How can ijk be equal to -1 if each is a square root?

Question

I'm learning about Quaternions for the first time and ran into something I can't quite understand. Supposedly i^2 = j^2 = k^2 = ijk = -1; but if i, j, and k are all square roots of -1, then shouldn't the product of all three be -1 times its square root, or -i? (or -j, or -k as the case may be).

Answer 1

Note that $3$ and $-3$ are both square roots of $9$ in usual real numbers. But the product of these two square roots is not $9$. Only self-multiplication of a square root of $x$ can give $x$.

Also in quaternions commutative law is not valid for multiplication. In such a case there cannot be a limit on number of square roots for any element.

We are merely looking at a binary operation that is associative, has an identity and is distributive over addition. For convenience we call it multiplication; it shares lot of properties of usual multiplication, and when restricted to a subset identifiable with real (or complex) numbers it is usual multiplication. Thats all.

Answer 2

Some people might argue that $i^2 = -1$ doesn't make sense because $i^2 = \sqrt{-1} \cdot \sqrt{-1} = \sqrt{-1 \cdot -1} = \sqrt{1} = 1$. Of course this argument is terrible in many ways, but the most important point is that it, if you are careful, you can define $i = \sqrt{-1}$ in a way that does not always behave in the way that real numbers behave but, rather, adds new structures to the real numbers in a consistent and very useful way. The same thing can be said for quaternions.

One way of looking at the set of quaternions, $\mathbb H$, is that they can be divided into a scalar part and a vector part. We make the following identification between $\mathbb H$ and $\mathbb R \times \mathbb R^3$

$$a + bi + cj + dk \longleftrightarrow (a, \vec{v}) \quad \text{(where $\vec v = (b,c,d))$}$$

Then \begin{align} (a_1, \vec v_1) + (a_2, \vec v_2) &= (a_1+a_2, \vec v_1 + \vec v_2) \\ (a_1, \vec v_1) (a_2, \vec v_2) &= (a_1 a_2 - \vec v_1 \cdot \vec v_2, a_1 \vec v_2 + a_2 \vec v_1 + \vec v_1 \times \vec v_2) \end{align}

It takes some effort to show that this is consistent with the usual definitions of quaternions, but, ultimately, it is.

If you make the associations

\begin{align} i &\leftrightarrow(0, (1, 0, 0)) \\ j &\leftrightarrow(0, (0, 1, 0)) \\ k &\leftrightarrow(0, (0, 0, 1)) \end{align}

then you will find that $i, j,$ and $k$ behave just as described.