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Definition
Proof
Here is a counting. That is, we list the elements
N 1 2 3 4 5 6 7 ... Z 0 1 -1 2 -2 3 -3 ...You can (if you wish) write down a formula: 1/2 n if n is even, 1/2 (n - 1) if n is odd.

N is countable.
Proof
Count the points with integer coefficients in the positive quadrant as shown. (The formula is now rather tricky to write down.)


Proof
Write down a positive rational x/y at the point (x, y) in the plane and count as in the last example, except that you can leave out a point if you have already counted the rational. You have to do this because, for example, 1/1 = 2/2 etc.
This will count all the positive rationals. Then use the same trick as in example 1 to count all the rationals.

Suppose we could write down all the decimal expansions of the reals in (0, 1) in a list:
0.a1a2a3a4a5...
0.b1b2b3b4b5...
0.c1c2c3c4c5...
0.d1d2d3d4d5...
...
and now define a decimal x = x1x2x3x4... by x1
a1 (or 9), x2
b2 (or 9), x3
c3 (or 9), etc.
Then the decimal expansion of x does not end in recurring 9's and it differs from the nth element of the list in the nth decimal place. Hence it represents an element of the interval (0, 1) which is not in our counting and so we do not have a counting of the reals in (0, 1).

A then count through the elements of A leaving out those which are not in B.
is finite or countable.
N is countable.
{1} to count A1 , N
{2} to count A2 , etc.
N which is thus countable by the last result.
n is a root of x2 - n = 0 and so is algebraic.
2 +
3 satisfies the equation (x2 - 5)2 = 24 x2 and so is also algebraic.
Q
...
Q (n+ 1 times). This is countable by an extension of the above result about N
N and so there are only countably many such polynomials. Such a polynomial can have at most n roots and so there are only countably many such roots.
0, 1) and x is irrational then ax is transcendental. This was proved by Gelfond in 1934.
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