Something I've always wondered...
There is an infinite amount of numbers between any two numbers. For example, there are an infinite amount of numbers between say, 1 and 2, but there are also an infinitite amount of numbers between 1 and 3. But for every number, n, between 1 and 2, there is the number n+1 between 1 and 3. So there are more numbers between 1 and 3 than there are between 1 and 2, and yet they are still both infinite. Why is this?
The short answer is: if you were to count, you'd find that there are the same amount of numbers in both sets.
Of course, we'd need to count them in a finite amount of time, though.
Which brings me to the definition of 'counting', and the long answer:
In mathematics, we say that a set contains the same amount of elements as another set if there exists a one-to-one onto function (bijective) that maps each element of the first set to an element of the second set. We can easily see, for example, that the sets {1,2,3,4} and {400,21,8,-24} contain the same number of elements by mapping, say, 1 -> 400, 2 -> 21, 3 -> 8 and 4 -> -24.
This becomes more difficult for infinite sets. Suppose we need to count how many non-negative even numbers are there. We can assign 0 -> 0, 1 -> 2, 2 -> 4, 3 -> 6, etc. This means that the set of non-negative even numbers contains the same amount of elements as the set {0,1,2,3,...}, which is usually called the set of
natural numbers. In mathematics, if a set contains the same amount of elements as the set of natural numbers, then it is called
countably infinite. This is considered the 'smallest' infinity. The number of elements of such a countably infinite set is sometimes called Aleph-0.
Now, in your example, the set of all numbers between 1 and 2 aren't even countably infinite, if you also consider numbers such as the square root of 2. There is an elegant proof by Cantor that explains why, which goes like this: Suppose the numbers between 1 and 2 are countably infinite. Then there must exist a bijective function that maps the natural numbers into the set of numbers between 1 and 2. A number between 1 and 2 can be written as an infinite decimal starting with 1 and a decimal point (1.99999999999... can be proved to be equal to 2, while 1.00000000000... is, of course, equal to 1). Hence we have, say:
0 -> 1.abcdefg...
1 -> 1.hijklmn...
2 -> 1.opqrstu...
etc. This should be a list that contains all numbers between 1 and 2.
Now create a number 1.ABCDEFG... such that
A = 0 if a != 0
A = 1 if a = 0
B = 0 if i != 0
B = 1 if i = 0
C = 0 if q != 0
C = 1 if q = 0
etc. This new number is distinct from each of the numbers in the bijective function, and is also a number between 1 and 2. This means that not all numbers were included in the bijective function after all! So the numbers between 1 and 2 are not countably infinite.
The set of numbers between 1 and 2 thus contain more numbers than the set of natural numbers. It can be shown that it contains as many numbers as the set of
real numbers (the set of real numbers being the set of all infinite decimals). We could just map them as follows, for example:
a.b -> 1.ab
where 'a' is the (finite) set of digits before the decimal point and 'b' is the (infinite) set of digits after the decimal point.
The amount of numbers in the set of real numbers is sometimes called Aleph-1. This illustrates the point that there are some 'infinities' that are 'bigger' than 'others'.
Using a similar mapping, the set of all numbers between 1 and 3 can also be shown to contain as many numbers as the set of real numbers, and thus the sets [1..2] and [1..3] are shown to have the same number of elements, Aleph-1. For example, one such mapping would be:
ab.c -> 1.ac, if b is the digit 0,1,2,3 or 4
ab.c -> 2.ac, if b is the digit 5,6,7,8 or 9
where 'a' is the (finite) set of digits before the decimal point except the last digit (this set could be empty), b is the digit before the decimal point, and 'c' is the (infinite) set of digits after the decimal point.
That's the long answer. If you didn't understand anything, don't blame me. You asked the question, I provided you with the exact answer!
