I recently read John Baez's paper on the octonions (http://\nmath.ucr.edu/home/baez/octonions/) and in particular the section on\nthe Cayley-Dickson Construction of the complex numbers, quaternions,\noctonions, sedenions, etc. ([URL]http://math.ucr.edu/home/baez/octonions/[/URL]\nnode5.html).\n\nIf the octonion (a0+a1e1+a2e2+a3e3+a4e4+a5e5+a6e6+a7e7) is expressed\nas the ordered set of real numbers (a0, a1, a2, a3, a4, a5, a6, a7),\nthen after expanding all the terms of octonion multiplication in the\nCayley-Dickson Construction, one gets the following result:\n\n(a0,a1,a2,a3,a4,a5,a6,a7)(b0,b1,b2,b3,b4,b5,b6,b7) =\n\n(a0b0 -\- a1b1 -\- a2b2 -\- a3b3 -\- a4b4 -\- a5b5 -\- a6b6 -\- a7b7,\na0b1 + a1b0 + a3b2 -\- a2b3 + a5b4 -\- a4b5 + a6b7 -\- a7b6,\na0b2 + a2b0 + a1b3 -\- a3b1 + a6b4 -\- a4b6 + a7b5 -\- b7a5,\na0b3 + a3b0 + a2b1 -\- a1b2 + a7b4 -\- a4b7 + a5b6 -\- a6b5,\na0b4 + a4b0 + a1b5 -\- a5b1 + a2b6 -\- a6b2 + a3b7 -\- a7b3,\na0b5 + a5b0 + a4b1 -\- a1b4 + a2b7 -\- a7b2 + a6b3 -\- a3b6,\na0b6 + a6b0 + a4b2 -\- a2b4 + a3b5 -\- a5b3 + a7b1 -\- a1b7,\na0b7 + a7b0 + a4b3 -\- a3b4 + a1b6 -\- a6b1 + a5b2 -\- a2b5)\n\nSince the sedenions and all higher-dimensional algebras created by the\nCayley-Dickson Construction have zero-divisors, for example in the\nsedenions (e1 + e10)(e15 -\- e4) = 0, I was looking for a way to\ndemonstrate that no such zero-divisors of the form\n(ex + ey)(ez -\- ew) = 0 exist in the octonions simply by looking at\nthe above expansion. In doing so I discovered that if the signs are\nswitched on a single pair of terms in the above expansion, for example\nif instead of + a2b1 -\- a1b2 we had + a1b2 -\- a2b1, then in this 8-\ndimensional algebra we would have zero-divisors, for example (e1 + e5)\n(e2 -\- e6) = + e4 -\- e4 -\- e7 + e7 = 0. What's more, if the signs are\nswitched on the terms a3b2, a2b3, a1b3, and a3b1 as well, we get an\nalgebra with still more zero-divisors of this form, and in which the\ncoefficients a1, b1, etc., still appear four times with positive sign\nand four times with negative sign in the expansion. Note that in this\nexample, the octonions e1, e2, e3 whose coefficients' signs have been\nswitched constitute together with 1 a subset of quaternions within the\noctonions.\n\nI call these octonions with one or more pairs of switched signs\nmutated octonions, because the transformation from octonions to them\nreminds me of a mutation of DNA in genetics. I have verified that\nthese mutated octonions, like the sedenions, are a *-algebra with a\nconjugation\n(a0, a1, a2, a3, a4, a5, a6, a7)* = (a0, -a1, -a2, -a3, -a4, -a5, -a6,\n-a7) which satisfies a**=a and (ab)* = b*a*, and that they are nicely\nnormed since a + a* is in the reals and aa* = a*a > 0 for all nonzero\na in the mutated octonion algebras. (See the section of Baez's paper\non the Cayley-Dickson Construction for these definitions.)\n\nWhile the octonions with three pairs of switched signs appear more\nsymmetrical than the octonions with a single pair of switched signs,\nin a certain way the single-pair mutation makes more sense. When the\nsigns of the coefficients of e1, e2, and e3 are all switched, products\nthat equal +/-2e1, +/-2e2, and +/-2e3 in the octonions all become 0.\nFor example (e1 + e4)(e3 -\- e6) = 2e2 in the octonions, 0 in the three-\nmutation octonions; (e3 + e7)(e2 -\- e6) = 2e1 in the octonions, 0 in\nthe three-mutation octonions; and (e2 + e6)(e1 -\- e5) = 2e3 in the\noctonions, 0 in the three-mutation octonions. But when only the signs\nof the products e1e2 and e2e1 are switched from -e3 and +e3\nrespectively to +e3 and -e3 respectively, then products that equal\n+/-2e3 in the octonions become 0, but products that equal +/-2e1 and\n+/-2e2 in the octonions are unaffected. Thinking in terms of the\ngenetic mutation concept, this is a mutation of e1 and e2 that\nproduces a relative increase in the amount of them by reducing the\namount of e3.\n\nQuaternions can also be mutated in this way. Mutating three pairs of\nterms in the expansion of the quaternion product would just create\nanother set of normal quaternions, but mutating one pair of terms\nworks the same way as in the case of octonions. The mutated quaternion\nmultiplication table is as follows:\n\n(1)(1) = 1 ; (1)(e1) = e1 ; (1)(e2) = e2 ; (1)(e3) = e3\n(e1)(1) = e1 ; (e1)(e1) = -1 ; (e1)(e2) = +e3 ; (e1)(e3) = +e2\n(e2)(1) = e2 ; (e2)(e1) = -e3 ; (e2)(e2) = -1 ; (e2)(e3) = -e1\n(e3)(1) = e3 ; (e3)(e1) = -e2 ; (e3)(e2) = +e1 ; (e3)(e3) = -1\n\nThe properties of mutated quaternions are rather remarkable. Like the\nsedenions and the mutated octonions, they are a *-algebra with a\nconjugation (a0, a1, a2, a3)* = (a0, -a1, -a2, -a3) which satisfies\na**=a and (ab)* = b*a*, and they are nicely normed. The really\ninteresting thing about them is the patterns formed by the products of\ntwo mutated quaternions with two nonzero +/-1 coefficients. In the\nnormed division algebras, the reals, complex numbers, quaternions, and\noctonions, the product of two n-ions each with two nonzero +/-1\ncoefficients always has either four nonzero coefficients or one\nnonzero coefficient. But in the mutated quaternions, as in the mutated\noctonions and in the sedenions, the product of two of them each with\ntwo nonzero +/-1 coefficients can have either zero, one, two, or four\nnonzero coefficients. What's remarkable about the mutated quaternions\nis not only their zero-divisors, the products with zero nonzero\ncoefficients, but also the products with two nonzero coefficients, and\nthe relationship between the two sets of products.\n\nThe first noteworthy observation is that in the mutated quaternions,\nin each pair of zero-divisors one of them must have a nonzero real\npart. Here are the products of two mutated quaternions with two\nnonzero +/-1 coefficients that have zero nonzero coefficients, i.e.,\nthe zero-divisors:\n\n(1 + e1)(e2 -\- e3) = 0 (e2 + e3)(1 + e1) = 0\n(1 -\- e1)(e2 + e3) = 0 (e2 -\- e3)(1 -\- e1) = 0\n\n(1 + e2)(e1 + e3) = 0 (e1 -\- e3)(1 + e2) = 0\n(1 -\- e2)(e1 -\- e3) = 0 (e1 + e3)(1 -\- e2) = 0\n\nNow the relationship between these products and the products with two\nnonzero coefficients: they are generated by switching the sign of e1\nor e2 in each of the (1 +/- e1) and (1 +/- e2) terms of the above zero-\ndivisors. But that's not all: in each case, the resulting product\nequals exactly two times the (e2 +/- e3) or (e1 +/- e3) term!\n\n(1 -\- e1)(e2 -\- e3) = 2e2 -\- 2e3 (e2 + e3)(1 -\- e1) = 2e2 + 2e3\n(1 + e1)(e2 + e3) = 2e2 + 2e3 (e2 -\- e3)(1 + e1) = 2e2 -\- 2e3\n\n(1 -\- e2)(e1 + e3) = 2e1 + 2e3 (e1 -\- e3)(1 -\- e2) = 2e1 -\- 2e3\n(1 + e2)(e1 -\- e3) = 2e1 -\- 2e3 (e1 + e3)(1 + e2) = 2e1 + 2e3\n\nThus, wherever (1 + e1) is a zero-divisor, (1 -\- e1) in effect\nmultiplies the other term by 2, and vice versa, and the same holds for\n(1 + e2) and (1 -\- e2). Thus we can say that in the (1 +/- e1) and (1\n+/- e2) terms of the above equations, e1 and e2 are functioning\nexactly like real numbers +/-1 ! Where (1 + e1) in effect multiplies\nby 0 and (1 -\- e1) multiplies by 2, e1 is functioning as the real\nnumber -1, and where (1 -\- e1) multiplies by 0 and (1 + e1) multiplies\nby 2, e1 is functioning as the real number 1.\n\nWe can define precisely where e1 and e2 function as 1 and -1 as\nfollows:\n\nIf (1 +/- e1) is the left-hand multiplier and e2 and e3 have the same\nsign, e1 functions as 1.\nIf (1 +/- e1) is the left-hand multiplier and e2 and e3 have opposite\nsigns, e1 functions as -1.\nIf (1 +/- e1) is the right-hand multiplier and e2 and e3 have the same\nsign, e1 functions as -1.\nIf (1 +/- e1) is the right-hand multiplier and e2 and e3 have opposite\nsigns, e1 functions as 1.\nIf (1 +/- e2) is the left-hand multiplier and e1 and e3 have the same\nsign, e2 functions as -1.\nIf (1 +/- e2) is the left-hand multiplier and e1 and e3 have opposite\nsigns, e2 functions as 1.\nIf (1 +/- e2) is the right-hand multiplier and e1 and e3 have the same\nsign, e2 functions as 1.\nIf (1 +/- e2) is the right-hand multiplier and e1 and e3 have opposite\nsigns, e2 functions as -1.