Existence of Irrational Number that has same $n$ digits of a given Rational Number.

Question

Problem: Any decimal number $z$ of $n$ digits after decimal point is given. Does there always exist an irrational number $q$ (or expression related to irrational number) which has same $n$ digits of $z$ after decimal point?

Example: Consider, rational $0.41421356237309504880168872420$ as $z$, then $q=\sqrt2$ since $\sqrt2$ has same $29$ digits of $z$ after decimal point.

Decimal Point is a point or dot used to separate the whole number part from the fractional part of a number.

Answer 1

Yes, and you can explicitly construct such number. Suppose your rational number is $$d = 0.d_1d_2d_3 \dots d_n$$ where $d_i \in \{0,1,\ldots,9\}$ stands for a digit .

Then $d$ is rational. Now take $d' = d + 10^{-n-1} \cdot \sqrt{2}$. This number has digits $$d' = 0.d_1d_2d_3 \dots d_n 141421356 \dots$$ where the tail is just the decimal expansion of $\sqrt{2}$. Clearly $d'$ is irrational and $d$ and $d'$ have the same first $n$ digits after the decimal point.

You can do something similar for any rational and with any irrational, not just $\sqrt{2}$.

Answer 2

Since the rationals are dense in the reals, so are the rationals times $\sqrt{2}$ (or any other irrational number), and these numbers are all irrational.

This does what is wanted.