Really, how many were in the hotel room? Our intuition complete misfit hidden in this issue. Think

Our perception is deceived us about eternity. The Turkish "forever," it has a mold. Endless marks ans spencers train tracks' also appears as if the invention. Though infinitely far away, but nothing marks ans spencers ever comes a time when we can not reach such a place. Or what happens when reached unknowable as a mystical place. 19. There was no agreement on the concept marks ans spencers of the infinite among mathematicians up to the end of the century, was defined as a mathematical concept of infinity. Defining the mathematical infinite Georg was with the development of Cantor's set theory [3, pages 238-239].
The purpose of this hypothetical German mathematician David Hilbert marks ans spencers created by our hotel full of everyday mathematical infinity infinite perception, to underline that the agreement with our intuition. A hotel with an infinite number marks ans spencers of rooms Hilbert Hotel. The rooms are numbered by natural numbers: 0,1,2,3,4,5, ...
Each room has a number, marks ans spencers a natural number of the number of each room; but it does not mean the end of the room. What you buy is natural numbers, numbers had a room with that number. This hotel, let's assume that in the same way that a busload of visitors come in an infinite number of seats. Seats are numbered with the natural numbers: 0,1,2,3,4,5, ... (There marks ans spencers is also a passenger marks ans spencers in every seat). There is no difficulty in place these passengers to the hotel.
Really, how many were in the hotel room? Our intuition complete misfit hidden in this issue. Think Like an infinite number of misleading us. All passengers need to do a mapping accomodation, and two passengers not to match the same accomodation.
In this way kodlayal all passengers of the bus 0. Overall (0, n), 0 bus of the code will be sitting in the passenger seat number. Kodlayal passengers in one bus in the same way. (1 m), are encoded in the passenger sitting in seat number 1 bus m.
Now let's put in the room for passengers. The number of passengers on the second bus let us place the code in the rooms numbered 0 through count natural doubled. That code (0,0) to be moved to the passenger room number 0, code (0,1) passenger room number 2, code (0,2) is the number 4 passenger room. Overall code (0, n) the number of passengers will be moved to 2n room. 0 bus of all passengers in this way, let us place to drop one passenger in each room. 0 all bus passengers in all rooms are number one number after they had settled in the room was still empty mind you. As you may have guessed in 1 bus passengers will be placed in the room with only the numbers marks ans spencers numbered. Overall code (1, m) 2m + 1 number will be placed in the passenger room. So we put all passengers marks ans spencers to be a passenger in every room.
Zorlaştıral further questions. Suppose that three bus passengers came this way. Passengers hotel room we can still put each one to be a passenger? Answer marks ans spencers yes again. Likewise, kodlayal all passengers. This time there's a bus, let him also 2 bus. 2 buses of the month so kodlayal passengers. Overall (2, k) code to the passenger sitting in seat number 2 bus k. Now let's placement:
Let's place the second number of rooms 0 bus three times in the code of passengers. So (0, n) 3n-coded number of passengers are settled in the room. 0 bus when everyone settles in a numbered only with the number of rooms that can be divided into 3 has expired. Let's put one bus of passengers as follows: code (1, m) 3m + 1 passenger numbers are settled in the room. So we put one of the passengers of the bus number 3 to the number of remaining portion of the room first. 2 buses will place the remaining number of course rooms numbered with section 2 of the 3. Code (2, k) the number of passengers 3k + 2 will be placed in the room. Again, all passengers to be one person in every room, we placed without marks ans spencers leaving anyone exposed.
Z = {..., -3, -2, -1,0,1,2,3, ...} Let's look to the clusters. Which one has more elements? At first glance just like the Z Hilbert hotel's marks ans spencers seems to have more elements. N wherein each element corresponds to Z 'As there are two elements. A similar argument Z and N's, we will show that it contains the same amount of staff. To be a bit more mathematical Z's elements N 'leads to the elements, and any natural number of publicly release the (covering) and two different integers leads to different natural numbers (one to one) will find a function f. F we find Z's and N's staff and will be exactly the same amount of elements that match the covers.
0 and define our function before the positive number, n is a positive marks ans spencers integer or 0, f (n) = 2n get. So the function f 0 and positive numbers leads to two floors. We will also define as negative numbers: n is a positive integer again f (n) = 2n - get one. So negative numbers were matched with an odd number. We define exactly the function f and clear that covering. Therefore, to be exact and natural marks ans spencers numbers covering the full number eşled