Inductive Reasoning and Conjecture 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 T F P Q PVQ T F

Truth Table (H) P Q R ~P ~PΛQ (~PΛQ)VR T F F 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 B Conjunction AΛB Blue Shade ~AΛ~B Disjunction AVB

Conditional Statements 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, then I will carry my umbrella. Hypothesis 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 Converse P → Q Q → P Write the conditional statement here. Switch the order of the Q and the P. Inverse Contrapositive ~P → ~Q ~Q → ~P Negate the conditional statement here. Negate the Converse statement here.

Inverse Contrapositive Example Conditional Converse If two angles are right angles, then they’re congruent. If two angles are congruent then they’re right angles. (P → Q) (Q → P) Inverse Contrapositive If two angles are not right angles, then they’re not congruent. If two angles are not congruent, then they’re not right angles. (~Q → ~P) (~P → ~Q)

Inverse Contrapositive Truth? Conditional Converse If two angles are right angles, then they’re congruent. If two angles are congruent then they’re right angles. (T) (F) Inverse Contrapositive If two angles are not right angles, then they’re not congruent. If two angles are not congruent, then they’re 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)

Postulates and Paragraph Proofs 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 Statement Reason x = 5 H Given 5 = x Symmetric Prop C The two column proof derives it’s name b/c it has two columns. Two columns are Statements and Reasons. Statement Reason x = 5 H Given 5 = x Symmetric Prop C These two statements satisfy the LOD of the If-Then form. Think of this in If-Then form.

Example #1 Given: 3(x – 2) = 42 Prove x = 16 Statement Reason H C 3x – 6 = 42 Distribution Prop H C 3x – 6 + 6 = 42 + 6 Add/Subt Prop H C 3x = 48 Substitution Prop H C 3x/3 = 48/3 Mult/Div Prop H C x = 16 Substitution Prop

Geometric Proofs Segments Angles Reflexive AB = AB Symmetric If AB = BC, then BC = AB Transitive If AB = BC and BC = CD then AB = CD Angles Reflexive m<ABC = m<ABC Symmetric If m<ABC = m<XYZ, then m<XYZ = m<ABC 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.

Proving Segment Relationships 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. B A C

Example of SAP Given: AB = CD Prove: AC = BD A B C D Statement Reason AB = CD Given BC = BC Reflexive Prop AB + BC = CD + BC Addition Prop AB + BC = AC BC + CD = BD SAP What do you need to do to AB to make it AC? What do you need to do to CD to make it BD? Substitution AC = BD

Another Example Given: AC = BD A B C D Prove: AB = CD Statement Reason AC = BD Given AB + BC = AC BC + CD = BD SAP AB + BC = BC + CD Substitution BC = BC Reflexive Prop AB = CD Add/Subt Prop

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 –

Proving Angle Relationships 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 m<ADC = m<ADB + m<BDC B D C

Same Process In Proof E C 1 Given: m<1 = m<2 Prove: m<EBD= m<CBA 3 D 2 A B Statement Reason m<1 = m<2 Given m<3 = m<3 Reflexive m<1 + m<3 = m<2 + m<3 Add/Subt Prop m<1 + m<3 = m<EBD m<2 + m<3 = m<CBA AAP m<EBD = m<CBA Substitution

Angle Theorems Supplement Theorem – If two angles are Linear Pair, then they are supplementary. Complement Theorem – If the non common sides of two adjacent angles are perpendicular, then the two adjacent angles are complementary. <1 & <2 are LP, <1 & <2 are Supp. 1 2 <3 & <4’s non-common sides are | so, <3 & <4 are comp. 3 4

Angle Theorems (Con’t) Angles Supplementary to the same angle (or congruent angles) are congruent. Angles Complementary 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.

Angle Theorems (Con’t) Vertical Angle Theorem – If two angles are vertical angles, then they’re congruent. 1 2 3 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.