Download Chapter 2.1

Chapter 2.1
Inductive Reasoning and
Conjecture
Vocabulary
Conjecture – A conjecture is an educated
guess based on known information.
Inductive Reasoning – This is reasoning
based on a number of examples to arrive
at a plausible generalization or prediction.
Counter Example – This is an example
that shows that a conjecture is false.
Inductive Reasoning
Let us look at this pattern:
What do you notice about the pattern?
What can you predict about the next picture?
It is going to have the same shape as the
last picture with…
another column of four boxes.
Keys
The key to inductive reasoning is to find a
pattern.
Look at example #1 in the book.
From the pattern, you can make a conjecture.
Sometimes the conjectures are wrong.
These conjectures are proved wrong by a
counterexample.
When this happens you need to look for
another pattern.
Counterexample
A counterexample is an example used to
prove a conjecture is false. (Wrong)
It only takes one counterexample to prove
a conjecture wrong.
So, good conjectures are “always true”
While bad conjectures can be “sometimes
true and sometimes false.”
Example
Let us look at some observations:
A Ford Mustang has two doors.
A Pontiac Solstice has two doors.
A Saturn Sky has two doors.
A BMW Z4 has two doors.
From this pattern I can make a conjecture
that “All cars have two doors”.
Is this ALWAYS true?
Counterexample: A Pontiac Grand Am is a
car that has four doors.
Chapter 2.2
Logic
Truth Values
A statement can have only two truth
values. That is the statement is either:
True (Implying always true) or
False (Implying not always true).
A statement can not be both True and
False.
Take the statement: All cars have two
doors. That is false b/c we found a
counterexample to refute it’s validity.
Negation
We can negate a statement by putting the
word “not” in it someplace.
The negation of a true statement makes is
now false.
The negation of a false statement makes it
now true.
We use ~ to indicate “not”
“All cars have two doors” becomes “Not all
cars have two doors”
The original statement was false, when we
negate it, it becomes true.
Compound Statement
Just like in English, you can put two
statements together and make one
compound statement.
You can do this by making the compound
statement either a conjunction or a
disjunction.
The compound statement also has a truth
value as a whole.
Conjunction (Λ)
The key word for a conjunction is and.
Take two statements:
All cars have two doors.
All birds fly.
To make a conjunction from these two
statements you simply put and in between them.
All cars have two doors and all birds fly.
In order for a conjunction to be true both
statements have to be true.
Disjunction (V)
The key word for disjunction is or.
Take the same two statements:
All cars have two doors.
All birds fly.
To make a disjunction from these two
statements you simply put or in between
them.
All cars have two doors or all birds fly.
In order for a disjunction to be true only one
statement has to be true.
Truth Value of Conjunctions
Just like statements, conjunctions also
have a truth value.
Conjunctions – both statements must be
true before the conjunction is true.
Raleigh is in NC and NYC is in New York.
Both statements are true so conjunction is
True.
Raleigh is in NC and NYC is in Michigan.
Only one statement is true so the
conjunction is false.
Truth Value of Disjunctions
Just like statements, disjunctions also have a
truth value.
Disjunctions – only one statement must be true
before the disjunction is true.
Raleigh is in NC or NYC is in New York.
Both statements are true so disjunction is True.
Raleigh is in NC or NYC is in Michigan.
Only one statement is true so the disjunction is
true.
Truth Tables
Conjunctions
Disjunctions
P
Q
PΛQ
P
Q
PVQ
T
T
T
T
T
T
T
F
F
T
F
T
F
T
F
F
T
T
F
F
F
F
F
F
Truth Table (H)
P
Q
R
~P
T
T
T
F
T
F
F
F
T
T
F
T
F
T
F
F
T
F
T
T
F
F
T
F
F
F
F
T
F
T
T
T
~PΛQ (~PΛQ)VR
F
F
F
T
F
T
F
F
T
F
T
T
F
T
T
F
Venn Diagrams
Venn Diagrams are diagrams with pictures
to portray Conjunctions and Disjunctions.
The overlapping portion or the two ovals is
your Conjunction.
The two ovals combined is your Disjunction.
A
Disjunction
B
AVB
Conjunction
AΛB
Blue Shade
~AΛ~B
Chapter 2.3
Conditional Statements
Conditional Statements
Conditional Statements can be written in
If-Then form.
Essentially, an If-Then statement says
“The If must be satisfied, before the Then
can happen.”
Example: If you pass this class, then you
can move on to Alg II.
What do you need to complete before you
move on to Alg II?
Conditional Statements
(Con’t)
Conditional Statements have two parts.
The part that follows the “IF” is called the
Hypothesis.
The part that follows the “Then” is called
the Conclusion.
If it is raining outside,
Hypothesis
then I will carry my umbrella.
Conclusion
Conditional Statements
(Con’t)
Conditional statements don’t always have
to have an If – Then in the statement.
It can be put in though.
Example: A Right Angle is an angle that
measures 90° (Definition of a Right Angle)
Does this have an If-Then in it?
Can we rewrite it to have an If-Then?
If an angle is a Right Angle, then the angle
has a measure of 90°
Converse, Inverse and
Contrapositive
Conditional
P→Q
Converse
Q→P
Write the conditional
statement here.
Switch the order of
the Q and the P.
Contrapositive
Inverse
~P → ~Q
Negate the conditional
statement here.
~Q → ~P
Negate the Converse
statement here.
Example
Conditional
Converse
If two angles are right If two angles are congruent
angles, then they’re
then they’re right angles.
congruent. (P → Q)
(Q → P)
Inverse
Contrapositive
If two angles are not If two angles are not
right angles, then
congruent, then they’re
they’re not congruent. not right angles.
(~Q → ~P)
(~P → ~Q)
Truth?
Conditional
Converse
If two angles are right If two angles are congruent
angles, then they’re
then they’re right angles.
congruent.
(F)
(T)
Inverse
Contrapositive
If two angles are not If two angles are not
right angles, then
congruent, then they’re
they’re not congruent. not right angles.
(T)
(F)
For all False statements – provide a Counter Ex
Truth?
If the conditional statement is a Definition
– then all four conditionals will be true.
If the conditional is true, then the
contrapositive is also true.
If the converse if false, then the inverse is
false.
Biconditional Statements (H)
The biconditional statement is the
conjunction of the conditional and
converse statements.
(P→Q)Λ(Q→P) gives you (P↔Q)
Biconditional statements have the key
term “if and only if” in it.
See pg 81.
Chapter 2.4
Deductive Reasoning
Deductive Reasoning
Deductive reasoning is very different than
Inductive reasoning.
In Inductive reasoning we used patterns to
predict an outcome.
In Deductive reasoning we use theorems,
definitions, postulates and corollaries to
reach a conclusion.
Two types of deductive reasoning is the Law
of Detachment and the Law of Syllogism.
Law of Detachment
The Law of Detachment follows a distinct
pattern.
Using the LOD we have three steps
1) Write the Conditional Statement – this
is usually a definition or theorem.
2) State a specific case of the Hypothesis
of the conditional statement being
satisfied.
3) State a specific case of the Conclusion
being satisfied.
LOD Example
1) If an angle is a right angle,
then the angle measures 90°
(This is the definition of a right angle)
2) Angle A is a right angle.
(This talks specifically about a certain
angle. Notice it satisfies the Hypothesis
of the conditional statement?)
3) m<A = 90°
(This talks about a specific angle. It
satisfies the conclusion of the conditional
statement.
LOD Example
1) If two angles are supplementary,
then the sum of their measures equals
180°
2) <A and <B are supplementary,
3) m<A + m<B = 180
Law of Syllogism
Just like LOD, LOS has a specific pattern.
Unlike LOD, LOS has THREE conditional
statements.
If P→Q
If Q→R
If P→R
Notice where the Q’s are?
Notice they are on the diagonal… it is as if
they were crossed out and only the P and
R remain.
Summary
LOD
Step 1
If – Then conditional
statement
Step 2
Specific Example of
Hypothesis being
satisfied.
Step 3
Specific example of
conclusion being satisfied
LOS
Step 1
If – Then conditional
statement
Step 2
If – Then conditional
statement (C of 1 is H of
2.
Step 3
If – Then conditional
statement (H of 1 is H of
3 and C of 2 is C of 3)
Chapter 2.5
Postulates and Paragraph Proofs
Postulates
Postulates are statements that describe a
fundamental relationship between basic
terms.
Postulates are accepted to be true.
Postulates are used in deductive
reasoning.
Basic Postulates
Through any two points there exists
exactly one line.
Through any three noncollinear points
there exists exactly one plane.
A line contains at least two points.
A plane contains at least three
noncollinear points.
If two points lie in a plane then the entire
line lies in the plane.
Basic Postulates (Con’t)
If two lines intersect then they intersect at
exactly one point.
If two planes intersect then they intersect
at exactly one line.
Proofs
A proof is a LOGICAL argument in which each
statement is supported by a postulate, theorem,
definition or corollary.
There are three types of proofs that we will deal
with in this class.
Two are “Formal” and one is “Informal”.
The informal proof is a paragraph proof.
You have done these in English.
The two formal proofs are “Two Column” and
“Flow” Proofs.
Key Elements
There are five key elements essential for a
good proof.
State the Theorem or Conjecture to be
proven.
List the Given information.
If possible, draw a figure or diagram to
illustrate the given information.
State what is to be proved.
Develop a System of deductive reasoning to
get you from the conjecture to the end.
Drive to the Mall?
If you were at JHS, what directions would you
give to a person (not from Jax) to get to the
Jacksonville Mall?
How detailed would it need to be?
Very detailed – can’t take anything for granted
Can you leave any steps out?
You can’t skip steps – if you don’t tell them to turn
on Western Extension – then what?
Do you need to follow traffic regulations?
You can’t break any laws b/c you’ll get a ticket or
worse – you’ll end up in jail.
Theorem
Once a statement of conjecture is proved true
then it can be called a Theorem.
Go to page R1 in the back of the book.
This section contains all the theorems,
postulates and corollaries that we will use in this
class.
I strongly suggest that you read them every night
so you can commit them to memory.
The more you know in this class the easier it will
be.
Chapter 2.6
Algebraic Proof
Algebraic Proof
Before we start on geometric proofs let us
practice algebraic proofs .
Building blocks of all proofs are theorems,
definitions, postulates and corollaries.
Algebraic building blocks are the
properties of equality for real numbers.
Properties if Equality for
Real Numbers
Reflexive Property – For every number a,
a = a.
Symmetric Property – For all numbers a
and b, if a = b, then b = a.
Transitive Property – For all numbers a, b
and c, if a = b and b = c, then a = c.
Addition/Subtraction Property – For all
numbers a, b and c, if a = b, then a + c = b
+ c and a – c = b – c.
Properties of Equality for
Real Numbers
Multiplication/Division Property – For all
numbers a, b and c, if a = b, then ac = bc
and a/c = b/c.
Substitution Property – For all numbers a
and b, if a = b then a may be replaced by b
in any equation and expression.
Distributive Property – For all numbers a,
b, and c, a(b + c) = ab + ac.
Two Column Proof
The two column proof derives it’s name b/c it
has two columns.
Two columns are Statements and Reasons.
Statement
x=5 H
5=x
C
These two statements
satisfy the LOD of the
If-Then form.
Reason
Given
Symmetric Prop
Think of this in If-Then
form.
Example #1
Given: 3(x – 2) = 42
Statement
3(x – 2) = 42 H
3x – 6 = 42 C
H
3x – 6 + 6 = 42 + 6
C
H
3x = 48
H
C
3x/3 = 48/3
C
H
x = 16
C
Prove x = 16
Reason
Given
Distribution Prop
Add/Subt Prop
Substitution Prop
Mult/Div Prop
Substitution Prop
Geometric Proofs
Segments
Angles
Reflexive
AB = AB
Reflexive
m<ABC = m<ABC
Symmetric
If AB = BC, then BC = AB
Symmetric
If m<ABC = m<XYZ,
then m<XYZ = m<ABC
Transitive
If AB = BC and BC = CD
then AB = CD
Transitive
If m<1 = m<2 and m<2 = m<3
then m<1 = m<3
Hints
Some important Hints for Proofs
Your first statement is always the given.
The last statement is always what you
want to prove.
Look for what changed from one
statement to another…. Think of a
Theorem, Postulate, Definition or Corollary
that would let you go from one to the other.
Chapter 2.7
Proving Segment Relationships
Segment Addition Postulate
The Segment Addition Postulate (SAP) is
a very important postulate b/c it allows you
to break one segment into two smaller
ones, or if the three points are collinear, it
allows you to make one big segment out of
two little ones.
If B is between A and C, then AB + BC =
AC.
A
B
C
Example of SAP
Given: AB = CD
Prove: AC = BD
Statement
A
B
C
D
Reason
AB = CD
Given
BC = BC
Reflexive Prop
AB + BC = CD + BC
Addition Prop
AB + BC = AC
SAP
BC
+
CD
=
BD
What do you need to do to AB to make it AC?
Substitution
AC
=
BD
What do you need to do to CD to make it BD?
Another Example
Given: AC = BD
Prove: AB = CD
Statement
AC = BD
AB + BC = AC
BC + CD = BD
AB + BC = BC + CD
BC = BC
AB = CD
A
B
C
Reason
Given
SAP
Substitution
Reflexive Prop
Add/Subt Prop
D
Big Difference
There is a big difference between the
Segment Addition Postulate (SAP) and the
Add/Subt Property.
The SAP takes two little segments and
makes one big segment from it.
Or, takes one big segment and breaks it
down into two little segments.
Add/Subt adds or subtracts the same thing
from each side of the equal sign.
Segment Congruence
Congruence of Segments is Reflexive,
Symmetric and Transitive.
Reflexive –
Symmetric –
Transitive –
AB  AB
if AB  CD
then CD  AB
if AB  CD
and CD  DE
then AB  DE
Chapter 2.8
Proving Angle Relationships
Angle Addition Postulate
The Angle Addition Postulate (AAP) is
exactly like the SAP except you’re using
angles.
The AAP takes two adjacent angles and
allows you to add them together.
Or, it allows you to divide one big angle
into two smaller angles.
A
D
C
m<ADC =
B m<ADB + m<BDC
Same Process In Proof
E
1
Given: m<1 = m<2
Prove: m<EBD= m<CBA B
Statement
m<1 = m<2
m<3 = m<3
m<1 + m<3 = m<2 + m<3
m<1 + m<3 = m<EBD
m<2 + m<3 = m<CBA
m<EBD = m<CBA
C
3
D
2
A
Reason
Given
Reflexive
Add/Subt Prop
AAP
Substitution
Angle Theorems
Supplement Theorem – If two angles are
Linear Pair, then they are supplementary.
2
1
<1 & <2 are LP,
<1 & <2 are Supp.
Complement Theorem – If the non
common sides of two adjacent angles are
perpendicular, then the two adjacent
angles are complementary.
3
4
<3 & <4’s non-common sides are
| so, <3 & <4 are comp.
Angle Theorems (Con’t)
Angles Supplementary to the same angle
(or congruent angles) are congruent.
1
2
3
<1 & <2 are Supp. <2 & <3 are Supp.
<1 is congruent to <3.
Angles Complementary to the same angle
(or congruent angles) are congruent.
Angle Theorems (Con’t)
Vertical Angle Theorem – If two angles are
vertical angles, then they’re congruent.
2
1
3
4
1  3
2  4
Right Angle Theorems
Perpendicular Lines intersect to form four
right angles.
All Right Angles are congruent.
Perpendicular Lines form congruent,
adjacent angles.
If two angles are congruent and
supplementary, then they’re right angles.
If two congruent angles form a linear pair,
then they’re right angles.