> You’re probably familiar with the idea that each piece has infinitesimal width, but what about the question of ‘how MANY pieces are there?’. The answer to that is a hypernatural number. Let’s call it N again.
Is that right? I thought there was an important theorem specifying that no matter the infinitesimal width of an integral slice, the total area will be in the neighborhood of (= infinitely close to) the same real number, which is the value of the integral. That's why we don't have to specify the value of dx when integrating over dx... right?
The number N in question will adjust with dx (up to infinitesimal error anyway). So if dx is halved, N will double. But both retain their character as infinitesimal and hyperfinite.
But they don't retain their status as hypernaturals! dx does not need to evenly divide the interval over which the integral is taken. Whenever it doesn't, the number of slices in the integral will fail to be a hypernatural number, because one of the slices will extend beyond the interval boundary.
The theorem tells us that the area of the extended interval that uses a hypernatural number of slices has the same real part as the area of the exact interval. It doesn't tell us that the exact interval contains a hypernatural number of slices.
Yes, but that's also the entire thing I was questioning. The essay says that an integral necessarily contains a hypernatural number of infinitesimal slices. I don't think that's true.
> You’re probably familiar with the idea that each piece has infinitesimal width, but what about the question of ‘how MANY pieces are there?’. The answer to that is a hypernatural number. Let’s call it N again.
Is that right? I thought there was an important theorem specifying that no matter the infinitesimal width of an integral slice, the total area will be in the neighborhood of (= infinitely close to) the same real number, which is the value of the integral. That's why we don't have to specify the value of dx when integrating over dx... right?