Download Strategies for Proofs

Strategies for Proofs
Examples
Strategies for Proofs
G. Carl Evans
University of Illinois
Summer 2013
Strategies for Proofs
Strategies for Proofs
Examples
Today
Practice with proofs
Become familiar with various strategies for proofs
Strategies for Proofs
Strategies for Proofs
Examples
Review: proving universal statements
Claim: For any integer a, if a is odd, then a2 is also odd.
Definition: integer a is odd iff a = 2m + 1 for some integer m
Suppose a ∈ Z is odd. Then by definition of odd, a = 2m + 1 for
some m ∈ Z. So
a2 = (2m + 1)2
= 4m2 + 4m + 1
= 2(2m2 + 2m) + 1
2m2 + 2m is an integer since m ∈ Z. So a2 is odd by definition
QED.
Strategies for Proofs
Strategies for Proofs
Examples
Proving Existential Statements
Claim: There exists a real number x, such that |x 3 | < x 2
Let x = 12 . Then |x 3 | = 81 and x 2 = 1/4. Since
an x where |x 3 | < x 2 QED.
Strategies for Proofs
1
8
<
1
4
there exists
Strategies for Proofs
Examples
Disproving Existential Statements
Claim to disprove: There exists a real x, x 2 − 2x + 1 < 0
Strategies for Proofs
Strategies for Proofs
Examples
Disproving Existential Statements
Claim to disprove: There exists a real x, x 2 − 2x + 1 < 0
In general, ¬(∃x, P(x)) ≡ ∀x, ¬P(x)
¬(∃x ∈ R, x 2 − 2x + 1 < 0)
∀x ∈ R, ¬(x 2 = 2x + 1 < 0)
∀x ∈ R, x − 2x + 1 ≥ 0
Let x be a real number. x 2 − 2x + 1 = (x − 1)2 . (x − 1)2 ≥ 0
since x − 1 is a real number and the square of a real number is
non-negative QED.
Strategies for Proofs
Strategies for Proofs
Examples
Disproving Universal Statements
Claim to disprove: For all real x, (x + 1)2 > 0
Strategies for Proofs
Strategies for Proofs
Examples
Disproving Universal Statements
Claim to disprove: For all real x, (x + 1)2 > 0
In general, ¬(∀x, P(x)) ≡ ∃x, ¬P(x)
¬(∀x ∈ R, (x + 1) > 0)
∃x ∈ R, (x + 1)2 ≤ 0)
If x = 1, (x + 1)2 = 0. So since 1 is a real there then exists a real
that proves the negation of the claim and thus the original claim is
disproved QED.
Strategies for Proofs
Strategies for Proofs
Examples
Proof By Cases
Claim: for every real x, if |x + 7| > 8, then |x| > 1
Suppose x ∈ R and |x + 7| > 8. Then there are two cases.
If x + 7 > 8, then x > 1, so |x| > 1.
If x + 7 < −8, then x < −15, so |x| > 1.
The conclusion holds for both cases QED.
Strategies for Proofs
Strategies for Proofs
Examples
Rephrasing Claims
Claim: There is no integer k, such that k is odd and k 2 is even.
¬(∃k ∈ Z, odd(k) ∧ even(k 2 )
∀k ∈ Z, ¬(odd(k) ∧ even(k 2 ))
∀k ∈ Z, ¬odd(k) ∨ ¬even(k 2 )
∀k ∈ Z, even(k) ∨ odd(k 2 )
At this point there is a proof by cases.
Strategies for Proofs
Strategies for Proofs
Examples
Proof By Contrapositive
Claim: For all integers a and b, (a + b ≥ 15) → (a ≥ ∨b ≥ 8)
Contrapositive: ∀a, b ∈ Z¬(a ≥ 8 ∨ b ≥ 8) → ¬(a + b ≥ 15)
∀a, b ∈ Z, a < 8 ∧ b < 8 → a + b < 15
To prove the claim we will prove the contrapositive,
∀a, b ∈ Z, a < 8 ∧ b < 8 → a + b < 15.
Suppose a ∈ Z, a < 8 and a ∈ Z, b < 8. Then a ≤ 7 and b ≤ 7.
So a + b ≤ 14 < 15 QED.
Strategies for Proofs
Strategies for Proofs
Examples
Proof strategies
Does this proof require showing that the claim holds for all
cases or just an example?
– Show all cases: prove universal, disprove existential
– Example: disprove universal, prove existential
Can you figure a straightforward solution?
– If so, sketch it and then write it out clearly, and youre done
If not, try to find an equivalent form that is easier
Divide into subcases that combine to account for all cases
– OR in hypothesis is a hint that this may be a good idea
Try the contrapositive
– OR in conclusion is a hint that this may be a good idea
More generally rephrase the claim: convert to propositional
logic and manipulate into something easier to solve
Strategies for Proofs
Strategies for Proofs
Examples
Claim: For integers j and k, if j is even or k is even, then jk is
even.
Definition: integer a is even iff a = 2m for some integer m
Strategies for Proofs
Strategies for Proofs
Examples
Claim: For all integers k, if 3k + 5 is even, then k is odd.
Strategies for Proofs
Strategies for Proofs
Examples
Disprove: For all real k, if k is rational, then
k3
k
Strategies for Proofs
is rational.
Strategies for Proofs
Examples
Alternate proof, by cases
Claim: For any real k, if k is rational, then k 2 is rational.
Definition: real k is rational iff k = m
n for some integers m and n
with n 6= 0
Strategies for Proofs
Strategies for Proofs
Examples
Claim: For all integers x, if x is odd, then x = 4k + 1 or
x = 4k − 1 for some integer k.
Strategies for Proofs