more about first order logic and methods of proof
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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 - PowerPoint PPT PresentationTRANSCRIPT
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
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 - 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 - 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 - 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 - 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 - 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
• 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
• 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