Dtjytyk

I have to calculate the center of the real quaternions, $mathbb{H}$.

So, I assumed two real quaternions, $q_n=a_n+b_ni+c_nj+d_nk$ and computed their products. I assume since we are dealing with rings, that to check was to check their commutative product under multiplication. So i'm looking at $q_1q_2=q_2q_1$. When I do this, I find that clearly the constant terms are identical, so it is clear that the subset $mathbb{R}$ is in the center. So, perhaps then that $mathbb{C}lemathbb{H}$. However i ended up, after direct calculation with the following system;

$$c_1d_2=c_2d_1$$
$$b_1d_2=b_2d_1$$
$$b_1c_2=b_2c_1$$

So the determination is then found by solving this system. Intuitively, I felt that this lead to $0$'s everywhere and thus the center of $mathbb{H}$, $Z(mathbb{H})=mathbb{R}$. I then checked online for some confirmation and indeed it seemed to validate my result. However, the proof method used is something I haven't seen. It was pretty straight forward and understandable, but again, I've never seen it. It goes like this;

Suppose $b_1,c_1,$ and $d_1$ are arbitrary real coefficients and $b_2, c_2,$ and $d_2$ are fixed. Considering the first equation, assume that $d_1=1$ (since it is arbitrary, it's value can be any real...). This leads to
$$c_1=frac{c_2}{d_2}$$
And that this is a contradiction, since $c_1$ is no longer arbitrary (it depends on $c_2$ and $d_2$)

I really like this proof method, although it is unfamiliar to me. I said earlier that for my own understanding, it seemed intuitively obvious, but that is obviously not proof:

1) What are some other proof methods for solving this system other than the method of contradiction used below? I was struggling with this and I feel I sholnd't be.

2) What other proofs can be found in elementary undergraduate courses that use this method of "assume arbitrary stuff", and "fix some other stuff" and get a contradiction? I found this method very clean and fun, but have never seen it used (as far as I know) in any elementary undergraduate courses thus far...

I am not sure where the contradiction lies exactly in your proof by contradiction. But here is another method.

An element $xin mathbb H$ belongs to the center if and only if $[x,y]=0$ for all $yin mathbb H$, where $[x,y]=xy-yx$ denotes the commutator of two elements.

We see immediately that $[x,1]=0$, whereas if $x=a+bi+cj+dk$ we have
$$
[x,i]=-2ck+2dj.
$$
Thus $[x,i]=0$ if and only if $c=d=0$. Similarly $[x,j]=0$ if and only if $b=d=0$. Thus the only elements $x$ which commute with both $i$ and $j$ are $xin mathbb R$; in particular, it follows that $Z(mathbb H)subset mathbb R$. Since it is clear that $mathbb Rsubset Z(mathbb H)$, the result follows.

Idea behind the proof: There are three special copies of the complex numbers sitting inside $mathbb H$: the subspaces
$$
mathbb C_i=mathbb R[i],qquad mathbb C_j=mathbb R[j],qquad mathbb C_k=mathbb R[k].
$$
Over $mathbb H$, all of these subspaces are their own centers: $Z_{mathbb H}(mathbb C_i)=mathbb C_i$ and so forth. Since $$mathbb H=mathbb C_i+ mathbb C_j+ mathbb C_k,$$
it follows that $Z(mathbb H)=Z(mathbb C_i)cap Z(mathbb C_j)cap Z(mathbb C_k)=mathbb R$.

If $a+bi+cj+dk$ is in center, then it should commute with generators
$$i,j,k,mbox{ and reals}.$$
For example, see what do we get for $(a+bi+cj+dk).i=i.(a+bi+cj+dk)$?