# Proving that a rational and irrational sequence converge to the same value

In order to prove that the Dirichlet function is not continuous for all real numbers, x, I first want to prove that there is a rational and irrational sequence that converge to the same value, $a$.

Let $a$ $\in$ $\mathbb{R}$. For any $n \in \mathbb{N}$, $a$ - $1\over n$< $a$ + $1\over n$. By a Theorem from class (If $x$ and $y$ are arbitrary, distinct real numbers, then there exists a rational number between $x$ and $y$), there exists a rational, $a_n$ between $a$ - $1\over n$ and $a$ + $1\over n$, i.e. $a$ - $1\over n$ < $a_n$ < $a$ + $1\over n$. Thus |$a_n - a$| < $1\over n$ and $a_n$ converges to $a$.

I think this takes care of the rational sequence, but I am having trouble coming up with an irrational sequence that converges to $a$. Is there a similar approach to proving that an irrational sequence exists that converges to the same value? Any hints would be greatly appreciated.

• You can put $b_n=a_n+\frac{\sqrt{2}}{n}$, as $a_n$ is rational, $b_n$ is irrrational. – Kelenner Apr 25 '17 at 15:58
• does $b_n$ converge to $a$ in this case? – Mathgirl Apr 25 '17 at 16:00
• Yes, as $a_n\to a$ and $\sqrt{2}/n \to 0$. – Kelenner Apr 25 '17 at 16:01
• One trivial argument is that between any two distinct reals there is an irrational number. Then you just use the same approach as with the rational sequence. To see that this is the case, let $x<y$. We know there is a rational $r$ between $x+\sqrt2$ and $y+\sqrt2$. Then $r-\sqrt2$ is irrational, and lies between $x$ and $y$. – Andrés E. Caicedo Apr 25 '17 at 16:07

Here's an example of this. Let $a$ be irrational. Then the sequence $a_n = a$ for all $a$ is an irrational sequence converging to $a$. Now take the decimal expansion for $a$ (every real number has one). A rational sequence converging to $a$ is given by the sequence of partial expansions, i.e $b_n$ is the first $n$ decimal places. So if $a$ is $\pi$, then $b_n$ is the sequence $3, 3.1, 3.14, 3.141, 3.1415, ...$ etc.