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N} = · i=1 ai1. 35 Set ki = |ai1| and aij = σj−1(ai1) then due to (16) we get t {1, . . , n} = · (17) {ai1, ai2, . . , aiki }. i=1 It remains to show that σ = σ1 ◦ · · · ◦ σt where σi = (ai1 · · · aiki ) is a ki-cycle. For this let b ∈ {1, . . , n} so that b = aij = σj−1(ai1) for some 1 ≤ i ≤ t and some 1 ≤ j ≤ ki. We now apply σ to b and get σ(b) = σ(aij) = σj(ai1) = aij+1, ai1, if j < ki, if j = ki = σi(b). Since the decomposition in (17) is disjoint and since b as well as σi(b) are contained in {ai1, .

Is conversely (i, i + 1) not an error pair of σ ′ then the composition with τ creates this error pair and σ has one error pair more than σ ′ . Thus sgn(σ) = − sgn(σ ′ ) = sgn(σ ′ ) · sgn(τ). 14 every permutation is a product of transpositions of consecutive integers. 40 Let σ1 = τ˜1 ◦ · · · ◦ τ˜r and σ2 = τ˜r+1 ◦ · · · ◦ τ˜r+s be given as products of such transpositions of consecutive numbers. By induction on r + s we see that sgn(σ1 ◦ σ2) = (−1)r+s = (−1)r · (−1)s = sgn(σ1) · sgn(σ2). , and b.

Also the rules for addition and multiplication of rational numbers can easily be formulated with the notion of equivalence classes. For (p, q), (r, s) ∈ M define: (p, q) + (r, s) := (ps + qr, qs), (p, q) · (r, s) := (pr, qs). There is, however, one problem one has to come about with this definition, namely, that it does not depend on the particular representatives that we have chosen. We say that we have to show that the operation is well defined (see also Footnote 22 on page 55). We demonstrate this for the addition of rational numbers.

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