Multiplication of Fractions

Theorem 161 (Landau 68). If A1/A2 ~ B1/B2 and C1/C2 ~ D1/D2, then A1/A2 • C1/C2 ~ B1/B2 • D1/D2.

So the product too belongs to what the equivalent fractions have in common.

Proof: A1•B2 = B1•A2 and C1•D2 = D1•C2. Multiplying, (A1•B2)(C1•D2) = (B1•A2)(D1•C2). Rearranging, (A1•C1)(B2•D2) = (B1•D1)(A2•C2).

Theorem 162 (Landau 69). A1/A2 • B1/B2 ~ B1/B2 • A1/A2. (Commutative Law of Multiplication)

Theorem 31 applied to antecedents and to consequents.

Theorem 163 (Landau 70). (A1/A2 • B1/B2) • C1/C2 ~ A1/A2 • (B1/B2 • C1/C2). (Associative Law of Multiplication)

Theorem 33 applied to antecedents and to consequents.

Theorem 164 (Landau 71). A1/A2 • (B1/B2 + C1/C2) ~ A1/A2 • B1/B2 + A1/A2 • C1/C2. (Distributive Law)

Proof: The left side is (A1•(B1•C2 + C1•B2))/(A2•(B2•C2)), which by Theorem 32 is (A1•(B1•C2) + A1•(C1•B2))/(A2•(B2•C2)). By Theorem 150 this is equivalent to the sum of ((A1•B1)•C2)/((A2•B2)•C2) and ((A1•C1)•B2)/((A2•C2)•B2), and by Theorem 133 these are equivalent to (A1•B1)/(A2•B2) and (A1•C1)/(A2•C2).

Theorem 165 (Landau 72). If A1/A2 > B1/B2, or ~, or <, then A1/A2 • C1/C2 > B1/B2 • C1/C2, or ~, or <, respectively.

If A1•B2 > B1•A2, then (A1•B2)(C1•C2) > (B1•A2)(C1•C2) by Theorem 34, that is, (A1•C1) (B2•C2) > (B1•C1)(A2•C2). The second case is Theorem 161. The third follows from the first by Theorems 136 and 135.

Theorem 166 (Landau 73). The converse of Theorem 165.

The three cases are mutually exclusive and exhaust all possibilities.

Theorem 167 (Landau 74). If A1/A2 > B1/B2 and C1/C2 > D1/D2, then A1/A2 • C1/C2 > B1/B2 • D1/D2.

By Theorem 165 twice, with Theorem 162, and then Theorem 143.

Theorem 168 (Landau 75). If A1/A2 ≳ B1/B2 and C1/C2 > D1/D2, or A1/A2 > B1/B2 and C1/C2 ≳ D1/D2, then A1/A2 • C1/C2 > B1/B2 • D1/D2.

By Theorems 161 and 165 if an equivalence holds; otherwise by Theorem 167.

Theorem 169 (Landau 76). If A1/A2 ≳ B1/B2 and C1/C2 ≳ D1/D2, then A1/A2 • C1/C2 ≳ B1/B2 • D1/D2.

By Theorem 161 if two equivalences hold; otherwise by Theorem 168.

Theorem 170 (Landau 77). For any A1/A2 and B1/B2 there is a fraction D1/D2 with B1/B2 • D1/D2 ~ A1/A2, and any two such are equivalent.

Proof: Any two such are equivalent by Theorem 166. To find one, let D1 = A1•B2 and D2 = A2•B1. Then B1/B2 • (A1•B2)/(A2•B1) ~ (A1•(B1•B2))/(A2•(B1•B2)) ~ A1/A2, by Theorems 31, 33 and 133.

Remark. Among fractions division is always possible, where subtraction is not (Theorem 160). This is the first place where a derived kind has an operation that number lacks.