2.5 The proofs of theorems 證明方法 p → q 與 p  q 之應用 Definition 2.8 Let n be an integer. We call n even if n is divisible by 2 - that is, if there exists an integer r so that n=2r. If n is not even, then we call n odd and find for this case that there exists an integer s where n=2s + 1.

Direct proofs ( 直接證法 ) Indirect proofs ( 間接證法 ) Proofs by contradictions ( 矛盾法或歸謬法 ) Assumption Result Derived Direct p(m) q(m) ( 直接證法 ) Contraposition ¬q(m) ¬p(m) ( 間接證法 ) Contradiction p(m) and ¬q(m) F 0 ( 矛盾證法 )

The implication p  q can be proved by showing that if p is true, then q must also be true. This shows that the combination p true and q false never occurs. Theorem 2.3 (p114) For all integers m and n, if m and n are both odd, then their product mn is also odd. Direct proofs ( 直接證法 )

If m is an even integer, then m + 7 is odd. Proof: ( 直接證法 ) Since m is even, we have m=2a for some integer a. Then m + 7 = 2a + 7=2a = 2(a + 3) + 1. Since a + 3 is an integer, we know that m + 7 is odd. THEOREM 2.4 p114

If m is an even integer, then m + 7 is odd. Proof: ( 直接證法 ) 1. 前提為真， m = 偶數, 即 m=2a ， a 為整數. ( 偶數之定義 ) 2. m + 7 = 2a + 7=2a = 2(a + 3) + 1. 其中 a + 3 為整數 3. ∴ m + 7 = 奇數 ( 奇數之定義 ) 結論為真 THEOREM 2.4 p114

If m is an even integer, then m + 7 is odd. Proof: ( 間接證法 ) Suppose that m + 7 is not odd, hence even. Then m + 7=2b for some integer b and m=2b − 7=2b − 8 + 1=2(b − 4) + 1, where b − 4 is an integer. Hence m is odd. [The result follows because the statements and are logically equivalent.] THEOREM 2.4 p114

If m is an even integer, then m + 7 is odd. Proof: ( 間接證法 ) 1. 若結論不為真，即 m + 7 不是奇數，而是偶數 2. m + 7=2b ， b 為整數 ( 定義 ) m=2b − 7=2b − 8 + 1=2(b − 4) + 1, 其中 b − 4 為整數 3. ∴ m 為奇數 ( 定義 ) 4. 因 與 等價 ∴原定理成立 THEOREM 2.4 p114

If m is an even integer, then m + 7 is odd. Proof: ( 矛盾證法 ) THEOREM 2.4 p114 Assumption Result Derived Contraposition ¬q(m) ¬p(m) ( 間接證法 ) Contradiction p(m) and ¬q(m) F 0 ( 矛盾證法 )

If m is an even integer, then m + 7 is odd. p(m) and ¬q(m) F 0 ( 矛盾證法 ) 1)Now assume that m is even and that m + 7 is also even. 2)Then m + 7 even implies that m + 7 = 2c for some integer c. And, consequently, m =2c−7=2c−8+1= 2(c − 4) + 1 with c − 4 an integer, so m is odd. 3)Now we have our contradiction. We started with m even and deduced m odd—an impossible situation, since no integer can be both even and odd.

If m is an even integer, then m + 7 is odd. p(m) and ¬q(m) F 0 ( 矛盾證法 ) 1)Now assume that m is even and that m + 7 is also even. ( 假設 m 是偶數, 且 m+7 亦是偶數 ) 2) Then m + 7 even implies that m + 7 = 2c for some integer c. And, consequently, m =2c−7=2c−8+1= 2(c − 4) + 1 with c − 4 an integer, so m is odd. m+7 是偶數, 故 m+7=2c, 其中 c 是整數 m=2c-7=2c-8+1=2(c-4)+1, (c-4) 是整數 依奇數定義, m=2(c-4)+1 是奇數 3) Now we have our contradiction. m 是偶數, 又是奇數, 相互矛盾 所以 m+7 不可能是偶數

續上 How did we arrive at this dilemma? ( 困境, 兩難推理 ) Simple - we made a mistake! This mistake is the false assumption - namely, m + 7 is even - that we wanted to believe at the start of the proof. Since the assumption is false, its negation is true, and so we now have m+7 odd.

For all integers m and n, if m and n are both odd, then their product mn is also odd. ( 如果 m 和 n 都是奇數時, mn 一定是奇數。 ) Proof: (1) Since m and n are both odd, we may write m=2a + 1 and n =2b + 1, for some integers a and b – because of Definition 2.8. (2) Then the product mn =(2a + 1)(2b + 1)=4ab + 2a + 2b + 1=2(2ab + a + b) + 1, where 2ab + a + b is an integer. (3) Therefore, by Definition 2.8 once again, it follows that mn is odd. THEOREM 2.3 p114

THEOREM 2.5 (p115) For all positive real numbers x and y, if the product xy exceeds 25, then x > 5 or y > 5. Proof: 解 1 – 直接證明法 若 xy > 25, 則 ??? 似乎無法繼續下去 解 2 – 間接證法 或 矛盾證法

THEOREM 2.5 (p115) For all positive real numbers x and y, if the product xy exceeds 25, then x > 5 or y > 5. p(x, y) → q(x, y) Proof: 利用間接證明法 (1) 假設 “x > 5 或 y > 5” 不為真, 即 0 < x ≦ 5 且 0 < y ≦ 5. (2) 故 0 =0 · 0 <x · y ≦ 5 · 5 =25 (3) 即 “xy > 25” 不為真 (4) 因 ┐q → ┐p 與 p → q 等價, 所以原式成立

THEOREM 2.5 (p115) For all positive real numbers x and y, if the product xy exceeds 25, then x > 5 or y > 5. Proof: Consider the negation of the conclusion—that is, suppose that 0 < x ≦ 5 and 0 < y ≦ 5. Under these circumstances we find that 0 =0 · 0 <x · y ≦ 5 · 5 =25, so the product xy does not exceed 25. (This indirect method of proof now establishes the given statement, since we know that an implication is logically equivalent to its contrapositive.)