[eli5] Why is imaginary number root -1? Why not root -n, since no number times each other would result in negative number?
In: Mathematics
>since no number times each other would result in negative number?
That is by definition not true, i ^2 = -1 by definition. There are no real numbers that squared become a negative number but real numbers do not include all numbers.
You could do the same and say there is no number you can add to 5 and get a smaller number as a result, so negative numbers do not exist. That is a true statement for natural numbers 0, 1, 2, 3, 4 and so on. But if you use integers that contain negative numbers too like -1 then it is not longer true. The idea of negative numbers have historically been rejected by many mathematicians because the did not think it made any sense. You clearly disagree with that idea. imaginary number is an exertion to a number just like negative numbers are.
The imaginary unit is not the root of -1. It is the square of the imaginary unit that is equal to -1. -1 has two square roots i and -i because the square of both are -1. It it just like the square root of 4 is 2 and -2 because 2*2= 4 and -2* -2 =4. The possible root is the principal root but not the only root
n usually stands for any integer in a series so not a specific number.
If you want the principal root of -n you can rewrite it as -1 * n and sqrt(-1 *n) = i* sqrt(n)
So defining i as the square root if -n do not make sense because it is not a single value but a lot of possible values. You could define i as i^2 =-2 or any other negative number. It would just be impractical compared to -1.
If i^2 was -2 then the principal root of -36 would be i3 but the principal root of 36 would be 6. That is a lot less practical then i6 and 6 . So there is a good reason to pick 1.
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