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 DerivedDirect p(m) q(m) ( 直接證法 )

Contraposition ¬q(m) ¬p(m) ( 間接證法 )

Contradiction p(m) and ¬q(m) F0 ( 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 + 6 + 1= 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 + 6 + 1= 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

)]()([ mqmpm )]()([ mpmqm

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

)]()([ mqmpm

THEOREM 2.4 p114

)]()([ mqmpm )]()([ mpmqm

If m is an even integer, then m + 7 is odd.

Proof: ( 矛盾證法 )

)]()([ mqmpm

THEOREM 2.4 p114

Assumption Result DerivedContraposition ¬q(m) ¬p(m) ( 間接證法 )

Contradiction p(m) and ¬q(m) F0 ( 矛盾證法 )

If m is an even integer, then m + 7 is odd.

p(m) and ¬q(m) F0 ( 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) F0 ( 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 bothodd, 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 m

n 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.)