Discrete Mathematics -- Chapter 4: Properties of the Ch t 4 P ti - - PowerPoint PPT Presentation
Discrete Mathematics -- Chapter 4: Properties of the Ch t 4 P ti f th Integers: Mathematical Induction Hung-Yu Kao ( ) Department of Computer Science and Information Engineering, N National Cheng Kung University l Ch K U
Hung-Yu Kao (高宏宇) Department of Computer Science and Information Engineering, N l Ch K U National Cheng Kung University
4.1 The Well-Ordering Principle: Mathematical
4.2 Recursive Definitions
4.3 The Division Algorithm: Prime Numbers 4.4 The Greatest Common Divisor: The Euclidean
4 5 The fundamental Theorem of Arithmetic 4.5 The fundamental Theorem of Arithmetic
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自然數要適合五點(Peano axioms)：
有一起始自然數 0。 有一起始自然數 0 任一自然數 a 必有後繼，記作 a +1。 0 並非任何自然數的後繼 0 並非任何自然數的後繼。 不同的自然數有不同的後繼。
（數學歸納公設）有一與自然數有關的命題。設此
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contains a smallest element. ( )
} 1 | { } | { ≥ ∈ = > ∈ =
+
x x x x Z Ζ Z
Z+ is well ordered
( )
f th i bl hi h t iti i t
} 1 | { } | { ≥ ∈ > ∈ x x x x Z Ζ Z
a)
If S(1) is true; and (basis step)
b)
If whenever S(k) is true, then S(k+1) is true. (inductive step)
b)
w e eve S(k) s ue, e S(k ) s
then S(n) is true for all
+
∈Z n
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P(n0) P(n0+1) P(n0+2) . . .
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More rigorously the validity of a proof by mathematical induction relies More rigorously, the validity of a proof by mathematical induction relies
Let S be a statement for which we have proven that S(1) holds and for
p ( ) all n∈ Z+ we have S(n)⇒S(n+1). Claim: S(n) holds for all n∈ Z+
Proof by contradiction: Define the set F⊆ Z+ of values for which S
does not hold: F = { m | S(m) does not hold}.
If F is non empt then F must have a smallest element ( ell ordering of
If F is non-empty, then F must have a smallest element (well-ordering of
Z+), let this number be z with ¬S(z). Because we know that S(1), it must hold that z>1. Because z is the smallest value, it must hold that S(z–1), hi h di f f ll
+
( ) ( ) which contradicts our proof for all n∈ Z+: S(n)⇒S(n+1).
Contradiction: F has to be empty: S holds for all Z+.
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2 ) 1 ( 1
3 2 1 ,
+ = +
= + ⋅ ⋅ ⋅ + + + = ∑ ∈
n n n i
n i n Z
) 1 ( 2 ) 1 1 ( 1 1 1 2 ) 1 ( 1
1 : ) 1 ( (i) 3 2 1 : ) ( Let
+ × = + =
= = ∑ = + ⋅ ⋅ ⋅ + + + = ∑
i n n n i
i S n i n S
1 2 ) 1 ( 1 2 1
) 1 ( 3 2 1 ) 1 ( Th (iii) i.e., true, is ) ( Assume (ii) ) ( ( )
+ + = =
∑ = ∑
k k k k i i
k k i k S i k S
2 ) 1 ( 1 1 1 1
) 1 ( ) 1 ( ) ( ) 1 ( 3 2 1 : ) 1 ( Then (iii)
+ = + =
+ + = + + ∑ = + + + ⋅ ⋅ ⋅ + + + = ∑ +
k k i k i
k k i k k i k S
2 ) 2 )( 1 ( + +
=
k k
2
2
3 2 1 t
+ + +
+ + + +
n n n i
Z
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2 2 1
3 2 1 , try
+ + = +
= + ⋅ ⋅ ⋅ + + + = ∈
n n i
n i n Z
integer is the same whether it is read from left to right or from right to left are called palindromes Without actually determining all of these left, are called palindromes. Without actually determining all of these three-digit palindromes, we would like to determine their sum.
) 10 101 ( 10 101 10 100 : palindrome typical The
9 9 9 9
∑ ∑ + ∑ = ⎞ ⎜ ⎜ ⎛ ∑ + = + + = b a aba b a a b a aba
[ ]
45 10 1010 10 ) 101 ( 10 ) (
9 9 9 1 1
∑ ⋅ + = ∑ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ∑ + = ∑ ∑ ∑ ⎠ ⎜ ⎜ ⎝ ∑
= = = = a b a b
a b a
[ ]
500 , 49 450 9 1010 ) (
9 1 1
= ⋅ + ∑ = ⎥ ⎦ ⎢ ⎣
= = = a a b
a
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1 = a
t =1+2+ +i= i(i+1)/2
2 2 1
1
1
⋅
= = t
2 3 2
3 2 1
2
⋅
= = + = t
2 4 3
6 3 2 1
3
⋅
= = + + = t
2 5 4
10 4 3 2 1
4
⋅
= = + + + = t
ti=1+2+…+i= i(i+1)/2
P f
] [ ] [ ) (
) 1 ( 1 ) 1 2 )( 1 ( 1 1 ) 1 (
2
+ = ∑ + = ∑ ∑ =
+ + + +
i i t
n n n n n n n i i n
] [ ] [ ) (
6 ) 2 )( 1 ( 2 2 6 2 1 2 1 2 1
= + = ∑ + = ∑ ∑ =
+ + = = =
i i t
n n n i i i i
Ex 4.4 prove it
700 , 171 ...
6 ) 102 )( 101 ( 100 6
100 2 1
= = + + + ∴ t t t
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P d 1 dditi d lti li ti ( dditi ll t i)
procedure SumOfSquares1 begin procedure SumOfSquares2 begin ( 1) (2 1)/6 begin sum:= 0 for i:= 1 to n do sum := sum + i2 sum:= n*(n+1)*(2*n+1)/6 end end
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4n n2-7 n 4n n2-7 1 4 -6 2 8
5 20 18 6 24 29
). 7 ( 4 , 6 For
2 −
< ≥ n n n
2 8 -3 3 12 2 4 16 9 6 24 29 7 28 42 8 32 57
). 7 ( 4 , 6 For < ≥ n n n
) 1 2 ( ) 7 ( 4 ) 7 ( 4 4 ) 1 ( 4 : ) 1 ( 6 ), 7 ( 4 : ) (
2 2 2
k k k k k k S k k k k S + + − < + − < + = + + ≥ − <
2*6+1=13 ≥4
. true is ) ( 7 ) 1 ( ) 1 2 ( ) 7 ( ) 1 ( 4 ) ( ) ( ) ( ) ( ) (
2 2
n S k k k k ∴ − + = + + − < + ⇒
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n
n
H H H
1 2 1 2 1
1 , , 1 , 1
2 1
+ ⋅ ⋅ ⋅ + + = ⋅⋅ ⋅ + = =
n H n H
n j
n
− + = ∑ ) 1 ( n, all For
Ex 4.9 : Harmonic numbers
1 ) 1 1 ( 1 : ) 1 ( Verify
1 1
1 1
H H H S
j
j
− + = = =
=
j
j=1
Proof
) 1 ( : true ) ( Assume
1 1
k H k H k S
k j
k k k j j
− + =
=
1) (k ] ) 1 ( [ : ) 1 ( Verify
1 1
1 1 1
H k H H k H k H H H k S
k k k j j
k j k j
+ − + = + − + = + = +
+ +
+ = =
) 1 ( 2) (k ] 1 1 1)[ (k 1) (k
1 1 1
k H H k k H H k H
k k k k
+ + = + − + − + = + +
+ + +
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. true is ) ( ) 1 ( 2) (k
1
n S k H k ∴ + − + =
+
Ex 4.10 :
we must compare r with no more than A r ∈
n+1 elements in An.
n
A r ∈
b r c c C b b B c c b b C B c c b b A S a a a a A S = = < < < ∪ = = ∴ < =
2 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1
with compare (i) } , { }, , { , , } , , , { : ) 2 ( Verify s comparison 2 most at }, , { : ) 1 ( Verify
1 1 1 1 2 1
n C B C r B r b r + = + ∴ ∴ = = ∈ ∈
2
s comparison 1 1 2 most at (iii) s comparison 2 most at 2 | | | | ,
, (ii) with compare (i)
1 1 1 1
k k k k
c c c b b b c c c b b b C B A k S k S
k k k
< ⋅ ⋅ ⋅ < < < < ⋅ ⋅ ⋅ < < ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = ∪ = +
+
2 1 2 1 2 1 2 1
}, , , , , , , , { Let : ) 1 ( Verify : true ) ( Assume
1
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k k
c c c b b b < < < < < < <
2 1 2 1
M th ti l i d ti l j l i i ifi ti
Ex 4.11 :
answer : true ) ( Assume answer : ) ( Verify xy k S xy x S
k
= = = find we now again loop while' ' the
top the return to and executed nsare instructio loop the , first time for the loop while' ' the
top the reach the program the 1, hen : ) 1 ( Verify k n w k S + = +
while n≠ 0 do begin x := x * y
1 ) 1 (k find we now again, loop while the
top the
1
k n xy x = − + =
end answer := x
true. is ) ( ) ( answer final The and with continue will loop while' ' the And
1 1 , 1
n S xy y xy y x k n y x
k k k
∴ = = = ∴ =
+
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) ( S
Ex 4.13 :
summands (e g 14 = 3 + 3+ 8) 14 ≥ n
: true ) ( Assume k S While : ) 1 ( Verify : true ) ( Assume = + k n k S k S 1 for s 8' with two s 3' five replace , s 3' five least at have sum the 14, sum, in the appears 8 no (ii) case 1 for s 3' e with thre 8 replace sum, in the appears 8
least at (i) case + = ∴ ≥ + = k n k n Q ) 1 ( ) ( 1 for s 8 with two s 3 five replace + ⇒ ∴ + = k S k S k n
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Consider the sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,…
Consider the sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,… which is defined by F1 = F2 = 1 and Fn+2 = Fn+1 + Fn
Clearly this sequence F1,F2,… will grow, but how fast?
C j t F ≥ (3/2)n f ll >10
Conjecture: Fn ≥ (3/2)n for all n>10 Proof by induction:
Base case: Indeed F11 = 89 ≥ (3/2)11 = 86.49756…
Inductive step for n>10:
Inductive step for n>10:
Assume Fn ≥ (3/2)n and Fn+1 ≥ (3/2)n+1, then indeed
Fn+2
= Fn + Fn+1 ≥ (3/2)n + (3/2)n+1 = (3/2)n(1 + (3/2))
2
By induction on n, the conjecture holds.
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We just saw a different kind of proof by induction where
the inductive step is ∀n>10: [P(n), P(n+1) ⇒ P(n+2)] This time the basis step is proving P(11) and P(12). p p g ( ) ( )
There are many variations of proof by induction:
Strong/ Complete induction: Here the inductive step is: Assume all of P(1),…,P(n), then prove P(n+1). The basis step is P(1) for this alternative form of induction. p ( )
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The sequence 1,1,2,3,5,8,13,21,34,… defined by
Fn+2= Fn+Fn+1 is the famous Fibonacci sequence.
A closed expression of Fn is
n n
n
For large n, it grows like Fn ≈ 0.447214 × 1.61803n. This (1+√5)/2 ≈ 1.61803 is the Golden Ratio.
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Fibonacci numbers often occur in the natural world. Shape of shells:
htt // h tt / j / /i d ht l
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http://www.research.att.com/~njas/sequences/index.html
Alternative Form: (or the Principle of Strong Mathematical I d ti ) Induction)
more occurrences of the variable n, which represents a positive p p integer.
a)
If S(n0), S(n0+1), …, S(n1-1), and S(n1) are true If h S( ) S( 1) S(k 1) d S(k) f
Basic step
b)
If whenever S(n0), S(n0+1), …, S(k-1), and S(k) are true for some , then S(k+1) is also true
1
where , n k k ≥ ∈
+
Z
inductive step
.
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Ex 4.14 :
(e g 14 = 3+3+8 15 = 3+3+3+3+3 16 = 8+8) 14 ≥ n (e.g., 14 3+3+8, 15 3+3+3+3+3, 16 8+8)
true are ) 16 ( and ) 15 ( ) 14 ( Verify S S S : ) 1 ( Verify 16 where true, are ) ( and ) 1 ( ), 2 ( , ), 15 ( ), 14 ( Assume true. are ) 16 ( and ), 15 ( ), 14 ( Verify + ≥ − − ⋅⋅ ⋅ k S k k S k S k S S S S S S true is ) 1 ( true is ) 2 ( , 2 14 and , 3 ) 2 ( 1 : ) 1 ( Verify + ∴ − ≤ − ≤ + − = + + k S k S k k k k k S Q ) (
14 15 16 17 18 19 20 21 22 …
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Ex 4.15 :
⎧ ≤ where , 3 a
n n
⎩ ⎨ ⎧ ≥ ∈ + + = = = =
+
− − −
3 where all for , and , 3 , 2 , 1
3 2 1 2 1
n n a a a a a a a
n n n n
Z
1 1
3 3 2 3 3 1 (i) = ≤ = ≤ = = a a
2 1 1 2 2
(ii) 3 9 3
− − +
+ + = = ≤ =
k k k k
a a a a a
1 2 1
3 ) 3 ( 3 3 3 3 3 3 3
+ − −
= = + + ≤ + + ≤
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Towers of Hanoi: (1883) Towers of Hanoi: (1883)
How to move the disks from one pole to another? How to move the disks from one pole to another?
Call this number M(n) Call this number M(n).
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To prove something by induction, you need a conjecture about the
general case M(n).
Here we have: M(1)=1, M(2)=3, M(3)=7,… Obvious conjecture… M(n) = 2n – 1 for all n > 0 Clearly, the basis step M(1)=1 holds. Feeling for general n case: Each additional disk
(almost) doubles the number of moves: M( +1) i t f t M( ) M(n+1) consists of two M(n) cases…
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How to move n+1 disks, using an n-disk ‘subroutine’?
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We just saw that for all n, we have M(n+1) = 2M(n)+1.
This enables our proof of the conjecture M(n)=2n–1.
Proof: Basis step: M(1) = 21–1 = 1 holds.
A th t M( ) 2n 1 h ld Th f t +1
Assume that M(n)=2n–1 holds. Then, for next n+1: M(n+1) = 2 M(n) + 1 = 2(2n–1) + 1 = 2n+1 – 1. Hence it holds for n+1 End of proof by induction on n Hence it holds for n+1. End of proof by induction on n.
With n=64 golden disks, and one move per second, this amounts to almost 600,000,000,000 years of work.
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1.
2.
your conjecture) your conjecture)
3.
4
4.
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Inductive proofs play an important role in computer science
Analyzing recursive algorithms often require the use of
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Th th d f i d ti l b li d t t t
The method of induction can also be applied to structures
and so on.
The crucial property that must hold is the well-ordered
principle: there has to be a notion of size such that all
smallest object.
Examples: vertex size of graphs, dimension of matrices,
depth of trees, length of sequences and so on.
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Take a 2n×2n square with one tile missing Can you tile it with L-shapes? y p
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Basis step of 2×2 squares is easy. Inductive step: Assuming that 2n×2n tiling is possible, tile a 2n+1×2n+1 square by looking at 4 sub-squares:
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1)
F0 = 0; F1 = 1;
2)
F = F + F 2 h ll f ≥
+
Z
2)
Fn = Fn-1 + Fn-2, 2 where all for ≥ ∈
+
n n Z
1 2 + = +
× = ∑ ∈ ∀
n n n i i
F F F n Z
= i
1 2 2 1 2
Assume 1 * 1 step, basis
+
× = = +
k k k i
F F F F F
2 1 2 1 2 1
Then Assume
+ + + =
+ = ×
k k i k i k k i i
F F F F F F
2 1 1
) (
+ + = =
+ × = + × =
k k k i i
F F F F F F
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2 1 1 1
) (
+ + + +
× = + × =
k k k k k
F F F F F
1)
2)
2 where all for ≥ ∈
+
n n Z
1 1 + − +
+ = ∈ ∀
n n n
F F L n Z
3 1 2 2 1
2 1 3 , 1 1 step, basis : Proof + = + = = + = + = = F F L F F L
2 1 1 1 k k 1 k 1 1 n
) ( ) ( Then 2 k where k, 1,
1,2,3,..., n for Assume
− + − − + + −
+ + + = + = ≥ = + =
k k k k n n
F F F F L L L F F L
2 1 2 1
) ( ) (
+ + − −
+ = + + + =
k k k k k k
F F F F F F
Fn-2 Fn-1 Fn Fn+1 Fn+2 Ln-1 Ln Ln+1
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1 ) 1 ( 1 ) 1 ( + + − +
+ =
k k
F F
E 4 21 E l i
b
1 , 1 , ) 1 ( ) (
, 1 1 , 1 ,
k k m k a k a k m a
k m k m k m
< ≥ − ≤ ≤ + + − =
− − −
m=1 m=2 m=3 1 1 1 1 4 1
Row sum
1=1! 2=2! 6=3!
! , , , , , 1
1 , , , ,
m a k a m k a a
m k k m k m k m
= ∑ < = ≥ = =
− =
m=3 m=4 m=5 1 4 1 1 11 11 1 1 26 66 26 1 6 3! 24=4! 120=5!
] 3 ) 1 [( ] 2 [ ] ) 1 [( ] ) 1 ( ) 1 [(
2 1 1 1 , 1 , , 1
⋅ ⋅ ⋅ + + − + + + + + = + + ∑ − + = ∑
= − = +
a a m a ma a a m a k a k m a
m m m m m m k m m k k m m k k m
] ) 1 ( 2 [ ] [ ] ) 1 ( [ ] 2 [ ] ) 1 ( 3 [ ] ) [( ] [ ] ) [(
1 , 1 , , , , 1 , 1 , 2 , 2 , 3 , 2 , 1 , 1 , , , 1 ,
⋅ ⋅ ⋅ + − + + + = + + + + + − + +
− − − − − −
a m a ma a a m a ma a a m a
m m m m m m m m m m m m m m m m m m m m m m
Induction on m
! ) 1 ( ) 1 ( ] [ ] 2 ) 1 [(
1 , 1 , 1 , 2 , 2 ,
+ = ∑ + = + + + − +
− = − − − −
m m a m a ma a a m
m k k m m m m m m m m m
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)! 1 ( ) ( + = m
D fi iti 4 1 If b di id d it b| if
d b Z b
there is an integer n such that a = bn, then b is divisor of a, or a is multiple of b.
, and , ≠ ∈ b Z b a
We also say for b nonzero: b divides a or b is a factor of a
, and , , integers three the
two divides and , If e) all for | | d) | )] | ( ) | [( c) )] | ( ) | [( b) 0) (a | and | 1 a) z y x a z y x x bx a b a c a c b b a b a a b b a a a + = ∈ ⇒ ⇒ ∧ ± = ⇒ ∧ ≠ Ζ ) ( | then each divides If 1 For g) ) ,
n combinatio linear called is y ( y), ( | )] | ( ) | [( f) integer. remaining the divides then , , , g , ) x c x c x c a c a c n i c b c bx c bx a c a b a a y y + + + ∈ ≤ ≤ + + ⇒ ∧ Ζ
) ( | then , each divides If . , 1 For g)
2 2 1 1 n n i i
x c x c x c a c a c n i + ⋅ ⋅ ⋅ + + ∈ ≤ ≤ Ζ
and | and | If f) an c am b c a b a = = ⇒ y) ( | ) ( ) ( ) ( y c bx a ny mx a y an x am c bx + ∴ + = + = + ∴
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Number theory is now an essential applicable tool in dealing
Primes are the positive integers that have only two positive
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Sacks spiral, 1994 Ulam spiral, 1963
+
Lemma 4.1: If and n is composite, then there is a prime
Proof
+
∈ Z n
Proof
If no such a prime Let S be the set of all composite integers that have no prime divisors. By Well-Ordering Principle, S has a least element m. If m is composite, then m=m1m2, 1<m1<m, 1<m2<m. Since
, m1 is prime or divisible by a prime, so exists a prime p,
S m1 ∉
,
1
p y p , p p, p|m1.
Since p|m1 and m=m1m2, so p|m. (contradiction, Theorem 4.3 (d))
S m1 ∉
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Th 4 4 Th i fi it l i (E lid B k IX)
Theorem 4.4: There are infinitely many primes. (Euclid, Book IX) Proof
If not If not Let p1 , p2, …, pk be the finite list of all primes. Let B = p1 p2 …pk+1.
Let B p1 p2 …pk 1.
Since B > pi , , B is not a prime. So B is composite, pj|B, . (Lemma 4.1)
k i ≤ ≤ 1
k j ≤ ≤ 1
j
Since pj|B and pj|p1 p2 …pk, so pj|1. (Contradiction, Theorem 4.3
(e))
integer. remaining the divides then , and , , integers three the
two divides and , If e) a z y x a z y x + =
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integer. remaining the divides then a
Z ∈ b a ,
unique
. , with , b r r qb a r q < ≤ + = ∈Z
r | (b) , i.e. , | (a) > / = a b r a b
We call r the remainder when a is divided by b, and q the quotient when a is divided by b.
(1) q, r exist
φ ≠ ∈ = > > − ∈ − = b b b b S S a t a tb a Z t tb a S ) 1 ( ) 1 ( h 1 l d If (ii) , then , and If (i) } , | { Let r , | (b) > / a b b r , | (c) < / a b φ ≠ > − ∴ ≤ ≤ − + − = − − = − − = ≤ S tb a a b b b a b a a tb a a t a , and 1 ) 1 ( ) 1 ( then , 1 let and If (ii) Q b S S P i i l ) O d i W ll ( l l h φ S S b q a c qb a c b r b r a b b q a b r qb a r r S S ∈ + − = ⇒ − = + = > / + = = − − = < ∴ ≠ f l t l t th i i t di t , ) 1 ( then , If (ii) . | ing contradict , ) 1 ( then , If (i) Principle) Ordering Well ( , element least a has φ Q
(2) q, r are unique
b r S r < ∴ .
element least the is ing contradict
2 2 1 1
if i t di t | | | let , and
are there If b b| r b q r b q a r's q's + = + =
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2 1 2 1 2 1 1 2 2 1
and if ing contradict , | | | r r q q q q b r r q b|q = = ∴ ≠ < − = − ⇒
Ex 4.25
15 and the remainder r = 5 (170=15*11+5) 15, and the remainder r = 5. (170=15*11+5)
and the remainder r = 0. (98=14*7) f h di id d d h di i b h h i
and the remainder r = 3. (-45=(-6)*8+3 or -45=(-5)*8-5?)
+
∈Z b a,
If a = qb, then -a = (-q)b. So, the quotient is -q, and the remainder is 0. If a = qb + r, then -a = (-q)b - r = (-q - 1)b + (b - r). So, the quotient
is -q -1, and the remainder is b - r. q ,
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Ex 4.27 : Write 6137 in the octal system (base 8)
Here we seek nonnegative integers r0 , r1 , r2, …, rk , 0< rk <8,
such that 6137= (rk… r2 r1 r0)8.
2 1
8 8 8 6137 ⋅ + ⋅ ⋅ ⋅ + ⋅ + ⋅ + =
k k
r r r r
2 3 4 1 2 1
1 8 7 8 7 8 3 8 1 ) 8 8 ( 8 + ⋅ + ⋅ + ⋅ + ⋅ = ⋅ + ⋅ ⋅ ⋅ + ⋅ + + =
− k k
r r r r
remainders
8 6137
8
) 13771 ( 1 8 7 8 7 8 3 8 1 = + + + +
8 767 1(r0) 8 95 7(r1) 8 11 7(r2)
2
8 1 3(r3) 0 1(r4)
How about base 2 and base 16?
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Ex 4.29 : Two’s Complement Method: binary representation of
negative integers
One’s Complement: interchange 0’s and 1’s. Add 1 to the prior result.
E l 6 1010 0001 1001 0110 6 ≡ → + → ≡
Example: Example: How do we perform the subtraction 33 – 15 in base 2
with patterns of 8-bits? 6 1010 0001 1001 0110 6 − ≡ → + → ≡ with patterns of 8 bits?
(-15) 11110001 (33) 00100001 +
00001111 (+15)
General formula: x – y = x + [(2n – 1) – y + 1] – 2n
100010010
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One’s complement of y
E 4 31 If
d i i h h i i h
+
Ex 4.31 : If and n is composite, then there exists a prime p such
that p|n and
Proof
+
∈Z n
. n p ≤
Composite n = n1n2 We claim that one of n1 , n2 must be less than or equal to If not then
d
, n
If not, then
tion) (contradic and
2 1 2 1
n n n n n n n n n n = > = > >
So, assume
(i) If n1 is prime, the statement is true.
.
1
n n ≤
( )
1
p , (ii) If n1 is not prime, there exists a prime p < n1 where p|n1. (Lemma 4.1) then
So p|n and
. n p ≤ .
1
n n p ≤ <
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p|
Definition 4.3: For , a positive integer c is called a greatest
common divisor of (最大公因數gcd(a,b) )
Z ∈ b a,
c d b a d b a c b c c|a | have we , and
divisor common any for (b) ) ,
divisor common a is ( | and (a)
| , y ( )
and b ?
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Theorem 4.6: For all , there exists a unique that is the
greatest common divisor of a and b.
f
+
∈Z b a,
} | { l t Gi bt t bt S b
+
Z Z
+
∈Z c
. ,
divisor common greast a is that claim We : Existence (i) Principle) Ordering
element least a has } , , | { let , , Given b a c c S bt as t s bt as S b a > + ∈ + = ∈
+
Z Z , , | If | 4.3(f)) (Theorem then , and if , , c r r qc a a c c d by d|ax d|b d|a by ax c S c < < + = / ∴ + + = ∈ Q | similarly, , |
element least the is ng contrdicti , ) ( ) 1 ( ) ( then b c a c S c S r b qy a qx by ax q a qc a r ∴ ∈ ∴ − + − = + − = − = 4 3(b)) (Theorem | and n the divisors common greast the are both , If : s Uniquenes (ii) | similarly, , |
2 1 1 | 2 2 1
c c c c c c c c b c a c ∴ ∴
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4.3(b)) (Theorem
2 1
c c = ∴
a a = ≠ ∈ 0) gcd(a then a if Z
b d b ll d l ti l i h d( b) 1
a a ≠ ∈ 0) gcd(a, then , a if , Z
. 1 with , = + ∈ by ax y x Z
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and apply the division algorithm n times as follows:
. ,
+
∈Z b a
and apply the division algorithm n times as follows:
r r r r q r r r r r q r < < + = < < + =
1 2 2 1 1
r r r r q r r r r r q r < < + = < < + =
⋅ ⋅ ⋅ 3 4 4 3 3 2 2 3 3 2 2 1 n n n n n n n n n
r q r r r r r q r = < < + =
− − − − − 1 1 1 1 2
Then rn, the last nonzero remainder, equals gcd(a, b)
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d ) ( | d | If ,
divisor any for , | Verify (i) ) , gcd( at verify th To b b a c r c b a r
n n =
| | and | Next ) (e) 4.3 Theorem ( | then and ), , ( | and | If
, 2 1 1 2 2 1 1 1 1
r c r c r c r r q r r c r r q r b r a r r c r c ⇒ + = + = = = Q
r r q r r r q r + = + =
3 2 2 1 2 1 1
| and | Verify (ii) | down Continuing | | and | Next
3 2 1
b r a r r c r c r c r c
n n n
⇒ ⇒
r r q r + + =
⋅ ⋅ ⋅ 4 3 3 2
) (e) 4.3 Theorem ( equation last the From
, 1 1 2 2 | 1 |
r r q r r r r r
n n n n n n n n
+ = ∴
− − − − −
Q
n n n n n n n
r q r r r q r = + =
− − − − 1 1 1 2
) gcd( ) , ( | [ ] | [ up, Continuing
1 | ] 1 2 | 1 | 2 3 |
b a r b r a r r r r r r r r r r r r r
n n n n n n
= ∴ = = ⇒ ∧ ⇒ ∧
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) , gcd( b a rn = ∴
Ex 4.34 : find the gcd(250,111), and express the results as a linear
combination of these integers.
< < + ⋅ = 111 28 28 111 2 50 2
< < + ⋅ = < < + ⋅ = < < + ⋅ = 27 1 1 27 1 28 28 27 27 28 3 111 111 28 28 111 2 50 2
= ∴ + ⋅ = 1 ) 111 , 250 gcd( 1 7 2 27
⋅ − + − = ⋅ + − = ⋅ − − = ⋅ − = ) ( ] 111 2 250 [ 4 111 ) 1 ( 28 4 111 ) 1 ( 28] 3 111 [ 1 28 27 1 28 1
Z ∈ + + = − ⋅ + ⋅ = k k k ] 250 9 [ 111 ] 111 250[4 1 unique not is it fact, In ) 9 ( 111 4 250
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Z ∈ + − + − = k k k ], 250 9 [ 111 ] 111 250[4 1 1. 111) gcd(250, 111)
111)
111) gcd(-250, = = = =
Ex 4.35 : the integers 8n+3 and 5n+2 are relatively prime.
1 3 1 2 ) 1 2 ( ) 1 3 ( 1 2 5 2 5 1 3 ) 1 3 ( ) 2 5 ( 1 3 8 + < + < + + + = + + < + < + + + = + n n n n n n n n n n 1 1 2 1 2 1 2 ) 1 2 ( 1 1 3 ) ( ) ( < < + ⋅ = + + < < + + = + n n n n n n n n 1 ) 2 5 , 3 8 gcd( 1 = + + + ⋅ = n n n n
(8n+3)(-5) + (5n+2)(8) = -15 + 16 = 1
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type of problem, not just one special case.
E 4 36 U
E lid l ith t d l d (i d d )
Ex 4.36 : Use Euclidean algorithm to develop a procedure (in pseudocode)
that will find gcd(a, b) for all a, b ∈ Z+
procedure gcd(a,b: positive integers) p g ( , p g ) begin r:= a mod b d:= b while r > 0 do c:= d d:= r r:= c mod d end return(d) d { d( b) d}
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end {gcd(a,b) = d}
Ex 4.37 : Griffin has two unmarked containers. One container holds 17
containers to measure exactly one ounce containers to measure exactly one ounce.
17 4 , 4 17 3 55 < < + ⋅ = ] 7 1 3 55 [ 4 17 4 4 17 1 4 1 , 1 4 4 17 ⋅ − − = ⋅ − = < < + ⋅ = 55 4 17 13 ] [ ⋅ − ⋅ =
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Ex 4.38 : On the average, Brian debug a Java program in six minutes, but
it takes 10 minutes to debug a C++ program. If he works for 104 minutes and doesn’t waste any time how many programs can be debug in each and doesn t waste any time, how many programs can be debug in each language.
= + ⇒ = + 52 5 3 104 10 6 y x y x + ≤ ≤ + − + − = − ⋅ + ⋅ = ⇒ − ⋅ + ⋅ = ⇒ = 3 52 5 104 ) 3 52 ( 5 ) 5 104 ( 3 ) 52 ( 5 104 3 52 ) 1 ( 5 2 3 1 1 ) 5 , 3 gcd( k y k x k k Q ⎧ = = = ≤ ≤ ∴ + − = ≤ − = ≤ 2 14 : 18 3 52 , 5 104
5 104 3 52
y x k k k y k x Q ⎪ ⎩ ⎪ ⎨ ⎧ = = = = = = = = = ⇒ 8 , 4 : 20 5 , 9 : 19 2 , 14 : 18 y x k y x k y x k
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⎩
integer solution x = x0, y = y0 if and only if gcd(a, b) divides c.
c is called a common multiple of a b
, , ,
+
∈Z c b a
+
∈Z c b a
if c is a multiple of both a and b. Furthermore, c is the least common multiple of a, b if it is the smallest of all positive integers that are common l i l f b W d b l ( b)
, , , ∈Z c b a
multiples of a, b. We denote c by lcm(a, b).
, , ,
+
∈Z c b a
, |
) , ( lcm , | If ma c b a c c r r qc d d c = ∴ = < < + = / Q | multiple common least the is that claim the contradict , but ) ( , also d c c c r r a qm n r qma na na d ∴ < < = − ⇒ + = =
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. | d c ∴
Theorem 4.10: For ab = lcm(a, b).gcd(a, b)
Ex 4.40 : By Theorem 4.10 we have
, ,
+
∈Z b a
then lcm(a, b) = ab.
, ,
+
∈Z b a
fact that gcd(168,458) = 24. As a result we find that lcm(a, b)?
192 , 3 ) 456 , 168 ( lcm 24 ) 456 , 168 gcd(
24 456 168
= = ∴ =
⋅
Q
24
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57
Lemma 4.2: If and p is prime, then
+
y) ( | )] | ( ) | [( c bx a c a b a + ⇒ ∧
Lemma 4.3:
. 1 some for | | and prime is If . Let
2 1
n i a p a a a p p a
i n i
≤ ≤ ⇒ ∈
⋅ ⋅ ⋅
+
Z
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2 1 i n i
Ex 4.41 : Show that is irrational.
Proof
2
4 2) (Lemma | 2 | 2 2 2 1 ) , gcd( and , , where , 2 not If
2 2 2 2
⇒ ⇒ = ⇒ = ∴ = ∈ = ⇒
+
a a a b b a b a
a b a
Z i ) ( di ) d( | | 4.2) (Lemma | 2 | 2 , 2 4 2 2 4.2) (Lemma | 2 | 2 2 2
2 2 2 2 2 2 2
⇒ = ⇒ = = ⇒ = ∴ ⇒ ⇒ = ⇒ = ∴ b b b b c b c a b c a a a a b
b
tion) (Contradic 2 ) , gcd( | 2 | 2 ≥ ∴ ∧ b a b a Q
p prime every for irrational is p
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y g p p uniquely, up to the order of the primes. ( )
k k
s s s
p p p n ⋅ ⋅ ⋅ =
2 2 1 1
existence (i) m m m m m m m m m m ) 1 ( primes
product as written be can , and , composite) is ( primes.
product a as e expressibl not integer smallest the is not, If existence (i)
2 1 2 1
< < = Q Q n- n m m m m true) are 1 , 2,3,4, for form, ve (alternati Induction al Mathematic use : uniqueness (ii) primes
product a as e expressibl be can ) , 1 (
2 1
⋅⋅ ⋅ = ∴ < < Q
j r t t t r k j i r t r t t k k
q p q q q p n p q q q p p p q p q q q p p p n
s s s
4.3) (Lemma | | | and , primes are , where Suppose
1 2 2 1 1 1 1 2 1 , 2 1 2 2 1 1 2 2 1 1
⇒ ⋅ ⋅ ⋅ ⇒ < ⋅ ⋅ ⋅ < < < ⋅ ⋅ ⋅ < < ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ = Q
. | | and prime is If
2 1 i n
a p a a a p p ⇒
⋅ ⋅ ⋅
i i j j j r
q p q p q p i if p q q p q p q p q q q p p
j
j find t can' we , 1 similarly, primes are and ) ( | | |
1 1 1 1 1 , 1 1 1 2 1 1 1
= ∴ = < = ∋ > = = ⇒ Q Q
. |
i
a p ⇒
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60
i i i i r t r t t k k
t s t s q p r k n n q q q p p p n
s s s p n
, , , , hypothesis induction By
1 1 1 2 2 1 1 1 2 2 1 1 1 1 1
= = = = ∴ < ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ = = ⇒
− −
Q
Ex 4.44 : How many positive divisors do 29,338,848,000 have? How
Ex 4.44 : How many positive divisors do 29,338,848,000 have? How
many of the positive divisors are multiples of 360? How many of the positive divisors are perfect squares?
For (i)
2 1
p p p n
k
s s s
11 7 5 3 2 000 , 848 , 338 , 29 ) 1 ( ) 1 )( 1 ( is
divisors positive
number The For (i)
3 3 5 8 2 1 2 2 1 1
= + ⋅ ⋅ ⋅ + + ⋅ ⋅ ⋅ = s s s n p p p n
k k k
76 5 1) 1)(1 1)(3 1 1)(3 2 5 )( 1 3 8 ( 5 3 2 360 (ii) 1728 1) 1)(1 1)(3 1)(3 5 )( 1 8 ( answer
2 3
= = + + + + + = ∴ (0,2,4) choices 3 have we , 5 For ) (0,2,4,6,8 choices 5 have we , 8 For (iii) 76 5 1) 1)(1 1)(3 1 1)(3 2 5 )( 1 3 8 ( answer
2 1
= = = + + + − + − + − = ∴ s s 60 1 2 2 3 5 answer (0) choices 1 have we , 1 For (0,2) choices 2 have we , 3 For
5 4 , 3
∴ = = s s s
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60 1 2 2 3 5 answer = ⋅ ⋅ ⋅ ⋅ = ∴
Ex 4.46 : Can we find three consecutive positive integers whose product is
a perfect square, i.e., m(m+1)(m+2) = n2 , ?
+
∈Z n m,
) 4 4 S ti f 21 E i ( ) 2 , 1 gcd( 1 ) 1 , ( gcd fact that the Use exist , such Suppose + + = = + m m m m n m square perfect a also is ) 1 ( square perfect a is | and ), 2 ( | , | then ), 1 ( | if , prime any For ) 4.4 Section
21 Exercise (
2 2
+ ∴ + / / + ∴ m n n p m p m p m p p Q ) 1 ( 1 2 ) 2 ( square perfect a also is ) 2 ( square perfect a also is ) 1 ( square perfect a is
2 2 2
+ = + + < + < + ∴ + ∴ m m m m m m m m m n Q Q integers. positive e consecutiv such three no are e that ther conclude we So, square perfect a be cannot ) 2 ( + ∴ m m
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4-1 :18, 24
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