math 409/409g history of mathematics
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Math 409/409G History of Mathematics. Perfect Numbers. What’s a perfect number?. A positive integer is a perfect number if it is equal to the sum of all its positive divisors except itself. - PowerPoint PPT PresentationTRANSCRIPT
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Math 409/409GHistory of Mathematics
Perfect Numbers
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What’s a perfect number?
A positive integer is a perfect number if it is equal to the sum of all its positive divisors except itself.
For example, 6 is a perfect number since the positive divisors of 6 are 1, 2, 3, and 6 and 6 1 + 2 + 3 (the sum of all positive divisors of 6 except 6).
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In fact, P1 6 is the smallest perfect number. The next in line is P2 28. (The positive divisors of 28 are 1, 2, 4, 7, 14, and 28 and 28 1 + 2 + 4 + 7 + 14.) The third and fourth perfect numbers are P3 496 and P4 8128.
As another example, no prime number p can be a perfect number since the only divisors of p are 1 and p and p ≠ 1.
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Why study perfect numbers?
Ancient philosophers thought that perfect numbers had mystical and religious significance: God created the world in 6 days and rested on the seventh; it takes 28 days for the moon to circle the earth.
We, of course, study perfect numbers for the sheer beauty of the mathematics.
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How do you find the positive divisors of a number?
Look at the prime factorization of the number.
The positive divisors of n are of the form
where
31 21 2 3
rkk k krn p p p p
31 21 2 3
raa a ard p p p p
0 1,2,3, , . for i ia k i r
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For example, 28 22 ·7. So the positive divisors of 28 are of the form d 2a ·7b where a 0, 1, or 2 and b 0 or 1. Since there are 3 choices for a and 2 for b, 28 has 3·2 6 positive divisors.
With b 0 we get the divisors 20, 21, and 22. And with b 1 we get 20 ·7, 21 ·7, and 22 ·7. So the six positive divisors of 28 are 1, 2, 4, 7, 14, and 28.
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Finding all positive divisors of a number can be an exhausting task.
For example, if n 25 ·33 ·72 ·11, then when determining the positive divisors of n we have: 6 choices for the exponent of two, 4 for the exponent of three,
3 for the exponent of seven, and 2 for the exponent of eleven.
So n has 6·4·3·2 144 positive divisors!
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To determine if n 25 ·33 ·72 ·11 is a perfect number we would have to find the 143 divisors other than n itself and then add them up.
There’s got to be a better way! We need a formula for the sum of these divisors.
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The sigma function
If n is a positive integer, then σ(n) is defined to be the sum of all the positive divisors of n.
Examples:
The positive divisors of 6 are 1, 2, 3, and 6. So σ(6) 1 + 2 + 3 + 6 12.
The positive divisors of 28 are 1, 2, 4, 7, 14, and 28. So σ(28) 56, the sum of all these positive divisors.
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Since σ(n) is the sum of all the positive divisors of n, σ(n) – n is the sum of all positive divisors except n. So n is a perfect number when σ(n) – n n. That is
n is a perfect number if σ(n) 2n.
Examples: 6 and 28 are perfect numbers since σ(6) 12 2·6 and σ(28) 56 2·28.
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Theorem:
Example:
1 21 2
rk k krn p p p
1 21 1 11 2
1 2
1 1 1( ) .
1 1 1
rk k kr
r
p p pn
p p p
3 22 1 2 1 7 1
28 2 7 (28) 56.2 1 7 1
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Proof that
For i 1, 2, 3, …, r, let
Ex. For n 28 22 ·71,
P1 1 + 2 + 22 and P2 1 + 7.
Consider the product P1P2P3···Pr .
1 2
1 2
1 1 11 2
1 21 2
1 1 1( ) .
1 1 1
r
r
k k kk k k r
rr
p p pn p p p n
p p p
2 31 .iki i i i iP p p p p
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Ex. For n 28,
P1P2 (1 + 2 + 4)(1 + 7)
1 + 2 + 4 + 7 + 14 + 28.
Each positive divisor of n appears exactly once in the expansion of this product, so σ(n) P1P2P3···Pr .
But is a
geometric series. Thus
2 31 iki i i i iP p p p p
1 1.
1
iki
ii
pP
p
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We now have our desired formula:
1 2
1 2
1 1 11 2
1 2
( )
1 1 1.
1 1 1
r
r
k k kr
r
n PP P
p p p
p p p
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Example: Is n 25 ·33 ·72 ·11 a perfect number?
Solution:
So n is not a perfect number.
6 4 3 2
5 4
2 1 3 1 7 1 11 1( )
2 1 3 1 7 1 11 1
2 3 5 7 19 2
n
n
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Can we generate perfect numbers?
Consider the first four perfect numbers
They are each of the form of 2k ·p where p is a prime number.
11
22
43
64
6 2 3
28 2 7
496 2 31
8128 2 127
P
P
P
P
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Will any number of the form 2k ·p where p is a prime number be a perfect number?
No. Consider 44 22 ·11.
So 44 is not a perfect number.
3 22 1 11 1(44) 84 2 44.
2 1 11 1
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What condition must be placed on the prime p and the exponent k of 2?
1
2
4
1
2
3
46
6 2
28 2
4
3
7
396 1
1
2
8128 2 27
P
P
P
P
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We now see that the first four perfect numbers are of the form 2k(2k + 1 – 1).
Is every number of this form a perfect number?
1 1
2 2
4 4
1
2
3
2
3
5
74
6 6
6 2
28 2
4
2 (2 1)
2 (2 1)
2 (2 1)
3
7
31
2
96
127 (2 1)
2
8128 2
P
P
P
P
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Consider n 28(29 – 1) 28 ·7·73 130,816.
So n is not perfect, and thus not all numbers of the form 2k(2k + 1 – 1) are perfect numbers.
Which numbers of the form 2k(2k + 1 – 1) are perfect numbers?
9 2 22 1 7 1 73 1( ) 302,512 2
2 1 7 1 73 1n n
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The pattern we found for the first four perfect numbers was that they were of the form 2k(2k + 1 – 1) where 2k + 1 – 1 is prime.
1 1
2 2
4 4
1
2
3
2
3
5
74
6 6
6 2
28 2
4
2 (2 1)
2 (2 1)
2 (2 1)
3
7
31
2
96
127 (2 1)
2
8128 2
P
P
P
P
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In the last example of a number that was not perfect, n 28(29 – 1) where 29 – 1 7·73 is not prime.
So will the numbers of the form
2k(2k + 1 – 1) where 2k + 1 – 1 is prime
generate perfect numbers?
Yes, as shown in the next theorem.
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Euclid’s Theorem: If 2k + 1 – 1 is prime, then 2k(2k + 1 – 1) is perfect.
Pf: Let p 2k + 1 – 1 be prime and set n 2k
·p.
Then
But p 2k + 1 – 1, and thus p + 1 2k + 1.
So n is perfect since
11
22 1 1( ) .
2 1 1(2 1 ( 1))k
k
pp
pn
1( ) 2 2( .2 ) 2k k p npn
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So Euclid’s theorem will generate even perfect numbers. Actually, it can be shown that all even perfect numbers are of the form 2k(2k + 1 – 1) where 2k + 1 – 1 is prime.
So Euclid’s theorem will generate all even perfect numbers. But generating these even perfect numbers not as easy as it looks since for each k, you have to determine if 2k + 1 – 1 is prime.
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If we look at the first four perfect numbers for inspiration to help us determine when 2k + 1 – 1 is prime, we suspect that k + 1 must be prime.
1 1
2 2
4 4
1
2
3
2
3
5
64
6 7
6
28
4
2 3 2 (2 1)
2 7 2 (2 1)
2 31 2 (2 1)
2 1
96
27 2 (812 28 1)
P
P
P
P
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Although I will not prove it here, it has been shown that if 2k + 1 – 1 is prime, then k + 1 must be prime.
Putting this together with Euclid’s theorem shows that the every even perfect numbers is of the form 2p - 1(2p – 1) where p is prime.
But we are still not guaranteed that every number of the form 2p - 1(2p – 1) where p is prime will be a perfect number.
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Ex: p 11 is prime, but 211 – 1 23·89 is not prime. So n 210(211 – 1) is not a perfect number.
The proof of this example is left as an exercise.
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This ends the lesson on
Perfect Numbers