Addition of Fractions

Theorem 149 (Landau 56). If A1/A2 ~ B1/B2 and C1/C2 ~ D1/D2, then A1/A2 + C1/C2 ~ B1/B2 + D1/D2.

So the sum belongs to what the equivalent fractions have in common.

Proof: A1•B2 = B1•A2 and C1•D2 = D1•C2. Multiplying the first by C2•D2 and the second by A2•B2, and rearranging, (A1•C2)(B2•D2) = (B1•D2)(A2•C2) and (C1•A2)(B2•D2) = (D1•B2) (A2•C2). Adding (Common Notion 2) and using Theorem 32, (A1•C2 + C1•A2)(B2•D2) = (B1•D2 + D1•B2)(A2•C2), which is the equivalence required.

Theorem 150 (Landau 57). A1/A + A2/A ~ (A1 + A2)/A.

By Definition 34 the sum is (A1•A + A2•A)/(A•A), which is ((A1 + A2)•A)/(A•A) by Theorem 32, and this is equivalent to (A1 + A2)/A by Theorem 133.

Theorem 151 (Landau 58). A1/A2 + B1/B2 ~ B1/B2 + A1/A2. (Commutative Law of Addition)

(A1•B2 + B1•A2)/(A2•B2) and (B1•A2 + A1•B2)/(B2•A2) have equal antecedents by Theorem 5 and equal consequents by Theorem 31.

Theorem 152 (Landau 59). (A1/A2 + B1/B2) + C1/C2 ~ A1/A2 + (B1/B2 + C1/C2). (Associative Law of Addition)

By Definition 34 and Theorems 31, 32 and 33, each side has the antecedent (A1•(B2•C2) + B1•(A2•C2)) + C1•(A2•B2) and the consequent A2•(B2•C2), the antecedents being grouped differently. They are equal by Theorem 4.

Theorem 153 (Landau 60). A1/A2 + B1/B2 > A1/A2.

This follows from Common Notion 4. In symbols: A1•B2 + B1•A2 > A1•B2, so (A1•B2 + B1•A2)•A2 > (A1•B2)•A2 = A1•(A2•B2), by Theorems 34, 31 and 33.

Theorem 154 (Landau 61). If A1/A2 > B1/B2, then A1/A2 + C1/C2 > B1/B2 + C1/C2.

Proof: A1•B2 > B1•A2, so (A1•B2)•C2 > (B1•A2)•C2, that is, (A1•C2)•B2 > (B1•C2)•A2. Also (C1•A2)•B2 = (C1•B2)•A2. Adding (Theorem 13), (A1•C2 + C1•A2)•B2 > (B1•C2 + C1•B2)•A2. Multiplying both by C2 and rearranging, (A1•C2 + C1•A2)(B2•C2) > (B1•C2 + C1•B2)(A2•C2), which is what is required.

Theorem 155 (Landau 62). If A1/A2 > B1/B2, or ~, or <, then A1/A2 + C1/C2 > B1/B2 + C1/C2, or ~, or <, respectively.

The first is Theorem 154. The second is in Theorem 149. The third follows from the first by Theorems 136 and 135.

Theorem 156 (Landau 63). The converse of Theorem 155.

The three cases are in both instances mutually exclusive and exhaust all possibilities (Theorem 134).

Theorem 157 (Landau 64). If A1/A2 > B1/B2 and C1/C2 > D1/D2, then A1/A2 + C1/C2 > B1/B2 + D1/D2.

By Theorem 154, A1/A2 + C1/C2 > B1/B2 + C1/C2. By Theorems 151 and 154, B1/B2 + C1/C2 > B1/B2 + D1/D2. Then Theorems 137 and 143.

Theorem 158 (Landau 65). 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 149 and 154 if an equivalence holds; otherwise by Theorem 157.

Theorem 159 (Landau 66). If A1/A2 ≳ B1/B2 and C1/C2 ≳ D1/D2, then A1/A2 + C1/C2 ≳ B1/B2 + D1/D2.

By Theorem 149 if two equivalences hold; otherwise by Theorem 158.

Theorem 160 (Landau 67). If A1/A2 > B1/B2, there is a fraction D1/D2 with B1/B2 + D1/D2 ~ A1/A2, and any two such are equivalent.

If A1/A2 ≲ B1/B2 there is none, by Theorem 153.

Proof: Any two such are equivalent by Theorem 156. To find one: A1•B2 > B1•A2, so A1•B2 = B1•A2 + D for exactly one D (Theorem 8). Let D1 = D and D2 = A2•B2. Then B1/B2 + D/(A2•B2) ~ (B1•A2)/(A2•B2) + D/(A2•B2) ~ (B1•A2 + D)/(A2•B2) ~ (A1•B2)/(A2•B2) ~ A1/A2, by Theorems 133, 149 and 150.