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July 12th, 2016, 12:53 PM  #1 
Newbie Joined: Oct 2014 From: 192.168.0.1 Posts: 13 Thanks: 2  Direct proof for this inequality theorem Theorem:$\displaystyle a\leqslant b+\epsilon, \forall \epsilon > 0 \Rightarrow a\leqslant b $ where $\displaystyle a,b\in \mathbb{R}$ I have seen proof by contradiction for this. But I tried to give a direct proof as follows; Let $\displaystyle a,b\in \mathbb{R} \wedge \epsilon > 0$ $\displaystyle a\leqslant b+\epsilon \Rightarrow ab\leqslant \epsilon$ That is $\displaystyle \epsilon_{min}=ab$ But since $\displaystyle \epsilon > 0$, $\displaystyle \epsilon_{min} > 0 \geqslant 0$ $\displaystyle \therefore ab\geqslant 0 \Rightarrow a\geqslant b$ Here I obtained a complete opposite of the theorem. Could someone please point me where I have gone wrong? Thanks 
July 12th, 2016, 01:05 PM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
You cannot write "$\displaystyle \epsilon_{min}= b a$" unless you first show that there is a minimum value for $\displaystyle \epsilon$ (and there isn't).
Last edited by Country Boy; July 12th, 2016 at 01:10 PM. 
July 12th, 2016, 10:53 PM  #3 
Newbie Joined: Oct 2014 From: 192.168.0.1 Posts: 13 Thanks: 2 
Thanks! Is that the condition "$\displaystyle \epsilon > 0$" which keeps us away from concluding that $\displaystyle \epsilon_{min}=ab$ ? And what if we didn't have that condition and the statement was as follows? $\displaystyle a\leqslant b+\epsilon, \forall \epsilon > ab \Rightarrow a\leqslant b $ where $\displaystyle a,b\in \mathbb{R}$ Is the above statement legit? 

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