Hilbert’s Hotel 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 ...

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

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

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

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 ... 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ... n 2n

There is no biggest prime. (i.e. the set of all primes is infinite)

There is no biggest prime. (i.e. the set of all primes is infinite) Proof by reduction ad absurdum:

There is no biggest prime. (i.e. the set of all primes is infinite) Proof by reduction 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.

There is no biggest prime. (i.e. the set of all primes is infinite) Proof by reduction 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.

There is no biggest prime. (i.e. the set of all primes is infinite) Proof by reduction 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

There is no biggest prime. (i.e. the set of all primes is infinite) Proof by reduction 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.

There is no biggest prime. (i.e. the set of all primes is infinite) Proof by reduction 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.

There is no biggest prime. (i.e. the set of all primes is infinite) Proof by reduction 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.

There is no biggest prime. (i.e. the set of all primes is infinite) Proof by reduction 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

1st bus: passenger n goes to the room 3n 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ... 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 ... 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ... 1st bus: passenger n goes to the room 3n

1st bus: passenger n goes to the room 3n 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ... b1p1 b1p2 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 ... b1p3 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 ... 1st bus: passenger n goes to the room 3n

2nd bus: passenger n goes to the room 5n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ... b1p1 b2p1 b1p2 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 ... b1p3 b2p2 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 ... 2nd bus: passenger n goes to the room 5n

3rd bus: passenger n goes to the room 7n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ... b1p1 b2p1 b3p1 b1p2 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 ... b1p3 b2p2 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 ... 3rd bus: passenger n goes to the room 7n

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ... b1p1 b2p1 b3p1 b1p2 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 ... b1p3 b2p2 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 ... 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

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ... 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 ... 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 ... 15 35 21 39 33 45

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ... 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 ... 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 ... 15 = 3  5 35 = 5  7 21 = 3  7 39 = 3  11 33 = 3  11 45 = 9  5 = 32  5

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ... 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 ... 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 ... 3n  5m 3n  13m 3n  7m 3n  17m 3n  11m

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ... 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 ... 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 ... 5n  7m 5n  11m 5n  13m 3n  5m 3n  13m 3n  7m 3n  17m 3n  11m

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ... 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 ... 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 ... 3n  5m  7p 3n  5m  7p 19q  43t 5n  7m 5n  11m 5n  13m 3n  5m 3n  13m 3n  7m 3n  17m 3n  11m

 5, if the nth digit of rn  5 the nth digit of r0 =   6, if the nth digit of rn  5

r0 0.  5, if the nth digit of rn  5 the nth digit of r0 = 