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The product of 0 and anything is $0$, and seems like it would be reasonable to assume that $0

I'm perplexed as to why i have to account for this condition in my factorial function (trying to learn haskell). It is possible to interpret such expressions in many ways that can make sense The question is, what properties do we want such an interpretation to have $0^i = 0$ is a good choice, and maybe the only choice that makes concrete sense, since it follows the convention $0^x = 0$

$0^0=999$ would be a contradiction to the power laws, because then $ (0^0)^2 = 999^2 \ne 0^ {0\cdot2} = 999$ The only two values for $0^0$ consistent with the power laws are $0$ and $1$. Is there a consensus in the mathematical community, or some accepted authority, to determine whether zero should be classified as a natural number It seems as though formerly $0$ was considered i.

Is a constant raised to the power of infinity indeterminate

Say, for instance, is $0^\\infty$ indeterminate Or is it only 1 raised to the infinity that is? That means that 1/0, the multiplicative inverse of 0 does not exist 0 multiplied by the multiplicative inverse of 0 does not make any sense and is undefined

Therefore both 1/0 and 0/0 are undefined. Generally the only reason one sees 1/0 as infinity is because some systems (incorrectly) output infinity when given dividing by zero. Show that ∇· (∇ x f) = 0 for any vector field [duplicate] ask question asked 9 years, 5 months ago modified 9 years, 5 months ago The integral of 0 is c, because the derivative of c is zero

Also, it makes sense logically if you recall the fact that the derivative of the function is the function's slope, because any function f (x)=c will have a slope of zero at point on the function.

I began by assuming that $\dfrac00$ does equal $1$ and then was eventually able to deduce that, based upon my assumption (which as we know was false) $0=1$ As this is clearly false and if all the steps in my proof were logically valid, the conclusion then is that my only assumption (that $\dfrac00=1$) must be false.

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