Proof Techniques There are some of the most common proof techniques. 1. Direct Proof 2. Proof by Contradiction 3. Proof by Contapositive 4. Proof by Cases

1. PROOF TECHNIQUES
Presented in The Course of Tutorial Fundamental First Semester in
Academic Year 2017/2018
By : Group 3
1. Nun Hafizah Nur/1711440005
2. Nurul Fadhilah Alimuddin/1711440008
3. Fadhillah Suhardi/1711441006
4. Nurmasita/1711441011
5. Meutiah Nahrisyah/1711441014
6. Elvira/1711442006
Program of ICP Mathematic Education Mathematic Department
UNIVERSITAS NEGERI MAKASSAR
2017

2. PROOF TECHNIQUES
There are some of the most common proof techniques. In this chapter, we will examine about :
1. Direct proof
This is the simplest and easiest method of proof avaible to us. There are only two steps to a direct
proof :
- Assume that P is true.
- Use P to show Q must be true.
Direct proof of 𝑃( 𝑥) → 𝑄(𝑥) for all 𝑥 ∈ 𝐷
 Assume that 𝑃(x) is true for an arbitrary 𝑥 ∈ 𝐷,
 And show that 𝑄(𝑥) is true for this 𝑥.
Properties of Integers : In order to illustrate this type of proof we need to know the following :
- The negative integers of an integers is an integer.
- The sum and difference of two integers is an integer.
- The product of two integers is an integer.
- An integer 𝑛 is even if 𝑛 = 2𝑘 for some integer 𝑘.
- An integers 𝑛 is odd if 𝑛 = 2𝑘 + 1 for some integer 𝑘
*Example :
Theorem 1 : If 𝒏 is an odd integer, then 𝟓𝒏 + 𝟑 is an even integer.
Proof : Assume that 𝑛 is an odd integer, then 𝑛 = 2𝑘 + 1 for 𝑘 ∈ ℤ
Then, 5𝑛 + 3
= 5(2𝑘 + 1) + 3
= 10𝑘 + 8
= 2(5𝑘 + 4)
= 2𝑚, where 𝑚 = (5𝑘 + 4) ∈ ℤ
It follows that 5𝑛 + 3 is an even integer when 𝑛 is an odd integer.
Theorem 2 : If 𝒏 is an odd integer, then 𝟒𝒏 𝟑 + 𝟐𝒏 − 𝟏 is odd.
Proof : Assume that 𝑛 is odd. So, 𝑛 = 2𝑘 + 1 for 𝑘 ∈ ℤ
Then, 4𝑛3 + 2𝑛 − 1
= 4(2𝑘 + 1)3 + 2(2𝑘 + 1) − 1
= 4(4𝑘3 + 12𝑘2 + 6𝑘 + 1) + 4𝑘 + 2
= 16𝑘3 + 48𝑘2 + 28𝑘 + 5

3. = 2(8𝑘3 + 24𝑘2 + 14𝑘 + 2) + 1
= 2𝑚 + 1, where 𝑚 = (8𝑘3 + 24𝑘2 + 14𝑘 + 2) ∈ ℤ
It follows that 4𝑛3 + 2𝑛 − 1 is odd when 𝑛 is an odd integers.
2. Proof by Contradiction
The proof by contradiction is grounded in the fact that any preposition must be either true or
false, but not both true and false at the same time.We arrive at a contradiction when we are able
to demonstrate that a statement is both simultaneously true and false.
The method of proof by contradiction.
- Assume that 𝑃 is true
- Assume that ¬𝑄 is true
- Use that 𝑃 and ¬𝑄 to demonstrate a contradiction
*Example :
Theorem 1 : Showthat if x is a prime number not equal to 3, then
𝒙
𝟑
is not an integer.
Proof : Assume that
𝑥
3
is an integer.
Then,
𝑥
3
= 𝑛
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.
Theorem 2 : √ 𝟐 is an irrational number.
Proof : Let us assume that √2 is not an irrational number. Then √2 is a rational number, so we
can write,
√2 =
𝑎
𝑏
Where 𝑎 and 𝑏 are integers and 𝑏 ≠ 0. Moreover, we may assume that the fraction
𝑎
𝑏
is in the
lowest term, that is 𝑎 and 𝑏 have no common factors other than 1. (If 𝑎 and 𝑏 have common
factors other than 1, then we can cancelthose common factors and get the fraction in which the
numerator and denominator has no common factors other than 1).
So,
√2 =
𝑎
𝑏

4. → (√2)2 = (
𝑎
𝑏
)2
→ 2 =
𝑎2
𝑏2
→ 𝑎2 = 2𝑏2
→ 𝑎2 is an even integers
Because 𝑎2is an even integer, we must have a an even integer.Therefore, we can write 𝑎 = 2𝑛 for
some integer 𝑛. This implies that 𝑎2 = 4𝑛2. We now subtitute this value of 𝑎2 into 𝑎2 = 2𝑏2 to
obtain
𝑎2 = 2𝑏2 = 4𝑛2
This implies that,
𝑏2 = 2𝑛2
And so 𝑏2 is even. We can now conclude that 𝑏 is even. Thus, we have proved that both 𝑎 and 𝑏
are even and so have 2 as a common factors. This contridacts our assumption that 𝑎 and 𝑏 have
no common factors other than 1. We have now arrivedat a contradiction. Consequently, we can
conclude that √2 is an irrational number.
3. Proof by contrapositive
The statement 𝑃 → 𝑄 is equivalent to ¬𝑄 → ¬𝑃
- Assume ¬𝑄 is true
- Show that ¬𝑃 must be true
- Observe that 𝑃 → 𝑄 by contraposition
Proof by contrapositive can be an effective approach when a traditional irect proof is tricky, or it
can be a different way to think about the substance of a problem.
*Example :
Theorem 1 : Let 𝒙 ∈ ℤ. Then 𝟏𝟏𝒙 − 𝟕is even if 𝒙 is odd.
Proof : Assume that 𝑥 is even. Then 𝑥 = 2𝑘 for some 𝑘 ∈ ℤ
Now, 11𝑥 − 7
=11(2𝑘) − 7
=22𝑘 − 7
=2(11𝑘 − 4) + 1

5. =2𝑝 + 1, where 𝑝 = 11𝑘 − 4 ∈ ℤ
Hence 11𝑘 − 7 is odd. Therefore if 11𝑥 − 7 is even then 𝑥 is odd is true by the technique of
proof by contrapositive.
Theorem 2 : Let 𝒙 ∈ ℤ. Then 𝒙 𝟐is even if 𝒙 is even.
Proof : Assume that 𝑥 is odd. Then 𝑥 = 2𝑘 + 1 for some 𝑘 ∈ ℤ
Now, 𝑥2 = (2𝑘 + 1)2
= 4𝑘2 + 4𝑘 + 1
= 2(2𝑘2 + 2) + 1
= 2𝑚 + 1, where 𝑚 = 2𝑘2 + 2 ∈ ℤ
Hence 𝑥2is odd. Therefor if 𝑥2is even then 𝑥 is even is true by the techniques of proof by
contrapositive.
4. Proof by cases
Sometimes it’s hard to prove the whole theorem at once so you split the proof into several cases,
and prove the theorem separately for each case.
Definition:
- Two integers 𝑥 and 𝑦 are of the same parity if 𝑥 and 𝑦 are both even or both odd
- Two integers 𝑥 and 𝑦 are of opposite parity if 𝑥 is even and 𝑦 is odd, or 𝑥 is odd and 𝑦 is
even.
*Example :
Theorem 1: Let 𝒙, 𝒚 ∈ ℤ. Then 𝒙 and 𝒚 are ofthe same parity if 𝒙 + 𝒚 is an even.
Proof strategy : There are two implications to prove here :
1). If 𝑥 and 𝑦 are of the same parity then 𝑥 + 𝑦 is an even.
2). If 𝑥 + 𝑦 is an even then 𝑥 and 𝑦 are of the same parity.
Proof 1: Assume that 𝑥 and 𝑦 are of the same parity. Consider two cases.
Case 1, 𝑥 and 𝑦 both even.
Then, 𝑥 = 2𝑘, 𝑦 = 2𝑙 for some 𝑘, 𝑙 ∈ ℤ.
So, 𝑥 + 𝑦
=2𝑘 + 2𝑙
=2( 𝑘 + 𝑙)

6. =2𝑝, where 𝑝 = 𝑘 + 𝑙 ∈ ℤ.
Therefore, 𝑥 + 𝑦 is even.
Case 2, x and y both odd.
Then, 𝑥 = 2𝑘 + 1, 𝑦 = 2𝑙 + 1 for some 𝑘, 𝑙 ∈ ℤ.
So, 𝑥 + 𝑦
=2𝑘 + 1 + 2𝑙 + 1
=2( 𝑘 + 𝑙 + 1)
=2𝑞, where 𝑞 = 𝑘 + 𝑙 + 1 ∈ ℤ.
Therefore, 𝑥 + 𝑦 is even.
Theorem 2 : Let 𝒙, 𝒚 ∈ ℤ. Then 𝒙 and 𝒚 are ofthe opposite parity if 𝒙 + 𝒚 is an odd.
Proof :Assume that 𝑥 and 𝑦 are of opposite parity. Consider two cases.
Case 1, 𝑥 is even and 𝑦 is odd.
Then 𝑥 = 2𝑘, 𝑦 = 2𝑙 + 1 for some 𝑘, 𝑙 ∈ ℤ.
So, 𝑥 + 𝑦
= 2𝑘 + 2𝑙 + 1
= 2( 𝑘 + 𝑙) + 1
= 2𝑚 + 1, where 𝑚 = (𝑘 + 𝑙) ∈ ℤ.
Hence, 𝑥 + 𝑦 is odd.
Case 2, 𝑥 is odd and 𝑦 is even.
Then, 𝑥 = 2𝑘 + 1, 𝑦 = 2𝑙 for some 𝑘, 𝑙 ∈ ℤ.
So, 𝑥 + 𝑦
= 2𝑘 + 1 + 2𝑙
= 2( 𝑘 + 𝑙) + 1
= 2𝑛 + 1, where 𝑛 = (𝑘 + 𝑙) ∈ ℤ.
Hence, 𝑥 + 𝑦 is odd.