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Definitions
- Even An integer n is even, iff n =2 k for some integer k. - n is even ↔ ∃ k such that n = 2 k
- Odd An integer n is odd, iff n =2 k +1 for some integer k. - n is odd ↔ ∃ an integer k such that n = 2 k +
- Divisible An integer n is divisible by m , iff n and m are integers such that that there is an integer k such that mk = n - n is divisible by m ↔ ∃ an integer k such that mk = n
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The sum of two even integers is even
the conclusion (end)
unwind definitions
an integer
1. Rewrite as a condition (using if, then)
2. Write the hypothesis (beginning) and
- For beginning: establish notation and
- For end: unwind definitions backwards
3. … and … wait …
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The sum of two even integers is even
- [conditional]
Proof:
- Let x and y be integers [hypothesis]
[conclusion]
y is even, so 2|y
There is an integer, e.g., b, with y=2b [as above]
There is an integer, e.g., a, with x=2a [def. of divisible]
x is even, so 2|x
(x+y) is even, so 2|(x+y)
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Direct Proof
simple combination of existing theorems
restated
the theorem
and beginning of the proof
your argument.
If x and y are even integers, then x+y is even
- x + y is even
[likewise for y]
(x+y)=2(a+b) so take c = (a+b) [add equations in 4 and 6] There is an integer, e.g, c, with (x+y)=2c [def. of divisible]
[def. of even]
[def. of even]
- Show that a given statement is true by
- With or without mathematical manipulations
- Template for Proof of an if-then theorem
- First sentence(s) of proof is the hypothesis
- Last sentence(s) of proof is the conclusion of
- Unwind the definitions, working from both end
- Try to forge a ‘link’ between the two halves of
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Concept Question
Given p → q; What is ¬q → ¬p?
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Proof of Implication
- pÆq contrapositive , ¬q → ¬p, is logically equivalent to pÆq
- prove ¬ q → ¬ p via the direct approach and then the original implication, p → q, is proven.
- first assume that q is false. Then use rules of inference, logical show that p must also be false
Note If q is true then the implication holds. We assume that q is false so that we can
well.
1. Negation
2. Implication
3. Contrapositive
4. I don’t know
Direct proof of
Indirect proof
equivalences, and previously proved theorems to
that it may not be the case that q is false.
explore this scenario and show that p must necessarily be false as
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Ex : Give an indirect proof of “If 3n + 2 is odd, then n is odd.” Recall again that this statement is implicitly a universal quantification “∀n(3n + 2 is odd → n is odd).”
Proof : We will prove the contrapositive , “If n is not odd, then 3n + 2 is not odd. That is, “If n is even, then 3n + 2 is even.”
[step 2: Translate assumptions into a form we can work with] Then n = 2k for some integer k. [Definition of even] [step 3: Work with it until it is in a form we need for concl.] So 3n + 2 = 3(2k) + 2 = 6k + 2 = 2(3k + 1). [step 4: Realize that you’re there and state your conclusion.]
So 3n + 2 = 2*m where m = (3k + 1), so 3n + 2 is even.
[step 1: Write assumptions] Let n be an even integer.
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Indirect Proofs
→ q:
¬q, ¬p follows
p → q ≡ ¬q → ¬p (the contrapositive)
Prove : “if a + b ≥ 15, then a ≥ 8 or b ≥ 8”.
Where a, b are integers
(a + b ≥ 15) → (a ≥ 8) ∨ (b ≥ 8)
∧ (b < 8)
⇒(a ≤ 7) ∧ (b ≤ 7)
⇒(a + b) ≤ 14
⇒ (a + b) < 15
- Prove the implication p
- Assume: (a < 8)
- Proof:
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Proof by Contradiction
logical negation of the result we wish
to prove, and then reach some kind of
contradiction.
¬Q.
conclusion contradicting one of the assumptions (or something obviously untrue like 1 = 0)
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Proof by Contradiction Example
4. ¬ H [ Assumption ]
7. ¬G
6,7 = Contradiction!
Rainy days make gardens grow.
Gardens don’t grow if it is not hot.
When it is cold outside, it rains.
1. R → G
2. ¬H → ¬G
3. ¬H → R
- Assume, along with the hypotheses, the
- To prove: "If P, then Q“
- assume P and
- the contradiction we arrive at could be some
- Prove that it is hot
5. R [Modus Ponens 3,4]
6. G [Modus Ponens 1,5]
[Modus Ponens 2,4]
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[Example]
Proof : Assume 5n+6 is odd and n is even
- 5n+6 = 5* 2k + 6 = 2 * (5k + 3)
- Since 5k+3 is an integer, 5n+6 is an even number, contradicting the assumption that it was odd
- Thus if 5n+6 is odd, then n is odd
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Proof by Induction
A proof by Induction has five basic parts:
Prove " if 5n+6 is odd, then n is odd" by contradiction
Then n = 2k for some integer k
1. State the proposition
2. Verify the base case
3. Formulate the inductive hypothesis
4. Prove the inductive step
5. State the conclusion of the proof
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Proof by Induction
F(0) = 1
Assume: F(n-1) = (n-1)(n-2)…2
multiply both sides by n,
= F(n)
Factorial(n) is the product of the first n positive integers
nF(n-1) = n(n-1)…32
Rule of Inference Tautology Name p ∴ p^ ∨ q
p → (p ∨ q) Addition
p ∧ q (p ∧ q) → p Simplification ∴ p p, q (^) (p ∧ q) → p ∧ q Conjunction ∴ p ∧ q p, p → q ∴ q
(p ∧ (p → q)) → q Modus Ponens
¬q, p → q ∴ ¬p
(¬q ∧ (p → q)) → ¬p Modus Tollens
p → q, q → r ∴ p → r
((p → q) ∧ (q → r)) → (p → r)
Hypothetical Syllogism
p ∨ q, ¬p ∴ q
((p ∨ q) ∧ ¬p) → q Disjunctive Syllogism
p ∨ q, ¬p ∨ r ∴ q ∨ r
(p ∨ q) ∧ (¬p ∨ r) → q ∨ r
Resolution
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