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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?

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    $\begingroup$ We know, by definition, that $$i = \sqrt{-1}.$$ $\endgroup$ Commented Oct 18, 2014 at 2:17
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    $\begingroup$ Alternatively, by definition, the two roots of $x^2+1=0$ are denoted $i$ and $-i$. Hence, by rearrangement, $i^2=-1$. $\endgroup$ Commented Oct 18, 2014 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. $\endgroup$ Commented Oct 18, 2014 at 2:21
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    $\begingroup$ Because two $90^\circ$ rotations in the same direction make a $180^\circ$ rotation? $\endgroup$ Commented Oct 18, 2014 at 2:31

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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 $a,b \in \mathbb{R}$.

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    $\begingroup$ 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!! $\endgroup$ Commented Oct 18, 2014 at 4:00
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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 $i^2=(-1,0)=-1$.

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  • $\begingroup$ This is very concise (+1) and to the point $\endgroup$ Commented Oct 18, 2014 at 4:01
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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.

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    $\begingroup$ 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. $\endgroup$ Commented Oct 18, 2014 at 4:11
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    $\begingroup$ This is not only black magic. (+1)The way how Jamil used the rule for square root multiplication with negative radicants is simply false! $\endgroup$ Commented Oct 18, 2014 at 4:53
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A note: Since $i^2 = -1$, there is no way to tell the difference between $i$ and $-i$.

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  • $\begingroup$ Yes there is, assume $i=-i$, by adding $i$ to both sides you get $2i=0$ which is a contradiction. So $i \neq -i$ $\endgroup$ Commented Jul 27, 2015 at 21:08

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