How do quaternions not show that $-1=1$? Where is the proof wrong?

Question

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?

Answer 1

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.

Answer 2

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}

Answer 3

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.

Answer 4

because the multiplication of Quaternions is noncommutative.

i.e.

(ijk)(ijk) is not the same as (ii)(jj)(kk), you cannot exchange places.