[eli5] Why is imaginary number root -1? Why not root -n, since no number times each other would result in negative number?

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[eli5] Why is imaginary number root -1? Why not root -n, since no number times each other would result in negative number?

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9 Answers

Anonymous 0 Comments

sqrt(-1) is the part that can’t be simplified but you can factor it out of anything. sqrt(-n) is simply sqrt(-1 * n) and sqrt(a * b) = sqrt(a) * sqrt(b) so you’d get sqrt(-1) * sqrt(n).

Anonymous 0 Comments

You can write any negative square root as i times a number. -1 is used to make things as simple as possible. For example take sqrt(-4):

sqrt(-4) = sqrt(-1)*sqrt(4) = 2i

Anonymous 0 Comments

Something has to be the “standard” unit. For real number, it is “1”, for imaginary number, it is “i”. The value of i is conventionally agreed to be sqrt of (-1). Of course if everyone agree, “i” could be sqrt of (-2) or (-3) or (-n), but why do that, it will look weird. Just make an xy graph and you will see why doing so is weird.

And no, it doesn’t solve the problem of “no number times each other would result in negative number”. Per your proposal, i x i = -n which is still a negative number.

But more importantly, there are numbers that when times each other (square), result in negative number. *They are imaginary numbers.*

Anonymous 0 Comments

All numbers that can be written as √(-n) where n is a real number are imaginary.

Only √(-1) is the imaginary *unit.*

Think of it like “north” or “°C” or “meter.” It’s not that √(-1) is special because it’s somehow any more or less like any other number that is the square root of a negative number. Instead, it’s that it is the *simplest* such number, just as 1°C is the simplest *unit* of Celsius temperature or 1m is the simplest *unit* of metric distance.

Anonymous 0 Comments

>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.

Anonymous 0 Comments

Because the purpose for imaginary numbers is that you can calculate with them in the same way as with real numbers. If i was just an abstract element sqrt(-n) then sqrt(-1) would be i but sqrt(-4) would also be i. But then we have i=sqrt(-4)=sqrt(4)*sqrt(-1)=2*i. This is only possible if i=0 and the whole thing falls apart.

Or to make a more simple argument, if i=sqrt(-n) then what is i² ? It’d have to be undefined and what use is an imaginary number if you can’t even square it?

Anonymous 0 Comments

That’s because people chose i = sqrt(-1) as a basis for imaginary numbers, like 1 = sqrt(1) is the basis for real ones.

Anonymous 0 Comments

all other numbers are made of ones.

One is the elementary particle that cannot be broken down into any more component parts (except, obviously, it can, but not for these purposes).

Anonymous 0 Comments

Math is about abstractions. In math, we are always interseted in properties of things, not what the things actually are. If in any context there are properties we care about and if two things have all those properties the same, we consider them essentially same. It’s like either is just a relabeling of the other.

For example, in the context of structures that are composed of a set with an operation on that set,

>the set of integers {1,-1} with the operation * (multiplication), that is

>1*1=1, 1*(-1)=-1, (-1)*1=-1, (-1)*(-1)=1

is considered to be the same as

>the set rotational symmetries of a nonsquare rectangle with the operation ∘ (of composition of maps)

because there are just two rotational symmetries of a nonsquare rectangle, rotation by 0 and rotation by straight angle, which if I label by symbols *S*, *F* (*S* for “stays” and *F* for “flips”), satisfy

>*S*∘*S=S, S*∘*F=F, F*∘*S=F, F*∘*F=S*

one can clearly see that it’s “the same” formulas as for the set {1,-1} with operation *. We have just relabeled the element 1 with *S*, the element -1 with *F*, and the operation * with ∘.

In the context of your question, defining imaginary numbers by adding a number *i* such that *i^2*=-1 to real numbers with addition and multiplication, and defining “different imaginary numbers” by adding *x* such that *x^2=-n* for some positive *n* other than 1, one finds that either is just a relabeling of the other, so *both* will be called complex numbers.