2.1 conjectures and counterexamples

Chapter 2 – Reasoning and ProofAnswer Key

2.1 Conjectures and Counterexamples

Answers

1. True

2. COULD NOT IDENTIFY THE CONJECTURE

3. False, maybe a raccoon ate the bread with peanut butter instead of the chipmunk

4. False, it is possible that the other students, that are not her friends, in her geometry class are at different grade levels.

5. Answers vary

6. n = 1 would be a counterexample because . Recall that a whole number is>112

0, 1, 2, 3, … n, so 0 could also be a counterexample.

7. Counterexamples include: 21, 51, 81, 121, and 151

8. is one counterexample. Any positive improper fraction (where the numerator is greater than the denominator) could be a counterexample.

9. A triangle is a counterexample.

10. A girl that doesn’t like ice cream would be a counterexample.

CK12 Geometry Concepts 1


11. Not everyone takes choir in high school.

12. Obtuse angles do not have complementary angles.

13. 13, 14, 15 yearolds are teenagers and cannot drive yet.

14. Any negative integer could be a counterexample.

15. Equations can have any real number as a solution.



2.2 IfThen Statements

Answers

1. Hypothesis : 5 divides evenly into x.

Conclusion : x ends in 0 or 5.

2. Hypothesis : A triangle has three congruent sides.

Conclusion : It is an equilateral triangle.

3. Here, the “if” is in the middle of the statement, making the hypothesis the second half.

Hypothesis : Three points lie in the same plane.

Conclusion : The three points are coplanar.

4. Hypothesis : x = 3.

Conclusion : .

5. Hypothesis : You take yoga.

Conclusion : You are relaxed.

6. Hypothesis : You are a baseball player.

Conclusion : You wear a hat.

7. Hypothesis : I am 16 years old.

Conclusion : I will learn how to drive.



8. Hypothesis : You do your homework.

Conclusion : You can watch TV.

9. Hypothesis : The lines are parallel.

Conclusion : Alternate interior angles are congruent.

10. Hypothesis : You are a kid.

Conclusion : You like ice cream.

11. If it is Thursday, then Susie is eating pizza.

12. If you are Raychel, then you completed your homework

13. If it is a weekday, then Alex is going to school

14. If you are a student, then you have a math class.

15. If a shape is a square, then it has right angles



2.3 Converse, Inverse, and Contrapositive

Answers

1. If B is the midpoint of , then AB = 5 and BC = 5. This could be true, but we don’t know the length of AC. AB = BC, but we cannot say they are 5 without knowing the length of AC.

2. If AB 5 and BC 5, then B is not the midpoint of . Again, this could be true, but we don’t know AC. Also, A, B and C might not be collinear.

3. A counterexample would be if the points A, B, and C are not collinear, thus not creating

the line segment .

4. If AB 5 and BC 5, then B is not the midpoint of .

5. →qp

6. Contrapositive

7. Contrapositive

8. →qp

9. If a U.S. citizen can vote, then he or she is 18 or more years old.



If he or she is 18 or more years old, then he or she is a U.S. citizen that can vote

10. If a whole number is prime, then it has exactly two distinct factors.

If a whole number has exactly two distinct factors, then it is prime

11. If the points are collinear, then there is a line that contains the points.

If the points are on the same line, then they are collinear.

12. If , then .x 82 = 1 x = 9

If , then .x = 9 x 82 = 1

13. a) True

b) False, OR x = 4 x =− 4

c) False, let , then x =− 4 6x2 = 1

d) True

14. a) True

b) True

c) True

d) True

15. a) True

b) True

c) True



d) True



16. a) True

b) False, Any angle that is less than 90° is an acute angle.

c) False, Any angle that is less than 90° is an acute angle.

d) True

17. Answers vary

18. Answers vary



2.4 Inductive Reasoning from Patterns

Answers

1. 4 th figure: 9 dots, 10 th figure: 21dots

2. 4 th figure: 20 dots, 10 th figure: 110 dots

3. a)

b) There are two more points in each star than its figure number.

4. a) 10 triangles

b) 48

c) , for any positive integer .n2 n

5. 1) 20, 23

2) 107



3) , for any positive integer .n3 + 2 n



6. 1) 19, 24

2) 164

3) , for any positive integer .1 n 1 − 5 n

7. 1) 64, 128

2) 34,359,738,370

3) for any positive integer .,2n n

8. 1) 12, 1

2) 307

3) , for any positive integer 8 1n 7 − 1 .n

9. 1) , .76 8

7

2) 3635

3) , for any positive integer nn+1 .n

10. 1) .,2312

2714

2) .70139

3) .2n4n 1−



11. 1) 36, 49

2) 1225

3) for any positive integer ,n2 .n

12. 1) 20, 27

2) 860

3) , for any positive integer 2n(n+3) .n

13. 1) 21, 28

2) 861

3) , for any positive integer 2(n+1)(n+2) .n

14. 1) 105, 168

2) 34440

3) , for any positive integer 2n(n+1)(n+2) .n



15.

The points are collinear. The equation of this graph is x y = 5 − 2

16. Answers vary.



2.5 Deductive Reasoning

Answers

1. Law of Detachment

2. No conclusion

3. Law of Syllogism.

4. Law of Contrapositive.

5. Law of Detachment.

6. Law of Contrapositive

7. Law of Contrapositive.

8. Law of Syllogism

9. This is a logical argument, but it doesn’t make sense because we know that circles exist.



10. →q p

p

q ∴

11. →qp

q

p ∴˜

12. If I need a new watch battery, then I go to the mall. If I don’t shop, then I didn’t go to the mall, but since I went to the mall, then I shop. If I shop, then I will buy shoes. Since I shop, then I conclude that I bought shoes.

13. If Anna’s teacher gives, notes, Anna writes them down. If Anna writes down the notes, she can do the homework. If Anna’s parents don’t buy her ice cream, then she didn’t get an A on her test. If Anna didn’t get an A on her test, then she couldn’t do the homework. But since she did do her homework, she got an A on her test. Since she got an A on her test, we conclude that her parents bought her ice cream.

14. Inductive Reasoning

15. Deductive Reasoning





2.6 Truth Tables

Answers

1.

p q r r ∧q p p∧q)∨ r ( ˜ T T T F T T T T F T T T T F T F F F T F F T F T F T T F F F F T F T F T F F T F F F F F F T F T

2.

p q r q q∨r ∨( q∨r) p ˜ T T T F T T T T F F F T T F T T T T T F F T T T F T T F T T F T F F F F F F T T T T F F F T T T

3.

p q r r ∨ r q ˜ ∧(q∨ r) p ˜ T T T F T T T T F T T T T F T F F F T F F T T T F T T F T F F T F T T F F F T F F F F F F T T F



4. There are a lot more false cells in #3 than #1

5. When at least one is true., q, or r p

6.

p q r ∨q∨r p T T T T T T F T T F T T T F F T F T T T F T F T F F T T F F F F

7.

p q r ∨q p r p∨q)∨ r ( ˜ T T T T F T T T F T T T T F T T F T T F F T T T F T T T F T F T F T T T F F T F F F F F F F T T

8.

p q r p q p∧ q ˜ p∧ q)∧r ( ˜ ˜ T T T F F F F T T F F F F F T F T F T F F T F F F T F F F T T T F F F F T F T F F F



F F T T T T T F F F T T T F



9.

p q r p q p∨ q ˜ p∨ q)∧r ( ˜ ˜ T T T F F F F T T F F F F F T F T F T T T T F F F T T F F T T T F T T F T F T F T F F F T T T T T F F F T T T F

10. True

11. False

12. True

13. False

14. True

15. False



2.7 Properties of Equality and Congruence

Answers

1. 3x + 11 = 16

3x = 27 Subtraction PoE (subtract 11 from both sides)

x = 9 Division PoE (divide both sides by 3)

2. 7x – 3 = 3x – 35

4x – 3 = 35 Subtraction PoE

4x = 32 Addition PoE

x = 8 Division PoE

3. g + 1 = 19

g = 18 Subtraction PoE

g = 27 Multiplication PoE

4. MN = 5

MN = 10 Multiplication PoE

5. 5 m ∠ABC = 540°

m ∠ABC = 108° Division PoE



6. 10b – 2(b + 3) = 5b

10b – 2b + 6 = 5b Distributive Property

8b + 6 = 5b Combine like terms

6 = 3b Subtraction PoE

2 = b Division PoE

b = 2 Symmetric PoE (this step is not necessary, but helpful to see how the Symmetric property works)

7.

Multiplication PoE (multiplied everything by 12)

3y = 6 Subtraction PoE

y = 2 Division PoE

*Students could have also found a common denominator and would have ended up with the same answer.*

8.

3AB + 4 AB = 144 + 6AB Multiplication PoE (multiplied everything by 12)

7AB = 144 + 6AB Combine like terms

AB = 144 Subtraction PoE

9. y + z = x + y 10. CD = 5 11. Substitute y – 7 for x in the second equation: y – 7 = z + 4 12. 6x – 12 = y 13. Yes, they are collinear.



14. No, they are not collinear. Since the points are not collinear then there exist no midpoint

because no line segment exists that contains those points. 15. Since the 124°, then that angle has to be obtuse since all obtuse angles have∠NOP m =

an angle measurement greater than 180°.



2.8 TwoColumn Proofs

Answers

1.

Statement Reason 1. ∠MLN ≅ ∠OLP Given 2. m∠MLN = m∠OLP ≅ angles have = measures 3. m∠MLO = m∠MLN + m∠NLO m∠NLP = m∠NLO + m∠OLP

Angle Addition Postulate

4. m∠NLP = m∠NLO + m∠MLN Substitution 5. m∠MLO = m∠NLP Substitution 6. ∠NLP ≅ ∠MLO ≅ angles have = measures

2.

Statement Reason

1. , Given

2. ∠BED is a right angle ∠AEC is a right angle

⊥ lines create right angles

3. m∠BED = 90° m∠AEC = 90°

Definition of a right angle

4. m∠BED = m∠2 + m∠3 m∠AEC = m∠1 + m∠3

Angle Addition Postulate

5. 90° = m∠2 + m∠3 90° = m∠1 + m∠3

Substitution

6. m∠2 + m∠3 = m∠1 + m∠3 Substitution 7. m∠2 = m∠1 Subtraction PoE 8. ∠2 ≅ ∠1 ≅ angles have = measures



3.

Statement Reason 1. ∠1 ≅ ∠4 Given 2. m∠1 = m∠4 ≅ angles have = measures 3. ∠1 and ∠2 are a linear pair ∠3 and ∠4 are a linear pair

Given (by looking at the picture) could also be Definition of a Linear Pair

4. ∠1 and ∠2 are supplementary ∠3 and ∠4 are supplementary

Linear Pair Postulate

5. m∠1 + m∠2 = 180° m∠3 + m∠4 = 180°

Definition of supplementary angles

6. m∠1 + m∠2 = m∠3 + m∠4 Substitution 7. m∠1 + m∠2 = m∠3 + m∠1 Substitution 8. m∠2 = m∠3 Subtraction PoE 9. ∠2 ≅ ∠3 ≅ angles have = measures

4.

Statement Reason 1. Given 2. ∠1 and ∠2 are right angles ⊥ lines create right angles. 3. ∠1 ≅ ∠2 Right Angles Theorem

5.

Statement Reason 1. Given 2. ∠1 and ∠2 make a right angle ⊥ lines create right angles 3. m∠1 + m∠2 = 90° Definition of a right angle OR

Substitution PoE 4. ∠1 and ∠2 are complementary Definition of complementary angles

ANSWERS MAY VARY FOR PROBLEMS 6 13

6. , AHG and ∠CHE ∠ GHE and ∠AHC ∠

7. , , , , ≅ HE AH ≅ HC GH ≅ HP MH ≅ MA GM ≅ PC EP

8. , .PMA and ∠PMG ∠ MPE and ∠MPC ∠



9. AHC ∠

10. , MHA and ∠AHC ∠ GHE and ∠EHP ∠

11. GC

12. and GC AE

13. GHM and ∠AHM ∠

14. AGC and ∠EGC ∠

15. NOT COMPLETE

Statement Reason 1. ACH≅∠ECH ∠

. is on the interior of CH ACE ∠ 1. Given

2. ∠ACH ∠ECH m = m 2. 3. 3. Angle Addition Postulate 4. 4. Substitution 5. ∠ACE m∠ACH m = 2 5. 6. 6. Division PoE 7. 7.


2.1 conjectures and counterexamples

Documents