02 Jpg The product of 0 and anything is $0$, and seems like it would be reasonable to assume that $0! = 0$. 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 $0^x = 0$. on the other hand, $0^ { 1} = 0$ is clearly false (well, almost —see the discussion on goblin's answer), and $0^0=0.
Image0 2 The Players Magazine 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? i am just curious. say, for instance, is $0^\\infty$ indeterminate? or is it only 1 raised to the infinity that is?. 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. 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.
02 0 Free Photo Download Freeimages 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. 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. Why is any number (other than zero) to the power of zero equal to one? please include in your answer an explanation of why $0^0$ should be undefined. @swivel but 0 does equal 0. even under ieee 754. the only reason ieee 754 makes a distinction between 0 and 0 at all is because of underflow, and for ∞, overflow. 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. Whether or not $0$ divides $0$ in any particular mathematical context depends on how divisibility is defined there. your reference to "nan" suggests that you're interested in whether some computer language allows $0 0$. that depends on the language. @arturo: i heartily disagree with your first sentence. here's why: 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.
05 02 01 Royalty Free Vector Image Vectorstock Why is any number (other than zero) to the power of zero equal to one? please include in your answer an explanation of why $0^0$ should be undefined. @swivel but 0 does equal 0. even under ieee 754. the only reason ieee 754 makes a distinction between 0 and 0 at all is because of underflow, and for ∞, overflow. 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. Whether or not $0$ divides $0$ in any particular mathematical context depends on how divisibility is defined there. your reference to "nan" suggests that you're interested in whether some computer language allows $0 0$. that depends on the language. @arturo: i heartily disagree with your first sentence. here's why: 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.
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