Lecture on 4 methods of Proof: Direct Proof, Proof by Contrapositive, Proof by Contradiction, Proof by Induction

1. Methods of Proof
Harshit Kumar

2. A Quick Review
• Direct proof
• Proof by contrapositive
• Proof by contradiction
• Proof by induction

3. Contrapositive
• If 3k+1 is even, then k is odd.
• Try to prove its contrapositive
• If k is even, 3k+1 is odd
» Let k = 2n
» 3k + 1 = 3(2n)+1 = 2(3n)+1 which is odd

4. Contrapositive
• For any integers a and b, a+b<=15 implies
that a<8 or b<8
• Try to prove its contrapositive
• If a>= 8 and b>=8, then a+b>15
» a + b >= 8 + 8 = 16 > 15

5. Contrapositive
• Prove by contrapositive
» If n2 is divisible by 3, then n is divisible by 3

6. Contrapositive - Answer
• Contrapositive:
» If n is not divisible by 3, then n2 is not divisible
by 3.
• Case 1: n=3k+1
– n2 = (3k+1)2 = 9k2 +6k+1 which is not divisible by 3
• Case 1: n=3k+2
– n2 = (3k+2)2 = 9k2+12k+4 which is not divisible by 3
» Hence n2 is not divisible by 3

7. Contradiction
• If 40 coins are distributed among 9 bags,
so that each bag contains at least one
coin. Then at least 2 bags contain the
same number of coins.

8. Contradiction - Answer
• Assume the contrary, every bag contain
different number of coins.
• Minimum number of coin required =
» 1 + 2 + 3 + …. + 9 = 45 > 40
» Contradiction!

9. Contradiction
Prove is irrational
Given: If n2 is divisible by 3, then n is divisible by 3

10. Contradiction - Answer
Assume is rational, then we can write
where p and q do not have common factor > 1
p2 is divisible by 3, so p is divisible by 3
q2 is divisible by 3, so q is divisible by 3
Since p and q are both divisible by 3, it contradict with our
assumption. Hence is irrational.

11. Contradiction
• For all prime numbers a, b and c,
a2 + b2 ≠ c2

12. Contradiction - Answer
• Assume the contrary, there exist prime
number a, b and c such that a2 + b2 = c2
• Then we get
» a2 = c2 – b2
» a2 = (c-b)(c+b)

13. Contradiction - Answer
• Since a is prime
• There are 3 cases
Case c+b c-b
1 a a Implies b = 0, Contradiction
Implies c = 3, b = 2, but 2+3=5 is not perfect
2 a2 1 square. Contradiction
3 1 a2 Implies b<=1, c<=1, Contradiction

14. Contradiction
here are no positive integer solution x and y
for
2
- y2 = 1

15. Contradiction - Answer
• Assume the contrary, there exist positive x
and y such that x2 – y2 = 1
• Then we get
» (x+y)(x-y) = 1
» (x+y)=1 or (x+y)=-1
» Contradiction!

16. Proof by Induction - 1
n +1
∑ i ⋅ 2i = n 2 n + 2 + 2
i =1
For all integer n >= 0
Base cases: n=0, L.S.= 2 =R.S.
Assume f(k) is true, i.e
When n = k+1

17. Proof by Induction - 2
7n – 1 is divisible by 6, for all n >= 1.
Base case: n=1 71-1=6 which is divisible by 6.
Assume f(k) is true, i.e. 7k-1=6m for some m
When n = k+1
7k+1-1 = 7(6m+1) - 1 = 42m + 6 = 6(7m+1)
So by induction 7n – 1 is divisible by 6, for all n >= 1.

18. Proof by Induction - 3
• For which positive integers n satisfy
2n+3≤2n? Prove your answer using
induction.

19. Proof by Induction - 3
• You can check the statement is wrong for
n=1,2,3.
• When n=4, 2(4)+3=11<16=24
• Assume 2k+3<2k (k≥4)
• When n=k+1
Hence by induction 2n+3<2n for all integer n≥4

20. Exercises
• Following exercises can be found in
textbook
• 3.1: 11
• 3.2: 15
• 3.7: 3 10 17 24
• 4.2: 6 13
• 4.3: 8 19

21. END