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- 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. ContradictionProve is irrational Given: If n2 is divisible by 3, then n is divisible by 3
- 10. Contradiction - AnswerAssume 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. Contradictionhere are no positive integer solution x and yfor 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 >= 0Base cases: n=0, L.S.= 2 =R.S.Assume f(k) is true, i.eWhen n = k+1
- 17. Proof by Induction - 27n – 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 mWhen 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+1Hence 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

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