To show that the set of real numbers is larger than the set of natural numbers we assume that the real numbers can be paired with the natural numbers and arrive at a contradiction. So suppose we can order the real numbers thus:

1 A.a1a2a3a4 ... 2 B.b1b2b3b4 ... 3 C.c1c2c3c4 ... 4 D.d1d2d3d4 ... 5 E.e1e2e3e4 ... 6 F.f1f2f3f4 ...

So the first number has an integer part *A* and the
decimal digits are *a1*, *a2*, *a3*,
and so on. (Of course, it does not matter that we run out
of letters, the notation is just due to the limitations of
the html language).

We'll now construct a number *z* that is not in the
above list. *z* is obtained as follows:

The integer part of *z* is zero. If *a1=2* we
make the first digit of *z* equal to 3, otherwise we
make it equal to 2. Similarly, we make the second digit of
*z* equal to 3 if *b2=2* and 2 otherwise. In
general, the *i* -th digit of *z* is 3 if the
*i* -th digit of the *i* -th number in the
list is 2 and 3 otherwise.

Now, *z* differs from every number in the list (in
fact, it differs from the *i* -th number in the *i
* -th digit), and so it cannot be in the list. The set
of real numbers is larger than the set of rational numbers.

**Note 1**: There's a small technical difficulty in that
the decimal representation of a real number is not
necessarily unique. For example,

However, this problem can be overcome easily (by picking that representation of a number that ends with an infinite string of zeros rather than nines).

** Note 2**: Cantor showed that the set of real numbers
is equivalent to the powerset of the natural numbers.

Fine print, your comments, more links, Peter Alfeld, PA1UM

[16-Aug-1996]