1. Methods of Proof Leo Cheung

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. 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.

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

7. Contradiction For all prime numbers a, b and c, a 2 + b 2 ≠ c 2

8. Contradiction - Answer Assume the contrary, there exist prime number a, b and c such that a 2 + b 2 = c 2 Then we get »a 2 = c 2 – b 2 »a 2 = (c-b)(c+b)

9. Contradiction - Answer Since a is prime There are 3 cases Casec+bc-b 1aa 2a2a2 1 31a2a2 Implies b = 0, Contradiction Implies b<=1, c<=1, Contradiction Implies c = 3, b = 2, but 2+3=5 is not perfect square. Contradiction

10. Proof by cases |x||y| = |xy| for all real numbers x, y Case 1: x>=0, y>=0 »|x||y| = xy, |xy| = xy Case 2: x>=0, y<0 »|x||y| = x(-y), |xy| = -(xy) Case 3: x =0 »|x||y| = (-x)y, |xy| = -(xy) Case 4: x<0, y<0 »|x||y| = (-x)(-y), |xy| = xy They are equal in every cases.

11. Proof by Induction 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

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

13. 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

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