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2012/12/01

irrational numbers 2

<exercises>

1. \sqrt{n^2+1} is an irrational number, in which n\in N.

[proof] If we put \sqrt{n^2+1}=p/q, n^2=p^2/q^2-1. We also know n<\sqrt{n^2+1}<n+1. Although n^2 is an integer number, obviously, p^2/q^2-1 is not an integer. These are a contradiction.

2. \log_{10}2 is an irrational number.

[proof] we set \log_{10}2=p/q, where p,q\in \mathbb{Z}, q\ne 0. Therefore 2^q=10^p. However, 10^p=(2^p)(5^p). Two equations can never exist at the same time, because uniqueness of the  factorization.  As these show that there cannot be the initial definition of \log_{10}2=p/q, \log_{10}2 is an irrational number.

3. e is an irrational number.

[proof] First of all, we accept the fact that
 e=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\cdots +\frac{1}{n!}+\cdots  

Next, as always, let e be p/q>0. If we make n in above equation q,
the equation is decomposed to
 e=\left( \frac{1}{0!}+\cdots +\frac{1}{q!}\right) +\left( \frac{1}{(q+1)!}+ \frac{1}{(q+2)!}+\cdots \right) 

Therefore we can get the equation
  q!\left( \frac{1}{(q+1)!}+\frac{1}{(q+2)!}+\cdots \right) =q!\left( \frac{p}{q}-\left( \frac{1}{0!}+\cdots +\frac{1}{q!}\right) \right) 

Obviously the right side of this equation is an integer number.

The other hand, the left side is, as we are able to assume q>1,
\begin{eqnarray*} \mbox{left side } &=& \frac{1}{(q+1)}+\frac{1}{(q+1)(q+2)}+\cdots    \\  &<& \frac{1}{2}+\frac{1}{2^2}+\cdots  =1 \end{eqnarray*}

Since the left side of the equation is not an integer number, a contradiction occurred.

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