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Methods of ProofMethods of Proof

• Proof techniques in this handout– Direct proof– Division into cases– Proof by contradiction

• In this handout, the proof techniques will be used to prove properties in number theory.

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Even and Odd IntegersEven and Odd Integers

Definition: An integer n ● is even iff

an integer k such that n=2k; ● is odd iff

an integer k such that n=2k+1.

Ex: If x and y are integers, is even or odd? 724 2 xyx

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Method of Direct ProofMethod of Direct Proof

To prove a statement: xD if P(x) then Q(x).

• Suppose x is a particular but arbitrarily chosen element of D

for which P(x) is true;

• Show the conclusion Q(x) is true by using ♦ definitions; ♦ previously established results;

♦ rules of logical inference.

Method of Direct Proof (Ex.)Method of Direct Proof (Ex.) Show xZ if x is odd

then 3x+9 is even.Proof: Suppose x is an arbitrarily chosen odd integer.

Then x=2k+1 for some integer k. (by definition) So 3x+9 = 3(2k+1)+9 (by substitution)

= 6k+3+9 (by distributive law) = 2(3k+6) (by factoring out a 2) (*)

3k+6 is an integer. (**) Hence 3x+9 is even based on (*), (**) and definition of even integers.

▀

(this is what we needed to show)

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Directions for writing proofsDirections for writing proofs

1) Write the theorem to be proved.

2) Clearly mark the beginning of your proof

with the word Proof.

3) Make your proof self-contained.

(Identify all variables used in the proof;

state the sources of outside facts).

4) Write proofs in complete English sentences.

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Common mistakes in proofsCommon mistakes in proofs

• Arguing from examples

• Using same letter

to mean two different things

• Jumping to a conclusion

(without adequate reasons)

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Types of Mathematical Types of Mathematical StatementsStatements

Theorems: Very important statements that have many and varied consequences.

Propositions: Less important and consequential.

Corollaries: The truth can be deduced almost immediately

from other statements.Lemmas: Don’t have much intrinsic interest

but help to prove other theorems.

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DivisibilityDivisibility

• Definition: For n,d Z and d≠0 we say that n is divisible by d

iff n=d·k for some k Z .• Alternative ways to say:

n is a multiple of d , d is a factor of n , d is a divisor of n , d divides n .

• Notation: d | n .• Examples: 6|48, 5|5, -4|8, 7|0, 1|9 .

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Properties of DivisibilityProperties of Divisibility

• For xZ, 1|x .

• For xZ s.t. x≠0, x|0 .

• For a,b,cZ, if a|b and a|c then a|(b+c) .

• Transitivity: For a,b,cZ,

if a|b and b|c then a|c .

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Quotient-Remainder Quotient-Remainder TheoremTheorem

Theorem: For nZ and dZ+ ! q,rZ such that

n=d·q+r and 0≤r<d.q is called quotient; r is called remainder.Notation: q = n div d; r = n mod d.Examples: 1) 53 = 8·6+5. Hence

53 div 8 = 6; 53 mod 8 = 5. 2) -29 = 7·(-5)+6. Hence

-29 div 7 = -5; -29 mod 7 = 6.

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Example of using div and Example of using div and modmod

• Last year Halloween was on Thursday.

Q.: What day is Halloween this year?

Solution: There are 365 days between

10/31/14 and 10/31/15.

365 mod 7 = 1.

Thus, if 10/31/14 was Friday

then 10/31/15 is Saturday.

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Proof Technique: Proof Technique: Division into CasesDivision into Cases

Suppose at some stage of a proof

● we know that

A1 or A2 or A3 or … or An is true;

● want to deduce a conclusion C.Use division into cases:

Show A1→C, A2→C, …, An→C.

Conclude that C is true.

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Division into Cases: Division into Cases: ExampleExample

Proposition: If nZ such that neither of 2 or 3 divide n, (1)then n2 mod 12 = 1. (2)

Proof: Suppose nZ s.t. neither of 2 or 3 divide n.By quotient-remainder theorem,

exactly one of the following is true: a) n=6k, b) n=6k+1, c) n=6k+2, d) n=6k+3, e) n=6k+4, f) n=6k+5 for some integer k. (3)n can’t be 6k, 6k+2 or 6k+4 because

in that case 2 | n (which contradicts (1) ). (4)n can’t be 6k+3 because in that case 3 | n

(which contradicts (1) ). (5)

Division into Cases: Division into Cases: Example(cont.)Example(cont.)

• Proof(cont.): Based on (3), (4) and (5), either n=6k+1 or n=6k+5.

Let’s show (2) for each of these two cases. Case 1: Suppose n=6k+1.

Then n2 = (6k+1)2=36k2+12k+1 (by basic algebra) = 12(3k2+k)+1 (6)

Let p=3k2+k. Then p is an integer. n2 = 12p+1 . ( by substitution in (6) ) Hence n2 mod 12 = 1 by quotient-remainder th-m.

Case 2: Suppose n=6k+5. (exercise) ■

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Method of Proof by Method of Proof by ContradictionContradiction

1. Suppose the statement to be proved is false.

2. Show that this supposition logically leads to a contradiction.

3. Conclude that the statement to be proved is true.

Example of proof by contradiction.

Theorem: There is no greatest integer.

The proof on the board.

We will see several contradiction proofs in graph theory.