Appetizer: True or False 1.2  2 2.If 2 = 3, then 5 = 6. 3.If 1 = 2, then I am the Pope. 4.Open the window. 5.If the proof is wrong, then the proposition.

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Appetizer: True or False 1.2  2 2.If 2 = 3, then 5 = 6. 3.If 1 = 2, then I am the Pope. 4.Open the window. 5.If the proof is wrong, then the proposition is wrong. 6.If 1=2 or 3 = 4, then 2 = 3. 7.If I come, then I come. 8.If it is a rhombus, then its diagonals are perp. 9.If the diagonals of a quad. are perp, then it is a rhombus. A2 Part A Mathematical Logic

Truth Tables (I) “ P and Q ”, denoted by P  Q PQPQPQ TT TF FT FF PQPQ T F F F

Truth Table for “ P or Q ”, denoted by P  Q PQP  Q TT TF FT FF T T T F

Truth Table for negation of P, denoted by ~P P~P~P T F ~P~P F T

Negate the following statements: (i) The sun is spherical and the plane can fly. (ii) London is not the capital of China or the house 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 values of their components.

De Morgan ’ s Law Let P, Q be two propositions, then (I) ~(P  Q)  (II) ~(P  Q) 

Proof of ~(P  Q)  (~P)  (~Q) PQPQPQ ~(P  Q)~P~P~Q(~P)  (~Q) TT TF FT FF

Proof of ~(P  Q)  (~P)  (~Q) PQP QP Q ~(P  Q)~P~P~Q(~P)  (~Q) TTTFFFF TFTFFTF FTTFTFF FFFTTTT

Section 3 Conditional Propositions: If P then Q, denoted by P  Q Determine the truth value of the following: 1.If Fun is a boy, then he is a human being. 2.If Fun is a human being, then Fun is a boy. 3.If x = 2 then x 2 = 4. 4.If x 2 = 4, then x = 2. 5.If a triangle is isosceles, then it is equilateral. 6.If a triangle is equilateral, then it is isosceles. 7.If |a|= -a, then a < 8.

Truth Table for P  Q PQP  Q TT TF FT FF T F T T

How many statements can be formed by using: P, Q, There are 16 equivalent statements can be formed. In general, there are 2 to the power 2 n equivalent statements formed by n statements.

Popular questions 1.If x = 2, then x 2 = 4. 2.If x 2 = 4, then x = 2. Are they true? 6.x = 2  x 2 = 4. 7. x 2 = 4  x = 2.

Definition 3.4 Let P  Q be a conditional proposition. This proposition has the following three derivatives( 衍生命題 ): 1.The converse( 逆命題 ) Q  P, 2.The inverse( 否命題 ) (~P)  (~Q) 3.The contrapositive( 逆反命題 ) (~Q)  (~P) 1.Make the truth tables for these four propositions. 2.Are they equivalent? 1.Make the truth tables for these four propositions. 2.Are they equivalent?

Proof by contrapositive( 反證法 ) P  Q  (~Q)  (~P) Example 1 If n 2 is an even integer then n is an even integer. Proof: If n is odd, then n = 2k + 1 and (2k + 1) 2 = 4k 2 + 4k + 1 = 2(2k 2 + 2k) + 1 is odd. Thus by contrapositive, the proposition is correct.

Example 2 Given that p and m are real numbers such that p 3 +m 3 =2, prove that p + m  2. Proof: Assume that p+m>2, then p 3 +m 3 >(2-m) 3 +m 3 =6m 2 -12m+8 =6(m-1) 2 +2  2. Thus, by contrapositive, 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 =p 2 /q 2, therefore 2q 2 =p 2. This implies that 2|p 2 and hence 2|p. So p=2k for some integer k. Putting it back to 2q 2 =p 2, 2q 2 =(2k) 2 i.e. q 2 =2k 2. Again, we have 2|q and 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 there are only n prime numbers, say p 1, p 2, p 3, …,p n. Now construct a new number p= p 1 p 2 p 3 … p n + 1, then p is a new prime number since p is not divisible by p i ’ s and p > p i ’ s. This leads to a contradiction that p 1, p 2, p 3, …,p n are the only prime numbers. So there are infinitely many prime numbers.

Examples of proof by contradiction Examples of proof by contradiction ~(P  Q)  ~( (~P)  Q)  P  (~Q)  F 1.If x=  2, then x is irrational. Proof: Assume that x=  2 and x is not rational, then … 2.If a + b  2 = c + d  2, then b = d. Proof: Assume that a + b  2 = c + d  2 but b  d, then  2 = (a – c)/(d – a), which is rational and hence is a contradiction. Thus b = d. P.65, Q.6

Illustrative Examples 3.If  ABC is a acute triangle and  A>  B>  C, prove that  B> 45 . Proof: Assume that  ABC is a acute triangle and  B  45 , then  C < 45 . But  A=180  -  B-  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 prime. 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 a 2 +b 2 +c 2 +d 2 +ab+cd  1. Proof: Assume that a, b,c and d are real numbers and ad-bc=1, but a 2 +b 2 +c 2 +d 2 +ab+cd=1, then a 2 +b 2 +c 2 +d 2 +ab+cd=ad-bc. Multiplying it by 2, we get 2a 2 +2b 2 +2c 2 +2d 2 +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 contradiction, a 2 +b 2 +c 2 +d 2 +ab+cd  1.

Section 4 Biconditional Propositions Definition 4.1 Let P and Q be two propositions. The biconditional proposition P  Q (read as “ P if and only if Q ” ) is defined as P  Q  (P  Q)  (Q  P)

Complete the Truth Table of P  Q  (P  Q)  (Q  P) PQPQPQQPQP(P  Q)  (Q  P) TT TF FT FF

Theorem 4.1 1.If (P  Q)  (Q  R) then P  R. 2.If (P  Q)  (Q  R) then P  R. 3.P  Q  Q  P Group discussion: Prove proposition 1-3

If (P  Q)  (Q  R) then P  R. Proof: PQRPQPQ  Q  R  P  R TTTTTTTT FTTTTTTT FFTTTTTT FFFTTTTT

Universal Quantifier :  for all Existential Quantifier:  for some 1.Some birds are white. In symbol, (  bird B)(B is white) 2.For any integer n, the equation x 2 -nx+1=0 must have a real solution. (  integer n)(x 2 -nx+1=0 has a real solution) 3.The equation x n +y n =z n has no integral solutions for all integers n  3. (  integer n  3)(x n +y n =z n has no integral solutions.) 4.For some real numbers n, if n 2 =4 then n = 2. (  real n)(n 2 =4  n = 2) Classwork:1.Translate the propositions on P.64 Q4 to symbols. 2.Negate the above Propositions.

Resources 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 x 2  4 and x > 0, then x  2.

Sometimes the proposition is conditional i.e. P  Q, We need to negate it in order to prove it by contradiction.i.e. ~ (P  Q)  F. But ~ (P  Q)  ? (Hint: Find an equivalent statement for P  Q which involves P, Q, ~ and .)

P  Q  (~P)  Q PQP  Q(~P )  Q TTT F T T TFF F F F FTT T T T FFT T T F

Write down the negation of 1.P  Q 2.If today is Sunday, you need not go to school. 3.If I can live without food, then I need not earn money. 4.P  (P  Q) 5.In the classroom, all students are girls. ~(P  Q)  ~( (~P)  Q)  P  (~Q)

Write the negation of: 7.Nobody can answer the question. 8.For any positive integer n, n + 8 > 0. 9.All students are clever and some of them are lazy. 10.For any even number x, if x is divisible by 3 then x is divisible by 6. 11.There exist natural numbers p and q such that  2 = p/q.

Definition 3.2 When the conditional proposition P  Q is always true, we write P  Q and read as P implies Q. For instance, it is correct to write “ x = 2  x 2 = 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 P.

Pick out the different one from the following statements: 1.If I receive a bonus, I shall have a holiday in Spain. 2.I shall have a holiday in Spain if I receive a bonus. 3.I shall have a holiday in Spain provided that I receive a bonus. 4.I receive a bonus only if I shall have a holiday in Spain. 5.Receiving a bonus is a sufficient condition for a holiday in Spain. 6.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 BP  AC, PA = m, PB = h and PC = n. Prove that h 2 = mn iff  ABC = 90 . h mn A B C

Exercise on Logic 1.Prove that if 3|n 2 then 3|n. 2.Prove that  3 is irrational. 3.Prove that  2+  3 is irrational. 4.Prove that for any real numbers a, b, c and d, if a + bi = c + di then a = c and b=d, where i 2 = -1. 5.Prove that log2 is irrational. 6.If 2 p -1 is a prime number, then p is prime. 7.Given that a > 0 and b > 0, prove that [(a+b)/2] 3  (a 3 +b 3 )/2 8.Given that b and c are odd integers, prove that x 2 + bx + c = 0 has no integral roots.

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