# Constructing quaternions - proof that square of each imaginary unit is -1

During construction of vector space of quaternions over real numbers I encountered a problem that I can't quite put my finger on.

For the context: Hamilton a multiplication in plane that keeps preservation of length. Now he's would like to find something similar in 3 dimension but instead concludes couple things.

Let $$\mathbb{R}^n$$ be a Euclidean space consisting of vectors $$\mathbf x=\left\langle x_0,\dots,x_{n-1}\right\rangle^T$$, with $$\mathbf x\in\mathbb{R}^n$$. This is a vector space, and as such has the standard addition of vectors and multiplication by scalar. It also has scalar product $$\mathbf {\color {red}a} \cdot \mathbf {\color {blue}b} =\sum _{i=0}^{n-1}{\color {red}a}_{i}{\color {blue}b}_{i}={\color {red}a}_{1}{\color {blue}b}_{1}+{\color {red}a}_{2}{\color {blue}b}_{2}+\cdots +{\color {red}a}_{n}{\color {blue}b}_{n},$$

euclidean norm, and a standard basis $$E=\{e_0,\dots,e_{n-1}\}$$ where $$e_i$$ has a $$1$$ as its $$i$$th entry and $$0$$s elsewhere. Note that by this definition, $$e_0$$ has all $$0$$ entries.

Now we define multiplication as follows:

Multiplication in $$\mathbb{R}^n$$ is a morphism $$m:\mathbb{R}^n\times \mathbb{R}^n\to \mathbb{R}^n$$ that satisfies:

(i) is bilinear

(ii) has multiplicative unit $$e_0$$ that $$e_0\mathbf x=\mathbf xe_0=\mathbf x$$ for any vector $$\mathbf x$$ of our vector space.

Finally my question:

If we were to add third property to multiplication, that is the preservation of distances mentioned in beginning defined as:

(iii) $$|xy|=|x||y|$$, where $$|x|=\sqrt{x^2_0 + x^2_1 +...+x^2_{n-1}}$$

Then following statements are equivalent:

(a) $$x^2=-|x|^2 \iff$$ (b) scalar product of $$\mathbf x$$ and $$e_0$$ is $$0$$.

What I concluded so far:

(1) from right to left premise is equivalent to $$x_0=0$$.

(2) from left to right premise is equivalent to $$\mathbf{xx}=\mathbf x^2=\langle a_0,\dots,a_{n-1}\rangle^T=-|x|^2e_0=\langle-|x|^2,0,\dots,0\rangle^T,$$ and therefore $$a_0=-|x|^2$$.

However now I'm at loss of how to continue. How will this statement help us to prove that square of imaginary unit is equal to $$-1$$? It is a weaker version of statement above, as imaginary unit equals $$e_i, i=\{1,\dots,n-1\}$$.

• Everything before "Finally my question" can be stated as "Let $e_0$ be the multiplicative identity of the vector space $\mathbb{R}^n$. You don't need to enumerate standard conventions and definitions in elementary linear algebra. Moreover, the multiplicative identity of a vector space is a scalar, not a vector that is part of the space's basis as your definition of $e_0$ seems to suggest. – Brian Mar 18 at 23:23
• Quaternions are not commutative, hence not a field. The word you want is "skew-field" or "division ring". – Somos Mar 19 at 0:02
• In defining the inner product, you write the sum first from $0$ to $n-1,$ then from $1$ to $n.$ A vector with all $0$ entries cannot be a basis vector, so it makes no sense to say $e_0$ is all $0$s, but perhaps this is a result of the confusion over whether to count from $0$ to $n-1$ or from $1$ to $n.$ – David K Mar 19 at 3:33
• Your approach seems to imply that you could build a system with any number of imaginary units you want, for example there could be exactly two, or exactly four. From what I understand of quaternions, this doesn't work. – David K Mar 19 at 3:40
• @Brian Hi, unfortunately your comment makes little to no sense to me. Now in definition of quaternions, there is explicitly stated that {1,i,j,k} is a basis. Now this basis can be constructed as mentioned above or not? If e$_0$ doesn't equal to 4x1 vector, how should I interpret the (b) statement? That is are you trying to say that scalar product of x and e$_0$ is equal to x$_1$ + x$_2$ + x$_3$ since x is vector 3x1? – Jan Lhoták Mar 20 at 4:10

There are several equally valid ways to describe the system of quaternions. One way is to work with couples where one part is a real number and the other part is a vector in $$\,\mathbb{R}^3.\,$$ Thus, let $$\, e_0 := \bf{1}\,$$ be the real unit and $$\,e_1 := (1,0,0) = \bf{i},\,$$ $$e_2 := (0,1,0) = \bf{j},\,$$ and $$\,e_3 := (0,0,1) = \bf{k}\,$$ be the three vector units. At this point, the construction is easy, except for product of two vectors. There are two products of vectors. One is the scalar or dot product. The other is the vector or cross product. Now we introduce the quaternion product as a quaternion couple with $$\, \bf{u}\, v := -u\cdot v + u \times v.\,$$
If we restrict to vectors with only one quaternion unit, then we essentially have a version of complex numbers. Hamilton knew that he had to have $$\, \bf{i}^2 = j^2 = k^2 = -1\,$$ for this reason. What was surprising to him was that for a consistent system he had to require that $$\, \bf{k} := i\, j = - j\, i.\,$$