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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).
In the context of natural numbers and finite combinatorics it is generally safe to adopt a convention that $0^0=1$ 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) $0=1$ Extending this to a complex arithmetic context is fraught with risks, as is the ambition to justify limits of this form generally by analogy to the value of a particular limit of this form
Inclusion of $0$ in the natural numbers is a definition for them that first occurred in the 19th century
The peano axioms for natural numbers take $0$ to be one though, so if you are working with these axioms (and a lot of natural number theory does) then you take $0$ to be a natural number. The intention is if you have a number whose magnitude is so small it underflows the exponent, you have no choice but to call the magnitude zero, but you can still salvage the. For example, $3^0$ equals 3/3, which equals $1$, but $0^0$ equals 0/0, which equals any number, which is why it's indeterminate Also, 0/0 is undefined because of what i just said.
I heartily disagree with your first sentence There's the binomial theorem (which you find too weak), and there's power series and polynomials (see also gadi's answer) For all this, $0^0=1$ is extremely convenient, and i wouldn't know how to do without it In my lectures, i always tell my students that whatever their teachers said in school about $0^0$ being undefined, we.
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? In the context of limits, $0/0$ is an indeterminate form (limit could be anything) while $1/0$ is not (limit either doesn't exist or is $\pm\infty$) This is a pretty reasonable way to think about why it is that $0/0$ is indeterminate and $1/0$ is not
However, as algebraic expressions, neither is defined Division requires multiplying by a multiplicative inverse, and $0$ doesn't have one. Show that ∇· (∇ x f) = 0 for any vector field [duplicate] ask question asked 9 years, 7 months ago modified 9 years, 7 months ago
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