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.