Question

I have a question about quarternions. These are an extension of complex numbers. Given that
i^2 = j^2 = k^2 = ijk=-1
show that
ij = k = -ji
jk = i = -kj
ki = j = -ki
I am having trouble with the negatives

Answer 1

It is easy to see that given ijk = -1
ijk *k = -1 *k
ij (kk) = -1 *k
ij (-1) = -1 *k
ij = k

Answer 2

now prove -ji = k , thats what im having trouble with

Answer 3

Are you not allowed to take right hand multiplication?

Answer 4

this is non-commutative. but you can do right multiplication or left multiplication. and it is associative

Answer 5

im not seeing algebraically how to prove ji = -k

Answer 6

oh i think i see it

Answer 7

ij = k
ij i = ki
i j i = j
i i j i = i j
-1 ji = ij

Answer 8

Oh I see, well that figures, we didn't cover that topic much yet in my university so you have the following identity \(ijk=-1\) multiply by \(ji\) from the let to get \[\Large (ji)ijk = -ji \]

Answer 9

oh thats simpler :)

Answer 10

the terms cancel

Answer 11

(ji) (ijk) = (ji)(-1)
j(ii)jk = -ji
j(-1)jk = -ji
(-1)jjk = -ji
(-1)(-1)k = -ji
k = -ji

Answer 12

it's really just a game between left hand and right hand multiplication, but not very obvious at first glance, however if I didn't contribute to any algebraic fallacy than that would turn out to be your expression \[\Large j(ii)jk=-ji \implies -jjk=-ji \implies k=-ji \]

Answer 13

right, i also did it another way

Answer 14

since we easily prove that ij = k , jk = i, ki = j, we can use these and argue:
ij = k
ij*i = k*i
iji = j
i*iji = i*j
(ii) ji = ij
- ji = ij

Answer 15

but your way is easier, i have to do left multiplication and right multiplication by i

Answer 16

Exactly, often Quaternions get introduced in Linear Algebra courses and as Matrix Multiplication to empathize on the topic of left hand and right hand multiplication. However, your way works just as fine, verifying these identities can be very tedious. So better to have them in an exercise rather than in a test (-;

Answer 17

yes :)

Answer 18

i was looking for something simple and you found it

Answer 19

\[\checkmark \]

Answer 20

simple and elegant

Answer 21

i gave up too quickly :/

Answer 22

thanks again

Answer 23

You're very welcome :-)

Answer 24

also we have to be careful when we do imaginary or quarternions, they are not the same as those coordinate basis vectors, like in calculus. r= ai + bj + ck. this is not the same i as z = a + bi , or h = a + bi + cj + ek (quarternions)

Answer 25

complex numbers is a + bi. , and i think theres also octonians , but im not sure

Answer 26

octonians are not even associative

Answer 27

so they are completely different, but quarternions find some interesting applications

Answer 28

this is interesting you can define quarternion multiplication using ordered pairs
http://en.wikipedia.org/wiki/Cayley%E2%80%93wingspanson_construction

Answer 29

Exactly, it is important to realize that \(i,j,k\) are not complex numbers in the manner of Quaternions.