# Are there more irrational than rational numbers?

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The proof goes in two steps. First you prove that Q (the set of rational numbers) is countable. This means that you can find a one-to-one relation to Z, the set of integers.First of all if a set is countable, then a subset of this set is also countable.

Secondly, it is sufficient to look at the positive rationals. Now make a 2-dimensional infinite matrix M in the following way:<BR/>

<BR/>

1/1 2/1 3/1 4/1 5/1 6/1 7/1 ...<BR/>

1/2 2/2 3/2 4/2 5/2 6/2 7/2 ..<BR/>

1/3 2/3 3/3 4/3 5/3 6/3 7/3 ...<BR/>

.<BR/>

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Note that all elements of Q are in M, lots of them even more than once.<BR/>

Next step is to make of this 2-dimensional matrix a 1-dimensional row. We do this in the following way:<BR/>

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1/1 2/1 1/2 3/1 2/2 1/3 4/1 3/2 2/3 1/4 ...in other words, we take every element of M by looking diagonally:<BR/>

(1/1) (2/1 1/2) (3/1 2/2 1/3) (4/1 3/2 2/3 1/4)<BR/>

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Thus we now made a one to one correspondence between the elements of M and the set of natural numbers (i.e. the index of the row). But this means that the set, consisting of the elements of M is countable. But now Q is also countable, because it is a subset.

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For the second step in the proof we have two options: options 1 is to use set-theory and this will take 12 pages. The other option is to go to the binary representation of R. This is a bit tricky, i agree, but the proof is very elegant then. Denote by E the set of all infinite sequences with 0's and 1's, for example 1010001110101010...<BR/>

If you put a dot somewhere, this is a binary representation of a real number. We will show that A is not countable. Suppose E is a countable( but infinite) subset and thus we can put the elements of E in some order (the one to one correspondence provides this). Make a new number s in the following way: we look at the nth place of the nth element of E. If it is a 1, we put a 0 on the nth place in s. If it is a 0, we put a 1 on the nth place of E. For example, if E=<BR/>

10100101010101...<BR/>

10010111101011..<BR/>

01011010111010..<BR/>

01010110101011..<BR/>

10101011100110..<BR/>

then s=01100...<BR/>

Thus, s is not an element of E, since every element in s differs at least at one point (on the nth place). But clearly we have that s is an element of our original set A. Thus every countable subset is a proper subset of A, since there is at least one element in A that is not in E. Finish the proof yourself. Why can't A be countable?

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The idea of the proof was first used by Cantor, and is called Cantor's diagonal process.

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<H2> The Euler equation </H2>

The Euler equation is not a theorem, it is a definition. The only thing left to prove is that it fits with the common function exp(x). To see this is not so trivial. What you can do is consider the Taylor expansion series (which is often taken as the definition of the complex exponential function): Let us assume that |z|=1 and thus z=ix with 0<=x<2pi.

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<BR/>

             ___            ___             ___            ___                     ___<BR/>

              \   <u>(ix)</u><sup>n</sup>      \     <u>(ix)</u><sup>n</sup>       \     <u>(ix)</u><sup>n</sup>      \     (-1)<sup>m</sup><u>x</u><sup>2m</sup>        \      (-1)<sup>m</sup><u>x</u><sup>2m+1</sup> <BR/>

exp(z) = /__ n!     = /__   n!     + /__   n!     = /__      (2m)!    + i /__     (2m+1)!    =cos(x)+isin(x)<BR/>

             n=0

           n=0             n=0

            m=0

                 m=0 <BR/>

          

          

      

n  even

     

n  odd

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I know the sum-symbol sucks, but it is not so easy to find a better one (not even guideML can help me there).

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