definition of quaternions division associative algebra
Topics
About: algebra
The table of contents of this article
Starting Context
- The reader knows a definition of algebra over field.
Target Context
- The reader will have a definition of quaternions division associative algebra.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\( \mathbb{R}\): \(= \text{ the real numbers field }\)
\(*\mathbb{H}\): \(= \{a + b i + c j + d k \vert a, b, c, d, \in \mathbb{R}\}\), where \(i, j, k\) are some symbols, \(\in \{ \text{ the division associative algebras } \}\), with the \(\mathbb{R}\) vectors space structure and the multiplication specified below
//
Conditions:
\(\forall a + b i + c j + d k, a' + b' i + c' j + d' k \in \mathbb{H} ((a + b i + c j + d k) + (a' + b' i + c' j + d' k) = (a + a') + (b + b') i + (c + c') j + (d + d') k) \land \forall r \in \mathbb{R} (r (a + b i + c j + d k) = (r a) + (r b) i + (r c) j + (r d) k)\)
\(\land\)
\((a + b i + c j + d k) (a' + b' i + c' j + d' k) = (a a' - b b' - c c' - d d') + (a b' + b a' + c d' - d c') i + (a c' - b d' + c a' + d b') j + (a d' + b c' - c b' + d a') k\)
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\(\overline{a + b i + c j + d k} := a - b i - c j - d k\) is called "quaternion conjugate of \(a + b i + c j + d k\)".
\(\forall h, h' \in \mathbb{H} (\overline{h + h'} = \overline{h} + \overline{h'} \land \overline{h h'} = \overline{h'}\text{ }\overline{h})\): note the order.
2: Note
Let us see that \(\mathbb{H}\) is indeed an \(\mathbb{R}\) vectors space.
1) \(\forall a_1 + b_1 i + c_1 j + d_1 k, a_2 + b_2 i + c_2 j + d_2 k \in \mathbb{H} ((a_1 + b_1 i + c_1 j + d_1 k) + (a_2 + b_2 i + c_2 j + d_2 k) \in \mathbb{H})\) (closed-ness under addition): \((a_1 + b_1 i + c_1 j + d_1 k) + (a_2 + b_2 i + c_2 j + d_2 k) = (a_1 + a_2) + (b_1 + b_2) i + (c_1 + c_2) j + (d_1 + d_2) k \in \mathbb{H}\).
2) \(\forall a_1 + b_1 i + c_1 j + d_1 k, a_2 + b_2 i + c_2 j + d_2 k \in \mathbb{H} ((a_1 + b_1 i + c_1 j + d_1 k) + (a_2 + b_2 i + c_2 j + d_2 k) = (a_2 + b_2 i + c_2 j + d_2 k) + (a_1 + b_1 i + c_1 j + d_1 k))\) (commutativity of addition): \((a_1 + b_1 i + c_1 j + d_1 k) + (a_2 + b_2 i + c_2 j + d_2 k) = (a_1 + a_2) + (b_1 + b_2) i + (c_1 + c_2) j + (d_1 + d_2) k = (a_2 + a_1) + (b_2 + b_1) i + (c_2 + c_1) j + (d_2 + d_1) k = (a_2 + b_2 i + c_2 j + d_2 k) + (a_1 + b_1 i + c_1 j + d_1 k))\).
3) \(\forall a_1 + b_1 i + c_1 j + d_1 k, a_2 + b_2 i + c_2 j + d_2 k, a_3 + b_3 i + c_3 j + d_3 k \in \mathbb{H} (((a_1 + b_1 i + c_1 j + d_1 k) + (a_2 + b_2 i + c_2 j + d_2 k)) + (a_3 + b_3 i + c_3 j + d_3 k) = (a_1 + b_1 i + c_1 j + d_1 k) + ((a_2 + b_2 i + c_2 j + d_2 k) + (a_3 + b_3 i + c_3 j + d_3 k)))\) (associativity of additions): \(((a_1 + b_1 i + c_1 j + d_1 k) + (a_2 + b_2 i + c_2 j + d_2 k)) + (a_3 + b_3 i + c_3 j + d_3 k) = ((a_1 + a_2) + (b_1 + b_2) i + (c_1 + c_2) j + (d_1 + d_2) k) + (a_3 + b_3 i + c_3 j + d_3 k) = ((a_1 + a_2) + a_3) + ((b_1 + b_2) + b_3) i + ((c_1 + c_2) + c_3) j + ((d_1 + d_2) + d_3) k = (a_1 + (a_2 + a_3)) + (b_1 + (b_2 + b_3)) i + (c_1 + (c_2 + c_3)) j + (d_1 + (d_2 + d_3)) k = (a_1 + b_1 i + c_1 j + d_1 k) + ((a_2 + a_3) + (b_2 + b_3) i + (c_2 + c_3) j + (d_2 + d_3) k) = (a_1 + b_1 i + c_1 j + d_1 k) + ((a_2 + b_2 i + c_2 j + d_2 k) + (a_3 + b_3 i + c_3 j + d_3 k))\).
4) \(\exists 0 \in \mathbb{H} (\forall a_1 + b_1 i + c_1 j + d_1 k \in \mathbb{H} (a_1 + b_1 i + c_1 j + d_1 k + 0 = a_1 + b_1 i + c_1 j + d_1 k))\) (existence of 0 element): \(0 := 0 + 0 i + 0 j + 0 k \in \mathbb{H}\), and \(a_1 + b_1 i + c_1 j + d_1 k + 0 = a_1 + b_1 i + c_1 j + d_1 k + 0 + 0 i + 0 j + 0 k = (a_1 + 0) + (b_1 + 0) i + (c_1 + 0) j + (d_1 + 0) k = a_1 + b_1 i + c_1 j + d_1 k\).
5) \(\forall a_1 + b_1 i + c_1 j + d_1 k \in \mathbb{H} (\exists v' \in \mathbb{H} (v' + a_1 + b_1 i + c_1 j + d_1 k = 0))\) (existence of inverse element): \(v' := - a_1 + (- b_1) i + (- c_1) j + (- d_1) k \in \mathbb{H}\), and \(v' + (a_1 + b_1 i + c_1 j + d_1 k) = - a_1 + (- b_1) i + (- c_1) j + (- d_1) k + (a_1 + b_1 i + c_1 j + d_1 k) = (- a_1 + a_1) + (- b_1 + b_1) i + (- c_1 + c_1) j + (- d_1 + d_1) k = 0 + 0 i + 0 j + 0 k = 0\).
6) \(\forall a_1 + b_1 i + c_1 j + d_1 k \in \mathbb{H}, \forall r \in \mathbb{R} (r . (a_1 + b_1 i + c_1 j + d_1 k) \in \mathbb{H})\) (closed-ness under scalar multiplication): \(r (a_1 + b_1 i + c_1 j + d_1 k) = (r a_1) + (r b_1) i + (r c_1) j + (r d_1) k \in \mathbb{H}\).
7) \(\forall a_1 + b_1 i + c_1 j + d_1 k \in \mathbb{H}, \forall r_1, r_2 \in \mathbb{R} ((r_1 + r_2) (a_1 + b_1 i + c_1 j + d_1 k) = r_1 (a_1 + b_1 i + c_1 j + d_1 k) + r_2 (a_1 + b_1 i + c_1 j + d_1 k))\) (scalar multiplication distributability for scalars addition): \((r_1 + r_2) (a_1 + b_1 i + c_1 j + d_1 k) = (r_1 + r_2) a_1 + ((r_1 + r_2) b_1) i + ((r_1 + r_2) c_1) j + ((r_1 + r_2) d_1) k = (r_1 a_1 + r_2 a_1) + (r_1 b_1 + r_2 b_1) i + (r_1 c_1 + r_2 c_1) j + (r_1 d_1 + r_2 d_1) k = (r_1 a_1 + (r_1 b_1) i + (r_1 c_1) j + (r_1 d_1) k) + (r_2 a_1 + (r_2 b_1) i + (r_2 c_1) j + (r_2 d_1) k) = r_1 (a_1 + b_1 i + c_1 j + d_1 k) + r_2 (a_1 + b_1 i + c_1 j + d_1 k)\).
8) \(\forall a_1 + b_1 i + c_1 j + d_1 k, a_2 + b_2 i + c_2 j + d_2 k \in \mathbb{H}, \forall r \in \mathbb{R} (r ((a_1 + b_1 i + c_1 j + d_1 k) + (a_2 + b_2 i + c_2 j + d_2 k)) = r (a_1 + b_1 i + c_1 j + d_1 k) + r (a_2 + b_2 i + c_2 j + d_2 k))\) (scalar multiplication distributability for vectors addition): \(r ((a_1 + b_1 i + c_1 j + d_1 k) + (a_2 + b_2 i + c_2 j + d_2 k)) = r ((a_1 + a_2) + (b_1 + b_2) i + (c_1 + c_2) j + (d_1 + d_2) k) = r (a_1 + a_2) + (r (b_1 + b_2)) i + (r (c_1 + c_2)) j + (r (d_1 + d_2)) k = (r a_1 + r a_2) + (r b_1 + r b_2) i + (r c_1 + r c_2) j + (r d_1 + r d_2) k = (r a_1 + (r b_1) i + (r c_1) j + (r d_1) k) + (r a_2 + (r b_2) i + (r c_2) j + (r d_2) k) = r (a_1 + b_1 i + c_1 j + d_1 k) + r (a_2 + b_2 i + c_2 j + d_2 k)\).
9) \(\forall a_1 + b_1 i + c_1 j + d_1 k \in \mathbb{H}, \forall r_1, r_2 \in \mathbb{R} ((r_1 r_2) (a_1 + b_1 i + c_1 j + d_1 k) = r_1 (r_2 (a_1 + b_1 i + c_1 j + d_1 k)))\) (associativity of scalar multiplications): \((r_1 r_2) (a_1 + b_1 i + c_1 j + d_1 k) = r_1 r_2 a_1 + (r_1 r_2 b_1) i + (r_1 r_2 c_1) j + (r_1 r_2 d_1) k = r_1 ((r_2 a_1) + (r_2 b_1) i + (r_2 c_1) j + (r_2 d_1) k) = r_1 (r_2 (a_1 + b_1 i + c_1 j + d_1 k))\).
10) \(\forall a_1 + b_1 i + c_1 j + d_1 k \in \mathbb{H} (1 (a_1 + b_1 i + c_1 j + d_1 k) = a_1 + b_1 i + c_1 j + d_1 k)\) (identity of 1 multiplication): \(1 (a_1 + b_1 i + c_1 j + d_1 k) = 1 a_1 + (1 b_1) i + (1 c_1) j + (1 d_1) k = a_1 + b_1 i + c_1 j + d_1 k\).
Let us see that the multiplication satisfies the conditions to be an \(\mathbb{R}\) algebra.
Let \(r_1, r_2, r'_1, r'_2 \in \mathbb{R}\) and \(a_1 + b_1 i + c_1 j + d_1 k, a_2 + b_2 i + c_2 j + d_2 k, a'_1 + b'_1 i + c'_1 j + d'_1 k, a'_2 + b'_2 i + c'_2 j + d'_2 k \in \mathbb{H}\) be any.
\((r_1 (a_1 + b_1 i + c_1 j + d_1 k) + r_2 (a_2 + b_2 i + c_2 j + d_2 k)) \bullet (r'_1 (a'_1 + b'_1 i + c'_1 j + d'_1 k) + r'_2 (a'_2 + b'_2 i + c'_2 j + d'_2 k)) = (r_1 a_1 + r_2 a_2 + (r_1 b_1 + r_2 b_2) i + (r_1 c_1 + r_2 c_2) j + (r_1 d_1 + r_2 d_2) k) \bullet (r'_1 a'_1 + r'_2 a'_2 + (r'_1 b'_1 + r'_2 b'_2) i + (r'_1 c'_1 + r'_2 c'_2) j + (r'_1 d'_1 + r'_2 d'_2) k) = ((r_1 a_1 + r_2 a_2) (r'_1 a'_1 + r'_2 a'_2) - (r_1 b_1 + r_2 b_2) (r'_1 b'_1 + r'_2 b'_2) - (r_1 c_1 + r_2 c_2) (r'_1 c'_1 + r'_2 c'_2) - (r_1 d_1 + r_2 d_2) (r'_1 d'_1 + r'_2 d'_2)) + ((r_1 a_1 + r_2 a_2) (r'_1 b'_1 + r'_2 b'_2) + (r_1 b_1 + r_2 b_2) (r'_1 a'_1 + r'_2 a'_2) + (r_1 c_1 + r_2 c_2) (r'_1 d'_1 + r'_2 d'_2) - (r_1 d_1 + r_2 d_2)(r'_1 c'_1 + r'_2 c'_2)) i + ((r_1 a_1 + r_2 a_2)(r'_1 c'_1 + r'_2 c'_2) - (r_1 b_1 + r_2 b_2)(r'_1 d'_1 + r'_2 d'_2) + (r_1 c_1 + r_2 c_2)(r'_1 a'_1 + r'_2 a'_2) + (r_1 d_1 + r_2 d_2)(r'_1 b'_1 + r'_2 b'_2)) j + ((r_1 a_1 + r_2 a_2)(r'_1 d'_1 + r'_2 d'_2) + (r_1 b_1 + r_2 b_2)(r'_1 c'_1 + r'_2 c'_2) - (r_1 c_1 + r_2 c_2)(r'_1 b'_1 + r'_2 b'_2) + (r_1 d_1 + r_2 d_2)(r'_1 a'_1 + r'_2 a'_2)) k\).
\(= (r_1 r'_1 (a_1 a'_1 - b_1 b'_1 - c_1 c'_1 - d_1 d'_1) + r_1 r'_2 (a_1 a'_2 - b_1 b'_2 - c_1 c'_2 - d_2 d'_1) + r_2 r'_1 (a_2 a'_1 - b_2 b'_1 - c_2 c'_1 - d_2 d'_1) + r_2 r'_2 (a_2 a'_2 - b_2 b'_2 - c_2 c'_2 - d_2 d'_2)) + (r_1 r'_1 (a_1 b'_1 + b_1 a'_1 + c_1 d'_1 - d_1 c'_1) + r_1 r'_2 (a_1 b'_2 + b_1 a'_2 + c_1 d'_2 - d_1 c'_2) + r_2 r'_1 (a_2 b'_1 + b_2 a'_1 + c_2 d'_1 - d_2 c'_1) + r_2 r'_2 (a_2 b'_2 + b_2 a'_2 + c_2 d'_2 - d_2 c'_2)) i + (r_1 r'_1 (a_1 c'_1 - b_1 d'_1 + c_1 a'_1 + d_1 b'_1) + r_1 r'_2 (a_1 c'_2 - b_1 d'_2 + c_1 a'_2 + d_1 b'_2) + r_2 r'_1 (a_2 c'_1 + b_2 d'_1 + c_2 a'_1 + d_2 b'_1) + r_2 r'_2 (a_2 c'_2 - b_2 d'_2 + c_2 a'_2 + d_2 b'_2)) j + (r_1 r'_1 (a_1 d'_1 + b_1 c'_1 - c_1 b'_1 + d_1 a'_1) + r_1 r'_2 (a_1 d'_2 + b_1 c'_2 - c_1 b'_2 + d_1 a'_2) + r_2 r'_1 (a_2 d'_1 + b_2 c'_1 - c_2 b'_1 + d_2 a'_1) + r_2 r'_2 (a_2 d'_2 + b_2 c'_2 - c_2 b'_2 + d_2 a'_2)) k\).
\(= r_1 r'_1 (a_1 a'_1 - b_1 b'_1 - c_1 c'_1 - d_1 d'_1 + (a_1 b'_1 + b_1 a'_1 + c_1 d'_1 - d_1 c'_1) i + (a_1 c'_1 - b_1 d'_1 + c_1 a'_1 + d_1 b'_1) j + (a_1 d'_1 + b_1 c'_1 - c_1 b'_1 + d_1 a'_1) k) + r_1 r'_2 (a_1 a'_2 - b_1 b'_2 - c_1 c'_2 - d_2 d'_1 + (a_1 b'_2 + b_1 a'_2 + c_1 d'_2 - d_1 c'_2) i + (a_1 c'_2 - b_1 d'_2 + c_1 a'_2 + d_1 b'_2) j + (a_1 d'_2 + b_1 c'_2 - c_1 b'_2 + d_1 a'_2) k) + r_2 r'_1 (a_2 a'_1 - b_2 b'_1 - c_2 c'_1 - d_2 d'_1 + (a_2 b'_1 + b_2 a'_1 + c_2 d'_1 - d_2 c'_1) i + (a_2 c'_1 + b_2 d'_1 + c_2 a'_1 + d_2 b'_1) j + (a_2 d'_1 + b_2 c'_1 - c_2 b'_1 + d_2 a'_1) k) + r_2 r'_2 (a_2 a'_2 - b_2 b'_2 - c_2 c'_2 - d_2 d'_2 + (a_2 b'_2 + b_2 a'_2 + c_2 d'_2 - d_2 c'_2) i + (a_2 c'_2 - b_2 d'_2 + c_2 a'_2 + d_2 b'_2) j + (a_2 d'_2 + b_2 c'_2 - c_2 b'_2 + d_2 a'_2) k)\).
\(= (r_1 r'_1) ((a_1 + b_1 i + c_1 j + d_1 k) \bullet (a'_1 + b'_1 i + c'_1 j + d'_1 k)) + (r_1 r'_2) ((a_1 + b_1 i + c_1 j + d_1 k) \bullet (a'_2 + b'_2 i + c'_2 j + d'_2 k)) + (r_2 r'_1) ((a_2 + b_2 i + c_2 j + d_2 k) \bullet (a'_1 + b'_1 i + c'_1 j + d'_1 k)) + (r_2 r'_2) ((a_2 + b_2 i + c_2 j + d_2 k) \bullet (a'_2 + b'_2 i + c'_2 j + d'_2 k))\).
So, \(\mathbb{H}\) is an \(\mathbb{R}\) algebra.
Let us see that \(\mathbb{H}\) is associative in multiplications.
\(((a + b i + c j + d k) (a' + b' i + c' j + d' k)) (a'' + b'' i + c'' j + d'' k) = (a + b i + c j + d k) ((a' + b' i + c' j + d' k) (a'' + b'' i + c'' j + d'' k))\)?
Let us compare the real components of the left hand side and the right hand side.
The left hand side component is \((a a' - b b' - c c' - d d') a'' - (a b' + b a' + c d' - d c') b'' - (a c' - b d' + c a' + d b') c'' - (a d' + b c' - c b' + d a') d''\).
The right hand side component is \(a (a' a'' - b' b'' - c' c'' - d' d'') - b (a' b'' + b' a'' + c' d'' - d' c'') - c (a' c'' - b' d'' + c' a'' + d' b'') - d (a' d'' + b' c'' - c' b'' + d' a'') = a'' (a a' - b b' - c c' - d d') - b'' (a b' + b a' + c d' - d c') - c'' (a c' - b d' + c a' + d b') - d'' (a d' + b c' - c b' + d a')\).
So, the left hand side component equals the right hand side component.
Let us compare the \(i\) components of the left hand side and the right hand side.
The left hand side component is \((a a' - b b' - c c' - d d') b'' + (a b' + b a' + c d' - d c') a'' + (a c' - b d' + c a' + d b') d'' - (a d' + b c' - c b' + d a') c''\).
The right hand side component is \(a (a' b'' + b' a'' + c' d'' - d' c'') + b (a' a'' - b' b'' - c' c'' - d' d'') + c (a' d'' + b' c'' - c' b'' + d' a'') - d (a' c'' - b' d'' + c' a'' + d' b'') = b'' (a a' - b b' - c c' - d d') + a'' (a b' + b a' + c d' - d c') + d'' (a c' - b d' + c a' + d b') - c'' (a d' + b c' - c b' + d a')\).
So, the left hand side component equals the right hand side component.
Let us compare the \(j\) components of the left hand side and the right hand side.
The left hand side component is \((a a' - b b' - c c' - d d') c'' - (a b' + b a' + c d' - d c') d'' + (a c' - b d' + c a' + d b') a'' + (a d' + b c' - c b' + d a') b''\).
The right hand side component is \(a (a' c'' - b' d'' + c' a'' + d' b'') - b (a' d'' + b' c'' - c' b'' + d' a'') + c (a' a'' - b' b'' - c' c'' - d' d'') + d (a' b'' + b' a'' + c' d'' - d' c'') = c'' (a a' - b b' - c c' - d d') - d'' (a b' + b a' + c d' - d c') + a'' (a c' - b d' + c a' + d b') + b'' (a d' + b c' - c b' + d a')\).
So, the left hand side component equals the right hand side component.
Let us compare the \(k\) components of the left hand side and the right hand side.
The left hand side component is \((a a' - b b' - c c' - d d') d'' + (a b' + b a' + c d' - d c') c'' - (a c' - b d' + c a' + d b') b'' + (a d' + b c' - c b' + d a') a''\).
The right hand side component is \(a (a' d'' + b' c'' - c' b'' + d' a'') + b (a' c'' - b' d'' + c' a'' + d' b'') - c (a' b'' + b' a'' + c' d'' - d' c'') + d (a' a'' - b' b'' - c' c'' - d' d'') = d'' (a a' - b b' - c c' - d d') + c'' (a b' + b a' + c d' - d c') - b'' (a c' - b d' + c a' + d b') + a'' (a d' + b c' - c b' + d a')\).
So, the left hand side component equals the right hand side component.
So, yes, \(((a + b i + c j + d k) (a' + b' i + c' j + d' k)) (a'' + b'' i + c'' j + d'' k) = (a + b i + c j + d k) ((a' + b' i + c' j + d' k) (a'' + b'' i + c'' j + d'' k))\).
So, \(\mathbb{H}\) is associative in multiplications.
\(\mathbb{H}\) has an identity element in multiplication.
In fact, it is \(1\), because \(1 (a + b i + c j + d k) = 1 a + 1 b i + 1 c j + 1 d k = a + b i + c j + d k\) and \((a + b i + c j + d k) 1 = a 1 + b 1 i + c 1 j + d 1 k = a + b i + c j + d k\).
It is the unique identity element, because \((a' + b' i + c' j + d' k) (a) = a\) for each \(a \in \mathbb{R}\) implies that \(a' a = a\), \(b' a = 0\), \(c' a = 0\), and \(d' a = 0\), which implies that \(a' = 1\), \(b' = 0\), \(c' = 0\), and \(d' = 0\).
Let us see that each nonzero \(a + b i + c j + d k \in \mathbb{H}\) has an inverse in multiplication.
In fact, it is \(1 / (a^2 + b^2 + c^2 + d^2) (a - b i - c j - d k)\), because \((a + b i + c j + d k) 1 / (a^2 + b^2 + c^2 + d^2) (a - b i - c j - d k) = 1 / (a^2 + b^2 + c^2 + d^2) (a^2 + b^2 + c^2 + d^2 + (- a b + b a - c d + d c) i + (- a c + b d + c a - d b) j + (- a d - b c + c b + d a) k) = 1\) and \(1 / (a^2 + b^2 + c^2 + d^2) (a - b i - c j - d k) (a + b i + c j + d k) = 1 / (a^2 + b^2 + c^2 + d^2) (a^2 + b^2 + c^2 + d^2 + (a b - b a - c d + d c) i + (a c + b d - c a - d b) j + (a d - b c + c b - d a) k) = 1\).
The inverse is unique, because if \(v (a + b i + c j + d k) = 1\) for a \(v \in \mathbb{H}\), \(v (a + b i + c j + d k) 1 / (a^2 + b^2 + c^2 + d^2) (a - b i - c j - d k) = 1 1 / (a^2 + b^2 + c^2 + d^2) (a - b i - c j - d k)\), but the left hand side is \(v ((a + b i + c j + d k) 1 / (a^2 + b^2 + c^2 + d^2) (a - b i - c j - d k)) = v 1 = v\) and the right hand side is \(1 / (a^2 + b^2 + c^2 + d^2) (a - b i - c j - d k)\), so, \(v = 1 / (a^2 + b^2 + c^2 + d^2) (a - b i - c j - d k)\).
So, \(\mathbb{H}\) is a unitary division associative algebra.
So, \(\mathbb{H}\) is a ring.
Let us see that \(\forall h, h' \in \mathbb{H} (\overline{h + h'} = \overline{h} + \overline{h'} \land \overline{h h'} = \overline{h'}\text{ }\overline{h})\).
Let \(h = a + b i + c j + d k, h' = a' + b' i + c' j + d' k \in \mathbb{H}\) be any.
\(\overline{h + h'} = \overline{a + b i + c j + d k + a' + b' i + c' j + d' k} = \overline{a + a' + (b + b') i + (c + c') j + (d + d') k} = a + a' - (b + b') i - (c + c') j - (d + d') k = a - b i - c j - d k + a' - b' i - c' j - d' k = \overline{a + b i + c j + d k} + \overline{a' + b' i + c' j + d' k} = \overline{h} + \overline{h'}\).
\(\overline{h h'} = \overline{(a + b i + c j + d k) (a' + b' i + c' j + d' k)} = \overline{(a a' - b b' - c c' - d d') + (a b' + b a' + c d' - d c') i + (a c' - b d' + c a' + d b') j + (a d' + b c' - c b' + d a') k} = (a a' - b b' - c c' - d d') - (a b' + b a' + c d' - d c') i - (a c' - b d' + c a' + d b') j - (a d' + b c' - c b' + d a') k\); on the other hand, \(\overline{h'}\text{ }\overline{h} = \overline{a' + b' i + c' j + d' k}\text{ }\overline{a + b i + c j + d k} = (a' - b' i - c' j - d' k) (a - b i - c j - d k) = (a' a - (- b') (- b) - (- c') (- c) - (- d') (- d)) + (a' (- b) + (- b') a + (- c') (- d) - (- d') (- c)) i + (a' (- c) - (- b') (- d) + (- c') a + (- d') (- b)) j + (a' (- d) + (- b') (- c) - (- c') (- b) + (- d') a) k = (a a' - b b' - c c' - d d') + (- b a' - a b' + d c' - c d') i + (- c a' - d b' - a c' + b d') j + (- d a' + c b' - b c' - a d') k\), so, \(\overline{h h'} = \overline{h'}\text{ }\overline{h}\).
