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Lecture 1.5: Proof Techniques
CS 250, Discrete Structures, Fall 2013
Nitesh Saxena
Adopted from previous lectures by Cinda Heeren
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04/19/23Lecture 1.5 - Proof
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Course Admin Slides from previous lectures all posted HW1 posted
Due at 11am on Sep 19 Please follow all instructions Questions?
Competency exams have been graded Will return this Thursday
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Outline
Proof Techniques
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Main techniques Direct proofs Proofs using contraposition Proofs using contradiction
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Direct Proofs
An example:
Prove that if n = 3 mod 4, then n2 = 1 mod 4.
HUH?
7 = 3 mod 4
37 = 1 mod 4
94 = 2 mod 4
16 = 0 mod 4
7 = 111 mod 4
37 = 61 mod 4
94 = 6 mod 4
16 = 1024 mod 4
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Proof:
If n = 3 mod 4, then n = 4k + 3 for some int k.
This means that:
Coming back to our Theorem: If n = 3 mod 4, then n2 = 1 mod 4.
Direct Proofs
= (4k + 3)(4k + 3)
n2
= 16k2 + 24k + 9= 16k2 + 24k + 8 + 1
= 4(4k2 + 6k + 2) + 1
= 4j + 1 for some int j
= 1 mod 4. 04/19/23
Lecture 1.5 - Proof Techniques
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Proof:
If n is odd, then n = 2k + 1 for some int k.
This means that:
Theorem: If n is an odd natural number, then n2 is also an odd natural number.
Direct Proofs: another example
= (2k+1)(2k+1)n2
= 4k2 + 4k + 1
= 2(2k2 + 2k) + 1
= 2j + 1 for some int j
n2 is odd04/19/23
Lecture 1.5 - Proof Techniques
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Proofs by ContrapositionRecall: Contrapositive: p q and q p
Ex. “If it is noon, then I am hungry.” “If I am not hungry, then it is not noon.”
We also know that: p q q p
Therefore, if establishing a direct proof (p q) is difficult for some reason, we can instead prove its contraposition (q p), which may be easier.
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Proof:
If n is not odd, then n = 2k for some int k.
This means that:
Theorem: If 3n + 2 is an odd natural number, then n is also an odd natural number.
Proofs by Contraposition: example
= 3(2k) + 23n+2
= 2(3k) + 2
= 2j for some int j
3n+2 is not odd
= 2(3k + 1)
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Proof:
If a > sqrt(N) AND b > sqrt(N), then by multiplying the two inequalities, we get
Theorem: If N = ab where a and b are natural numbers, then
a <= sqrt(N) or b <= sqrt(N).
Proofs by Contraposition: another example
> N
This negates the proposition N=ab
ab
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Proofs by Contradiction
Recall: Contradiction is a proposition that is always False
To prove that a proposition p is True, we try to find a contradiction q such that (p q) is True.
If (p q) is True and q is False, it must be the case that p is True.
We suppose that p is False and use this to find a contradiction of the form r r
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Proof:
A: n is prime
B: n is odd
We need to show that A B is true
We need to find a contradiction q such that: (A B) q
We know: (A B) (A B) A B
This means that we suppose (n is prime) AND (n is even) is True
But, if n is even, it means n has 2 as its factor, and this means that n is not prime.
This is a contradiction because (n is prime) AND (n is not prime) is True
Theorem: Every prime number is an odd number
Proofs by Contradiction: example
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Proof:
A: 3n + 2 is odd
B: n is odd
We need to show that A B is true
We need to find a contradiction q such that: (A B) q
We know: (A B) (A B) A B
This means that we suppose that (3n + 2 is odd) AND (n is even) is True.
But, if n is even, it means n = 2k for some int k, and this means that 3n + 2 = 6K+2 = 2(3K+1) even.
This is a contradiction: (3n + 2 is odd) AND (3n +2 is even)
Theorem: If 3n + 2 is an odd natural number, then n is also an odd natural number.
Proofs by Contradiction: example
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04/19/23Lecture 1.5 - Proof
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Disproving something: counterexamples
If we are asked to show that a proposition is False, then we just need to provide one counter-example for which the proposition is False
In other words, to show that x P(x) is False, we can just show x P(x) = x P(x) to be True
Example: “Every positive integer is the sum of the squares of two integers” is False.
Proof: counter-examples: 3, 6,…
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Today’s Reading Rosen 1.7 Please start solving the exercises at the
end of each chapter section. They are fun.