proof. imagine that the exponent is absolutely like an element of carbon molecule. exponent is like a box that contain something, copy the box and paste then imagine that they will combine and has more energy, it will be two thing that are same like x*x in mathematic.

so easy. it means that there are no element, like an empty box. because the box still present in universe, it mean there must be a symbol to represent it, that is the 1 or one.

- 5 years, 3 months ago

If 0^0=1 what's the proof and if 0^0= undefined what's the proof ???

- 5 years, 3 months ago

really nice question....... i think so it is undefined.

- 5 years, 3 months ago

I agree with that. You can easily prove that it is an indeterminate form in mathematics,

- 5 years, 3 months ago

in laws of exponent they never mention that base a is not equal to 0

- 5 years, 3 months ago

undefined. bt nt equal 2 1.

- 5 years, 3 months ago

The way I was taught, even though 0^0 is indeterminate, mathematicians will consider it to be 1.

- 5 years, 3 months ago

considering is different thing...sometimes in physics also we consider that 1/0 = infinity...but that is wrong.. 0^0 is undefined

- 5 years, 3 months ago

it should be zero..

- 5 years, 3 months ago

But many take it to be 1.

- 5 years, 3 months ago

- 5 years, 3 months ago

Yes.

- 5 years, 3 months ago

only google calculator is showing 1! but other scientific calculators finds math error.

- 5 years, 3 months ago

I know that.

- 5 years, 3 months ago

0^0 is neither 1 nor indeterminate, it is undefined... there is a difference between indeterminate and undefined. e.g. app.0/app.0 is indeterminate where as 0/0 is undefined similarly, 0^0 is undefined as 0^0 = e^(0log0) where log 0 is undefined, but (app. 0)^app.0 is indeterminate whose value is 1...if both functions which are approaching 0 used in (app0)^app0 are the same. if both functions are x. lim (x app0) x^x can be found by equating it to e^xlogx where if xapp. 0 xlogx approaches 0[this can be found by using LH rule] and hence x^x approaches 1(as e^0 = 1)

- 5 years, 3 months ago

Just a consideration

- 5 years, 3 months ago