4 ms·

Great article! I've always intuited the rules with which we operate on imaginary numbers as a hack - we don't know what to do with i other than squaring it, so

Great article! I've always intuited the rules with which we operate on imaginary numbers as a hack - we don't know what to do with i other than squaring it, so we avoid doing anything with it and algebra our way around the issue. Example: (i + 3) + (i + 3) == i^2 + 6i + 9 == 6i + 8 is about as logical to me as (x + 3) + (x + 3) == x^2 + 6x + 9. Sure, if we knew what the heck x was this operation would have been easier, but since we don't, we just use FOIL to work around it. In the former, somehow we know what i^2 is but not i, so we can reduce a little further but not completely. Of course, I don't know how sound this gutfeel impression is.

dbaupp 12y ago
Very sound. One formal model of complex numbers is R[i]/(i^2 + 1), which means all real polynomials in the variable i (R[i]) after "simplifying" (/) by taking i^2 + 1 = 0 (equivalently i^2 = -1)... which is exactly the process you are describing.
DanielRibeiro 12y ago
It is very sound. You can think of i as being a variable 'x', and therefore, you can see Complex numbers are polynomials over real numbers (denoted as R[x]). But you can simplify the polynomials by using the fact that x2 = 1 for this construction (since x = i on this situation). This is the exact construction of Complex numbers as an algebraic extension over the field of Real numbers[1] What is more amazing: there is no such way to do the same on the complex numbers. This is because the Complex numbers form an Algebraically closed field[2] [1] http://en.wikipedia.org/wiki/Field_extension#Examples http://en.wikipedia.org/wiki/Field_extension#Examples [2] http://en.wikipedia.org/wiki/Algebraically_closed_field http://en.wikipedia.org/wiki/Algebraically_closed_field
acjohnson55 12y ago
i's irreducible simply because it's a unit-sized basis vector for how much of a quantity is in the "i direction". It's just something to hang a coefficient on so that it doesn't mix with the real part. We could just as well invent a similar thing for the reals, and call it g. Then your result is 6i + 8g. Or we could use tuples. It doesn't matter, as long as you do proper bookkeeping of the real and imaginary parts.

dbaupp 12y ago

Very sound. One formal model of complex numbers is R[i]/(i^2 + 1), which means all real polynomials in the variable i (R[i]) after "simplifying" (/) by taking i^2 + 1 = 0 (equivalently i^2 = -1)... which is exactly the process you are describing.

DanielRibeiro 12y ago

It is very sound. You can think of i as being a variable 'x', and therefore, you can see Complex numbers are polynomials over real numbers (denoted as R[x]). But you can simplify the polynomials by using the fact that x2 = 1 for this construction (since x = i on this situation). This is the exact construction of Complex numbers as an algebraic extension over the field of Real numbers[1] What is more amazing: there is no such way to do the same on the complex numbers. This is because the Complex numbers form an Algebraically closed field[2] [1] http://en.wikipedia.org/wiki/Field_extension#Examples http://en.wikipedia.org/wiki/Field_extension#Examples [2] http://en.wikipedia.org/wiki/Algebraically_closed_field http://en.wikipedia.org/wiki/Algebraically_closed_field

acjohnson55 12y ago

i's irreducible simply because it's a unit-sized basis vector for how much of a quantity is in the "i direction". It's just something to hang a coefficient on so that it doesn't mix with the real part. We could just as well invent a similar thing for the reals, and call it g. Then your result is 6i + 8g. Or we could use tuples. It doesn't matter, as long as you do proper bookkeeping of the real and imaginary parts.