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According to Wikipedia,

$$x+y = (x_0+y_0)+(x_1+y_1) i+(x_2+y_2) j+(x_3+y_3) k$$

and

$$\begin{align} x y &=( x_0 y_0 - x_1 y_1 - x_2 y_2 - x_3 y_3)\\ &+( x_0 y_1 + x_1 y_0 + x_2 y_3 - x_3 y_2) \mathrm i\\ &+( x_0 y_2 - x_1 y_3 + x_2 y_0 + x_3 y_1) \mathrm j\\ &+( x_0 y_3 + x_1 y_2 - x_2 y_1 + x_3 y_0) \mathrm k \end{align}$$

It's clear how to get the addition, but how do you get the multiplication? Especially, why is

$x_1 \cdot \mathrm i \cdot y_0 = x_1 \cdot y_0 \cdot \mathrm i$ for $x_1, y_0 \in \mathbb{R}$

It might be a dumb question, but I really don't get it. I would appreciate if you use only what I know for your answer, if possible.

What I know

$$i^2 = j^2 = k^2 = -1$$ $$ijk=-1$$

which implies

$$\begin{align} ij &= k\\ ji &= -k\\ jk &= i\\ kj &= -i\\ ki &= j\\ ik &= -j \end{align}$$

and

$$\begin{align} (-1) \cdot \mathrm i &= \mathrm i \cdot (-1)\\ (-1) \cdot \mathrm j &= \mathrm j \cdot (-1)\\ (-1) \cdot \mathrm k &= \mathrm k \cdot (-1) \end{align}$$

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    $\begingroup$ The basis is $1,i,j,k$ and $1\cdot i=i\cdot 1$ so that $r\cdot i=i\cdot r$ for real $r$. $\endgroup$ – Dietrich Burde Sep 9 '13 at 9:56
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    $\begingroup$ It is kind of defined that way. Multiplication by an element of $\mathbb R$ is special. That is why it is called an $R$-algebra. That means, I think you should add to the list of things you know that $rx = xr$ for any $r \in \mathbb R$. $\endgroup$ – Tunococ Sep 9 '13 at 10:10
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$ix_1=x_1i$ is only one contract and if we define $x=x_0+ix_1+jx_2+kx_3$ and $y=y_0+iy_1+jy_2+ky_3$(there is no reason that define $x=x_0+x_1i+x_2j+x_3k$ or $y=y_0+y_1i+y_2j+y_3k$ and one of them can be used), then we can define $xy$ as follows $$\begin{align} x y &=( x_0 y_0 - x_1 y_1 - x_2 y_2 - x_3 y_3)\\ &+i( x_0 y_1 + x_1 y_0 + x_2 y_3 - x_3 y_2) \\ &+j( x_0 y_2 - x_1 y_3 + x_2 y_0 + x_3 y_1) \\ &+k( x_0 y_3 + x_1 y_2 - x_2 y_1 + x_3 y_0) \end{align}$$ and also note that $$ix_1=(0,x_1,0,0)=x_1i,(x_1=(x_1,0,0,0),i=(0,1,0,0),j=(0,0,1,0),k=(0,0,0,1))$$ if we define $$xy=(x_0 y_0 - x_1 y_1 - x_2 y_2 - x_3 y_3,x_0 y_1 + x_1 y_0 + x_2 y_3 - x_3 y_2,x_0 y_2 - x_1 y_3 + x_2 y_0 + x_3 y_1,x_0 y_3 + x_1 y_2 - x_2 y_1 + x_3 y_0).$$

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    $\begingroup$ This is just making the position of $i^{-1}$ ambiguous by putting it under the fraction... (I could have believed that this was true even when $x$ is a quaternion.) $\endgroup$ – Tunococ Sep 9 '13 at 10:07
  • $\begingroup$ I've understood that $x \mathrm i = \mathrm ix$ with $x \in \mathbb{R}$, but why is $x \mathrm i y = xy \mathrm i$? Wait a second... Ah, I understand! Lets say $x = ab$ with $a,b \in \mathbb{R}$. Then I can use your argument to get $ab \mathrm i = a \mathrm i b$. Thanks! (Now I'll check if I get the same for general multiplication, then I'll accept or ask in comment my question.) $\endgroup$ – Martin Thoma Sep 9 '13 at 10:09
  • $\begingroup$ No, you don't need that argument to show $xyi = xiy$ with $x, y \in \mathbb R$ if you know $xi = ix$. This is because $xyi = x(yi) = x(iy) = xiy$. (I assume associativity, since you've written $xyi$ without parentheses.) $\endgroup$ – Tunococ Sep 9 '13 at 10:15
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    $\begingroup$ This post is very misleading. It is not at all clear why you aren't showing that $i$ commutes with any quaternion $x$ (not just $x \in \mathbb{R}$). In particular, you cannot move the denominator around as you do. $xi^{-1} \neq i^{-1}x$ in general! $\endgroup$ – Alexander Sep 19 '13 at 14:08
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One thing that should be said (that Tunococ has mentioned in the comments) is that

elements of $\Bbb F$ in any associative $\Bbb F$-algebra are essentially assumed to be central.

(The "elements of $\Bbb F$" are of course $\{\lambda 1\mid \lambda\in \Bbb F\}$, their usual embedding.)

This is hiding in the axiom of compatibility with scalars. For any $\Bbb F$ algebra $A$, and any $x\in A$, that axiom gives $\lambda x=((\lambda 1)x)1=(1(\lambda x))1=(\lambda x)1=x(\lambda 1)=x\lambda$.

The quaternions are, after all, an $\Bbb R$-algebra generated by $i,j$ with special relations. The historical perspective at the quaternion article does not really do a good job of explaining that point.

A common alternative definition of algebras makes this even more obvious. It goes something like "an $\Bbb F$-algebra is a ring $A$ and a ring homomorphism $r:\Bbb F\to Z(A)$" where $Z(A)$ is the center of $A$. Assuming this definition, one easily sees the "regular" algebra axioms are satsified by $r(\Bbb F)$, which is (isomorphic to, hence) identified with $\Bbb F$.

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