Ordering of Fractions

Theorem 134 (Landau 41).

If A1/A2, B1/B2 are arbitrary, then exactly one of the following is the case: A1/A2 ~ B1/B2, A1/A2 > B1/B2, A1/A2 < B1/B2

Proof: For the arbitrary A1, A2, B1, B2, exactly one of the following holds: A1•B2 > B1•A2, A1•B2 = B1•A2, A1•B2 < B1•A2.

Theorem 135 (Landau 42). If A1/A2 > B1/B2, then B1/B2 < A1/A2.

This follows from the meaning of greater and lesser. In symbols: A1•B2 > B1•A2 means B1•A2 < A1•B2, by Theorem 10.

Theorem 136 (Landau 43). If A1/A2 < B1/B2, then B1/B2 > A1/A2.

Likewise, by Theorem 10.

Theorem 137 (Landau 44). If A1/A2 > B1/B2, A1/A2 ~ C1/C2, and B1/B2 ~ D1/D2, then C1/C2 > D1/D2.

So greater and lesser belong to what equivalent fractions have in common, not to the particular pair of numbers.

Proof: B1•D2 = D1•B2, C1•A2 = A1•C2, and A1•B2 > B1•A2. Multiplying the two equalities, (B1•D2)(C1•A2) = (D1•B2)(A1•C2). Rearranging by Theorems 31 and 33, (B1•A2)(C1•D2) = (D1•C2)(A1•B2) > (D1•C2)(B1•A2), by Theorem 34. Hence C1•D2 > D1•C2, by Theorem 35.

Theorem 138 (Landau 45). If A1/A2 < B1/B2, A1/A2 ~ C1/C2, and B1/B2 ~ D1/D2, then C1/C2 < D1/D2.

By Theorem 136, B1/B2 > A1/A2. By Theorem 137, D1/D2 > C1/C2. By Theorem 135, C1/C2 < D1/D2.

Theorem 139 (Landau 46). If A1/A2 ≳ B1/B2, A1/A2 ~ C1/C2, and B1/B2 ~ D1/D2, then C1/C2 ≳ D1/D2.

By Theorem 137 if > holds. Otherwise C1/C2 ~ A1/A2 ~ B1/B2 ~ D1/D2, by Theorems 131 and 132.

Theorem 140 (Landau 47). The same with ≲ in place of ≳.

By Theorem 138 if < holds. Otherwise as in Theorem 139.

Theorem 141 (Landau 48). If A1/A2 ≳ B1/B2, then B1/B2 ≲ A1/A2.

Theorems 131 and 135.

Theorem 142 (Landau 49). If A1/A2 ≲ B1/B2, then B1/B2 ≳ A1/A2.

Theorems 131 and 136.

Theorem 143 (Landau 50). If A1/A2 < B1/B2 and B1/B2 < C1/C2, then A1/A2 < C1/C2. (Transitivity of Ordering)

Proof: A1•B2 < B1•A2 and B1•C2 < C1•B2. By Theorem 36, (A1•B2)(B1•C2) < (B1•A2)(C1•B2). Rearranging, (A1•C2)(B1•B2) < (C1•A2)(B1•B2), so A1•C2 < C1•A2, by Theorem 35.

Theorem 144 (Landau 51). If A1/A2 ≲ B1/B2 and B1/B2 < C1/C2, or A1/A2 < B1/B2 and B1/B2 ≲ C1/C2, then A1/A2 < C1/C2.

By Theorem 138 if an equivalence holds; otherwise by Theorem 143.

Theorem 145 (Landau 52). If A1/A2 ≲ B1/B2 and B1/B2 ≲ C1/C2, then A1/A2 ≲ C1/C2.

By Theorem 132 if two equivalences hold; otherwise by Theorem 144.

Theorem 146 (Landau 53). For any A1/A2 there is a greater fraction.

(A1 + A1)•A2 = A1•A2 + A1•A2 > A1•A2. So (A1 + A1)/A2 > A1/A2.

Theorem 147 (Landau 54). For any A1/A2 there is a lesser fraction.

A1•A2 < A1•A2 + A1•A2 = A1•(A2 + A2). So A1/(A2 + A2) < A1/A2.

Theorem 148 (Landau 55). If A1/A2 < B1/B2, there is a C1/C2 with A1/A2 < C1/C2 < B1/B2.

Proof: A1•B2 < B1•A2. Adding A1•A2 to both sides, and separately B1•B2 to both sides (Theorem 13), A1•A2 + A1•B2 < A1•A2 + B1•A2 and A1•B2 + B1•B2 < B1•A2 + B1•B2. By Theorem 32, A1•(A2 + B2) < (A1 + B1)•A2 and (A1 + B1)•B2 < B1•(A2 + B2). So A1/A2 < (A1 + B1)/(A2 + B2) < B1/B2.

Remark. The fraction found here is made by adding antecedents and adding consequents. Applied again and again, it yields fractions ever nearer to a given one from either side. This is the instrument for relative maximum and minimum later.