Appetizer- True or False2 - 2If 2 = 3, then 5 = 6.If 1 = 2, then .ppt

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1、Appetizer: True or False 2 2 If 2 = 3, then 5 = 6. If 1 = 2, then I am the Pope. Open the window. If the proof is wrong, then the proposition is wrong. If 1=2 or 3 = 4, then 2 = 3. If I come, then I come. If it is a rhombus, then its diagonals are perp. If the diagonals of a quad. are perp, then it

2、is a rhombus.,A2 Part A Mathematical Logic,Truth Tables (I) “P and Q”, denoted by PQ,Truth Table for “P or Q”, denoted by P Q,Truth Table for negation of P, denoted by P,Negate the following statements:,(i) The sun is spherical and the plane can fly. (ii) London is not the capital of China or the ho

3、use is made of wood. (iii) All birds are white. (iv) Some men are tall. (v) All women are kind and helpful.,Section 2 Equivalence of Two Propositions,Two propositions with the same components P, Q, R, are said to be logically equivalent(or equivalent) if they have the same truth value for any truth

4、values of their components.,De Morgans Law,Let P, Q be two propositions, then(I) (PQ) (II) (PQ) ,Proof of (PQ) (P)(Q),Proof of (P Q) (P) (Q),Section 3 Conditional Propositions: If P then Q, denoted by P Q,Determine the truth value of the following: If Fun is a boy, then he is a human being. If Fun i

5、s a human being, then Fun is a boy. If x = 2 then x2 = 4. 4. If x2 = 4, then x = 2. If a triangle is isosceles, then it is equilateral. If a triangle is equilateral, then it is isosceles. 7. If |a|= -a, then a 8.,Truth Table for P Q,How many statements can be formed by using:,P, Q,There are 16 equiv

6、alent statements can be formed. In general, there are 2 to the power 2n equivalent statements formed by n statements.,Popular questions,If x = 2, then x2 = 4. If x2= 4, then x = 2. Are they true?,x = 2 x2 = 4. 7. x2 = 4 x = 2.,Definition 3.4,Let P Q be a conditional proposition. This proposition has

7、 the following three derivatives(衍生命題): The converse(逆命題) Q P, The inverse(否命題) (P) (Q) The contrapositive(逆反命題) (Q) (P),Make the truth tables for these four propositions. Are they equivalent?,Proof by contrapositive(反證法) P Q (Q) (P),Example 1If n2 is an even integer then n is an even integer.Proof:

8、If n is odd, then n = 2k + 1 and(2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1 is odd. Thus by contrapositive, the proposition is correct.,Example 2,Given that p and m are real numbers such that p3+m3=2, prove thatp + m 2. Proof:Assume that p+m2, then p3+m3(2-m)3+m3=6m2-12m+8=6(m-1)2+2 2. Thus, by contr

9、apositive, p + m 2.,Proof by contradiction(歸謬法) (P) F,Example 3.3 Use the method of contradiction to prove that 2 is irrational. Proof: Suppose that 2 is not irrational, then 2 = p/q for some natural numbers p, q where (p, q) = 1. Since 2 =p2/q2, therefore 2q2=p2. This implies that 2|p2 and hence 2|

10、p. So p=2k for some integer k. Putting it back to 2q2=p2, 2q2=(2k)2 i.e. q2=2k2. Again, we have 2|qand 2|(p, q) , which is a contradiction.,Class work : Use the method of contradiction to prove that 3 is irrational.,Theorem (proved by Euclid): There are infinitely many prime numbers.,Proof: Assume t

11、here are only n prime numbers, say p1, p2, p3,pn. Now construct a new number p= p1p2p3pn + 1, then p is a new prime number since p is not divisible by pis and p pis. This leads to a contradiction that p1, p2, p3,pn are the only prime numbers. So there are infinitely many prime numbers.,Examples of p

12、roof by contradiction (PQ) ( (P)Q) P(Q) F,If x=2, then x is irrational. Proof: Assume that x= 2 and x is not rational, then 2. If a + b2 = c + d2, then b = d. Proof: Assume that a + b2 = c + d2 but bd, then 2 = (a c)/(d a), which is rational and hence is a contradiction. Thus b = d.,P.65, Q.6,Illust

13、rative Examples,If ABC is a acute triangle and ABC, prove that B 45.Proof:Assume that ABC is a acute triangle and B 45, then C 90 leads to a contradiction that ABC is a acute triangle. Thus,by the method of contradiction, B 45.,Illustrative Examples,4. If x=n and y=n+1, then x and y are relatively p

14、rime. Proof: Assume that x=n and y=n+1 and x and y have common factor other than 1, say f, then n=fg and n+1=fh. So 1 = f(h-g) and hence f=1, which is a contradiction. Thus the proposition is true. .,5. Given that a, b,c and d are real numbers and ad-bc=1, prove that a2+b2+c2+d2+ab+cd1.,Proof: Assum

15、e that a, b,c and d are real numbers and ad-bc=1, but a2+b2+c2+d2+ab+cd=1, then a2+b2+c2+d2+ab+cd=ad-bc. Multiplying it by 2, we get 2a2+2b2+2c2+2d2+2ab+2cd-2ad+2bc=0 i.e.(a+b)2+(b+c)2+(c+d)2+(a-d)2=0 a+b=b+c=c+d=a-d=0 i.e.a=b=c=d=0, which contradict to that ad-cd=1. Thus, by the method of contradic

16、tion, a2+b2+c2+d2+ab+cd 1.,Section 4 Biconditional Propositions,Definition 4.1 Let P and Q be two propositions. The biconditional proposition PQ (read as “P if and only if Q”) is defined as P Q (PQ) (QP),Complete the Truth Table of P Q (PQ) (QP),Theorem 4.1,If (PQ)(Q R) then P R.If (PQ)(Q R) then P

17、R.P Q Q P Group discussion: Prove proposition 1-3,If (PQ)(Q R) then P R. Proof:,Universal Quantifier : for all Existential Quantifier: for some,Some birds are white. In symbol, (bird B)(B is white) 2. For any integer n , the equation x2-nx+1=0 must have a real solution.(integer n)(x2-nx+1=0 has a re

18、al solution) The equation xn+yn=zn has no integral solutions for all integers n 3.(integer n 3)(xn+yn=zn has no integral solutions.) For some real numbers n, if n2=4 then n = 2.(real n)(n2=4 n = 2),Classwork:1.Translate the propositions on P.64 Q4 to symbols. 2.Negate the above Propositions.,Resourc

19、es,Write down the contrapositive of the following propositions: If you pass both Physics and Chemistry, then you are able to promote to F.7.If x2 4 and x 0, then x 2.,Sometimes the proposition is conditional i.e. PQ,We need to negate it in order to prove it by contradiction.i.e. (PQ) F.But (PQ) ?(Hi

20、nt: Find an equivalent statement for PQ which involves P, Q, and .),PQ (P)Q,Write down the negation of P Q If today is Sunday, you need not go to school. If I can live without food, then I need not earn money. P (P Q) In the classroom, all students are girls.,(PQ) ( (P)Q) P(Q),Write the negation of:

21、,Nobody can answer the question. For any positive integer n, n + 8 0. All students are clever and some of them are lazy. For any even number x, if x is divisible by 3 then x is divisible by 6. There exist natural numbers p and q such that 2 = p/q.,Definition 3.2,When the conditional proposition P Q

22、is always true, we write P Q and read as P implies Q. For instance, it is correct to write“x = 2 x2 = 4”, but incorrect to write“x + a = b x = a + b”,Definition 3.3,Let P Q be a conditional proposition. Then P is called the sufficient condition (充分條件) for Q,and Q is the necessary condition(必要條件) for

23、 P.,Pick out the different one from the following statements:,If I receive a bonus, I shall have a holiday in Spain. I shall have a holiday in Spain if I receive a bonus. I shall have a holiday in Spain provided that I receive a bonus. I receive a bonus only if I shall have a holiday in Spain. Recei

24、ving a bonus is a sufficient condition for a holiday in Spain. Having a holiday in Spain is a necessary condition for receiving a bonus.,Example 4.1,In the Figure, P is a point on AC such that BPAC, PA = m, PB = h and PC = n. Prove that h2= mn iff ABC = 90.,h,m,n,A,B,C,Exercise on Logic,Prove that i

25、f 3|n2 then 3|n. Prove that 3 is irrational. Prove that 2+ 3 is irrational. Prove that for any real numbers a, b, c and d, if a + bi = c + di then a = c and b=d, where i2= -1. Prove that log2 is irrational. If 2p-1 is a prime number, then p is prime. Given that a 0 and b 0, prove that (a+b)/23 (a3+b3)/2 8. Given that b and c are odd integers, prove that x2 + bx + c = 0 has no integral roots.,