Menu Statement, Not Statement and Open Sentence
Compound SentencesA. Negation
B. Conjunction
C. Disjunction
D. Implication
E. Biimplication

I. Statement, Not Statement and Open Sentence :

I. Statement, Not Statement and Open Sentence Statement / Preposition
Example :
1. Determine the truth value of the following sentences, the true or false:
The sum of angle of triangle is 180°
( )
b. Sin2 x =cos2x-1
( )

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c. Square is rectangle with same length
( )
d. 4 isn’t perfect quadratic
( )
e. The diagonals of Rhombus are perpendicular
( )
f. Log 10 + log 0,1 = 1 ( )

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That Example above are called Statement or Proposition
So, We can say that
The preposition is the sentence that has true or false value but neither.
The symbols that statement by using small letter, among other : p, q, r, …etc.

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B. Not Statement:
is the sentences which can’t be determined of their truth.
Not Statement consist of
a. Factual Sentence:
is the sentences which must be observed their truth is true or false.
Examples:
1. Bravura is black sweet boy
2. Lika is tall girl
3. Angga is Nurani’s boyfriend.
4. There is a thief in my class.

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b. Open Sentences:
is the sentences that still consist of variable.
Open Sentences can become the true statement if that variables are changed with constant that fulfill it.
Examples:
1. a) 2x+5=10
b) x is even number
c) x2-x < x+3

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2.Find of x, in order that the open sentence became the true statement!
a). Sec x = 2 ; 0< x < 3600
b). X2-4x-5> 0 ;x € R
c). 2logx + 2log (2X-1) =0

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SOLUTION
1. a) 2x+5 =10
2x=10-5
x=2,5
b) x is even number
x={2,4,6,8,…}
c) x2-x < x+3
x2-x –x-3 =0 + - +
x2 - 2x-3 = 0 -1 3
(x-3) (x+1) = 0
x=3 V x=-1 HP={xl -1 =x =3 x€R}

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Solution:
2 a). Sec x =2
x=600,3000
b). X2- 4x- 5 > 0
(X-5) (X+1) > 0
X=5 V X=-1
-1 5
HP ={xlx =-1 V x =5 x€ R}

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c). 2logx + 2log(2x-1) = 0
<=>2logx . (2x-1) =0
<=>2x2 - x =20 =1
<=>2x2 - x - 1 =0
<=>(2x+1) (x-1) =0
<=>x=-1/2 v x=1
(TM)
HP {1}

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Note:
Not statement generally in the form :
Command Sentences
Interrogative Sentences
Expectation Sentences
Exclamation Sentences
Example
Determine the roots of equation x2-2x = 0
Watch out dog!
I hope you will get better soon..

II. Compound Sentences :

II. Compound Sentences Compound Sentence is the union of two or more sentences with linking logic word, among other are:
1. And the symbol is “?” , “&”
2. Or the symbol is “v”
3. If …..then….symbol is “=>”
4. If and only if the symbol is “?”

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NEGATION (Ingkaran)
Negation of p statement can be written –p, ~p
Negation is the new statement which is gotten by adding the word ”Not “at that previous statement.
The truth of Negation :
Relation negation and Complement of the set. p
~p

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Example :
p : Jakarta is the capital city of Indonesia.
~p : Jakarta isn’t the capital city of Indonesia.
p : sin 300 – sin 900 < cos 00
~p : sin 300 – sin 900 = cos 00
p : The all prime number is odd number.
~p : There is prime number isn’t odd number
or Some prime number isn’t odd number

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B. CONJUNCTION
Conjunction is the compound sentences with the linking word “and”
If given that 2 statements p and q that conjunction p & q can be written “p ? q”
The linking “and” can be change the other words, among other :
But, although, nevertheless, etc.

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The truth of Conjunction
Conclusion :
The Conjunction is TRUE if the both statements are true.

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The relation between Conjunction and Set can be draw :
( p ? q) = ( p n q) p q

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EXAMPLE
A. Determine the truth value of the following Conjunction :
2 is even number and prime number ( )
v3v3v3=3 and 3 is composite number ( )
3log 27 = -3 and v12+v12+…. =3 ( )
9 isn’t root form and 50 is the prime number ( )
Ronal is stupid boy and 5 is composite number ( )

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Determine the value of x from sentences p(x) and q :
To solve the value of x is sentence p(x) ? q based on the Conjunction table
If p(x) ? q True, so q is true, and p(x) must TRUE
If p(x) ? q False, so q is true and p(x) must FALSE

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EXAMPLE
Determine the value of in order that conjunction is true :
a). x2 – 4x = 0 and 8 ½ is irrational number
b). 2log (x+1) = 3 and 9log 3 = 1/2
2. Determine the value of x in order that conjunction is false
a). x2 – 36 = 0 and Baghdad is capital city of Iraq
b). x2 – 2x – 3 > 0 and 2log10– 2log 5 =1
c). sin 2x =0, and < x < 270 and cotan 600 =v3

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SOLUTION
a) v8 is irrational number. So x2 – 4x = 0 must be true
? x(x-4)
? x =0 V x=4
HP {0,4}
b) 9log3 = ½. So 2log (x+1) = 3 must be true
x+1 = 23
x =8-1
x = 7

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2. a) x2-36 =0 and Baghdad is capital city of Irak
x2 ?v36
x ? ±6
b) x2- 2x -3 >0 and 2log10 – 2log5 =1
(x+1) (x-3) = 0
x = -1 V x= -3
+ - +
-1 3
HP ={x I -1 = x = 3 x€ R}
q = 2log – 2log5 =1
2log(10/5) = 2log2
2 = 2 (T)

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c) sin 2x =0, and < x < 2700 and cotan 600 =v3
p(x) ? q = F
q= cotan 600 = v3 (F)
p=sin 2x = 0 , 0 = x = 2700 must be false
2x = k. 1800
x = 900
HP={900, 1800, 2700}

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C. DISJUNCTION
is the compound sentence with the linking word “or”
If given that 2 statements p and q, so disjunction of it can be written “pVq”
Disjunction consist of :
1. Disjunction Inclusive (v)
* Disjunction is false if the both statements are false.

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2. Disjunction Exclusive (V )
* Disjunction is false if the both statements has the same value.

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The Relation Between Disjunction Inclusive and set
(pVq) = (pUq) P q

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Example :
Determine the truth of Disjunction
sin2x+cos2x = 2log2 or 1+ tan2x=sec2x
( )
2. y=mx+c has the slope m or cut in (0,c)
( )
3. The sunrise from west or the earth goes arround the moon.
( )
4. 2log2 ¼ = -4 or a Sin B=b Sin A
( )

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Determine The value of x from sentence p(x)Vq
1. In order p(x)Vq is true if q false so p(x) must be true.
2. In order p(x)Vq is false if q false so p(x) must be false.

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Example
Determine the value of x in order to Disjunction p(x)Vq is true.
a). x2+3x =4 or sin 600=1/2
? x2+3x-4=0
?(x+4)(x-1)
? x=-4 V x=1 HP={-4,1}
b). All of x € A,(2x+1) is even number or log (x-1) +log(x+2) =log 10
q:log (x-1)+ log(x+2)=log10

2.Like no.1 but Disjunction is false
a). 22x-2x-2 =0 or 2 is the root form
[F]
p: 22x-2x-2 =0
2x =p=> p2-p-2?0
(p-2)(p+1) ?0
p?2 V p? -1 [™]
b). 2 2log5 = 25 or vx2-4x = 0
q: vx2-4x < 0
(i) x2-4x <0 (ii) x2-4x = 0
x(x-4) <0 x(x-4) = 0
x=0 V x=4 x=0 V x=4
+ - +
0 4
+ - +
0 4

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D. IMPLICATION
is the command statement with the linking word “if….then…”
If given that p and q that implication of it can be written “p=>q”
p:antecedent
q:consequence
p=>q can be read:
1. If p then q
2. q if p
3. p is adequate condition of q
4. q is necessary condition of p

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The truth of Implication :
P ? Q = (p=>q)
The Relation Implication and set q P

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The characteristic of Implication :
I. Implication is false if:
Antecedent is true and Consequence is false
II. Implication is true if:
(i) Antecedent is false
(ii) Consequence is true

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EXAMPLE
If (42 + 32)1/2 =7 then 7 is irrational number
[ ]
2. If a3-b3 = (a-b)(a2+ab+b2) then x=-1 is the root of x2-x+2=0
[ ]
3. If -2 = 3 then -32 =-9
[ ]
4. If Adi is precident then 2 is even number
[ ]

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5. If log(x-y) =log x/y then Toni loves Nia
[ ]
6. If today is here after then the sunset in east
[ ]
7. If x2=16 then x=4
[ ]
8. If x2-x >0 then x<0 ( )
x(x-1)>0
x=0 V x=1
+ - +
0 1

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Conclusion: Implication is false if we give one example so that Implication is FALSE
Determine the value of x from the sentence
p(x) => q V p =>q(x) :
p(x) =>q is TRUE, if q is false so p(x) must FALSE
p(X) =>q is FALSE if q is false so p(x) must TRUE
p(x) =>q is FALSE if p is true so q(x) must FALSE

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EXAMPLE
Determine the x value in order that the Implication is TRUE!
a). 2log(x-1)=3 => am.an =amn
b). Sin 2700=-1 => vx-2 = v2x-3
2. Determine the x value in order to the Implication is FALSE!
a). v5+2 vx = v3 + v2 =>the roots of x2+3x+5 =0 are the real number
b). v6-2 v8 =2- v2 => ½ logx =-1

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SOLUTION
a) 2log (x-1) ? 3 => am.an =amn
x-1 ?23
x ? 9
so the value is TRUE
b) Sin 2700=-1 [F]
=> vx-2 = v2x-3
x-2 = 2x-3
-x =-1
x =1

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2. a). p : [T]
v5+ 2 vx = v3 + v2
v(3+2)+2 v3.2 = v3 + v2
x = 6 [T]
q : [F]
b). v6-2 v8 =2- v2 [T]
½ logx = -1
x= ½
x ? 2

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LOGICAL IMPLICATION
The implication p(x)=> q(x) is called the logical implication if each of x fulfills p(x) also q(x)
EXAMPLE
Please observe that implication include Logical Implication or not
a). x2-1 = 0 => x2+x-6 < 0 V x €R
b). x2+5 = x+7 => x2+x =0
c). x2 = 25 => x=-5
d). x= v3 => x2 =3

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SOLUTION
a) x2-1 = 0
(x-1)(x+1) = 0 + - +
x=1 V x=-1 -1 1 HP={xl-1 = x = 1x €R}
x2+x-6 < 0
(x+3)(x-2)<0 + - +
x=-3 V x=2 -3 2
HP={xl-3<x<2,x €R}
so its Logical Imp.

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b) x2+5 = x+7 => x2+x =0
x2 -x-2=0 x(x+1)=0
(x-2)(x+1)=0 x=0 V x=-1
x=2 V x=-1 HP={-1,2}
so it isn’t Logical Imp.
x=-1
c). X2 =25
x=±5=>Not Logical Imp.
d). X=v3
X2=(v3)2
X2=3 so, it Logical Imp.

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E. BIIMPLICATION
If the compound statement with the linking word “If and only if”
p biimplication q can be written “p?q”
Biimplication is often called Biimplication 2 direction and can be stated :
p?q ˜ (p=>q) ? (q=>p)

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The Truth table of Biimplication :
The characteristic of Biimplication :
* Biimplication is true if the both statements are the same truth value.

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The Relation between Biimplication and set :
(P=Q)
p?q can be read :
p if and only if q
If p then q and if q then p
p is necessary and enough condition to q
q is necessary and enough condition to p p = q

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EXAMPLE
Determine that truth value
1+ cotan2x = cosec2x if and only if ax log ay=y/x
( )
2. a sin B= b sinC if and only only if a2=b2+c2+2bc.cosA
( )
3. x2-6x +9=0 has twin roots if and only if y= x2+4 doesn’t cut x axis at 2 points
( )
4. l x l= -x for x > 0 if and only if l2x-1l =3
So the value of x= 2. or x=-1
( )
5. cos750 = -sin 150 if and only if sec (2700-x) =-sec x
( )

Determine the x value from the sentence p ?q or p?q(x)
EXAMPLE
Determine the value of x that Biimplication TRUE
a). 2x-5=9 ? 9 is the Composite number
b). Ln e=-1 ? x2-5x+4=0
c). v40-16(1+ v2) =4-2v2?lx+3l <2
d). vx2-2x = v3? a2log ½=4

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Like no.1 but that Biimplication FALSE
a). Sin2700=1? log2x=3
b) P(x,5) is culmination point of y=2x2+12x-13?1+ tan2 x=sec2x
c) Y=2x+5 has the slope 2 ? lx2+2xl =1
d). lxl=x for x>0 ? vx2+x >2

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SOLUTION
a). 2x-5=9 b).x2-5x+4 ?0
2x=14 (x-4)(x-1) ?0
x=7 x ?4 v x ?1
c). v40-16(1+ v2) =4-2v3 [T]
lx+3l <2
(x+3)2-22 <0
(x+3-2)(x+3+2)<0
(x+1)(x+5)<0
x=-1 V x=-5
+ - + HP={xl-5<x<-1,x€R}
-5 -1

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d). vx2-2x = v3
(i) x2 – 2x ‹3 (ii) x2 – 2x =0
x2 – 2x -3 <0 x(x-2) =0
(x-3)(x+1) <0 x=0 V x=2
x=3 V x=-1
+ - +
-1 3
+ - +
0 2
-1 0 2 3
HP={xl -1< x = 0 V 2 = x <3 x€ R}

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2. a) Sin2700=1 [F]
log2x=3
2x=103
x=500
b). P(x,5) is culmination point of y=2x2+12x-13
x? -b
2a
x ? -12 =>x?-3
2.2
?1+ tan2 x=sec2x [T]

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c). Y=2x+5 has the slope 2 [T]
lx2+2xl =1
(X2+2X-1) (x2+2x+1)<1
(-2± v22 -4.1-1) V (x+1)(x+1)
2 x=-1
-2 ± v8
2
x1=-2- 2v2 = -1- v2 x2= -2+2v2 =-1+ v2
2 2
+ - - +
-1- v2 -1 -1+ v2
HP={xl-1- v2 < x < -1+v2, x?1 x € R }

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LOGICAL BIIMPLICATION
Biimplication p(x)?q(x) is called Logical Biimplication each x fulfills p(x) also q(x) or vice versa
Example
Determine that biimplication include Logical Biimplication or not!
a). alogb=c ? ac=b
b). X2-4=0 ? x± 2
c). X2-4x-5>0 ? x<-1
d). lx+3l =2 ?x=-1 V x=-5
e). 2x+1=5 ?x2-4=0
2. Determine the x value so 3x-4=8 ? x2-6x+8=0 is called Logical Biimplication

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SOLUTION
1. a). Logical Biimp
b). Logical Bimp
c). X2-4x-5>0 => x<-1[F] ? x<-1 =>X2-4x-5>0[T]
(x-5)(x+1)>0 + - +
x=5 V x=-1 -1 5
Not Logical Biimp.
d). lx+3l =2 => x=-1 V x=-5 ? x=-1 V lx+3l =2
(x+3)2=22
(x+3)2-22=0
(x+3-2) (x+3+2)=0
x=-1 V x=-5
Logical Biimp

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e). 2x+1=5 =>x2-4=0 ? x2-4=0 => 2x+1=5
2x=4 (x-2)(x+2)=0
x=2 x=2 V x=-2
Not Logical Biimp.
2). 3x-4=8 => x2-6x+8=0 ? x2-6x+8=0 => 3x-4=8
3x=12 (x-4)(x-2)
x=4 x=4 V x=2
so x=4

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1. ~ ( p ? q ) = ( ~ p ? ~q )
2. ~ ( p ? q ) = ( ~ p ? ~q)
3. ~ ( p ? q ) = ( p ? ~q )
4. ~ ( p ? q ) = ( p ? ~q ) ? (q ? ~p)
~ ( p ? q ) = ( ~q ? p )
= ( ~p ? q ) III. NEGATION OF CONJUNCTION, DISJUNCTON, IMPLICATION & BIIMPLICATION NEXT SLIDE

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EXAMPLE :
Determine the negation of the statement below ;
1. 5 x 4 < 2 + 17 and 4 is composite number
2. A ? B = A or A ? B = B ? A
(read ; A subset B = A or A union B = B union A)
3. If Simon is handsome then Paula loves him.
4. ABCD is rectangle if and only if AC ? BD ANSWER :
5 x 4 = 2 + 17 or 4 is not composite number
A ? B ? A and A ? B ? B ? A
Simon is handsome and Paula doesn’t love him.
ABCD is rectangle and AC not perpendicular of BD or AC ? BD and ABCD is not rectangle NEXT SLIDE

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THE EQUIVALENCE OF THE COMPOUND STATEMENTS
To know the compound statement whether equivalence or not, we must prove the truth table of it equivalence or not.
If it has same the truth value then the compound statements is equivalence.
Formula of truth value is 2n, where n=the amount of statements. For example ;
? ~ ( p? q) = ( ~p ? ~q )
have 2 statements there are p & q
Truth values of p are T T F F
Truth values of q are T F T F NEXT SLIDE

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EXAMPLE :
Prove by using the truth table so that compound statements are equivalence!
1. ~ ( p ? q) = ( ~p ? ~q )
2. ~ ( p ? q ) = (~ p ? ~q )
3. ~ ( p ? q ) = ( p ? ~q )
4. ~ ( p ? q ) = ( p ? ~q ) ? (q ? ~p)
5. ~ ( p ? q ) = ( ~q ? p )
6. ( p ? q ) = ( ~p ? q )
7. ( p ? q ) = ~( p ? ~q )
8. ( p ? q ) = (~q? ~p )
9. ( q ? p ) = (~p ? ~q )
10. { p ? ( q ? r ) } = { ( p ? q ) ? r } NEXT SLIDE

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1.) ~ ( p ? q ) = ( ~p ? ~q ) SAME BACK

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2.) ~ ( p ? q ) = ( ~p ? ~q ) SAME BACK

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3.) ~ ( p ? q ) = ( p ? ~q ) SAME BACK

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4.) ~ ( p ? p ) = ( p ? ~q ) ? (q ? ~p) SAME BACK

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5.) ~ ( p ? q) = ( ~q ? p ) SAME BACK

6. ( p ? q ) = (~q ? p) :

6. ( p ? q ) = (~q ? p) SAME BACK

7. ( p ? q ) = ~( p? ~q) :

7. ( p ? q ) = ~( p? ~q) SAME BACK

8. ( p ? q ) = (~q ? ~p ) :

8. ( p ? q ) = (~q ? ~p ) SAME BACK

9. ( q ? p ) = (~p ? ~q) :

9. ( q ? p ) = (~p ? ~q) SAME BACK

10. { p ? ( q ? r ) } = { ( p ? q ) ? r } :

10. { p ? ( q ? r ) } = { ( p ? q ) ? r } SAME BACK

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V. CONVERSE, INVERS AND CONTRAPOSITION FROM THE STATEMENT “ p ? q”
Can be made :
(q ? p) is called Converse
(?p ? ?q) is called Inverse
(?q ? ?p) is called Contraposition
Example :
Find the converse, inverse and contraposition of the statement below ;
1. If Shoju is beautiful then Chooy loves her
2. If the line g // h then g and h haven’t intersection point Answer :
1. a) Converse : If Chooy loves Shoju then she is beautiful
b) Inverse : If Shoju is not beautiful then Chooy does not love her
c) Contraposition : If Chooy does not love Shoju then she is not beautiful
2. a) Converse : If g and h have no intersection point then the line g // h
b) Inverse : If the line g is not // h then g or h has intersection point
c) Contraposition : If g or h has intersection point then the line g is not // h NEXT SLIDE

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Make the truth table Conclusion :
( p ? q ) = (~q ? ~p )
( q ? p ) = (~p ? ~q ) NEXT SLIDE

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VI. TAUTOLOGY AND CONTRADICTION
Tautology is the compound statement which always has TRUE value
Contradiction is the compound statement which always has FALSE value
If the compound statement is not include tautology or contradiction then that compound statement is called CONTINGENTION NEXT SLIDE

Example : :

Example : Please observe this compound statements below whether Tautology, Contradiction or Contingention
{( p ? q ) ? ~q } ? ~p
( p ? q) ? (~p? ~q )
( p ? ~q) ? ( p ? q )
{( p ? q ) ? p } ? q
( p ? ~q ) ? r
( p ? ~q ) ? (q ? r) NEXT SLIDE

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a. {( p ? q ) ? ~q } ? ~p Tautology BACK

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( p ? q ) ? (~p ? ~q ) Contradiction BACK

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( p? ~q ) ? ( p ? q ) Contradiction BACK

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{( p ? q ) ? p } ? q Tautology BACK

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( p ? ~q ) ? r Tautology BACK

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f. ( p ? ~q ) ? (q ? r) Contingention BACK

VII. QUANTIFIER :

VII. QUANTIFIER Quantifier consist of 2 kinds :
Universal Quantifier
Existential Quantifier
Universal Quantifier
Is the statement which is using word “all” or “every” with the symbol ?
?x . p(x) can be read :
Every x are p(x)
All x are p(x) NEXT SLIDE

The relations between universal quantifier with set : :

The relations between universal quantifier with set : A B ?x ? A ? x ? B = A ? B Example :
The all human will be dead (True)
Every Monday held raising red and white flag ceremony at school (False)
Every x elements positive number x2 > x can be written
(?x > 0, x2 > x) (False) NEXT SLIDE

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The truth value of Universal Quantifier is true, if there was no answer (penyangkal) which make the statement false.
Meanwhile that quantifier statement is false if there is one minimum answer which make that statement false.
Example :
Determine the truth value of the quantifier below :
?x ? B, x ? R (True, because every whole number is real number)
?x > 0, x2 > x (False, because if we substitute X= ½ then (½)2 is not > ½. But ¼ < ½) NEXT SLIDE

2. Existential Quantifier :

2. Existential Quantifier Is the statement which is using word “there are” or “some” with the symbol ?
?x . p(x) can be read :
There x are p(x)
Some x are p(x)
The relations between universal quantifier with set : A B ?x ? A ? x ? B = A ? B NEXT SLIDE

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The truth value of Existential Quantifier is true, if there was minimum value which make the statement true.
Meanwhile that quantifier statement is false if there was no answer which make that statement false.
Example :
Determine the truth value of the quantifier below :
There are the odd numbers which divisible by 2 (False, because all of odd numbers are not divisible by 2)
Some students are not responsible (True)
?x . x2-8x+16=0 (True) NEXT SLIDE

VIII. NEGATION OF QUANTIFIER :

VIII. NEGATION OF QUANTIFIER Statement Negation
All A are B ? Not All A are B
? Some A are not B
Some A are B ? All A are not B
? There is no A which B 1. ?(?x . p(x)) = ?x . ?p(x) 2. ?(?x . p(x)) = ?x . ?p(x) The Formula are : NEXT SLIDE

Determine the negation of the statements below : :

Determine the negation of the statements below : All Rational numbers include Real numbers.
Some students in SMA 1 Kudus are smart.
There are no even prime number.
?x > 0, x2 > 0
?x ? C . x > 0
?x ? A . x2 + x = 0
?x ? Q . x2 - 4x + 4 = 0
Answer :
? Not All Rational numbers include Real numbers.
? Some rational numbers are not include real numbers. NEXT SLIDE

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2. ? All students in SMA 1 Kudus are not smart.
? There are no students in SMA 1 Kudus which smart
3. ? Some even number is prime number.
? All even number is prime number.
4. ? ?[ ?x > 0, x2 > 0 ]
? ?x > 0, x2 ? 0
5. ? ?[ ?x ? C . x > 0 ]
? ?x ? C . x ? 0
6. ? ?[ ?x ? A . x2 + x = 0 ]
? ?x ? A . x2 + x = 0
7. ? ?[ ?x ? Q . x2 - 4x + 4 = 0 ]
? ?x ? Q . x2 - 4x + 4 = 0 NEXT SLIDE

IX. DRAWING CONCLUSION :

IX. DRAWING CONCLUSION Drawing conclusion consists of 3 kinds :
Modus Ponens
p ? q (Premise 1)
p (Premise 2)
... q (Conclusion)
Modus Tollens
p ? q (Premise 1)
~q (Premise 2)
... ~p (Conclusion)
Syllogism
p ? q (Premise 1)
q? r (Premise 2)
... p ? r (Conclusion) NOTE :
The drawing conclusion beside is
called Valid If it’s include Tautology NEXT SLIDE

1. Let’s check the arguments at the slide above, by using the truth table ! :

1. Let’s check the arguments at the slide above, by using the truth table ! Modus Ponens
p ? q (Premise 1)
p (Premise 2)
... q (Conclusion) Valid Promise 1 connected with promise 2 by using CONJUNCTION The promises connected with the conclusion by using IMPLICATION Tautology

Slide 94:

Modus Tollens
p ? q (Premise 1)
~q (Premise 2)
... ~p (Conclusion) Tautology Valid

Slide 95:

Sylogism
p ? q (Premise 1)
q ? r (Premise 2)
... p ?r (Conclusion) Tautology Valid

Example !Determine Valid or Invalid that argument below : :

Example !Determine Valid or Invalid that argument below : ~p ? q
~p
... q
p ? q
p ? r
... ~r ? q
If they are pregnant then they are woman
Yanti is not woman
... Yanti is not pregnant
If x is an odd number then x2 is an odd number
If x2 is an odd number then x2+1 is an even number
... If x is an odd number then x2+1 is an even number NEXT SLIDE

Answer : :

Answer : ~p ? q
~p Include modus PONENS
... q (Valid)
p ? q = ~q ? p
p ? r p ? r Include SYLOGISM
... ~r ? q ~q ? r = ~r ? q (Valid)
p ? q
~q Include modus TOLLENS
... ~p (Valid)
p ? q
q ? r Include SYLOGISM
... p ? r (Valid) NEXT SLIDE

X. APPLICATION IN ELECTRIC CURRENT :

X. APPLICATION IN ELECTRIC CURRENT Disjunction as parallel connection
Conjunction as series connection q p p ? q = = p ? q NEXT SLIDE

Exercise !Draw the electric current or give the statement from the figure below : :

Exercise !Draw the electric current or give the statement from the figure below : ( p ? q ) ? (p v q v r)
{(p v q ? (~r v q} v (p v r) p q ~q ~r p p NEXT SLIDE

Answer : :

Answer : p p q r 1. 2. 3. r p p q ~r q {(~r ? p) v (p v ~q)} v (q ? p) NEXT SLIDE

XI. MATHEMATICAL INDUCTION :

Example !
1+2+3+…….+n = ½ n (n+1)
1.2+2.3+3.4+…..+n(n+1) =
………………………………..
5n – 1 is divisible by 4 XI. MATHEMATICAL INDUCTION How to solve the mathematical induction
(i) Prove that formula is true for n = 1
(ii) Let if the formula is true for n=k, prove that formula will be true for n = k+1
(iii) If (i) & (ii) is proven so that formula will be valid for n ? A NEXT SLIDE

Answer : :

Answer : 1+2+3+…….+n = ½ n (n+1)
(i) n = 1
1 = ½ .1 (1+1)
1 = 1 (true)
(ii) n = k ? 1+2+3+…….+k = ½ k (k+1) (true)
(iii) n = k+1 ? 1+2+3+…….+ k + (k+1) =½k(k+1)(k+2)
RKi = 1+2+3+…….+ k+ (k+1)
= ½ k (k+1)+(k+1)
= ½k(k+1){k+2}
= Rka (Valid)

Slide 103:

b) 1.2+2.3+3.4+…..+n(n+1) =
(i) n = 1
(ii) n = k ? 1.2+2.3+3.4+…..+k(k+1) =
(iii) n = k+1 ? 1.2+2.3+3.4+…..+(k+1) (k+2) =
RKi = + (k+1)(k+2)
=
= = RKa (true) (true) (Valid)

Slide 104:

1 + 31 + 32 + …. + 3n-1 = ½ (3n -1)
(i) n = 1
1 = ½ (31-1)
1 = 1
(ii) n = k ? 1 + 31 + 32 + …. + 3k-1 = ½ (3k -1)
(iii) n = k+1 ? 1 + 31 + 32 + …. + 3k-1+ 3k-1+1 = ½ (3k+1 -1)
RKi = ½ (3k -1) + 3k
= {½ 3k – ½} + 3k
= 1½ 3k – ½
= ½ 3k+1 – ½
= ½ (3k+1 -1)
= RKa (Valid) (true) (true)

Slide 105:

5n – 1 is divisible by 4
(i) n = 1 ? 51 – 1 = 4 {is divisible by 4 (true)}
(ii) n = k ? 5k – 1 = 4m {is divisible by 4 (true)}
(iii) n = k+1 ? 5k+1 – 1
5k+1 – 1 = 5k . 51 – 1
= 4 . 5k + (5k – 1)
= 4 . 5k + 4m
= is divisible by 4 + is divisible by 4
= is divisible by 4 (Valid)

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By: AjengCrossev (115 month(s) ago)

very difficult