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VALERIA IBARGUEN 9-1
CONDITIONAL STATEMENT




a)
b)
c)
This is a type of statement that can be written in a
form of “if p, then q”
P=Hypothesis
Q=conclusion
EXAMPLES:
If m<A=195°, then <A is obtuse
If an insect is a butterfly, then it has four wings
If an angle is obtuse, then it has a measure of
100°
COUNTER-EXAMPLES
A counter-example is a type of example that proves if a conjecture
or statement is false. This could be a drawing, a statement or a
number.
 EXAMPLES:
a)
For any real number x, x2 > x
5, 52 > 5
5, 25 > 5
b)
Supplementary angles are adjecent

c)
The radius of every planet in the solar system is less than
50,000 km.
Mercury
Venus
Earth
Mars
Jupiter
Saturn
Uranus
4880
12,100
12,800
6790
143,00
121,00 51
Neptune
49,500
DEFINITION


a)
b)
c)
This is a statement that tells or discribes a
mathematical object and can be written as a true
biconditional statement. A definition includes “if and
only if”
EXAMPLES:
A figure is a triangle if and only if it is a threesided polygon.
A ray, segment or line is a segment bisector if and
only if it divides a segment into two congruent
segments.
A traingle is straight if and only if it measures
180°.
BI-CONDITIONAL STATEMENTS


a)
b)
c)
This is a statment that is written in the form “p if and only
if q”. They are important. This is used when a conditional
statement and its converse are combined together.
EXAMPLES:
Converse: If x=3, then 2x+5=11
Biconditional: 2x+5=11 if and only if x=3
Converse: If a point divides a segment into two
congruent segments, then the point is a midpoint.
Biconditional: A point is a midpoint if and only if it
divides the segments into two congruent segments.
Converse: If the dates is July 40th, then it Independence
day. Biconditional: It is Independence day if and only if
it is July 40th.
DEDUCTIVE REASONING


a)
b)
c)
This is the type of process in which we use logic to draw conclusions
of something.
EXAMPLES:
If a team wins 10 games, the they play in the finals. If a team
plays in the finals they they travel to Boston. The Reavens won 10
games.
CONCLUSION:The Reavens will travel to Boston.
If two angles form a linear pair, then they are adjecent. If two
angles are adjecent, then they share a side. <1 and <2 form a
linear pair.
CONCLUSION: <1 and<2 share a side.
If a polygon is a triangle, then it has three sides. If a polygon has
three sides then it is not a quadrilateral. Polygon is a P triangle.
CONCLUSION: A polygon is not a quadrilateral because ithas
three sides.
LAWS OF LOGIC

-

-
Law of detachment:
If p-q is true we should assume if P is true then Q
must also be true
Law of Syllogism:
If P-Q is true and Q then R is true then if P is true
are must be true P and R is true.
LAW OF DETACHMENT
Given: If two segments are congruent then they have the same length.
AB≅XY
Conjecture: AB=XY
hypothesis: two segments are congruent
conclusion: they have the same lenght
The given AB≅XY statements does match the hypothesis so the conjecture
IS true.
 Given: If you are 3 times tardy, you must go to detention. John is in
detention.
Conjecture: John was tardy at least 3 times.
hypothesis: you are tardy 3 times
conclusion: you must go to detention.
The statement given to us matches the conclusion of a true conditiona, but
the hypothesis is not true since John can be in detention for another
reason so the conjecture is NOT valid.

LAW OF SYLLOGISM
GIVEN: If m<A 90°, then <A is acute. If <A is acute then it is not a right
angle.
p= the measure of an angle is less then 90°
q= the angle is acute
r= the angle is not a right angle.
-This is trying to explain us that pq and qr is the conclusion of the first
conditional and the hypothesis of the second conditional you can tell that at
the en pr. So IT IS VALID
 Given: If a number is divisible by 4 then it is divisible by 2. If a number is
even, then it is divisible by 2.
Conjecture: If a number is divisible by 4, then it is even.
p= A number is divisible by 4
q= A number is divisible by 2
r= A number is even
-What this means is that pq and rq. The Law of Syllogism cannot be used
to draw conclusions since q is the conlcusion of both conditional statements,
even though pr is true the logic used to dra the conclusion is NOT VALID.

ALGEBRAIC PROOF
An algebraic proof is an argument that uses logic,
definitions, properties. To do one, you have to do a
2 colum proof.
STATEMENT
REASON
 EXAMPLES:
2x-6=4x-10
Given
a)Prove: x=2 if
+10
+10
Addition Property
Given: 2x-6=4x-10 2x+4=4x
Simplify

-2x
-2x
Subtraction Property
4x=2x
Simplify
2 2
Divsion Property
2=x
Simplify
X=2
Symetric property
ALGEBRAIC PROOF
b)-5=3n+1

STATEMENT
c)sr=3.6
REASON
-5=3n+1
Given
-1
Subtraction
Property
-1
-6=3n
Simplify
3 3
Division Property
-2=n
Symplify
N=-2
Symmetric
Property
STATEMENT
REASON
sr=3.6
S=75 km/h
R=6 pixels per meter
GIVEN
(75)(6)=3.6p
Substitution property
450=3.6p
Simplify
3.6
Division Property
3.6
125=p
Simplify
P=125 pixels
Symmetric Property
SEGMENT AND ANGLE PROPERTIES
OF CONGRUENCE AND EQUALITY

PROPERTY OF EQUALITY:
ADDITION PROPERTY OF EQUALITY
If a=b, then a+c=b+c
SUBTRACTION PROPERTY OF EQUALITY If a=b, then a-c=b-c
MULTIPLICATION PROPERTY OF
EQUALITY
If a=b, then ac=bc
DIVISION PROPERTY OF EQUALITY
If a=b then c≠= then a/c=b/c
REFLEXIVE PROPERTY OF EQUALITY
A=A
SYMMETRIC PROPERTY OF EQUALITY
If a=b, then b=a
TRANSITIVE PROPERTY OF EQUALITY
If a=b and b=c, then a=c
SUBSTITUTION PROPERTY OF EQUALITY If a=b then b can be substituted for a in
any expression.
SEGMENT AND ANGLE PROPERTIES
OF CONGRUENCE AND EQUALITY

PROPERTIES OF CONGRUENCE:
Reflexive Property of Congruence
Figure A≅ figure A
− −
EF≅EF
Symmetric Property of Congruence
If figure A≅ figure B then figure B≅A
IF <1 ≅ <2 then <2 ≅ <1
Transitive Property of Congruence
If figure A ≅ figure B and figure B ≅
figure C then figure A ≅ figure C
If <1 ≅ <2 and <2 ≅ <3 then <1 ≅
<3
TWO-COLUM PROOFS


To do a two colum proofs you have to list each step
of how you found your answer.
EXAMPLES:
STATEMENT
REASON
C=9f+90
C=102
Given
102=9f+90
Substitution
-90
Subtraction
-90
12=9f
Simplify
9 9
Division
1.3=f
Simplify
F=1.3
Symmetric Property
TWO-COLUM PROOFS
STATEMENT
C=$5.75+$0.89
m
C=$11.98
REASON
Given
$11.98=$5.75+$ Substitution
0.89m
-$5.75
-$5.75
Subtraction
Property
$6.23=$0.89m
Simplify
$0.89m $0.89m
Division
7=m
Simplification
M=7
Symmetric
Property
STATEMENTE
REASON
Y+1=5
GIVEN
-1
Substitution
-1
Y=4
Simplify
Y=4
Symmetric
Property
LINEAR PAIR POSTULATE (LPP)


This is when all linear pairs are linear postulates,
SUPPLEMENTARY
EXAMPLES:
Given: angle<1 and < 2 are linear pair
Prove: <1 and <2 supplementary.
<1 and <2 are linear pair
Given
<1 and <2 form a linear pair
Definition of linear pair
M<1+M<2= 180°
Striaght angle’s definition
<1 and <2 are supplementary
Deffintion of supplementary
LINEAR PAIR POSTULATE
Given: <1 and <2 are supplementary <3 and <4
are supplementary.
Prove:<1≅<4
STATEMENT
REASON
<1 and <2 are supplementary <3 and
<4 are supplemtary
GIVEN
<1 + <2 = 180°, <3 + <4=180°
Deffinition of supplementary angles
M<1+M<2=M<3+M<4
Substitution
<2≅<3
Given
M<2=M<3
Deffinition of congruent andgles
M<1=M<4
Subtraction property of steps 3 and 5
<1≅<4
Definition of Congruent angles
LINEAR PAIR POSTULATE
Given: BE ≅ CE, DE ≅ AE
Prove: AB ≅ CD
SATEMENT
REASON
BE ≅ CE, DE ≅ AB
Given
BE + AE= AB
Segment Addition Postulate
CE + DE= CD
Segment Addition Postulate
CE + DE= AB
Substitution
CD=AB
Substitution
CONGRUENT COMPLEMENTS AND
SUPPLEMENTS THEOREMS

CONGRUENT COMPLEMENT THEOREM:
THEOREM
If two angles are
complementary to
the same angle (or
to two congruent
angles) then the two
angles are
congruent
HYPOTHESIS
<1 and <2 are
complementary
<2 and <3 are
complementary
CONCLUSION
<1 ≅ <3
CONGRUENT COMPLEMENTS AND
SUPPLEMENTS THEOREMS

CONGRUENT SUPPLEMENT THEOREM:
THEOREM
If two angles are
supplemtnary to the
same angle (or two
congruent angle)
then the two angles
are congruent
HYPOTHESIS
<1 and <2 are
supplementary
<2 and <3 are
supplementary
CONCLUSION
<1 ≅ <3
VERTICAL ANGLES THEOREM

VERTICAL ANGLE THEOREM:
THEOREM
Vertical angles are
congruent
HYPOTHESIS
<A and <B are
verical angles
CONCLUSION
<A ≅ <B
COMMON SEGMENTS THEOREM

COMMON SEGMENTS THEOREM:
THEOREM
Given a collinear
points A,B, C and
D arranged as
showns if AB≅CD
then AC≅BD
HYPOTHESIS
AB≅CD
CONCLUSION
AC≅BD