【学习笔记】离散数学（Discrete Math） - 证明 Proof 3
离散数学（Discrete Math）
- 证明proof
目录
离散数学（Discrete Math）
- 证明proof
Terminology 术语
How to prove conditional statements?
Direct proof直接证明法
Proof by contraposition对位证明法
Proof by contradiction 矛盾法/反证法
Vacaous and Trival proofs
Proof by cases 列举法
Uniqueness proofs 唯一性证明
名词补充
参考课程
Terminology 术语
Theorem 定理 ：A theorem is a statement that can be shown to be true.
Proposition 命题：Less important theorems are called propositions.
Proof 证明：A proof is a valid argument that established the truth of a theorem.
Axiom 公理：Statements that we assume to be true.
Lemma 引理：A less important throrem that is helpful in the proof of other results.
Corollary 推论：A theorem that can be established directly from a theorem that has been proved.
Conjecture 推测（猜想）：A statement that is being proposed to be a true statement.
How to prove conditional statements?
p->q
- Direct proof 直接证明法
- Proof by contraposition 对位证明法
- Proof by contradiction 矛盾法/反证法
- Vacaous and Trival proofs 空真和平凡证明
Direct proof 直接证明法
例：
P(n): n is odd.
Q(n): n^2 is odd.
∀n∈Z, P(n)-> Q(n)
证：
设n=2k+1，则有
n^2 = 2(2k^2+2k)+1
证毕
Proof by contraposition 对位证明法
p->q ≡┐q->┐p
Assume q is false by applying rule of inference, we conclude that p is false.
例：
Prove that if n is an integer and 3n+2 is odd, then n is odd.
证：
Assume n is not odd.
Then, n is even
There exists an integer k such that n=2k.
3n+2 = 3(2k)+2 = 2(3k+1)
Since 3k+1 is an integer 3n+2 is even.
So, the premise is not true.
Therefore, the theorem is proved.
例2：
Definition
The real number r is rational if there exists integers p and q with q≠0 such that r=p/q. Areal number that is not rational is called irrational.
Example
Prove that the sum of two rational numbers is rational.
证：
∀x,y,R(x), R(y)-> R(x+y)
Assume R(x) and R(y) are true. (Premise前提)
R(x)-> ∃p,q≠0 such that x = p/q.(def)
R(y)-> ∃r,s≠0 such that y = r/s.(def)
x+y = p/q + r/s = (ps+rq)/qs
ps+rq and qs are integer and qs≠0, x+y is rational.(def)
证毕
例3：
If n is an integer and n^2 is odd, then n is odd. ∀n∈Z, odd(n^2)-> odd(n).
证：
We prove ┐odd(n)-> ┐odd(n^2).
Assume ┐odd(n) is true.
i.e. n is even.
There exists k such that n=2k.
n^2 = 2(2k^2)
Since 2k^2 is an integer, n^2 is even. Therefore ┐odd(n^2) is true.
Proof by contradiction 矛盾法/反证法
If we want to proof p is true, we first assume ┐p(i.e. p is false). Then , by applying rules of inference we get some contradiction.(i.e. ┐p->F). Since ┐p->F is true ,we can get p is T.
例：
Prove that ‘√2 is irrational’ by contradiction.
证：
Assume √2 is not irrational.
There exists integers p,q≠0, such that √2 = p/q( Beside, we can assume p and q are both composite.)
P^2 = 2q^2
Since p^2 is even ,q is even (E(p^2)->E(P))
Assume ,p = 2k, (2k)^2 = 2q^2.
We get
q^2 = 2k^2
since q^2 is even ,q is even .
2|p(代表2为p的一个因子) and 2|q , imply p,q are not composite.
This is a contradiction.
Therefore ‘√2 is rational’ is false
And we prove the theorem.
Vacaous and Trival proofs 空真和平凡证明
虚假前件蕴含任何后件
To prove p->q is true ,we can
show p is false (vacuous proof) or
show q is true (trivial proof).
例1
Let P(n) =”if n>1,then n^2>n”. Prove P(0) is true.
证：
P(0) =”if 0>1,then 0^2>0.” in the conditional statement.
The premise “0>1” is false , so P(0) is true.
例2
Let P(n) =”If a and b are positive integers with a ≥ b, then a^n ≥ b^n”. Show that P(0) is true.
证：
P(0) =”If a and b are positive integers with a ≥ b, then a^0 ≥ b^0.”
Since a^0=1 and b^0=1,
a^0 ≥ b^0 is true.
So, P(0) is true.
Proof by cases 列举法
(p1∪p2∪p3 ∪…∪pn)->q ó p ≡ p1∪p2∪p3 ∪…∪pn and p->q
例：
Show that there are no solutions in integers x and y of x^2+3y^2 = 8
证：
以y的定义域分段进行判断即可
Uniqueness proofs 唯一性证明
∃!xp(x) ≡ ∃x0(p(x0)) ∧ ∀x ≠x0 ┐p(x)
So, we need to
- Find an x0 such that P(x0) is true;
- Show that ∀ x≠x0 ┐p(x) is true.
名词补充
WLOG 不失一般性原则 without loss of generality
Set 集合
Set builder 结构式式集合
Subsets 子集
参考课程
易志伟老师 – 离散数学 Discrete Mathematics