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This question was asked before for proof by contradiction and which got me into thinking whether i could prove it without using a contradiction

Original problem statement is here Prove by contradiction that a real number that is less than every positive real number cannot be posisitve

Question

There exists a real number less than every positive real number that is positive

With using the method of contradiction it can be easily proved. But without?

Here's what i tried

$$ \text{let} \space \exists \space a \in \Re \text{ such that } \forall \epsilon > 0 \space a < \epsilon $$

now we can write it as

$$ \epsilon \, - a > 0 \\ but \, \epsilon - a \in \Re+ \\ hence \, (\epsilon - a) - a > 0 \rightarrow \epsilon - 2a > 0 \\ \; \rightarrow \frac{\epsilon}{2} > a \\ similarly \, \frac{\epsilon}{n} > a \, where \, n \in \Re+ $$

from here onwards, i showed that when n goes to positive infinity, the whole term goes to 0 which makes $ 0 > a $

but i'm not sure how correct that is. Any ideas? :)

And also my apologies for poor formatting of equations, it's my first time with the tex

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    $\begingroup$ Well first, only nonstrict inequalities are preserved under limits, so taking limits gives only $a \leq 0$, not $a < 0$. Second, you may need to justify the preservation of nonstrict inequalities. This depends on who you are writing this for. Shorter proofs might exist depending on how you are defining $\mathbb{R}$ (for instance, you might be able to get a short proof by using the Archimedean property of $\mathbb{Q}$). $\endgroup$
    – Ian
    Commented Feb 3, 2015 at 13:50
  • $\begingroup$ yes i got the $ a < 0 $ result and stuck from there :/ this was a university assigment, well the original question was the thread i have posted as a link above, this is just for my curiosity whether it's possible $\endgroup$
    – ManZzup
    Commented Feb 3, 2015 at 13:51
  • $\begingroup$ Even if right, it still seems to be a proof by contradiction. $\endgroup$
    – coffeemath
    Commented Feb 3, 2015 at 13:51
  • $\begingroup$ @coffeemath since i didnt contradict any of the original facts or premises i thought this would be a varying approach $\endgroup$
    – ManZzup
    Commented Feb 3, 2015 at 13:53
  • $\begingroup$ Your final conclusion, which should be that $a \le 0$ rather than $a<0$ [see Ian's comment] is to the initial assumption that $a>0$ and so is a contradiction to that assumption. Actually in your initial start you only assumed for all $epsilon>0$ that $a<\epsilon,$ without the assumption that $a>0$ But if the assumption $a>0$ is dropped, you can't get a contradiction anyway since maybe for example the initial $a$ is $-1$. $\endgroup$
    – coffeemath
    Commented Feb 3, 2015 at 14:04

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