How can $i^2 = k^2 = j^2 = ijk = -1$ be true?

Question

I have just started to learn the basics of quaternions, but I immediately run into a wall.

Litteraly the first equation on Wikipedia is as follows

$i^2 = k^2 = j^2 = ijk = -1$

This implies

$i = \sqrt{-1}$

$j = \sqrt{-1}$

$k = \sqrt{-1}$

but now $ijk = -1$ also need to be true

$\sqrt{-1} * \sqrt{-1} * \sqrt{-1} = -1$

$-1 * \sqrt{-1} = -1$

$\sqrt{-1} = 1$

This can not be true. What am I missing here?

How do you conclude that $i = \sqrt{-1}$? These are not complex numbers. — , Jul 02 '16 at 23:18
@T.Bongers If $i^2 = -1$ then $i = \sqrt{-1}$ or am I not allowed to do that? — Maik Klein, Jul 02 '16 at 23:20
*Which* square root do you mean by $\sqrt{-1}$? Note that $-1$ has *six* square roots: $\pm i,\ \pm j,\ \pm k.$ And it happens that $ij=k,$ so that $ijk=kk=-1.$ — bof, Jul 02 '16 at 23:22
@MaikKlein well it is a square root of $-1$. Also, $-i, -j, -k$ are square roots. — , Jul 02 '16 at 23:23
You don't need *quaternions* for that. You run into the same wall when you learn about *integers: $$1^2=(-1)^2=1$$ so $$1=\sqrt1\sqrt1=1*(-1)=-1$$ — bof, Jul 02 '16 at 23:26
@bof Actually there are infinitely many square roots of $-1$ in the quaternions. — egreg, Jul 02 '16 at 23:34
@MaikKlein For $\sqrt a\sqrt a=a$ to hold, I think both square root symbols have to denote the same* squart root. — bof, Jul 02 '16 at 23:59

score 7 · Accepted Answer · answered Jul 02 '16 at 23:49

There is no such thing as $\sqrt{-1}$ in the complex numbers. Don't use that symbol that's not well defined and your understanding of the complex numbers will improve.

While it is possible to define a square root function over the nonnegative real numbers, satisfying the property $\sqrt{a}\sqrt{b}=\sqrt{ab}$ for $a,b\ge0$, there is no function $f$ defined over the complex numbers satisfying

$f(1)=1$;
$(f(z))^2=z$, for all $z\in\mathbb{C}$;
$f(z_1z_2)=f(z_1)f(z_2)$, for all $z_1,z_2\in\mathbb{C}$.

(here $f$ should be the square root).

Thus you can't use the relation $\sqrt{-1}\, \sqrt{-1}=\sqrt{(-1)^2}=\sqrt{1}=1$: you see well this gives an immediate contradiction. But it is only apparent: since no function satisfies the requirements above, you can't use it. Actually, this contradiction is a proof that the above function cannot exist.

Over the quaternions the situation is even more complicated. There are infinitely many quaternions $h$ such that $h^2=-1$.

To wit, consider $h=a+bi+cj+dk$; then \begin{align} h^2 &=(a+bi+cj+dk)(a+bi+cj+dk) \\ &=a^2-b^2-c^2-d^2+2abi+2acj+2adk \end{align} so we get $$ \begin{cases} a=0 \\[4px] b^2+c^2+d^2=1 \end{cases} $$ and the second equation has infinitely many solutions (imagine the unit sphere in three-space). Among these there are indeed $\pm i$, $\pm j$, and $\pm k$.

Don't forget that the quaternions are not commutative, so seemingly mysterious things can happen. They're not mysterious, though: follow the given rules, not those that you think apply.

(Although some sources choose to define the square root function on the negative reals even if the function no longer satisfies the nice properties.) — anon, Jul 03 '16 at 00:10
@arctictern Yes, but then $z^2=-1$ does not mean that $z = \sqrt{-1}$ if you define such a function. Because then we would have that since $i^2 = (-i)^2=-1$ then $i = \sqrt{-1} = -i$. — skyking, Aug 21 '17 at 08:16
@skyking That's nothing new; $z^2=+1$ already doesn't imply $z=1$, since both $+1$ and $-1$ are square roots of $1$. (Note that your "then we would have that since $i^2=(-i)^2$ then $i=-i$" proof by contradiction is nonsense.) — anon, Sep 22 '17 at 18:51

score 0 · Answer 2 · answered Jul 02 '16 at 23:45

I think the best way to understand this in my opinion is by looking at the matrix form formulas (4,5,6,7) http://mathworld.wolfram.com/Quaternion.html,

1

I

J

k

think about if $i^2=-1 iijk=-i$ so $jk=i$ further $jiijk=-ij$ so $k=-ji$, and other properties such as $jiijkj=-jij=-kj$ so $jk=i$ and $kj=-i$ and so on.

Arguably the simplest system with non commuting elements is the 2X2 matrix group which as is shown in the mathworld article is isomorphic to 1,i,j,k

score 0 · Answer 3 · answered Feb 01 '20 at 16:08

The way to think about this is not to think of these as normal multiplication, but rather rotation. To rotate by i means to take the point at 1 and sort of move it 90 degrees up to i. Rotation by j and k is completely similar. All other numbers on the unit circles of i, j, and k for their respective multiplications follow the same 90-degree rotations. What happens to the other quaternions? Well, the unit circle of j and k, for the rotation of i, follows a similar motion, where j goes to k, and k goes to -j. And, for multiplications by other quaternions, you can probably guess what happens by now. Obviously, 180 degrees of rotation starting at 1 gives -1. Now we can tackle ijk. As I already said, ij gives k. This means that ijk = k^2. Obviously, this is -1. Problem solved!

How can $i^2 = k^2 = j^2 = ijk = -1$ be true?

3 Answers3