Methods of Proof

Harshit Kumar

A Quick Review

• Direct proof

• Proof by contrapositive

• Proof by contradiction

• Proof by induction

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

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

Contrapositive

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

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

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.

Contradiction - Answer

• Assume the contrary, every bag contain different number of coins.

• Minimum number of coin required =» 1 + 2 + 3 + …. + 9 = 45 > 40» Contradiction!

Contradiction

Prove is irrational

Given: If n2 is divisible by 3, then n is divisible by 3

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.

Contradiction

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

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)

Contradiction - Answer

• Since a is prime• There are 3 cases

Case c+b c-b

1 a a

2 a2 1

3 1 a2

Implies b = 0, Contradiction

Implies b<=1, c<=1, Contradiction

Implies c = 3, b = 2, but 2+3=5 is not perfect square. Contradiction

Contradiction

There are no positive integer solution x and y for

x2 - y2 = 1

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!

Proof by Induction - 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

222 21

1

nn

i

i ni

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.

Proof by Induction - 3

• For which positive integers n satisfy 2n+3≤2n? Prove your answer using induction.

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

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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