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**Sample text**

Pi La n L + dim(Z/! is 2) o L 2 ) = dim dim L2 . A - B 4 V + V2 and = 1 + dim(F + F2 ). empty. Then o + L, La) 1 r Let us illustrate this in a few special cases. then L! = A, L 2 = B, F! = F2 = 0; if L v If dim L L2 ? , = dim L 2 = 0, then they do not meet and dim(L! o Lj o L2 is L 2) = the unique "line" through A 1 ; and B. Suppose n = dim F = 2, dim L^ = dim L 2 empty we have dim(Li o L 2 ) = 1 + dim(F! dim(F! If +F = Lj C\ 2) L2 is 1, V = F l 2 = 1. + F (parallel lines). not empty, then dim(L 1 n L + dim^ o L a) a) = 2.

27) We shall derive a contradiction from this assumption. 28) (a(b < left side is (ab + ax)' 9 = a'b' + aV or ((b The 9 x) x: + *) V= = a'b' = a'x'. + b'a' + V. (ax). 26), that (ax) = a'x'. 28) . GEOMETRIC ALGEBRA 40 = (ax)' + (c(d a'x'. The same method is used on and x)) ((c + x)d)'. 32) (xd)' = d'x'. 31). 30) obtain = and d'a' c'6" = 6V. 34) ((a + c)(b + = d)Y \ab) ~\~ \ad) ~\~ = a'b' + a'd' + c'b' + c'f or + d)'(o computation of the direct d)' < ((b A + c)'(b + (co) b'a' + d'a' + b'c' + dV. left side gives: (ca) ~\~ + c)' = = a b ~{~ ad ~f* c b ~\~ d c .

A field fc of this type is called a field of characteristic 0. Suppose, on the other hand, that the map n>ne has a non-zero kernel H, a subgroup consists of all multiples of the additive group Z. Such a subgroup 5* k GEOMETRIC ALGEBRA 34 pv of the smallest positive integer p in H. As remarked earlier p j* 1. If p were not a prime number, then p = db with positive a, & which < are p. Since a true since k is field of this type and The only < v < = p-e field. we would This number p 0, a prime number.