More about First Order Logic and Methods of Proof

Leo Cheung

A Quick Review

• First Order Logic

• Proof by contrapositive

• Proof by contradiction

• Proof by cases

Formulating statements

• Give the predicate» prime(x): x is prime

• Write the predicate in first order logic» odd(x): x is odd

• Write the following in first order logic» All odd numbers greater than 7 are the sum of

two odd primes.» There are infinite # of solutions for sin(x)=1

Formulating statements

• odd(x):

• All odd numbers greater than 7 are the sum of two odd primes.

• There are infinite # of solutions for sin(x)=1

Quantifiers

• Loves(x,y): x loves y

• Are they equivalent?

vs

vs

vs

Quantifiers

• Loves(x,y): x loves y

• Are they equivalent?

≠

≡

≡

Prove or Disprove

Prove or Disprove

Counterexample: a=20, b=2

Counterexample: n=11

Argument with quantifiers

• Nothing intelligible ever puzzles me.

• Logic puzzles me.

Argument with quantifiers

Logic is not intelligible

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

• Consider 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

Prove n2-2 is not divisible by 4 for all integer n.

Contradiction - Answer

• Assume the contrary, then» There exists x2 – 2 = 4m» x must be even, write x=2k» 4k2-2 = 4m» Contradiction!

• 4k2-2 is not divisible by 4• 4m is divisible by 4

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!

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

Proof by cases

• |x||y| = |xy| for all real numbers x, y

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, y>=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.

Proof by cases

Prove max(x,y) + min(x,y) = x+y

Proof by cases

Prove max(x,y) + min(x,y) = x+y

• Case 1: x ≥ y» max(x,y) + min(x,y) = x + y

• Case 2: x < y» max(x,y) + min(x,y) = y + x = x + y

Exercises in textbook

• 2.3: 11 18 20 22 28 36 53

• 2.4: 3 5 9 10 31 32

• 3.1: 11 17 50• 3.2: 13 33 34• 3.3: 17 19 29• 3.5: 15• 3.6: 3 5 10 21 27• 3.7: 7 9

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