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EXAMPLE 54 The equivalence class [(4, 12)] contains all of the pairs (4, 12), (1, 3), ( 2, 6) (Of course it contains in nitely many other pairs as well) This equivalence class represents the fraction 4/12 which we sometimes also write as 1/3 or ( 2)/( 6) If [(a, b)] and [(c, d)] are rational numbers then we de ne their product to be the rational number [(a c, b d)] This is well de ned (unambiguous), for the following reason Suppose that (a, b) is related to (a, b) and (c, d) is related to (c, d) We would like to know that [(a, b)] [(c, d)] = [(a c, b b)] is the same equivalence class as [(a, b)] [(c, d)] = [(a c, b d)] In other words we need to know that (a c) (b d) = (a c) (b d) But our hypothesis is that a b =a b and c d =c d (52)
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Multiplying together the left sides and the right sides we obtain (a b) (c d) = (a b) (c d) Rearranging, we have (a c) (b d) = (a c) (b d) But this is just Eq 52 So multiplication is unambiguous EXAMPLE 55 The product of the two rational numbers [(3, 8)] and [( 2, 5)] is [(3 ( 2), 8 5)] = [( 6, 40)] = [( 3, 20)] This is what we expect: the product of 3/8 and 2/5 is 3/20 If q = [(a, b)] and r = [(c, d)] are rational numbers and if r is not zero (that is, [(c, d)] is not the equivalence class zero in other words, c = 0) then we de ne
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the quotient q/r to be the equivalence class [(ad, bc)] We leave it to you to check that this operation is well de ned EXAMPLE 56 The quotient of the rational number [(4, 7)] by the rational number [(3, 2)] is, by de nition, the rational number [(4 ( 2), 7 3)] = [( 8, 21)] This is what we expect: the quotient of 4/7 by 3/2 is 8/21 How should we add two rational numbers We could try declaring [(a, b)] + [(c, d)] to be [(a + c, b + d)], but this will not work (think about the way that we usually add fractions) Instead we de ne [(a, b)] + [(c, d)] = [(a d + b c, b d)] That this de nition is well de ned (unambiguous) is left for the exercises We turn instead to an example EXAMPLE 57 The sum of the rational numbers [(3, 14)] and [(9, 4)] is given by [(3 4 + ( 14) 9, ( 14) 4)] = [( 114, 56)] = [(57, 28)] This coincides with the usual way that we add fractions: 9 57 3 + = 14 4 28
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Notice that the equivalence class [(0, 1)] is the rational number that we usually denote by 0 It is the additive identity, for if [(a, b)] is another rational number then [(0, 1)] + [(a, b)] = [(0 b + 1 a, 1 b)] = [(a, b)] A similar argument shows that [(0, 1)] times any rational number gives [(0, 1)] or 0 By the same token the rational number [(1, 1)] is the multiplicative identity We leave the details for you Of course the concept of subtraction is really just a special case of addition [that is, a b is the same as a + ( b)] So we shall say nothing further about subtraction
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