4. Vocabulary
1. Indirect Reasoning: Assuming that the conclusion
of a conditional is false, then showing this
assumption leads to a contradiction
2. Indirect Proof:
3. Proof by Contradiction:
5. Vocabulary
1. Indirect Reasoning: Assuming that the conclusion
of a conditional is false, then showing this
assumption leads to a contradiction
2. Indirect Proof: Prove a statement to be true by
assuming that which you are trying to prove to
be false, then show that it leads to a
contradiction
3. Proof by Contradiction:
6. Vocabulary
1. Indirect Reasoning: Assuming that the conclusion
of a conditional is false, then showing this
assumption leads to a contradiction
2. Indirect Proof: Prove a statement to be true by
assuming that which you are trying to prove to
be false, then show that it leads to a
contradiction
3. Proof by Contradiction: Another name for indirect
proof
8. Steps to Write an Indirect
Proof
1. Identify what you are trying to prove, then
assume the conclusion to be false (opposite).
9. Steps to Write an Indirect
Proof
1. Identify what you are trying to prove, then
assume the conclusion to be false (opposite).
2. Follow a logical argument based on your false
assumption to find a contradiction in your new
assumption.
10. Steps to Write an Indirect
Proof
1. Identify what you are trying to prove, then
assume the conclusion to be false (opposite).
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.
11. Example 1
State 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.
12. Example 1
State the assumption necessary to start an indirect
proof for each statement.
a. EF is not a perpendicular bisector.
EF is a perpendicular bisector.
b. 3x = 4y + 1
c. If B is the midpoint of LH and LH = 26, then BH is
congruent to LB.
13. Example 1
State the assumption necessary to start an indirect
proof for each statement.
a. EF is not a perpendicular bisector.
EF is a perpendicular bisector.
b. 3x = 4y + 1 3x ≠ 4y + 1
c. If B is the midpoint of LH and LH = 26, then BH is
congruent to LB.
14. Example 1
State the assumption necessary to start an indirect
proof for each statement.
a. EF is not a perpendicular bisector.
EF is a perpendicular bisector.
b. 3x = 4y + 1 3x ≠ 4y + 1
c. If B is the midpoint of LH and LH = 26, then BH is
congruent to LB.
BH is not congruent to LB.
15. Example 2
Write an indirect proof to show that if −2x + 11 < 7,
then x > 2.
16. Example 2
Write an indirect proof to show that if −2x + 11 < 7,
then x > 2.
Assume that x ≤ 2.
17. Example 2
Write an indirect proof to show that if −2x + 11 < 7,
then x > 2.
Assume that x ≤ 2.
Then −2(2) + 11 < 7
18. Example 2
Write an indirect proof to show that if −2x + 11 < 7,
then x > 2.
Assume that x ≤ 2.
Then −2(2) + 11 < 7
−4 + 11 < 7
19. Example 2
Write an indirect proof to show that if −2x + 11 < 7,
then x > 2.
Assume that x ≤ 2.
Then −2(2) + 11 < 7
−4 + 11 < 7
7 < 7
20. Example 2
Write an indirect proof to show that if −2x + 11 < 7,
then x > 2.
Assume that x ≤ 2.
Then −2(2) + 11 < 7
−4 + 11 < 7
7 < 7
Not true
21. Example 2
Write an indirect proof to show that if −2x + 11 < 7,
then x > 2.
Assume that x ≤ 2.
Then −2(2) + 11 < 7
−4 + 11 < 7
7 < 7
Not true
Also, −2(1) + 11 < 7
22. Example 2
Write an indirect proof to show that if −2x + 11 < 7,
then x > 2.
Assume that x ≤ 2.
Then −2(2) + 11 < 7
−4 + 11 < 7
7 < 7
Not true
Also, −2(1) + 11 < 7
−2 + 11 < 7
23. Example 2
Write an indirect proof to show that if −2x + 11 < 7,
then x > 2.
Assume that x ≤ 2.
Then −2(2) + 11 < 7
−4 + 11 < 7
7 < 7
Not true
Also, −2(1) + 11 < 7
−2 + 11 < 7
9 < 7
24. Example 2
Write an indirect proof to show that if −2x + 11 < 7,
then x > 2.
Assume that 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
25. Example 2
Write an indirect proof to show that if −2x + 11 < 7,
then x > 2.
Assume that 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.
26. Example 3
Write an indirect proof to show that if x is a prime
number not equal to 3, then x/3 is not an integer.
27. Example 3
Write an indirect proof to show that if x is a prime
number not equal to 3, then x/3 is not an integer.
Assume x/3 is an integer.
28. Example 3
Write an indirect proof to show that if x is a prime
number not equal to 3, then x/3 is not an integer.
Assume x/3 is an integer.
Then,
x
3
= n.
29. Example 3
Write an indirect proof to show that if x is a prime
number not equal to 3, then x/3 is not an integer.
Assume x/3 is an integer.
Then,
x
3
= n.
This means x = 3n.
30. Example 3
Write an indirect proof to show that if x is a prime
number not equal to 3, then x/3 is not an integer.
Assume x/3 is an integer.
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. By contradiction, x/3 is not an integer.
31. Example 4
Prove by contradiction.
K
J L
8 5
7
Given: ∆JKL with side lengths 5, 7, and 8.
Prove: m∠K < m∠L
32. Example 4
Prove by contradiction.
K
J L
8 5
7
Given: ∆JKL with side lengths 5, 7, and 8.
Prove: m∠K < m∠L
Assume m∠K ≥ m∠L.
33. Example 4
Prove by contradiction.
K
J L
8 5
7
Given: ∆JKL with side lengths 5, 7, and 8.
Prove: m∠K < m∠L
Assume m∠K ≥ m∠L.
Then, by angle-side relationships, JL ≥ JK.
34. Example 4
Prove by contradiction.
K
J L
8 5
7
Given: ∆JKL with side lengths 5, 7, and 8.
Prove: m∠K < m∠L
Assume m∠K ≥ m∠L.
Then, by angle-side relationships, JL ≥ JK.
This mean that 7 ≥ 8, which is not true.
35. Example 4
Prove by contradiction.
K
J L
8 5
7
Given: ∆JKL with side lengths 5, 7, and 8.
Prove: m∠K < m∠L
Assume m∠K ≥ m∠L.
Then, by angle-side relationships, JL ≥ JK.
This mean that 7 ≥ 8, which is not true.
So the assumption m∠K ≥ m∠L is false, so by
contradiction, m∠K < m∠L
