I am trying to prove that if $x\in \mathbb{R}$,$y\in \mathbb{R}$ and $x
Here is what I tried to do:
I am trying to show that given $x$,$y$,there exist $m$ and $n$ such that $x
By the Archimedean property, as $x
$y-x>0$ so $n(y-x)>1$ for some $n\in \mathbb{N}$.So, $ny>1+nx$.
There must also be an integer $m$ between $nx$ and $1+nx$ so that $nx
Which gives $nx
Is this proof correct?
I just read Arkeet's comment and I think the statement "There must also be an integer $m$ between $nx$ and $1+nx$" needs a formal proof.I don't think I have one at hand.