102 Week 1

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    Hilberts Hotel

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    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 ...

    n 2n

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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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    There is no biggest prime.

    (i.e. the set of all primes is infinite)

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    There is no biggest prime.

    (i.e. the set of all primes is infinite)

    Proof byreduction ad absurdum:

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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.

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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.

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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

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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.

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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.

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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.

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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 ..

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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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    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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    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.