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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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ManZzup
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    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 $\mathbb{Q}$). – Ian Feb 03 '15 at 13:50
  • 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 – ManZzup Feb 03 '15 at 13:51
  • Even if right, it still seems to be a proof by contradiction. – coffeemath Feb 03 '15 at 13:51
  • @coffeemath since i didnt contradict any of the original facts or premises i thought this would be a varying approach – ManZzup Feb 03 '15 at 13:53
  • 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 $-1$. – coffeemath Feb 03 '15 at 14:04
  • You are asking for a real number which is less than itself. No epsilons needed to see that such a number does not exist. – Christian Blatter Feb 03 '15 at 14:04
  • @coffeemath sorry but i didnt get the last part, because i didnot specifically assumed $ a > 0 $ in the beginning, and only got $ a < 0 $ as a result. Will that still be a contradiction? – ManZzup Feb 03 '15 at 14:15
  • OK you have $\epsilon>0$ and $a<\epsilon$ so you do get $\epsilon-a>0.$ In the next line the step is to subtract $a$ from each side, and you conclude that $\epsilon-2a>0.$ However from $\epsilon-a>0$ after subtracting the $a$ the result may be negative, not positive. Consider the case of $\epsilon=1,\ a=0.9.$ Then $\epsilon-a=0.1$ but $\epsilon-2a=-0.8$ is negative, going against your conclusion that $\epsilon-2a>0$ at that point. – coffeemath Feb 03 '15 at 14:39
  • @coffeemath ahh got it, thankxx :) so there's no way other than contradiction to prove this? – ManZzup Feb 03 '15 at 14:46
  • Manzup: I don't think without contradiction creeping in. Even using the least upper bound property, some kind of contradiction argument is used to show that an element of a set is less than or equal to its least upper bound. – coffeemath Feb 03 '15 at 15:28
  • @coffeemath since ManZzup assumed $a$ was the 'smallest' positive real number , so $(\epsilon-a)-a=\epsilon-2a>0$ is reasonable . – iMath Nov 14 '16 at 09:23
  • @ManZzup If $\lim {x\rightarrow \infty }a{n}=a$ and $\lim {x\rightarrow \infty }b{n}=b$, and there exists a number $N > 0$ such that $a_{n}\leq b_{n}$ for all $n>N$, then $a\leq b$ . You should know this in order to get $a \le 0$ rather than $a<0$. – iMath Nov 14 '16 at 11:07

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