Quaternions: why does ijk = -1 and ij=k and -ji=k Currently i am studying quaternions.
I do understand that i, j and k are imaginairy numbers. so $i^2 = j^2 =k^2 = -1$. 
But I could not understand this:
$$\begin{matrix}ij=k,&ji=-k,\\jk=i,&kj=-i,\\ki=j,&ik=-j\end{matrix}$$ 
Why is this? There seems no explanation why that is true. I would like to understand why that is true instead of just assuming that that is true.
Could somebody provide me some help?
 A: 
It occurred to Hamilton during a walk to scratch into a bridge the following:
$$i^2 = j^2 = k^2 = ijk = -1.$$
From this we deduce by basic algebra:


*

*$ij = -ijkk = k$

*$jk = -iijk = i$

*$jki = -1$

*$ki = -jjki = j$

*$i^{-1} = -i^{-1}ijk = -jk = -i$

*$j^{-1} = -j^{-1}jki = -ki = -j$

*$k^{-1} = -ijkk^{-1} = -ij = -k$

*$ji = j^{-1} i^{-1} = (ij)^{-1} = k^{-1} = -k$

*$kj = -i$ similarly

*$ik = -j$ similarly


So we can provide complete arithmetic operations for the set numbers of the form $a + i b + j c + k d$:
$(a + i b + j c + k d) + (\alpha + i \beta + j \gamma + k \delta) = (a + \alpha) + i (b + \beta) + j (c + \gamma) + k (d + \delta)$
$\begin{array}?
(a + i b + j c + k d) \cdot (\alpha + i \beta + j \gamma + k \delta) &=&
a (\alpha + i \beta + j \gamma + k \delta) + i b (\alpha + i \beta + j \gamma + k \delta) + j c (\alpha + i \beta + j \gamma + k \delta) + k d (\alpha + i \beta + j \gamma + k \delta) \\
&=& a \alpha + i a \beta + j a \gamma + k a \delta +
i b \alpha + i i b \beta + i j b \gamma + i k b \delta +
j c \alpha + j i c \beta + j j c \gamma + j k c \delta +
k d \alpha + k i d \beta + k j d \gamma + k k \delta \\
&=& (a \alpha - b \beta - c \gamma - d \delta) +
i (a \beta + b \alpha + c \delta - d \gamma) +
j (a \gamma - b \delta + c \alpha + d \beta) +
k (a \delta + b \gamma - c \beta + d \alpha)
\end{array}$
$$\frac{1}{a + i b + j c + k d} = \frac{a - i b - j c - k d}{(a + i b + j c + k d)(a - i b - j c - k d)} = \frac{a - i b - j c - k d}{a^2 + b^2 + c^2 + d^2}$$
A: You can derive the properties of quaternions through clifford algebra and the geometric product of vectors.
The geometric product works like so:
$$e_a e_b = \begin{cases} 1, & a = b \\ -e_b e_a, & a \neq b \end{cases}$$
where $a, b$ can be $x$, $y$, or $z$ as usual.  This captures both the work of the cross product and the dot product in one product of basis vectors.
You can then identify
$$\begin{align*}
i &= -e_y e_z \\
j &= -e_z e_x \\
k &= -e_x e_y
\end{align*}$$
And then the properties of quaternions naturally follow.
$$i^2 = (-e_y e_z)(-e_y e_z) = (e_y e_z)(e_y e_z) = -e_y (e_z e_z) e_y = -e_y e_y = -1$$
And similarly for $j^2$ and $k^2$, as well as the $ijk$ product:
$$ijk = (-e_y e_z)(-e_z e_x)(-e_x e_y) = - e_y e_z e_z e_x e_x e_y = -e_y (e_z e_z)(e_x e_x) e_y = -e_y e_y = -1$$
This allows you to interpret quaternions in a very geometric way:  the $i,j,k$ do not represent vectors, but rather oriented planes.  It's just that in 3d each plane has a unique normal vector, so we often abuse this duality.
A: $i,j,k$ are not imaginary numbers. Imaginary numbers arise only when you are talking about the complex plane $\mathbb{C}$, which has a very simple one to one mapping with the 2-D plane $\mathbb{R}^2$. Quaternions arise when you are talking about three dimensions, i.e. looking for solutions to $x^2+1 = 0$  in 3-D. 
If you want to get a physical picture, consider $i$ as rotating a vector or a line segment in 3-D by $90^{\circ}$ taking X-axis as the axis of rotation. Similarly, $j,k$ correspond to rotations about Y and Z axes respectively. This is similar to imaginary number $i$, which corresponds to a right angle rotation in the complex plane. Since in 3-D there are more than one independent axes of rotation possible, 3 to be precise, there are 3 quaternions.
Now, two $90^{\circ}$ rotations about X, Y or Z axis will take the vector $\mathbf{x}$ to its mirror image $\mathbf{(-x)}$. S0, $i^2=j^2=k^2 =-1$.
A right angle rotation about $X$ axis followed by an equal amount of rotation about Y corresponds to an overall effective rotation of $90^{\circ}$ about Z axis. So, $ij = k$. Similarly, you can physically verify the quaternion multiplication laws.
Do not try to think of their multiplication as arithmetic. They are compositions of rotation operations. If this looks too confusing, a little background in group theory will bring you enough mathematical maturity to be comfortable with these.
A: The defining properties relating $i, j $ and $k$ are
$$
i^2 = -1 \\ j^2 = -1 \\ k^2 = -1 \\ ijk = -1.
$$
From these you get for example that 
$$\begin{align}
i(ijk) &= -i \Rightarrow \\
i^2jk &= -i \Rightarrow\\
jk &= i.
\end{align}
$$
Likewise you can derive the other identities.
