52
II - Convergence: Discrete variables
(0.3)
(a, b).(e, d) = (ae - bd, ad + be),
and to prove that conditions (C 1) to (C 4) above are satisfied 8 . This requires
only the simplest algebraic calculations and, for beginners, may even be a
good exercise in elementary algebra.
Suppose for instance we want to check the associativity of multiplication
(1.5), which in C is the identity
[(a, b)(a', b')](a", b") = (a, b)[(a', b')(a", b")].
Applying (0.3) blindly, this first reduces to
. (aa' - bb', ab' + ba')( a", b") = ( a, b) (a' a" - b'b", a'b" + b' a"),
then, again by (0.3), to
((aa' - bb')a" - (ab' + ba')b", (aa' - bb')b" + (ab' + ba')a") =
= (a(a'a" - b'b") - b(a'b" + b'a"),a(a'b" + b'a") + b(a'a" - b'b")).
Applying rule (1.9) of nO 1 for real numbers, we are reduced to proving that
(aa')a" - (bb')a" - (ab'W' - (ba')b", (aa')b" - (bb')b" + (ab')a" + (ba')a") =
= (a(a'a") - a(b'b") - b(a'b") - b(b'a"), a(a'b") + a(b'a") + b(a'a") - b(b'b")),
and the result then follows from the associativity rules (1.5) for real numbers.
Rules (1.1), (1.2), (1.6) and (1.9) for complex numbers are proved in similar
ways.
The existence of complex numbers "zero" and "one" is clear: they are the
pairs (0,0) and (1,0). The "opposite" of (a, b) is obviously (-a, -b). To prove
(1.8), i.e. that we can solve (a, b)(x, y) = (1,0) for any pair (a, b) =J. (0,0), we
write this as
(0.4)
ax - by = 1, ay + bx = 0;
if b = 0, in which case a =J. 0, the pair (l/a,O) is a solution by (1.8) for lR. If
b =J. 0, the second relation is equivalent to x = -ay Ib, hence the first to
(a 2 + b 2 )y = -b :
and in JR this can be solved in one and only one way provided that a 2 +b 2 =J. O.
But since b =J. 0, we may write a = be for some e E JR, and the relation
a 2 + b 2 = 0 then is equivalent to
8 These are easy to understand when one sees that the pair ( a, b) must after all
represent the sought-for symbol a + ib:
(a + ib) + (c + id)
a + b + i(c + d),
(a + ib)(c + id) = ac + ibc + aid + i 2 bd
ac - bd + i(ad + be).
II - Convergence: Discrete variables
(0.3)
(a, b).(e, d) = (ae - bd, ad + be),
and to prove that conditions (C 1) to (C 4) above are satisfied 8 . This requires
only the simplest algebraic calculations and, for beginners, may even be a
good exercise in elementary algebra.
Suppose for instance we want to check the associativity of multiplication
(1.5), which in C is the identity
[(a, b)(a', b')](a", b") = (a, b)[(a', b')(a", b")].
Applying (0.3) blindly, this first reduces to
. (aa' - bb', ab' + ba')( a", b") = ( a, b) (a' a" - b'b", a'b" + b' a"),
then, again by (0.3), to
((aa' - bb')a" - (ab' + ba')b", (aa' - bb')b" + (ab' + ba')a") =
= (a(a'a" - b'b") - b(a'b" + b'a"),a(a'b" + b'a") + b(a'a" - b'b")).
Applying rule (1.9) of nO 1 for real numbers, we are reduced to proving that
(aa')a" - (bb')a" - (ab'W' - (ba')b", (aa')b" - (bb')b" + (ab')a" + (ba')a") =
= (a(a'a") - a(b'b") - b(a'b") - b(b'a"), a(a'b") + a(b'a") + b(a'a") - b(b'b")),
and the result then follows from the associativity rules (1.5) for real numbers.
Rules (1.1), (1.2), (1.6) and (1.9) for complex numbers are proved in similar
ways.
The existence of complex numbers "zero" and "one" is clear: they are the
pairs (0,0) and (1,0). The "opposite" of (a, b) is obviously (-a, -b). To prove
(1.8), i.e. that we can solve (a, b)(x, y) = (1,0) for any pair (a, b) =J. (0,0), we
write this as
(0.4)
ax - by = 1, ay + bx = 0;
if b = 0, in which case a =J. 0, the pair (l/a,O) is a solution by (1.8) for lR. If
b =J. 0, the second relation is equivalent to x = -ay Ib, hence the first to
(a 2 + b 2 )y = -b :
and in JR this can be solved in one and only one way provided that a 2 +b 2 =J. O.
But since b =J. 0, we may write a = be for some e E JR, and the relation
a 2 + b 2 = 0 then is equivalent to
8 These are easy to understand when one sees that the pair ( a, b) must after all
represent the sought-for symbol a + ib:
(a + ib) + (c + id)
a + b + i(c + d),
(a + ib)(c + id) = ac + ibc + aid + i 2 bd
ac - bd + i(ad + be).