geometry section 5-4 1112

Section 5-4Indirect Proof

Tuesday, March 13, 2012

Essential Questions

How do you write indirect algebraic proofs?

How do you write indirect geometric proofs?


Vocabulary

1. Indirect Reasoning:

2. Indirect Proof:

3. Proof by Contradiction:


Vocabulary

1. Indirect Reasoning: Assuming that the conclusion of a conditional is false, then showing this assumption leads to a contradiction

2. Indirect Proof:



Vocabulary


2. Indirect Proof: Prove a statement to be true by assuming that which you are trying to prove to be false, then show that is leads to a contradiction



Vocabulary


2. Indirect Proof: Prove a statement to be true by assuming that which you are trying to prove to be false, then show that is leads to a contradiction

3. Proof by Contradiction: Another name for indirect proof


Steps to Write an Indirect Proof


Steps to Write an Indirect Proof1. Identify what you are trying to prove, then assume the conclusion to be false.



2. Follow a logical argument based on your false assumption to find a contradiction in your new assumption.



2. Follow a logical argument based on your false assumption to find a contradiction in your new assumption.

3. Identify the contradiction, thus proving that the original conclusion must be true.


Example 1State the assumption necessary to start an indirect

proof for each statement.

a. EF is not a perpendicular bisector.

b. 3x = 4y + 1

c. If B is the midpoint of LH and LH = 26, then BH is congruent to LB.





EF is a perpendicular bisector.

b. 3x = 4y + 1







b. 3x = 4y + 1 3x ≠ 4y + 1







b. 3x = 4y + 1 3x ≠ 4y + 1


BH is not congruent to LB.


Example 2Write an indirect proof to show that if −2x + 11 < 7,

then x > 2.



then x > 2.

Assume that x ≤ 2.



then x > 2.


Then −2(2) + 11 < 7



then x > 2.


Then −2(2) + 11 < 7

−4 + 11 < 7



then x > 2.


Then −2(2) + 11 < 7

−4 + 11 < 7

7 < 7



then x > 2.


Then −2(2) + 11 < 7

−4 + 11 < 7

7 < 7

Not true



then x > 2.


Then −2(2) + 11 < 7

−4 + 11 < 7

7 < 7

Not true

Also, −2(1) + 11 < 7



then x > 2.


Then −2(2) + 11 < 7

−4 + 11 < 7

7 < 7

Not true

Also, −2(1) + 11 < 7

−2 + 11 < 7



then x > 2.


Then −2(2) + 11 < 7

−4 + 11 < 7

7 < 7

Not true

Also, −2(1) + 11 < 7

−2 + 11 < 7

9 < 7



then x > 2.


Then −2(2) + 11 < 7

−4 + 11 < 7

7 < 7

Not true

Also, −2(1) + 11 < 7

−2 + 11 < 7

9 < 7

Also not true



then x > 2.


Then −2(2) + 11 < 7

−4 + 11 < 7

7 < 7

Not true

Also, −2(1) + 11 < 7

−2 + 11 < 7

9 < 7

Also not true

Both cases provide a contradiction, so therefore x > 2 must be true.


Example 3Write an indirect proof to show that if x is a prime

number, then x/3 is not an integer.




Assume x/3 is an integer.





Then,

x

3= n.





Then,

x

3= n.

This means x = 3n.





Then,

x

3= n.

This means x = 3n.

This cannot be true, because if x is prime, then 3 cannot be a factor of x as shown in the equation

x = 3n. Therefore, x/3 is not an integer.


Example 4Prove by contradiction.

K

J L

8 5

7Given: ∆JKL with side lengths 5, 7, and 8.

Prove: m∠K < m∠L



K

J L

8 5



Assume m∠K ≥ m∠L.



K

J L

8 5




Then, by angle-side relationships, JL ≥ JK.



K

J L

8 5





This mean that 7 ≥ 8, which is not true.



K

J L

8 5





This mean that 7 ≥ 8, which is not true.

So the assumption m∠K ≥ m∠L is false, so therefore m∠K < m∠L


Check Your Understanding

Explore problems #1-10 on page 354


Problem Set


Problem Set

p. 355 #11-29 odd, 43, 51, 53

“It's not that I'm so smart, it's just that I stay with problems longer.” – Albert Einstein


geometry section 5-4 1112

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