Is \(0^0=1\) and can be applied everywhere in mathematics?

Note by Aditya Parson
5 years ago

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

Ashutosh Mittal - 5 years ago

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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/

Saurabh Dubey - 5 years ago

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

Tan Wei Sheng - 5 years ago

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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....

Raja Metronetizen - 5 years ago

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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 . :(

Zubayet Zico - 5 years ago

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it is indeterminate form in chapter limits....

Riya Gupta - 5 years ago

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

Aditya Parson - 5 years ago

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

Riya Gupta - 5 years ago

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@Riya Gupta read the article posted by Saurabh.

Aditya Parson - 5 years ago

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@Aditya Parson ok thanks....i'll do that...

Riya Gupta - 5 years ago

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

Djordje Marjanovic - 5 years ago

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

Diksha Verma - 5 years ago

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

Attain k Gupta - 5 years ago

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

Sreehari Vp - 5 years ago

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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.

how about x^0?

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.

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

Zubayet Zico - 5 years ago

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

Diksha Verma - 5 years ago

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

Aditya Parson - 5 years ago

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

Diksha Verma - 5 years ago

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undefined. bt nt equal 2 1.

Shubham Malik - 5 years ago

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

Kenneth Chan - 5 years ago

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

Arushit Mudgal - 5 years ago

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it should be zero..

Diksha Verma - 5 years ago

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But many take it to be 1.

Aditya Parson - 5 years ago

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LIke google! https://www.google.com/search?q=0^0

Kenneth Chan - 5 years ago

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@Kenneth Chan Yes.

Aditya Parson - 5 years ago

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

Diksha Verma - 5 years ago

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@Diksha Verma I know that.

Aditya Parson - 5 years ago

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@Aditya Parson 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)

Jatin Yadav - 5 years ago

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Just a consideration

Arushit Mudgal - 5 years ago

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