CH.1The Logic Language of Proofs 1-1.Proposition(命題)
Def:A proposition (or statement) is a sentence that is true or false. The connect of proposition
:”And” :”or” :”Not”
If P and Q are proposition,then P Q:P and Q,P Q:P or Q,P:The denial of proposition P. (1) T T T T F F F T F F F F (2) T T T T F T F T T F F F (3) T F F T
Def:The proposition of “P Q” is referred to as the conjuction of P and Q. Remark:
(1)”P Q” as “P and Q”,that is P Q is true if and only if (a)P is true and also Q is true.
Or
(b)Each of P and Q is true. Or
(c)P and Q are both true.
(2)The conjuction of P and Q is false iff at least one of P or Q is false. Def:The proposition “P Q” referred to as the disjuction of P and Q.
Def:If P is a proposition,the assertion that P is false is commonly known as denial or negation of P.
Remark:”P” as “not P”
ex.P:”5 is an odd integer” The P is true.Hence,P:” 5 is not an odd integer”P is false.
Def:The two connective “” and “” called the conditional and the biconditional are defined by the fallowing truth tables
T T T
T F F
F T T
F F T
Def:The proposition “PQ” referred to as a conditional proposition. Remark:”PQ” is mean (1)If P,then Q
(2)If P is true,then Qis also true. (3)P is true only if Q is true. (4)P implies Q.
(5)Q is true whenever P is true.
(6)For P to be true,it is necessary for Q to be true. (7) For Q to be true,it is sufficient for P to be true.
Def:In the conditional proposition “If P,then Q.”P is called hypothesis or condition and Q is called the conculsion.
ex.P:If >3,then 7is a positive integer.(T)
Q:If a robin is a bird ,then 4 is a prime number.(F) R:If a cauliflower is a bird,then a robin is a bird.(T)
S:If 5 is an even integer,then no one will pass this course.(T) Def:If P and Q are propositions,then the converse of PQ is QP. P Q PQ QP
T T T T
T F F T
F T T F
F F T T
Def:If P and Q are propositions,then the contraposition of PQ is Q P. ex.The contrapositive of “If 4<3,then moon is made of cheese.” is “If moon is not made of cheese,then 4 3.”
Def:The proposition “PQ” is referred to as a biconditional proposition. RK:”PQ” is mean(1)P if and only if Q.(iff,)
(2)For P to be true,it’s necessary and sufficient for Q to be true. (3)P implies Q and Q implies P.
(4)P is equivalent to Q. 1-2 Expressions and Tautology
RK: (1)A expression either true or false.
(2)A expression has no meaning until its variables are replaced by proposition. (3)Each expression has a truth table,but a proposition has no truth table. ex.(X Y) (X(YZ)) is an expression.
Def:Two expressions are equivalent,provide that they have the same true values,for all possible values of true or false for all variables appearing in either expression.Two expressions X and Y are equivalent,we write X Y.(函數等價 or logically equivalent)
Ex.Are the expressions X Y and XY equivalent.(Yes) X Y XY T T T T F F F T T F F T X Y X X Y T T F T T F F F F T T T F F T T
Ex.Are the expressions W X and Y Z equivalent?(No) XYW W X Y X . . FTF . . F T
Ex.Which of the fallowing statements are propositions and which are expressions? (a) If 2=3,then 2 is a rational number.
(b) P and Q are solutions of the equation. x -5x+6=0 2 Analysis:The statement(a)is true it is a proposition.
(b)has no mean until it is a expression. If P:x=3 and Q:x=2 then the resulting proposition is true.
If P:x=4 and Q:x=5 then the resulting proposition is false. Ex. P P PP T F T F T T P P PP T F F F T F
Def: (1)A proposition expression is called a tautology(contradiction),if it yields a true(false) proposition regardless of what propositions keplace it variables. Ex.(PQ)(Q P) is a tautology. P Q PQ P Q Q P (PQ)(Q P) T T T F F T T T F F F T F T F T T T F T T F F T T T T T Proposition1-1.
Each of the following propositional expressions is a tautology. (a) P Q(P Q) (b) (PQ)P Q (c) (P Q)P Q (d) (P Q)P Q (e) (P)P Pf)(d) P Q P Q P Q (P Q) P Q (P Q) PQ T T F F T F F T T F F T F T T T F T T F F T T T F F T T F T T T
應用:Prove that if x,y
H:if x,y
X:x=0 Y:y=0
Suppose that x0(P),then 1
x
1 1
(xy)= 0=0 (Q) P Q
↑
1 y=(1 x)y=1(xy)
x x
Proposition 1-2:
Each of the following propositional expressions is a tautology. (1) PP (2) (PQ)(QP) (3) (PQ)(P Q) (4) [(PQ) (QP)](PQ) (5) [(PQ) (QR)](PR) 遞移性 (6) [(PQ) (QR)](PR) 遞移性 (7) [P (Q R)][(P Q) (P R)] 分配性 (8) [P (Q R)][(P Q) (P R)] 分配性 (9) [(P Q) R][P (Q R)] 結合性 (10) [(P Q) R][P (Q R)] 結合性 (11) {(P)[Q (Q)]}P (12) (PQ)(R PR Q) (13) (PQ)(R PR Q) Proposition 1.4:
Each of the following expression is a contradiction. (1)(PQ) (P Q) (2)[(P Q) P] (Q) (3)(P Q) (P) Pf)(1)By proposition 1.1(2)(PQ)PQ (PQ) (P Q)(PQ)(PQ),By definition (PQ)(PQ) is a contradiction (PQ) (P Q) is a contradiction. 1-3.Quantifiers(量詞 , )
In this section,we discuss two type of mathematical sign. Variable and Constants
Ex.A sentence such as “x<4” is not proposition,because it contain a variable. Rk: (1)sentence “x<4” is called a propositional function.
(2)The set of object that can replace the variable is called the universe or set of meanings.
Def: (1)The word or sign in a propositional function to be replaced is called a variable”.
(2)Any object in the set of meanings is called a ”constant”.
(3)The callections of objects in the set of meanings that can be substituted to make a propositional function a true proposition is called the true set of the propositional function.
Ex.The truth set of “x<4” the set of all real numbers that are less than 4. Ex.(1)x +2x+16=0 :The truth set is empty. 2
(2)sin x+ cos x=1 : The truth set is the set of all real numbers. 2 2 (3)x -5x+6=0 : The truth set is the set of 2 and 3. 2
Rk: (1)A propositional function may have more than one variable. ex.”x+y=2”The truth set is the live with slope-1 and y-intercept 2
(2)The sentence “For all real number x,x+2=2+x.” is not a propositional function, because this sentence is a true sentence.(x is called apparent or pound variable)
(3)If variable occurs is an expression in such a way that in order to turn the expression into a proposition,it is necessary to replace with a constant,then the variable occurs in the expression as an actual or free variable.
Ex.In the propositional function x 2
2 y dy=39
x:actual variable y:apparent variable
A:Phrases of existential quantifiers( ) 1. for some x
2. there exists a element x 3. there are x and y
ex.(1)There is a real number x such that x+2=5.
(2)There are real number x and y such that x+y=7 and x-1=1. (3)x +1=0 for some real number x 3
(4)There exists a real number x such that x +1=0 . 3 B: Phrases of universal quantifiers( )
1. for each x 2. for any x 3. for all x
ex.(1)For each(Every) real number x,x2 x (2)Any even integer is a multiple of f. (3)x+1>x for all real number x.
Notations:If p(x) (p(x,y)) is a propositional function,then (1) For each x, p(x) is denoted by “( x)p(x)”
(2) There exists x such that p(x) is denoted by ( x)p(x).
(3) For each x, there is a y such that p(x,y) is denoted by ( x)( y)p(x,y) Rk:Sometime the universal and existential quantifiers can be tucked away in a statement(hidden quantifiers)
Ex.If n is a positive integer,then n is the sum of 4 perfect square.
change sentence:If n is any positive integer, then there are four perfect square whose sum is n.
( n)(n is positive integer), (P ,P ,P ,P12 22 32 42)(n=P +P +P +P ) 12 22 32 42
其他(universal): 2
x >0, whenever(for all) x>0.If n is even, then 2
n is even.
The negation of propositions
Ex. P:”For each real number x,x2 ”(true) P: ( x)(0 x2 ) 0 P:”There is a real number x such that x <0 ” 2 P: ( x)(x <0) 2 Ex.Q:”There is a real number such that x +2=0” (2 x)(x +2=0) 2 Q:“For each real number x,x +22 0” ( x)(x +22 0) Theorem1.5:If p(x) is a propositional function with variable x, then (a)[( x)p(x)] is equivalent to ( x)[p(x)]
(b)[( x)p(x)] is equivalent to ( x)[p(x)]
Pf)(a) suppose that proposition [( x)p(x)] is true. ( x)p(x) is false so that truth set of p(x) is not universe.Therefore the truth set of [p(x)] is nonempty. ( x)[p(x)] is true.
改寫: [( x)p(x)] is true. ( x)p(x) is false The truth set of p(x) is not universe. The truth set of [p(x)] is nonempty. ( x)[p(x)] is true. Counterexample 反例
Ex.P:”If x is any real number,then x 0” false Counterexample:x=-3<0
Def:If p(x) is a propositional function with variable x,then a counterexample to ( x)p(x) is a object t in the set of meaning such that p(t) is false.
Def:An integer m divides an integer n if there is an integer q such that n=qm multiple divisor Def:An integer n is a perfect square provied there is an integer k such that n=k 2 Ex.(a)Either prove or find a counterexample to the statement;if n is prime( n, prime),then is n
2 -1 is prime Pf)
n n 2 -1 2 3 3 7 5 31 7 127 11 2047=2389 反例 1-4.Methods of proof (PQ) 1.Direct method (PQ) 2.contrapositive method (Q P) 3.contradiction method (PQ) P Q Review.PQ has three conditions is true. (1) Each of P and Q is true.
(2) P is false and Q is true. (3) Each of P and Q is false.
Def:An integer n is said to be odd,if there is an integer k such that n=2k+1 and n is said to be even,if there is an integer q such that n=2q.
Theorem1.6 If n is odd integer,then n is odd. 2 Pf)The hypethesis:”n is odd”
The conclusion:”n is odd” 2
Since n is odd,there is a integer k such that n=2k+1. Since 2 n = 2 2 2 (2k+1) =4k +4k+1=2(2k +2k)+1 and 2 2k +2k is integer, thus 2 n =2k+1 is odd. Contraposition (QP)
Theorem1.7 [Suppose that m and b are real number with m0 and that f is a linear function defined by f(x)=mx+b.]非命題
If xy(命題),then f(x)f(y)(conclusion). Pf)Q P
If f(x)=f(y),then x=y.Suppose that f(x)=f(y) mx+b=my+b mx=my x=y. Theorem1.8:Suppose a,b and c are integers.If a and b are even and c is odd,then the equation ax+by=c does not have an integral solution for x and y.
has an integral solution for x and y.Then if x and 0 y are integral solution of 0 equation,and so. ax +by =c (odd) since a and b are even,0 0 ax ,by are even.Hence 0 0 ax +b0 y is even.This is a contradiction. 0
Def)A real number(
n0,such that r=m
n and r is irrational(
c
Theorem.If r is a real number such that r =2 ,then r is irrational. 2
Pf)Suppose that there exists a rational r such that r =2 ,since r is rational,there are 2 integer m and n such that r=m
n ,we can assume that m and n has no common divisors greater than 1,so
2 2 2 m r = =2 n ,hence 2 2 m =2n and so 2 m is even. By theorem 1.6,m is even.Therefore,there exists a integer p such that
m=2p.So,2n =m =(2p) =4p =2(2p ) ,and hence 2 2 2 2 2 n =2p ,so 2 2 n is even.Again by 2 theorem 1.6,n is even.The conclusion,m and n are even.
Theorem 1.11:If a,b and c are real number,then a +b +c2 2 2 ab+bc+ac Pf)Soppose that a,b and c are real
number.(a-b) +(b-c) +(c-a)2 2 2 (0 a -2ab+b )+(2 2 b -2bc+c )+(2 2 c -2ac+a )2 2 0 2(
2 2 2
a +b +c )-2(ab+bc+ac)0a +b +c2 2 2 ab+bc+ac Theorem1.12:An integer n is odd iff n2 is odd. Pf)HW
Let n=2k+1,then n2=(2k+1)2=4k2+4k+1=2(2k2+2k)+1. 2k2+2k 2(2k2+2k)+1 is even.Then we can say that n is even. 2
Theorem1.13:Let x be a real number.The x=1 if and only if Pf)x -3x +4x-2 =0 3 2
( )trival
( )x -3x +4x-2 =(x-1)(x3 2 2-2x+2)=0.Then either x-1=0 or x2-2x+2=0.But x2-2x+2=(x-1)2+1>0.Therefore x-1=0 x=1
1-5.The contradiction method of proof when do you use the contradiction method of proof?
Theorem1.1:There do not prime numbers a,b and c such that a3+b3=c3. 1. Direct method(universe):(PQ)( x)p(x)
3. contradiction method:we want to show that,if a,b and c are primes,then a3+ b3c3.Suppose otherwise,there are a,b and c are primes,such that a3+b3=c3. case 1.If a and b are odd,then a3 and b3 are odd.Thus c3 is even and so.c is even.Therefore c=2,this contradicts to a3+ b3>8=c3.
case 2.At least one of a and b is even,say b=2.Therefore, c3-
a3=8 (c-a)(c2+ac+a2)=8.Since a and c are primes,c2,ac,a2 4.This implies c2+ac+a2 12.Thus,c -a3 3 12.( 3 3 2 2 c -a = c-a c +ac+a 12 1 12 ) This contradicts to c3- a3=8
Theorem1.15:If a,b and c are odd integers,then ax2+bx+c=0 does not have a rational solution.
Pf)Suppose a,b and c are odd integers and p and q are integers such that p q is a solution ax2+bx+c=0.we can assume p and q have no common divisors greater than 1.Then a( 2 2 p q )+b( p q)+c=0. ap 2
+bpq+cq2=0.we will first prove that p and q are both odd.Suppose that p is even.Then ap2+bpq is even and so cq2 must be even.this contradicts to c and q are odd.
Suppose that q is even.Then bpq+cq2 is even and so ap2 must be even. this contradicts to.Thus p and q both odd and so ap2,bpq,cq2 are odd.Hence,their sum is odd,this is contradiction.
1-6.More proofs(element-chasing method)
Theorem1.16:If x is say real number such that x2-x-2<0 then -1<x<2 Pf)since x2-x-2=(x-2)(x+1)<0
i.x-2<0 and x+1>0 ii.x-2>0 and x+1<0
case i:x<2 and x>-1 -1<x<2 case ii:x>2 and x<-1 x
Theorem1.17:There do not exist natural number m and n such that 7 = 1 +1
17 m n
Since 1 1,
m n>0,we have neither m nor n can be 2,since
1 1 7 + > 2 n 17 for all n Since 1+ <1 7 5 5 17 and 1 1 1 1 + < +
be less than 5.Therefore one of m and n must be 3 or 4.But 7 1- = 4 1
17 3 51 for all n n and 7 - =1 1 1
17 4 68 for all nn .Therefore,since we have examinal all passibilities,there do not exist m,n 使得___.
