Respuesta :
A decimal expansion of a rational number either terminates or follows a repeating pattern of a *finite* sequence of digits.
Neither of these is the case here. The decimal expansion is obviously non-terminating, nor is the sequence of digits finite.
Writing the number as
[tex]2.010010001\ldots=1+1.010010001\ldots[/tex]
and only considering the second number, you have the following sequence of digits: [tex]\{1,0,1,0,0,1,0,0,0,1,\ldots\}[/tex], where the [tex]n[/tex]th term, starting with [tex]n=0[/tex] corresponds to the number [tex]10^{-n}[/tex]. The sequence can be described recursively by the recurrence
[tex]\begin{cases}a_0=1\\a_{n+1}=a_n+n+2&\text{for }n\ge1\end{cases}[/tex]
and explicitly by [tex]a_n=\dfrac{n(n+3)}2[/tex].
This sequence is not periodic, and indeed diverges to [tex]\infty[/tex] as [tex]n\to\infty[/tex]. This means the number cannot be rational.
Neither of these is the case here. The decimal expansion is obviously non-terminating, nor is the sequence of digits finite.
Writing the number as
[tex]2.010010001\ldots=1+1.010010001\ldots[/tex]
and only considering the second number, you have the following sequence of digits: [tex]\{1,0,1,0,0,1,0,0,0,1,\ldots\}[/tex], where the [tex]n[/tex]th term, starting with [tex]n=0[/tex] corresponds to the number [tex]10^{-n}[/tex]. The sequence can be described recursively by the recurrence
[tex]\begin{cases}a_0=1\\a_{n+1}=a_n+n+2&\text{for }n\ge1\end{cases}[/tex]
and explicitly by [tex]a_n=\dfrac{n(n+3)}2[/tex].
This sequence is not periodic, and indeed diverges to [tex]\infty[/tex] as [tex]n\to\infty[/tex]. This means the number cannot be rational.
