1. Hilbert’s Hotel 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
...

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

3. 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

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

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

6. 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

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

8. 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

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

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

11. 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.

12. 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.

13. 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

14. 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.

15. 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.

16. 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.

17. 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

18. 1st bus: passenger n goes to the room 3n2 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

19. 1st bus: passenger n goes to the room 3n2 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

20. 2nd bus: passenger n goes to the room 5n1 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

21. 3rd bus: passenger n goes to the room 7n1 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

22. 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 Bus Bus
Bus Bus Bus
Bus Bus Bus

23. 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
...

24. 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  7
21 = 3  = 3  11
33 = 3  = 9  5 = 32  5

25. 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

26. 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

27. 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

42.  5, if the nth digit of rn  5
the nth digit of r0 = 
 6, if the nth digit of rn  5

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