Basic Proof Methods.pptx

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Basic Proof Methods

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Some Terminology

Theorem – statement that can be shown to be true

Proposition – less important true statement

Proof – valid argument that establishes the truth of theorem

Axiom – statements which are supposed to be true without proof

Corollary – statement following directly from a theorem

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Direct Proof

Direct proof

– To prove conditional

statement

p



q

we start with assumption

that

p

is true, then using axioms,

definitions, proved theorems, inference

rules, we show that also

q

is true.

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Direct Proof Example

• Definition: An integer n is called odd iff n=2k+1 for some integer k; n is even iff n=2k for some k.

• Axiom: Every integer is either odd or even.

• Theorem: (For all numbers n) If n is an odd integer, then n2 is an odd integer.

• Proof: If n is odd, then n = 2k+1 for some integer k. Thus, n2 = (2k+1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1. Therefore n2 is of the form 2j + 1 (with j the integer 2k2 + 2k), thus n2 is odd. □

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Another Example

• Definition: A real number r is rational if there

exist integers p and q ≠ 0, with no common factors other than 1 (i.e., gcd(p,q)=1), such that r=p/q. A real number that is not rational is called irrational.

• Theorem: Prove that the sum of two rational numbers is rational. Let r and s are rational then So r+s is also a rational number

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Proof by Contraposition

• _Proof

_{by contraposition}

_or

_{indirect proof}

_is

based on the equivalency

p



q





q



p.

• To prove the implication

p



q

we suppose

that



q

is true, and we prove



p

in a

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Indirect Proof Example -1

• Theorem: (For all integers n)

If 3n+2 is odd, then n is odd.

• Proof: Suppose that the conclusion is false, i.e., that

n is even. Then n=2k for some integer k. Then 3n+2 = 3(2k)+2 = 6k+2 = 2(3k+1). Thus 3n+2 is even,

because it equals 2j for integer j = 3k+1. So 3n+2 is not odd. We have shown that ¬(n is odd)→¬(3n+2 is odd), thus its contra-positive (3n+2 is odd) → (n is odd) is also true. □

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Indirect Proof Example - 2

• Theorem: Prove that if n=ab, where a and b are positive integers, then .

• Proof: Suppose that the conclusion is false, i.e., that means . Multiplying a and b we obtain . So we

proved that assumption is also false, so the implication follows □

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Proof by Contradiction

• _{The proof by contradiction of proposition}

_p

_is

based on assumption that



p is true

and then

obtain a false proposition.

• Proving

p

– Assume p, and prove that p (r  r)

– (r  r) is a trivial contradiction, equal to F

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Contradiction Proof Example

• Theorem: Prove that is irrational.

• Proof: let where gcd(p,q)=1. Then

2= hence 2q2= p2 and p is an even number p=2k. Then 2q2= 4k2 and q is also an even number, and

gcd(p,q)1, and we have a contradiction.

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Proof by Cases

To prove

we need to prove

Example: Prove the implication “If n is an integer not divisible by 3 then, then n2  1 (mod 3)

Proof: If n  1(mod 3) then n=3k+1 and n2=9k2+6k+1,

and n2 1 (mod 3) . If n  2(mod 3) then n=3k-1 and

n2=9k2-6k+1, and n2 1 (mod 3). 1 2

(

p



p

 

...

p

_n

)



q

1 2

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Proof of Equivalence

• To prove of equivalence pq we use tautology

(pq)  (pq) (qp)

Theorem: If n is an integer, then n is odd if and only

n2 is odd.

Proof: we prove

a) If n is odd then n2 is odd ( pq)

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Equivalence of a group of propositions

To prove

we need to prove

1 2

[

p



p

 

...

p

_n

]

1 2 2 3 1

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Example

• _{Show that the statements below are equivalent:}

p

₁

:

n is even

p

₂

:

n-1 is odd

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Counterexamples

• When we are presented with a statement of the form

xP(x) and we believe that it is false, then we look for a counterexample.

• Example

– Is it true that “every positive integer is the sum of the squares of three integers?”

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Proving Existence

• _{A proof of a statement of the form}

_

_x

_P

₍

_x

_{) is}

called an

existence proof

.

• If the proof demonstrates how to actually find

or construct a specific element

a

such that

P

(

a

)

is true, then it is called a

constructive

proof.

• Otherwise, it is called a

non-constructive

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Constructive Existence Proof

• Theorem: There exists a positive integer n that is the sum of two perfect cubes in two different ways:

– equal to j 3 + k 3 and l 3 + m 3 where j, k, l, m are positive integers, and {j,k} ≠ {l,m}

• Proof: Consider n = 1729, j = 9, k = 10,

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Non- constructive Existence Proof

• Theorem: For each integer n there exists a prime

number p greater than n.

• Proof: Suppose that such a prime number does not exist. Then all primes are less or equal than n. Let

P={p1,p2,p3, …, pk} be set of all primes  n. Number

p=p_1*p_2*p_3* …_* p_k+1 is greater than n and not divisible by any p1,p2,p3, …, pk it means that there exists at least