MAT 2720 Discrete Mathematics Section 2.2 More Methods of Proof Part II

MAT 2720Discrete Mathematics

Section 2.2More Methods of Proof

Part II

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Goals Indirect Proofs

• Contrapositive• Contradiction• Proof by Contrapositive is considered as a

special case of proof by contradiction• Proof by cases• Existence proofs

Proof by Contradiction Proof by Contrapositive

Proof by Contradiction

If then P Q

Example 2

Analysis Proof

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

Suppose 3 2 is even.n

Proof by Contradiction

Analysis Proof by Contradiction of If-then Theorem• Suppose the negation of the conclusion is true. • Find a contradiction.• State the conclusion.

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

Proof by Contradiction The method also work with statements other

then If P then Q

Example 3

Analysis Proof

2 is an irrational number.

Proof by Cases

Example 4

Analysis Proof

, ,xy x y x y

if 0if 0

x xx

x x

Proof by Cases

Analysis Proof by Cases of If-then Theorem• Split the domain of interest into cases.• Prove each case separately.• State the conclusion.

•Note that the cases do not have to be mutually exclusive. They just have to cover all elements in the domain.

, ,xy x y x y

Existence Proofs

Example 5

Analysis Proof

Suppose , such that .Show that such that .

a b a bx a x b

Existence Proofs

Analysis Existence Proof• Prove the statement by exhibiting an element in the domain of interest that satisfies the given conditions.•State the conclusion.

Suppose , such that .Show that such that .

a b a bx a x b

MAT 2720Discrete Mathematics

Section 2.4 Mathematical Induction

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Preview Review Mathematical Induction

• Why?• How?

The Needs…

( 1)1 2 ; 2

n n n Nn

Theorems involve infinitely many, yet countable, number of statements.

Principle of Mathematical Induction (PMI)

PMI: It suffices to show 1. P(1) is true.2. If P(k) is true, then P(k+1) is also

true, for all k.

(1) (2) (3) (66) (67) (68)P P P P P P

Principle of Mathematical Induction (PMI)

PMI: It suffices to show 1. P(1) is true. (Basic Step)2. If P(k) is true, then P(k+1) is also

true, for all k (Inductive Step)

(1) (2) (3) (66) (67) (68)P P P P P P

Format of Solutions In this course, it is extremely important

for you to follow the exact solution format of using mathematical induction.

Do not skip steps.

Format of Solutions In this course, it is extremely important

for you to follow the exact solution format of using mathematical induction.

Do not skip steps.

Example 1Use mathematical induction to prove that

whenever n is a nonnegative integer.

2 2 2 2 ( 1)(2 1)(2 3)1 3 5 (2 1)3

n n nn

Checklist A

2 2 2 2 ( 1)(2 1)(2 3)1 3 5 (2 1)3

n n nn

0,

Checklist B

2 2 2 2 ( 1)(2 1)(2 3)1 3 5 (2 1)3

n n nn

Checklist C

2 2 2 2 ( 1)(2 1)(2 3)1 3 5 (2 1)3

n n nn

Checklist C

2 2 2 2 ( 1)(2 1)(2 3)1 3 5 (2 1)3

n n nn

Checklist D

Proof by Mathematical InductionDeclare P(n) and the domain of n.• Basic Step

• Write down the statement of the first case. • Do not do any simplifications or algebra on the statement of the first case.• Explain why it is true.

• For A=B type, simplify and/ or manipulate each side and see that they are the same.

• Inductive Step• Write down the k-th case. This is the inductive hypothesis.• Write down the (k+1)-th case. This is what you need to prove to be true.• For A=B type, we usually start form one side of the equation and show that it equals to the other side.

• In the process, you need to use the inductive hypothesis.• Conclude that p(k+1) is true.

• Make the formal conclusion by quoting the PMI

Example 2

1

Show that for 1, 2,3,...

! 2nn

n

Example 3

1

0

Show that for real number 1,

1 , for

1

nni

i

r

a rar n

r

N