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In this lectureIn this lecture

Number Theory

● Quotient and Remainder

● Floor and CeilingProofs

● Direct proofs and Counterexamples

(cont.)

● Division into cases

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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 Sunday.

Q.: What day is Halloween this year?

Solution: There are 365 days between

10/31/10 and 10/31/11.

365 mod 7 = 1.

Thus, if 10/31/10 is Sunday

then 10/31/11 is Monday.

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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 s.t. 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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Floor and CeilingFloor and Ceiling

Definition: For any real number x,

● the floor of x:

= the unique integer n s.t. n ≤ x < n+1;

● the ceiling of x:

= the unique integer n s.t. n-1 < x ≤ n.

Examples:

x

x .37.3;47.3

.53/14;43/14

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Properties of Floor and Properties of Floor and CeilingCeiling

• For xR and mZ , .

• is false.Counterexample: For x=1.7, y=2.8,

Note: If x,y>0 and the sum of their fractional parts is <1

then

mxmx

",," yxyxRyxFor

8.27.1348.27.1

yxyx

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Properties of Floor and Properties of Floor and CeilingCeiling

• Theorem: For nZ ,

• Proof: Case 1: Suppose n is odd.

Then n=2k+1 for some integer k. (1)

oddisnif

evenisnifn

nn

21

22

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Properties of Floor and Properties of Floor and CeilingCeiling

• Proof (cont.): By substitution from (1),

(2)

because kZ and k ≤ k+1/2 < k+1.

On the other hand,

n=2k+1 → k=(n-1)/2. (by basic algebra) (3)

Based on (2) and (3),

Case 2: n is even (left as exercise). ■

kkkkn 21

21

22

212

2

21

2nn