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