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Prove that , for every real number $c$, there exists a sequence ${a_{n}}$ of rational numbers and a sequence ${b_{n}}$ of irrational numbers such that $\lim a_{n} = \lim b_{n} = c$, could anyone give me a hint please?

My attempt: one part is answered here For every real number $a$ there exists a sequence $r_n$ of rational numbers such that $r_n$ approaches $a$.

Intuition
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    Decimal expansion of $c $? Density or rationals/irrationals (answer to your previous question) is likely useful too. – AnyAD Sep 27 '18 at 10:37
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    Take $r_n$ as in the cited answer and use $s_n:=r_n+\frac{\sqrt{2}}n$. – Jens Schwaiger Sep 27 '18 at 11:17
  • @JensSchwaiger I could not understand .... could you please explain it in details ..... also even the cited answer I did not understand .... it seems like it lacks alot of details. – Intuition Sep 27 '18 at 22:34
  • As mentioned in the problem cited in the present question the existence of a sequence $(r_n)$ of rationals converging to $c$ is more or less a basic property of the field of reals. To get a sequence $(s_n)$ of irrational numbers converging to $c$ it is enough to put $s_n:=r_n+t_n$ where all $t_n$ are irrational and where $\lim_{n\to\infty}t_n=0$. (Note that all $s_n$ are irrational since otherwise the difference $t_n=s_n-r_n$ would be rational too.) Finally $\sqrt{2}$ is irrational and so is $t_n:=\frac{\sqrt{2}}{n}$. Also note that this sequence converges to $0$. – Jens Schwaiger Sep 28 '18 at 04:17

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