• Cantor's first diagonal argument.

• "Godel counting" argument.
• when a >= 0, b>0 , ... a/b  corresponds to  2^b3^a
• when a<0, b>0, ... a/b corresponds to 2^b5^a

Definition 1. An algebraic number is the solution to a polynomial equation with integer coefficients.
 

Fact 2. A polynomial equation of degree N can have at most N distinct real number solutions.
 

Fact 3. There are a finite number of integer polynomial equations of degree N where the sum of the absolute values of the coefficients is less than a given number K.
 

Proof Procedure: (modified slightly from Cantor's)
First:
List algebraic number roots from integer polynomials of degree 1 with sum of the absolute values of the coefficients equal to 1. (This is just 1x=0 and -1x=0.)

Then:
use degree 1, sum 2;
(This is 2x=0, -2x=0, 1+1x=0, -1+1x=0, 1-1x=0, -1-1x=0.)
then degree 2, sum 1;
then degree 3, sum 1;
then degree 2, sum 2;
then degree 1, sum 3;. ...
 

Consider a sequence of integers, with each integer between 0 and 9.
If we let kth integer of the sequence determine the 2k+1 th decimal place of a number then we can determine a unique real number by specifying that the 2k th decimal be a 5.

Let S be the set of of all real numbers that are specified in this fashion.
S is a subset of [0,1].
 

Suppose there is a function,  f : N ->  S.
Then let b=.b1 5 b3 5 b5 5 b7 5 b9 5 ... where we choose  b2k+1 so that it is not equal to the 2k+1st digit of f(k).

Thus b is a member of s but b not equal to f(k) for any k, so f is not "onto".
Hence S is not a countable set, and consequently neither is [0,1].
 

Note: The set of {0,1} - valued  sequences is also "uncountable."
 

Proof: Suppose f: R -> P(R).
Let  B = {x such that x is not an element of f(x)}.
Suppose B = f(b) for some b.
If b is in B then b is not in f(b)=B, which is a contradiction.
So b is not in B, but then b is not in f(b), so b is in B!

Thus B is not f(b) for any b, and f is not onto.
 

Thus: There are sets which are larger than the reals.
 

Footnote  on measuring sets of real numbers:

The rational numbers between 0 and 1 have  "measure" zero.
Proof:  List the rationals a₁,a₂,a₃,...  and for any number s,  use an interval of length s/2^k about a_k. Then the set of rational numbers has measure less than s. Since s can be any number, the set of rationals between 0 and 1 has measure 0.

Generalization:  Any countable set of real numbers has "measure" zero.