Prove $\alpha i=i\alpha$ iff $c=d=0$. Let $\alpha \in \mathbb H$ and $\alpha=a+bi+cj+dk, a,b,c,d \in \mathbb Q$.

Question

Prove $\alpha i=i\alpha$ iff $c=d=0$. Let $\alpha \in \mathbb H$ and $\alpha=a+bi+cj+dk, a,b,c,d \in \mathbb Q$.

My Attempt:

$(\rightarrow):$ $$\alpha i=ai-b-ck+dk \Rightarrow -b+ai+dj-ck$$

$$i\alpha =ai-b+ck-dk \Rightarrow -b+ai-dj+ck$$

Thus, $\alpha i =i \alpha$ because $d=-d$ and $c=-c$ implies $c=d=0$.

$(\leftarrow):$ If $c=d=0$ then,

$$(a+bi+0j+0k)i=i(a+bi+0j+0k)$$ $$(a+bi)i=i(a+bi)$$ $$ai-b=ai-b$$ Thus, $\alpha i = i \alpha$.

I'm not positive if I'm allowed to begin the second part with the equivalence:$(a+bi+0j+0k)i=i(a+bi+0j+0k)$. I know when you're trying to prove two things equal you can't start the proof with stating the equality as true. Does it work in this instance because we are stating $c=d=0$ first, then plugging them in?

Answer 1

Call $[\alpha,\beta]$ the commutator of two quaternions; then it is not hard to see that this is a bilinear product. Moreover $[1,i]=0$ and $[i,i]=0$. Knowing this, calculate $[\alpha,i]$, and make use of the fact that $j$ and $k$ are linearly independent.