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8/3/2019 102 Week 1
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ...
Hilberts Hotel
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ...
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ...
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ...
n n+1
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ...
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ...
n 2n
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ...
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ...
n 2n
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ...
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 .
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 .
n 2n
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There is no biggest prime.
(i.e. the set of all primes is infinite)
8/3/2019 102 Week 1
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There is no biggest prime.
(i.e. the set of all primes is infinite)
Proof byreduction ad absurdum:
8/3/2019 102 Week 1
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There is no biggest prime.
(i.e. the set of all primes is infinite)
Proof byreduction ad absurdum:
Suppose the opposite. i.e. that there is a natural number which is a
prime and is bigger than any other prime. Call it p.
8/3/2019 102 Week 1
12/43
There is no biggest prime.
(i.e. the set of all primes is infinite)
Proof byreduction ad absurdum:
Suppose the opposite. i.e. that there is a natural number which is a
prime and is bigger than any other prime. Call it p.
Let p1, p2, ..., pn be all the primes smaller than p.
8/3/2019 102 Week 1
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There is no biggest prime.
(i.e. the set of all primes is infinite)
Proof byreduction ad absurdum:
Suppose the opposite. i.e. that there is a natural number which is a
prime and is bigger than any other prime. Call it p.
Let p1, p2, ..., pn be all the primes smaller than p.
Consider the number q = p1 p2 ... pn + 1
8/3/2019 102 Week 1
14/43
There is no biggest prime.
(i.e. the set of all primes is infinite)
Proof byreduction ad absurdum:
Suppose the opposite. i.e. that there is a natural number which is a
prime and is bigger than any other prime. Call it p.
Let p1, p2, ..., pn be all the primes smaller than p.
Consider the number q = p1 p2 ... pn + 1
q is not divisible by any natural number except 1 and q itself.
Therefore, q is a prime.
8/3/2019 102 Week 1
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There is no biggest prime.
(i.e. the set of all primes is infinite)
Proof byreduction ad absurdum:
Suppose the opposite. i.e. that there is a natural number which is a
prime and is bigger than any other prime. Call it p.
Let p1, p2, ..., pn be all the primes smaller than p.
Consider the number q = p1 p2 ... pn + 1
q is not divisible by any natural number except 1 and q itself.
Therefore, q is a prime.
q is also bigger than p.
8/3/2019 102 Week 1
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There is no biggest prime.
(i.e. the set of all primes is infinite)
Proof byreduction ad absurdum:
Suppose the opposite. i.e. that there is a natural number which is a
prime and is bigger than any other prime. Call it p.
Let p1, p2, ..., pn be all the primes smaller than p.
Consider the number q = p1 p2 ... pn + 1
q is not divisible by any natural number except 1 and q itself.
Therefore, q is a prime.
q is also bigger than p.
Therefore, q is a prime bigger than p, which contradicts our
assumption that p is the biggest prime.
8/3/2019 102 Week 1
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There is no biggest prime.
(i.e. the set of all primes is infinite)
Proof byreduction ad absurdum:
Suppose the opposite. i.e. that there is a natural number which is a
prime and is bigger than any other prime. Call it p.
Let p1, p2, ..., pn be all the primes smaller than p.
Consider the number q = p1 p2 ... pn + 1
q is not divisible by any natural number except 1 and q itself.
Therefore, q is a prime.
q is also bigger than p.
Therefore, q is a prime bigger than p, which contradicts our
assumption that p is the biggest prime.
Conclusion: There is no biggest prime. QED
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ..
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ..
1st bus: passenger n goes to the room 3n
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 .
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31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
1st bus: passenger n goes to the room 3n
b1
p1
b1
p2
b1p3
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
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31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
2nd bus: passenger n goes to the room 5n
b1
p1
b1
p2
b1
p3
b2
p1
b2
p2
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
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31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
3rd bus: passenger n goes to the room 7n
b1
p1
b1
p2
b1
p3
b2
p1
b2
p2
b3
p1
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
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31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 ..
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ..
Bus 4 11 Bus 7 19 Bus 10 31
Bus 5 13 Bus 8 23 Bus 11 37
Bus 6 17 Bus 9 29 Bus 12 41
b1
p1
b1
p2
b1p3
b2
p1
b2
p2
b3
p1
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 ..
8/3/2019 102 Week 1
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31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
15 35
21 39
33 45
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 .
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31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
15 = 3 5 35 = 5 7
21 = 3 7 39 = 3 11
33 = 3 11 45 = 9 5 = 32 5
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 .
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31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
3n 5m 3n 13m
3n 7m 3n 17m
3n 11m
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 .
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31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
3n 5m 3n 13m
3n 7m 3n 17m
3n 11m
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
5n 7m
5n 11m
5n
13m
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31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
3n 5m 3n 13m
3n 7m 3n 17m
3n 11m
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
5n 7m
5n 11m
5n 13m
3n 5m 7p
3n 5m 7p 19q 43t
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5, if the nth digit of rn 5
the nth digit of r0 = 6, if the nth digit of rn 5
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5, if the nth digit of rn 5
the nth digit of r0 = 6, if the nth digit of rn 5r0 0.