1
$\begingroup$

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?

$\endgroup$
18
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ $i j=k....j k=i...k i=j,$ But $j i=-k...k j=-i...i k=-j$. $\endgroup$ – DanielWainfleet May 11 '16 at 20:55
2
$\begingroup$

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}

$\endgroup$
  • 3
    $\begingroup$ Well, $(ijk)(ijk)=(-1)^2=1$. No need to do so lengthy a computation. $\endgroup$ – egreg May 11 '16 at 20:24
  • 3
    $\begingroup$ @egreg : This was to be an exercise for those who need to get used to the non-commutativity of quaternion multiplication. That probably includes anyone who asks a question like this one. $\qquad$ $\endgroup$ – Michael Hardy May 11 '16 at 20:39
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

because the multiplication of Quaternions is noncommutative.

i.e.

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.