1. Section 2.1 Proof Techniques Introduce proof techniques:   o        Exhaustive Proof: to prove all possible cases, Only if it is about a finite collection   o        Direct Proof: to prove PQ is true, Assume P is true, and deduce Q (like what we have done in Chap1)   o        Contraposition: to prove P Q is true By proving Q P  
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2. k[ P(k) true  P(k+1) true] P(n) true,n
o       Contradiction: to prove PQ is true, (indirect proof) Assume both the hypothesis and the negation of the conclusion are true, then try to deduce some contradiction from this assumption.   o        Counterexample: to disprove something   o        Induction: to prove that P(n) is true, n Use the principle of mathematical induction:   P(1) is true
k[ P(k) true  P(k+1) true] P(n) true,n
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3. n n! 2n 1 2 4 3 6 8 · Proof by Exhaustion:
eg: “For any positive integer n less than or equal to 3, n! < 2n” n n! 2n 1 2 4 3 6 8
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4. eg:. “If an integer between 5and 15 is divisible by 6, then it
eg: “If an integer between 5and 15 is divisible by 6, then it is also divisible by 3”
(Note: If the above problem is about all integers, then we cannot use exhaustive proof)
Number
Divisible by 6
Divisible by 3 5 No 6
Yes: 6 = 1X6
Yes: 6 = 2X3 7 8 9
Yes 10 11 12
Yes: 12 = 2X6
Yes: 12 = 4X3 13 14 15
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5. eg: “Every integer less than 10 is bigger than 5”
Proof by Counterexample:
eg: “Every integer less than 10 is bigger than 5”
To prove the statement is not true,
find a counterexample,
integer = 4 < 10, but not > 5.
eg: disprove that “the sum of any three consecutive integers is even.”
counterexample : 2+3+4=9
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6. eg: “for all x [x is divisible by 6  x is divisible by 3]”
·        Direct Proof:
eg: “for all x [x is divisible by 6  x is divisible by 3]”
X is divisible by 6
 x = k*6, for some integer k (hypothesis)
 x = k*2*3
 x = (k*2)*3
(k*2) is some integer x is divisible by 3
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7. eg: “the product of two even integers is even”
Let x = 2m, y = 2n for some integer m, n (hypothesis)
 xy = (2m)(2n) = 2(2mn), which is even.
eg: “the sum of two odd integers is even”
Let x = 2m+1, y = 2n+1 for some integer m,n (hypothesis)
 x + y = 2m + 2n+ 2 = 2(m+n+1),
where m+n+1 is an integer
x+y is even.
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8. Prove: n even  n2 even (P  Q  Q  P)
·        Contraposition:
eg: “If the square of an integer is odd, then the integer must be odd.”
“n2 odd  n odd”
Prove: n even  n2 even (P  Q  Q  P)
Let n = 2m for some integer m (hypothesis)
 n2 = n * n = 2m*2m = 2(2m2)
 n2 is even.
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9. · eg: “xy is odd iff both x and y are odd.”
() by direct proof:
Let x = 2m+1, y=2n+1 for some m, n  integers
 xy = (2m+1)(2n+1) = 4mn + 2m + 2n +1
= 2(2mn+m+n) +1
 since 2mn + m + n is an integer, xy is odd.
() by contraposition:
to prove that : x even or y even, then xy even
case1 x even, y odd: Let x = 2m, y = 2n+1
xy = 2(2mn + m), which is even
case2 x odd, y even: similar to case1.
case3 x even, y even: Let x = 2m, y = 2n
xy = 2(2mn), which is even
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10. · Proof by Contradiction: Since (P  Q F)  (P  Q) is a tautology,
It is sufficient to prove P  Q F, then P  Q is true.
eg: “If a number added to itself gives itself, then the number is 0.”
i.e. “x + x = x, then x = 0”:
Assume x + x = x and x  0
 2x = x and x  0
 2 = 1, which is a contradiction
the assumption must be wrong
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11.  4 + 4 – 3 = 0 or 5 = 0, which is a contradiction.
·eg: “if x2 + 2x – 3 = 0, then x  2”
Let x2 + 2x – 3 = 0 and assume x = 2,
 – 3 = 0 or 5 = 0, which is a contradiction.
( by direct proof: if x2 + 2x – 3 = 0
 (x + 3)(x –1)= 0
 x = -3 or x = 1
 x  2 )
(by contraposition: if x = 2
 x2 + 2x – 3 = 5  0 )
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