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The product of 0 and anything is $0$, and seems like it would be reasonable to assume that $0 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. I'm perplexed as to why i have to account for this condition in my factorial function (trying to learn haskell).

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It is possible to interpret such expressions in many ways that can make sense 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) $0=1$ 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.