Concept of Imaginary numbers

Question

Why does $i = \sqrt{-1}$? Why does this concept work? Why did someone think it should be this? I'm not fully grasping this concept.

Answer 1

Remove the 'does'. We define $\sqrt{-1}$ to be $i$ simply because $i$ is a more compact notation; otherwise, there has been no change.

Similar notational conventions exist in other parts of algebra. Consider, for example, subtraction on the nonnegative integers. In that system, $2-5$, say, has no answer. To make such questions answerable, we accept that forms like $0-3=2-2-3$ are also numbers, and we shorten them as $-3$, in this case. Thus we have the negative integers. Something similar occurred for integer division. The set of integers is not closed with respect to this operation, so we extend the number system to have forms like $a/b$, where $a$ is any integer and $b$ is a positive integer. This is the only exception to the compression proclivity of mathematicians, perhaps because people could immediately connect the positive rationals with practical problems about some measurements. It took long before the negative integers were accepted, and also before the imaginary numbers were accepted because people couldn't similarly immediately connect them with things they were familiar with. Eventually when they found such interpretations, they were gladly accepted, but contracted to make them more wieldy.

However in modern algebra, we do not need to wait for intuitive interpretations of new forms that may be discovered, so long as they are well-defined and they can be viewed as extensions of previous systems.

Answer 2

If $i^2 = -1$, then you get a square root for every real.

It might be less mystical to think about extending the rationals with any square root. Take $d^2 =5$. Then $(a_1 + b_1d)(a_2 + b_2d)=(a_1a_2 + b_1b_2)+(a_1b_2 + a_2b_1)d $. This is the same pattern as for $d^2=-1$ with signs changed.

This all works because the sets we are defining occur as the images of homomorphisms of a polynomial ring.

Answer 3

Well, $i $ shouldn’t be defined as $\sqrt{-1}$. Nevertheless it’s one solution of $z^2=-1$.