1
$\begingroup$

This question already has an answer here:

In this KhanAcademy link at 2:25, Sal (the narrator) says that $i^2$ is negative 1 and he didn't explain why.

Why is this so? What is the intuition behind it?

$\endgroup$

marked as duplicate by apnorton, hardmath, Ross Millikan, user147263, Claude Leibovici Oct 18 '14 at 5:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 4
    $\begingroup$ We know, by definition, that $$i = \sqrt{-1}.$$ $\endgroup$ – user795305 Oct 18 '14 at 2:17
  • 1
    $\begingroup$ Alternatively, by definition, the two roots of $x^2+1=0$ are denoted $i$ and $-i$. Hence, by rearrangement, $i^2=-1$. $\endgroup$ – Fengyang Wang Oct 18 '14 at 2:20
  • $\begingroup$ Yeah, it is by definition. We just label $\sqrt{-1}$ as $i$. It is much easier to write, takes up less space, and also reminds us that $\sqrt{-1}$ is not a real number. $\endgroup$ – layman Oct 18 '14 at 2:21
  • 1
    $\begingroup$ Because two $90^\circ$ rotations in the same direction make a $180^\circ$ rotation? $\endgroup$ – bof Oct 18 '14 at 2:31
11
$\begingroup$

It is defined that way.

I believe the proper definition of $i$ is $i^2 = -1$ and not $i = \sqrt{-1}$ as is commonly stated.

It is useful in many deep areas of mathematics, but as a starting point, it serves as a solution to $x^2 + 1 = 0$.

I recommend looking deeper into the theory of complex numbers, particularly the polar form.

To help you understand further. Realize that we have many sets of numbers such as the natural numbers, the rationals, and the reals. The complex number system is just another set and each complex number can be defined as an ordered pair $a + ib$ or $(a,b)$ where $a,b \in \mathbb{R}$.

$\endgroup$
  • 1
    $\begingroup$ Yes (+1) The better definition is that $i$ is the imaginary unit for which its square is $-1$ The introduction of $i=\sqrt{-1}$ is a rather poor definition found in a lot of high school books, but you won't find that in complex analysis books. Thanks for pointing that out!! $\endgroup$ – imranfat Oct 18 '14 at 4:00
7
$\begingroup$

We define two mappings of the sets $\mathbb{R}^2\times\mathbb{R}^2\to \mathbb{R}^2$ by

$$+:\langle(a,b),(c,d)\rangle\mapsto(a+c,b+d)$$ $$\cdot:\langle(a,b),(c,d)\rangle\mapsto(ac-bd,ad+bc)$$

They're respectively addition and multiplication. For these two mappings the axioms of a field are satisfied (why?). The field thus written is what is called the field of the complex numbers $\mathbb{C}$ (you can find other definition but all are equivalents).

The mappping $x\to (x,0)$ from the reals to the complex is clearly injective and preserve addition and product in $\mathbb{R}$, for that reason we can consider $\mathbb{R}$ as a subfield of $\mathbb{C}$ and abusing the notation when we say the real $x$ in $\mathbb{C}$, really is $(x,0)$. Let $a+ib:=(a,b)$. The element $i=(0,1)$ is such that $i^2=(-1,0)=-1$.

$\endgroup$
  • $\begingroup$ This is very concise (+1) and to the point $\endgroup$ – imranfat Oct 18 '14 at 4:01
0
$\begingroup$

From a purely computational point of view, we can show why $i^2=-1$. By definition, $i=\sqrt{-1}$. So what does $i^2$ mean? Well, if we are squaring something, then we multiply that object by itself. So $i^2=i \cdot i = \sqrt{-1} \cdot \sqrt{-1} = \sqrt{(-1)^2} = -1$.

We can proceed this way to find $i^3$ and $i^4$ as well.

$\endgroup$
  • 2
    $\begingroup$ There's some black magic there. It's not true in general that $\sqrt{a}\sqrt{b} = \sqrt{ab}$, unless you select a branch. For example, you could do $-1 = \sqrt{-1}\sqrt{-1} = \sqrt{1} = 1$ if you are not careful. $\endgroup$ – Chantry Cargill Oct 18 '14 at 4:11
  • 2
    $\begingroup$ This is not only black magic. (+1)The way how Jamil used the rule for square root multiplication with negative radicants is simply false! $\endgroup$ – imranfat Oct 18 '14 at 4:53
-1
$\begingroup$

A note: Since $i^2 = -1$, there is no way to tell the difference between $i$ and $-i$.

$\endgroup$
  • $\begingroup$ Yes there is, assume $i=-i$, by adding $i$ to both sides you get $2i=0$ which is a contradiction. So $i \neq -i$ $\endgroup$ – JacksonFitzsimmons Jul 27 '15 at 21:08

Not the answer you're looking for? Browse other questions tagged or ask your own question.