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.
0 件のコメント:
コメントを投稿