Given that any r in R has a rational sequence that converges to it I want to show that between every 2 reals there is a rational. I want to show this just using the convergence fact only and not Archimedian property, I get stuck with then inequalities any help please.
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Let $a,b \in \mathbb{R}$, take a sequence of rationals converging to $a$ and use your assumption with $\epsilon=\frac{|a-b|}{2}$. You get rationals between $a$ and $b$.
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$\begingroup$ Hi, thanks but I did do that and get stuck with the inequalities. so wlog, take a<b then by assumption can fine a sequence of rationals {x_n} -> a, therefore for the choice (b-a)/2 we have for some m in N |x_n-a|<(b-a)/2 for all n>m. So then (3a-b)<2x_n<(b+a). Can similarly get an inequality using b and with different choices epsilon, taking epsilon to be b-a yields (2a-b)<x_n<b etc but the problem I have is manipulating the inequalities to get a rational between a and b, its probably some simply algebra but please can anyone offer some guidance. Thank you. $\endgroup$– JensCommented Feb 3, 2014 at 14:50