# There isn't a product operation that is commmutative on $\mathbb{R}^{n}$ that satisfies all the field axioms for $n \geq 3$.

This proof is broken down into simple easy algebra and vector questions. I would like to discuss different answers and approaches.

There are 5 questions from 7.6.3 - 7.6.7. You can read the paragraph above 7.6.3. You can also read the first part on quarternions. Exclude the "rotations of ijk space section.

Here is what I have tried.

Q1: I used the Pythagorean Theorem to get the norm equal to $\sqrt{2}$. Then I used the property $\text{Norm}(uv) = \text{Norm}(u) \text{Norm}(v)$ to show that that $2 = \text{Norm}(1 - i^{2})$. Am I right?

Q2: I used the property $\text{Norm}(uv) = \text{Norm}(u) \text{Norm}(v)$ since $\text{Norm}(i) = 1$. But I don’t know how to show $i^{2} = -1$. One says to use the Triangle Inequality. I guess the equality implies that $i^{2}$ and $1$ are collinear.

Q3: And thus I don’t know how to do this.

Q4: The map $p \longmapsto pi$ multiplies all distances in $\mathbb{R}^{n}$ by $|i| = 1$, since $|pi| = |p||i|$. For any points $p_{1}$ and $p_{2}$ in $\mathbb{R}^{n}$, $|p_{1} * i - p_{2} * i| = |(p_{1} - p_{2})i| = |p_{1} - p_{2}||i|$. Therefore, the distance $|p_{1} - p_{2}|$ between any two points is multiplied by $|i| = 1$. Therefore, the map is an isometry of $\mathbb{R}^{n}$. Therefore, since $i$ and $j$ are perpendicular directions, $i * i$ and $i * j$ are still perpendicular by the isometry. (An isometry preserves the distance between points.)

Still not sure why $\mathbf{1}$ and $ij$ are perpendicular.

Q5: From $jiij= j i^{2} j = jj i^{2} = j^{2} i^{2} = - \mathbf{1} * - \mathbf{1} = \mathbf{1}$, therefore $1 = -1$, which is a contradiction.

-
Please use MathJax to write all mathematical expressions. I’m pretty sure that loving math is just as important as loving typography. –  Haskell Curry Mar 7 '13 at 4:00

-