How do quaternions not show that $-1=1$? Where is the proof wrong? Given the rules of quaternions:
$$ i^2=j^2=k^2=ijk=-1$$
could it not be used to show that $-1=1$? As follows:
$$ijk=-1$$
$$ijk\cdot ijk=i^2\cdot j^2\cdot k^2=(-1)(-1)=1$$
$$i^2=-1$$
$$j^2=-1$$
$$k^2=-1$$
$$i^2\cdot j^2\cdot k^2 = (-1)(-1)(-1)=-1$$
thus $i^2\cdot j^2\cdot k^2$ both equals $1$ and $-1$.
What is wrong with this reasoning, and what does $i^2\cdot j^2\cdot k^2$ actually equal?
 A: The answer from "lisyarus" is good, but here's a slower version that reminds us of three basic identities: $$\begin{align}ij & =-ji, \\ jk & =-kj, \\ ki & =-ik.\end{align}$$
We have:
\begin{align}
ijk \cdot ijk & = ij(ki)jk \\[8pt]
& = ij(-ik)jk & & \text{since }ki=-ik \\[8pt]
& = -ij(ik)jk \\[8pt]
& = -i(ji)kjk \\[8pt]
& = -i(-ij)kjk & & \text{since }ji = -ij \\[8pt]
& = i(ij)kjk \\[8pt]
& = iijk(jk) \\[8pt]
& = iijk(-kj) & & \text{since }jk = -kj \\[8pt]
& = -iij(kk)j \\[8pt]
& = -iij(-1)j \\[8pt]
& = iijj \\[8pt]
& = (-1)(-1) \\[8pt]
& = 1.
\end{align}
A: You have an error in your proof.
When you say $ijk \cdot ijk = i^2j^2k^2$, you assume that quaternion multiplication is commutative, which is false.
A: The rule $ij=k$ can be derived from $i^2=j^2=k^2=ijk=-1$; indeed
$$
(ijk)k=(-1)k
$$
so $ij(-1)=(-1)k$. Now $kij=k^2=-1$, so we similarly get $ki=j$ and therefore also $jk=i$.
But if we try $i(ijk)=i(-1)$ we get $i^2jk=-i$ and therefore $jk=-i$. Similarly we also get $ik=-j$ and $kj=-i$.
So the identity $(ijk)(ijk)=i^2j^2k^2$ cannot hold and it is not a contradiction.
A: because the multiplication of Quaternions is noncommutative.
i.e.
(ijk)(ijk) is not the same as (ii)(jj)(kk), you cannot exchange places.
