1 discrete math methods of proof. proofs a proof is a valid argument that establishes the truth of...
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Discrete Math
Methods of proof
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Proofs A proof is a valid argument that establishes the
truth of a theorem. The statements used in a proof include
axioms (or postulates) which are statements we assume to be true
Premises of the theorem, Previous proven theorems, and lemmas (sometimes
prove a part of the theorem) Rules of inference Definitions, All terms used in a proof must be defined
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Theorems A theorem is a statement that can be shown to
be true (using a proof) The statement of a theorem is a conjecture, a
statement that is proposed to be true (usually based on some evidence)
A conjecture becomes a theorem after it is formally proven to be true
A theorem may be stated as a quantification of a conditional statement
A statement that is somewhat important will become a theorem when it is proven, Simpler statements (propositions) will become facts, or results
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Types of Proof Direct Proof Indirect Proof or Proof by Contraposition Proof by contradiction Proof using cases Exhaustive proofs Proof by mathematical induction Other methods of proof
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Definition (to be used in proofs) Consider any two integers a and b. We
say that a divides b if and only if there exists an integer q such that b=qa. We write this symbolically as a|b When a divides b then a is a factor of b When a divides b then b is a multiple of a When a divides b then a is divisible by b If no such q exists then a does not divide b
a b
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Another Definition (used in proofs)
Let p>1 be an integer. p is prime when it is divisible by only 1 and
itself (by 1 and p) Otherwise p is called a composite integer
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Disproving a proposition Sometimes it is easy to disprove a
proposition or theorem by finding a counter example for which the statement is not true.
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Example: Disproving a proposition Proposition
Every positive integer is a prime number x universe of discourse positive integers P(x) x is a prime number ∀x P(x)
It is easy to find a counterexample that will disprove the proposition ∀x P(x) Consider the positive integer 8 2|8, the definition of a prime number states that a prime
number is divisible only by 1 and by itself 2 is not 1 or 8, there is a contradiction and the proposition
is disproved
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Example: Proving a proposition? Proposition
Every positive integer is a prime number x universe of discourse positive integers P(x) x is a prime number ∀x P(x)
It is easy to find examples that will satisfy the proposition ∀x P(x) Consider the positive integers 2, 3, and 11 We can show that 2, 3 and 11 are all prime We cannot say that all positive integers are prime based
on those particular examples Proof by example is no more convincing when the
proposition is true !!!
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Direct Proofs A direct proof shows that a conditional
statement p → q is true. Proving the theorem or proposition requires that We begin with p and show step by step that q is true We show the case P true Q false never occurs We justify each step of our proof with an axiom or
proven proposition (like a rule of inference) We have already seen direct proofs of a
number of propositions while we were studying the rules of inference
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Example of Direct Proof Proposition (Theorem) to be proven
If a|b and b|c then it follows that a|c a, b, c are positive integers P(a,b) a|b Q(b,c) b|c R(a,c) a|c ∀a, ∃b, ∃c [ ( P(a,b)^Q(b,c) ) → R(a,c) ]
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Example of Direct Proof Proposition (Theorem) to be proven
∀a, ∃b, ∃c ( P(a,b)^Q(b,c) ) → R(a,c) ) Replace a, by arbitrary positive integers x x, is arbitrarily chosen from the universe of
positive integers (may represent any positive integer)
(P(x,b)^Q(b,c) ) → R(x,c) This is an example of Universal Instantiation
of x
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Example of Direct Proof1. ∀a, ∃b, ∃c ( P(a,b)^Q(b,c)) → R(a,c))
2. ∃b, ∃c ( P(x,b)^Q(b,c) )→ R(x,c)) using Universal Instantiation
3. ( P(x,y)^Q(y,z) → R(x,z) ) using Existential Instantiation
4. Using the definition of divisible1. P(x,y) → ∃q y=qx where q is a positive integer
2. Q(y,z) → ∃r z=ry where r is a positive integer
5. z=ry=r(qx)=(rq)x where r and q are positive integers
6. z=ry=r(qx)=(rq)x rq is a positive integer (because the product of two positive integers is a positive integer)
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Example of Direct Proof1. Using the definition of divisible
x|z means ∃s=rq z=sx where s is a positive integer
So if P(x,y)^Q(y,z) is true then R(x,z) is true P(x,y)^Q(y,z) → R(x,z) is true
We have constructed a y and a z for which this is true we can use Existential Generalization to give ∃b, ∃c ( P(a,b)^Q(b,c) ) → R(a,c) )
Since x is arbitrary (any element of the universe) P(a,b)^Q(b,c) ) → R(a,c) by Universal Generalization
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Indirect Proof Also called Proof by Contraposition The contrapositive of a proposition or theorem is
proven rather than the proposition or theorem itself The contrapositive is logically equivalent to the
original statement, so the same thing is being proved
Sometimes the contrapositive is easier to prove than the original statement
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Example Problem Proposition (Theorem) to be proven
if a ≠ b then it follows that b ∤ a a, b are positive integers for which a|b Q(a,b) a ≠ b R(a,b) b ∤ a Q(a,b) → R(a,b) ¬ R(a,b) → ¬Q(a,b) Contrapositive
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Example Problem Proposition (Theorem) to be proven
If a|b and a ≠ b then it follows that b ∤ a ¬ R(a,b) → ¬ Q(a,b) Contrapositive a, b are positive integers for which a|b ¬ Q(a,b) a = b ¬ R(a,b) b∣a
Prove the contrapositive: demonstrate that when a|b and b∣a then a = b
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Example of Indirect Proof ¬R(a,b) → ¬Q(a,b)
1. ¬R(a,b) Premise
2. ¬ R(x,y) Universal Instantiation3. x|y from 2 4. y|x definition of universe5. y=rx and x=qy definition of divisible6. y=rx=r(qy)=(rq)y rq=1 so r=q= 1 and y=x 7. ¬Q(x,y)8. ¬Q(a,b) Universal Generalization
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Proof by contradiction We want to prove a statement p is true We can prove p is true if we can
demonstrate that for some proposition r that contradiction q
Because ¬p → q is true we conclude that ¬p is false (p is true)
This approach to proving p is true is called proof by contradiction
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Example: proof by contradiction Prove that is irrational
P(x) is irrational
Suppose that ¬P(x) is true (p is rational) Then there exist two integers, with no common factors, a and b
such the a/b = Squaring both sides gives 2 = a2/b2
2b2 = a2 so a2 is even by the definition of even By the definition of even a=2c So 2b2 = 4c2 or b2 = 2c2, so b2 is even, and b is even But both a and b are even so they have a common factor 2 This contradiction shows ¬P(x) is false so P(x) must be true
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Proof by cases Sometimes cannot prove a theorem or
proposition using a single argument. In these situations you can often divide
the problem into cases, then demonstrate the validity of the proposition or theorem using a different argument for each case
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Proof by cases If P ↔ P1 v P2 v … v Pk then
P → Q iff P1 v P2 v … v Pk → Q
or
P → Q iff P1 → Q ^ P2 → Q ^ … ^ Pk → Q
So we can prove by demonstrating that each of the following statements is true
P1 → Q (one case)
P2 → Q (another case)
… Pk → Q (last case
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Proof by cases: Example Prove ∀n ∈ Z [n/2] [n/2] = [n2/4] Case 1: n is even n= a*2
[2a/2] [2a/2] = [4a2/4]= a2
[n2/4] = [(2a*2a)/4] = a2
Case 2: n is odd n =a*2+1 [(2a+1)/2] [(2a+1)/2] = ([a+ ½ ] [a+ ½ ])
= a(a+1) + ¼ [n2/4] = [(2a+1)2/4] = [(4a2+4a+1)/4]
= a(a+1) + ¼ 23
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Your turn Prove by cases
If x,y ∈ ℝ and x+y >= 100 then x >= 50 or y >= 50
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Your turn: If x,y ∈ ℝ and x+y >= 100 then
x >= 50 or y >= 50
Case 1: for x,y ∈ ℝ y<50 x+y >= 100
x >= 100 – y If y>=50 then y = 50 + r for r an arbitrary non negative real
number x >= 100 – (50+r) = 50 – r x >= 50-r So x may be <50 , 50, or >50, all three possibilities are
consistent with the conclusion x >= 50 or y >= 5025
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Your turn: If x,y ∈ ℝ and x+y >= 100 then
x >= 50 or y >= 50
Case 1: for x,y ∈ ℝ y>=50 x+y >= 100
x >= 100 – y If y<50 then y = 50 - r where r is some positive real number x >= 100 – (50 - r) = 50 + r x >= 50+r So x must be >50, If x>50 then x>=50 So if y<50 then x>=50 and the conclusion is true
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Your turn: Indirect proof
If x,y ∈ ℝ and x+y >= 100 then x >= 50 or y >= 50
Contrapositive for x,y ∈ ℝ If x<50 and y<50 then x+y < 100
If x<50 then x = 50 – q where q is some positive real number
If y<50 then y = 50 – r where r is some positive real number
Since q and r are positive number q+r is a positive number
So x+y = 100 – (q+r) = 100 – positive real # < 100 27
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Proof by Exhaustion When we prove by cases for all possible
cases. Usually break down into a few cases and
prove them all Sometimes will need to prove for many
different cases
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Existence proofs Many theorems are assertions that some
particular type of object exists We can prove these types of theorems by
constructive proofs in which we construct an example of the object, since we have an example we know the type of object exists
We can also use non-constructive proofs that show the type of object exists without demonstrating a particular example of that type of object
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Example: Constructive Prove that there exists an integer solution
to the equation x * y = z2
Proof: The integers x=2 y=8 and z=4 satisfy the
equation. We have shown that an integer solution to
the equation exists
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Another example: constructive There exists a positive rational number less than
1/100. By the definition of a rational number x, the rational
number can be expressed as the ratio of two integers a/b
Let a=1 and b=100 To construct a smaller positive rational number we
must increase b or decrease a. But a is already the smallest possible positive rational number so we must increase b
Thus a smaller rational number is 1/10131
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Uniqueness proofs Theorems often assert that there is
exactly one object with particular properties
Uniqueness proofs show That an object with the particular properties
exists We show that only one object with the
particular properties exists
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