Prove that \{a + bi : a,\ b \in R\}

Step 1
Note:
${\displaystyle \to H=\{a_0+a_1\hat{i}+a_i\hat{j}+a_3\hat{k}\mid ai\in \mathbb{R}\}}$
where ${\displaystyle i^2=j^2=k^2=-1}$, Called Quaternions ring
Now, ${\displaystyle Z\left(H\right)=\left\{\text{Set of those element of H which are Conmuty with every elements of H}\right\}}$
${\displaystyle Z\left(H\right)=\{a_0\mid a_0\in \mathbb{R}\}=\mathbb{R}}$
Now, let ${\displaystyle S=\{a+bi,a,b\in \mathbb{R}\}}$
${\displaystyle S\subseteq H}$ Clearly
Now ${\displaystyle \mathrm{\forall }x,y\in S,x=a+bi}$
${\displaystyle y-x+di}$
${\displaystyle x-y=a-c+(b-a)i\in S}$
and ${\displaystyle xy=(a+bi)\times (c+di)=(ac-bd)+(ad+bc)i\in S}$
${\displaystyle \Rightarrow S}$ is Subring of H
$\to (a+bi)\times (1+0\times i)=(a+bi)\Rightarrow 1$is unity
${\displaystyle \to {(a+bi)}^{-1}=\frac{a}{a^2+b^2}-\frac{bi}{a^2+b^2}}$ for any element
${\displaystyle \to }$ S is Commotative
${\displaystyle \to }$ S is field any ${\displaystyle S=\{a+bi\mid a,b\in \mathbb{R}\}\stackrel{\sim }{=}\subset Z\left(H\right)=\mathbb{R}}$

Following your suggestion:
Let ${\displaystyle h=a+bi+cj+dk}$
Then investigate conditions for i and h to commute!
${\displaystyle i\times h=i\times (a+bi+cj+dk)}$
${\displaystyle =ai+b{i}^2+cij+dik}$
${\displaystyle =ai-b+ck-dj}$
${\displaystyle h\times i=(a+bi+cj+dk)\times i}$
${\displaystyle =ai+b{i}^2+cji+dki}$
${\displaystyle =ai-b-ck+dj}$
Thus i and h commute only if ${\displaystyle c=d=0}$
Proceeding similarly we find that
j and h commute only if ${\displaystyle b=d=0}$
and
k and h commute only if ${\displaystyle b=c=0}$
Thus it seems as if I am being driven to the conclusion that the only Hamilton Quaternions that commute with every element of the ring of Hamilton Quaternions are elements of the form ${\displaystyle a+0i+0j+0k}$
But I am unsure of how to formally and validly argue from the facts established above to conclude this!

Step 1Proven that only quaternions of the form:$a+bi$ commute with i (this makes sense if you think about it, as neither j nor k commute with i). well if $q=a+bi+cj+dk$ is in Z(H), it has to commute with i, since i is a quaternion. so that alone narrows down the possibilities right there. what you've done with j and k is fine, although you could also show that if $a+bi$ commutes with j, then $b=0$, and likewise for k. this means that only real quaternions $(b=c=d=0)$ can possibly be in the center. it's not hard to show that all real quaternions are indeed central, which settles the matter. EDIT: a little reflection should convince you that the center of the quaternions has to be a field. the only possibilities for the dimension (as a vector space over R) of this field is either 1 or 2 (if it was 2, it would have to be an isomorph of the complex numbers). the fact that i doesn't commute with j or k, effectively kills this possibility. it actually makes more sense to think of i,j and k being ''3 identical copies of $\sqrt{-1}$'', because there's no real way to tell them apart from one another. this is why H is sometimes viewed as ''scalars+vectors'', the ''pure quaternion (non-real)'' part, acts very much like a vector in $R^3$, which was actually Hamilton's original goal-to find an ''algebra'' for 3-vectors. the fact that $Z(Q_8)$, the center of the group of quaternion units, is equal to $\{-1,1\}$ should reassure you that your conclusion is correct.