How would you show that if we have a sequence { a n } of irrational numbers then for any number a in the reals, the | a n − a | &lt; ϵ for ϵ &gt; 0.

Question

How would you show that if we have a sequence   {      a    n    } of irrational numbers then for any number   a in the reals, the       |        a    n    −  a      |    &amp;lt;  ϵ for   ϵ  &amp;gt;  0.

hitturn35 · Accepted Answer

We know that the every real number is the limit of a sequence of rationals.Let   x be real, and let   y  =  x  −      2  . Then there is a sequence   (      r    n    ) of rationals with limit y. It follows that the sequence   (      2    +      r    n    ) of irrationals has limit   x.

Alfredo Cooley · Accepted Answer

Hints: show first there is an irrational number in   (  x  ,  y  ) for   x  &amp;lt;  y. Then, apply this to   (  a  −      1    n    ,  a  ) to obtain a sequence that converges to a.

How would you show that if we have a sequence {a_n} of irrational numbers then for any number a in the reals, the |a_n−a|<epsilon for epsilon>0.

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